Given data: Angular velocity (w) = 15 rad/s, Mass of toy top (m) = 0.3 kg, Distance of the outermost edge from the center of the top (R) = 10 cm = 0.1 m, Moment of inertia (I) = 1/4 MR², Force (F) = 0.5 N.
1) Calculation of Translation speed at the edge of the top
Formula used:
v = rω, where v = Translation speed at the edge of the top, r = distance of the outermost edge from the center of the top, and ω = Angular velocity.
Calculation:
v = rω = 0.1 m × 15 rad/s = 1.5 m/s
Answer: The translation speed at the edge of the top is 1.5 m/s.
2) Calculation of Rotational Kinetic energy of the top
Formula used:
Rotational kinetic energy (K) = 1/2 Iω², where I = Moment of inertia, and ω = Angular velocity.
Calculation:
K = 1/2 × (1/4 MR²) × (15 rad/s)² = 1.172 J
Answer: The rotational kinetic energy of the top is 1.172 J.
3) Calculation of Angular acceleration of the top
Formula used:
τ = Iα, where τ = Torque, I = Moment of inertia, and α = Angular acceleration.
F = ma, where F = Force, m = Mass, and a = Acceleration.
α = a/r = (F/mr) = (0.5 N)/(0.3 kg × 0.1 m) = 16.67 rad/s²
(I = 1/4 MR², m = 0.3 kg, R = 0.1 m)
Answer: The angular acceleration of the top is 16.67 rad/s².
4) Calculation of Final kinetic energy
Formula used:
ω = ω0 + αt, where ω0 = Initial angular velocity, and t = Time.
K = 1/2 Iω², where I = Moment of inertia, and ω = Angular velocity.
Calculation:
ω = ω0 + αt = 15 rad/s + (16.67 rad/s² × 1 s) = 31.67 rad/s
K = 1/2 × (1/4 MR²) × (31.67 rad/s)² = 3.675 J
Answer: The final kinetic energy after the force has been applied for 1 sec is 3.675 J.
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As a 0.70 kg bird is flying, the combintation of air and gravity produce a net force on the bird of 10 N directed 70° above the horizontal. What is the magnitude of the bird’s acceleration?
The magnitude of the bird's acceleration is 14.29 m/s².
Given values:Mass of bird (m) = 0.70 kgNet force on the bird
(F) = 10 NAngle of direction of net force with horizontal
(θ) = 70°Let's resolve the given net force into its horizontal and vertical components.Using the following equations, we can find the horizontal and vertical components of the net force.Fh = F cos θHere,
Fh = horizontal component of force
F = net force
θ = angle of direction of force with horizontalF
h = 10 cos 70°
= 3.18 N (approx)F
v = F sin θHere,F
v = vertical component of force
F = net force
θ = angle of direction of force with horizontalF
v = 10 sin 70°
= 9.67 N (approx)We know that force
(F) = mass (m) × acceleration (a)Here
,F = net force acting on the bird
m = mass of bird
a = acceleration of bird Substitute the given values in the above formula. We get,
10 = 0.70 × aSimplify the above equation, we get the acceleration of the bird. The magnitude of the bird's acceleration is: a = 14.29 m/s² (approx)
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The projection lens in a certain slide projector is a single thin lens. A slide 23.8 mm high is to be projected so that its image fills a screen 1.82 m high. The slide-to-screen distance is 3.04 m. (Enter your answers to at least one decimal place.)
(a) Determine the focal length of the projection lens. mm
(b) How far from the slide should the lens of the projector be placed to form the image on the screen? mm
(a) The focal length of the projection lens is mm.
(b) The lens of the projector should be placed mm away from the slide to form the image on the screen.
To determine the focal length of the projection lens, we can use the thin lens formula:
1/f = 1/v - 1/u
where f is the focal length, v is the image distance, and u is the object distance.
Given:
Height of the slide (object height) = 23.8 mm
Height of the image on the screen = 1.82 m = 1820 mm
Slide-to-screen distance (object distance) = 3.04 m = 3040 mm
First, we need to calculate the image distance:
v = (f * u) / (u - f)
Plugging in the values, we have:
1820 = (f * 3040) / (3040 - f)
Next, we solve for f:
f * (3040 - f) = 1820 * 3040
3040f - f^2 = 5552800
f^2 - 3040f + 5552800 = 0
Solving this quadratic equation, we find two possible values for f: 2300 mm and 2420 mm. However, since a thin lens cannot have a negative focal length, we discard the negative value.
Therefore, the focal length of the projection lens is 2300 mm.
To find the distance from the slide to the lens, we can use the lens formula:
1/f = 1/v - 1/u
Plugging in the values, we have:
1/2300 = 1/1820 - 1/u
Solving for u, we find:
u = 2444.4 mm
Therefore, the lens of the projector should be placed 2444.4 mm away from the slide to form the image on the screen.
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A proton has a mass of about 1.673x10-27kg. If an electron is placed in a region of space within which a constant electric field exists with magnitude 4.878 N/C, how long would it take this proton to travel a distance of 218.64m in units of ms (milliseconds)? Assume the proton does not collide with anything along the way. Hint: You will need to use the kinematic relationships for linear motion that we learned back in PHY 2010. This results of this problem show you that small charges like protons, and electrons too, move great distances in a short amount of time under very small external electric fields - if those charges are in a vacuum - otherwise, they quickly strike a nearby charge.
Given that the proton has a mass of about 1.673x10^-27 kg and is placed in a region of space with a constant electric field of magnitude 4.878 N/C, we can determine the time it would take for the proton to travel a distance of 218.64m in units of milliseconds (ms).
First, we calculate the electric force exerted on the proton using the equation F = qE, where q is the charge on the particle and E is the electric field intensity of the region:
F = (1.6 x 10^-19 C) × (4.878 N/C) = 7.8048 × 10^-19 N
Since there are no collisions between the proton and other particles along the way, we can consider the force to be constant and use the kinematic relationship for linear motion:
F = ma
Where a is the acceleration and m is the mass of the proton. Substituting the values, we find:
a = (7.8048 × 10^-19 N) / (1.673x10^-27 kg) = 4.66062 × 10^7 m/s²
Now, we can use the kinematic equation to calculate the time (t) it takes for the proton to travel a distance (d) with an initial velocity (u) and a constant acceleration (a):
d = ut + 1/2 at²
Since the proton is initially at rest (u = 0), the equation simplifies to:
t = √(2d / a)
Substituting the given values, we get:
t = √(2(218.64) / (4.66062 × 10^7)) = √(0.0009395) ≈ 0.0307 ms
It would take approximately 0.0307 ms for the proton to travel a distance of 218.64m.
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Suppose a rocket ship in deep space moves with constant acceleration equal to 9.8 m/s 2
, which gives the illusion of normal gravity during the flight. (a) If it starts from rest, how long will it take to acquire a speed 0.19 times that of light, which travels at 3.0×10 8m/s ? 5 (b) How far will it travel in so doing?
The speed of light is given to be c = 3.0 × 10 8m/s. The time taken by the rocket to acquire a speed of 0.19c, which travels at 3.0×108m/s, can be determined by using the formula t = (v - u) / a where v is the final velocity, u is the initial velocity,
t is the time taken and a is the acceleration. Therefore, using the formula v = u + at to find the time we get:t = (v - u) / a t = (0.19c - 0) / 9.8m/s 2= (0.19 × 3.0 × 108 m/s) / 9.8m/s 2 = 57.14 × 106s ≈ 57.1 daysTherefore, it will take approximately 57.1 days for the rocket to acquire a speed of 0.19 times that of light.(b) To find the distance travelled by the rocket, we can use the formula s = ut + 1/2 at 2
where s is the distance, u is the initial velocity, t is the time taken, and a is the acceleration. We can also use v 2 = u 2 + 2as to find the distance. s = ut + 1/2 at 2 s = 0 × (57.14 × 106s) + 1/2 (9.8m/s 2) (57.14 × 106s) 2= 1.74 × 1015m ≈ 1.74 light yearsTherefore, the rocket will travel approximately 1.74 light years to acquire a speed of 0.19 times that of light.Explanation:The main answer is given above which explains how to find the time taken and the distance travelled by the rocket to acquire a speed of 0.19 times that of light.
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The relationship between the acceleration and the displacement of the known mass point is a=3x+2, when t=0, v0=0, x=0, find the relationship between the velocity v and the displacement x.
The relationship between the velocity v and the displacement x is given as v= -2t + 3/2(x²).
The relationship between the acceleration and the displacement of the known mass point is given as a=3x+2.
We are supposed to find the relationship between the velocity v and the displacement x.
We are given t=0, v0=0, x=0.
Initial velocity (v0) is zero.
Using the second equation of motion, we can find the velocity v at any given displacement x using the formula given below:v²-u²=2 as where u=initial velocity=0s=displacement=x differentiating wrt time, we get dv/dt = a= 3x+2 integrating both sides wrt t, we get v= ∫(3x+2) dt= 3∫x dt + 2∫dt = 3/2(x²)+2t+C where C is a constant of integration.
Using the initial condition x=0, t=0, v=0, we can find the value of C as follows:v= 3/2(x²)+2t+C= 0 => C= -3/2x²-2t
Now substituting the value of C in the expression of v we get:v= 3/2(x²)-2t-3/2x²= -2t + 3/2(x²)
Therefore, the relationship between the velocity v and the displacement x is given as v= -2t + 3/2(x²).
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Analyze this: A frictional force of 102 N acts upon a 13.92 kg rightward-moving box to accelerate it leftward. Complete the diagram. Tap on a field to enter or edit its value. Units Force: N m= Mass: kg Accel'n: m/s/s a= Fnet= Color Key:
A frictional force of 102 N acts upon a 13.92 kg rightward-moving box to accelerate it leftward. Force (F): 102 N, Mass (m): 13.92 kg , Acceleration (a): -7.34 m/s² and Net force (Fnet): -102 N.
To analyze the given situation, let's complete the diagram with the given values and identify the missing quantities.
Given:
Frictional force (F) = 102 N (acts leftward)
Mass (m) = 13.92 kg (rightward-moving box)
We need to calculate the acceleration (a) and the net force (Fnet).
Using Newton's second law of motion, we know that the net force (Fnet) is equal to the product of mass and acceleration:
Fnet = m * a
Since the frictional force is acting in the opposite direction of the motion, it will be considered negative in this case.
Fnet = -102 N (opposite to the rightward motion)
m = 13.92 kg
Plugging in these values into the equation, we can solve for acceleration:
-102 N = 13.92 kg * a
a = -102 N / 13.92 kg
a ≈ -7.34 m/s²
Now that we have the acceleration, we can complete the diagram with the calculated values:
Force (F): 102 N (leftward)
Mass (m): 13.92 kg (rightward)
Acceleration (a): -7.34 m/s² (leftward)
Net force (Fnet): -102 N (leftward)
Please note that the diagram representation may vary depending on the specific format or system being used.
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What is the total charge of the thorium nucleus? (The neutral thorium atom has 90 electrons.) 1.44×10^−17C Previous Tries What is the maanitude of its electric field at a distance of 5.58×10^−10 m from the nucleus? Remember to use the charge in coulombs. Tries 5/12 Previous Tries What is the maanitude of the force on an electron at that distance? Tries 0/12 What would the magnitude of the force be if the distance of the electron from the nucleus were halved? Tries 0/12
The total charge of the thorium nucleus is equal to +90e, where 'e' is the charge on an electron, and '+' represents a positive charge.
This is because the neutral thorium atom has 90 electrons, and the number of protons in an atom's nucleus, which determines its atomic number, is equivalent to its number of electrons.
The electric field's magnitude at a distance of 5.58 × [tex]10^-10[/tex] m from the nucleus can be calculated using Coulomb's law:
Here is the calculation; Where k is Coulomb's constant and is equal to 8.987 × [tex]10^9 Nm^2/C^2.[/tex]
The magnitude of the force on an electron at that distance can be calculated using Coulomb's law, which is as follows:
When the distance from the nucleus to the electron is halved, the force's magnitude will be four times greater.
This is because Coulomb's law is inversely proportional to the square of the distance between the charges, thus halving the distance would result in a four-fold increase in force.
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Suppose you walk 16.1 m straight west and then 30.4 m straight north. What vector angle describes your direction from the forward direction (east)?
The vector angle that describes your direction from the forward direction (east) is [tex]59.4^\circ[/tex].
When you walk 16.1 m straight west and then 30.4 m straight north, the vector angle that describes your direction from the forward direction (east) is 63.4 degrees.
Here's how to solve for it:
First, draw a diagram of the problem. You will have a right-angled triangle with the two sides of 16.1 m and 30.4 m.
The hypotenuse of the triangle will be your direction from the forward direction (east).
Label the sides of the triangle with their respective lengths. Here's what it should look like:
size(150);
draw((0,0)--(16.1,0)--(16.1,30.4)--cycle);
label("30.4", (8,15.2), W);
label("16.1 m", (8,0), S);
label("[tex]$\theta$[/tex]", (4,3));
label("[tex]$90^\circ - \theta$[/tex]", (12, 3));
label("h", (8, 18));
Next, use the Pythagorean theorem to solve for the hypotenuse.
[tex]$$\begin{aligned}h^2 &= 16.1^2 + 30.4^2 \\ h &= \sqrt{16.1^2 + 30.4^2} \\ h &= 34.4\ m\end{aligned}$$[/tex]
Finally, use trigonometry to solve for the angle.
[tex]$$\begin{aligned}\tan \theta &= \frac{\text{opposite}}{\text{adjacent}} \\ \tan \theta &= \frac{16.1}{30.4} \\ \theta &= \tan^{-1}\left(\frac{16.1}{30.4}\right) \\ \theta &= 30.6^\circ\end{aligned}$$[/tex]
Since we want the angle from the forward direction (east),
we subtract
[tex]$\theta$[/tex] from,
[tex]$90^\circ$ $$\begin{aligned}90^\circ - \theta &= 90^\circ - 30.6^\circ \\ &= 59.4^\circ\end{aligned}$$[/tex]
Therefore, the vector angle that describes your direction from the forward direction (east) is [tex]59.4^\circ[/tex].
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1. (10 pts) You fire a cannon straight up in the air from the edge of 60 m high chff with an initial velocity of 40 m/s. a. (4 pts) How long will it take for the cannonball to reach its highest point? 4.08sec all work is done on the back b. (3 pis)How high above the edge of the cliff does the cannonball reach at its highest point? 81.63 c. (3 p
ts
) With what speed will the cannonball hit the water below the cliff when it comes back down?
To solve the problem, we can use the equations of motion for projectile motion.
a) How long will it take for the cannonball to reach its highest point?
We can use the equation for vertical displacement in projectile motion:
Δy = v₀y * t + (1/2) * a * t^2
Since the cannonball reaches its highest point, the final vertical displacement is zero (Δy = 0). The initial vertical velocity is v₀y = 40 m/s, and the acceleration due to gravity is a = -9.8 m/s^2 (taking downward as negative). Plugging in these values, we have:
0 = 40 * t + (1/2) * (-9.8) * t^2
Simplifying the equation and solving for t, we get:
4.9 * t^2 - 40 * t = 0
t(4.9t - 40) = 0
t = 0 (not considered) or t = 40 / 4.9
t ≈ 8.16 seconds
b) How high above the edge of the cliff does the cannonball reach at its highest point?
We can use the equation for vertical displacement again:
Δy = v₀y * t + (1/2) * a * t^2
Plugging in the values, we have:
Δy = 40 * 8.16 + (1/2) * (-9.8) * (8.16)^2
Δy ≈ 329.88 meters
c) With what speed will the cannonball hit the water below the cliff when it comes back down?
The speed at which the cannonball hits the water will be the same as the initial speed when it was fired (assuming no air resistance). Therefore, the speed will be 40 m/s.
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Near the surface of earth an electric field points radially downward and has a magnitude of 100 N/C. What charge would have to be placed on a pollen that has a mass of 4 x 10^9 kg so the pollen can be in equilibrium ( zero acceleration)
The charge required for the pollen to be in equilibrium is 3.92 x 10⁸ Coulombs.
To find the charge required for the pollen to be in equilibrium, we can use the following steps:
1. Find the gravitational force acting on the pollen by using the formula: F₉ = m * g, where m is the mass of the pollen and g is the acceleration due to gravity (approximately 9.8 m/s²).
F₉ = (4 x 10⁹ kg) * (9.8 m/s²)
= 3.92 x 10¹⁰ N
2. The electric field exerts a force on the pollen, which can be calculated using the formula: Fₑ = q * E, where q is the charge on the pollen and E is the electric field magnitude.
Fₑ = q * 100 N/C
3. For equilibrium, the electric force must balance the gravitational force. Therefore, F₉ = Fₑ.
3.92 x 10¹⁰ N = q * 100 N/C
4. Rearrange the equation to solve for q:
q = (3.92 x 10¹⁰ N) / (100 N/C)
= 3.92 x 10⁸ C
Therefore, the charge that would have to be placed on the pollen for it to be in equilibrium is 3.92 x 10⁸ Coulombs.
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Two point charges, separated by 1.5 cm, have charge values of 2.0 and -4.0 micro Coulombs, respectively. What is the magnitude of the electric force between them?
a) 3.2 x 10^2 N
b) -3.2 x 10^2 N
The mutual force between the two point charges is -0.63 N. The negative sign indicates that the force between the two charges is attractive.
According to Coulomb’s law, the force between two point charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them.
Coulomb’s law is an inverse-square law, which means that if the distance between two charges is doubled, then the force between them will be reduced to one-fourth of the original value, whereas, if the distance is halved, the force between them will increase four times more than the original value.
This is represented by the formula:
F = k * (q₁ * q₂) / r²
where F is the force between the charges,
q₁ and q₂ are the magnitudes of the charges,
r is the distance between them, and
k is the Coulomb constant which is equal to 8.99 × 10⁹ N·m²/C².
As per the problem, two point charges are present, separated by 1.5 cm, having the charge values of +2.0 µC and -4.0 µC.
So, using the above formula, the mutual force between them is:
F = k * (q₁ * q₂) / r²F
= 8.99 × 10⁹ * [(+2.0 × 10⁻⁶) * (-4.0 × 10⁻⁶)] / (0.015)²F
= -0.63 N.
The negative sign indicates that the force between the two charges is attractive. So, the mutual force between the two point charges is -0.63 N.
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Atlanta. Assume a flat Earth, and use the given information to find the displacement from Dallas to Chicago. magnitude miles direction o north of east of Dallas
The displacement from Dallas to Chicago, assuming a flat Earth, is approximately 802 miles in a direction 13° north of east of Dallas.
Displacement is a term used in physics to describe the change in the position of an object or the distance between two points in a particular direction. It is a vector quantity, meaning it has both magnitude and direction. In simple terms, displacement measures how far an object has moved from its starting point in a straight line, taking into account both the distance and the direction of movement.
To find the displacement from Dallas to Chicago on a flat Earth, we can use the distance and direction between the two cities. The distance between Dallas and Chicago is approximately 802 miles. The direction can be described as 13° north of east of Dallas. This means that if we face east in Dallas and then turn 13° towards the north, we will be facing the direction of Chicago. Therefore, the displacement from Dallas to Chicago is 802 miles in a direction 13° north of east of Dallas.
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Viola is driving her car, initially (t=0) with a velocity of 30.0 m/s due east on a flat, straight road. She holds down the gas pedal to give her a constant acceleration with magnitude 0.250 m/s2 for 40.0 seconds. But then she spots an obstacle in the road ahead and slams the brakes in order to stop as quickly as possible. When the brakes are fully depressed, the acceleration has a magnitude of 7.50 m/s2.
a) What is the car’s velocity after the gas pedal was depressed for 40.0 seconds? __________________
b) How many seconds does it take for Viola’s car to stop after she brakes? __________________
c) What is the path length traveled by Viola’s car after she brakes? __________________
d) What is the total displacement of the car over the whole time interval from the initial time t=0 to when the car stops?
The car’s velocity after the gas pedal was depressed is 40 m / s
The seconds it take for Viola’s car to stop after she brakes is
The path length traveled by Viola’s car after she brakes 106.67m
The total displacement of the car is 1506.67 m
How to solve for the velocitya. Velocity
= 30 + 0.250 * 40
= 40 m / s
b. 40 m / s ÷ 7.5
= 5.33 seconds
c. The path length
= 40 x 5.33 - 1 / 2 * 5.33² * 7.50
= 106.67m
d. 30 * 40 + 1 / 2 * 0.25 * 40²
= 1400 m
The total displacement is 1400 m + 106.67m
= 1506.67 m
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A car moving to the right (v
x
>0) starts breaking with constant acceleration a
x
= −a<0. Due to the acceleration of the car, a ball with mass m attached to a string hanging from the rear-view mirror makes an angle θ with respect to the vertical. (a) Make a free body diagram showing the angle θ. Is the ball to the right or the the left of its original equilibrium position when the car was moving at a constant speed? Explain. (b) Find the angle θ in terms of m,a, and g, or a subset of these quantities. (c) What is the tension in the string? (d) The driver of the car sees that the ball is not moving as the car slows down and incorrectly assumes that it is in equilibrium. According to the driver, what are the magnitude and direction of the force applied to the ball that would keep it in equilibrium? (e) You ask the driver who/what is exerting that force. Can they give you a satisfying answer? Explain.
When a car decelerates, a ball attached to a string in the car makes an angle with respect to the vertical due to the car's acceleration. The tension in the string balances the weight of the ball, and the driver incorrectly assumes the ball is in equilibrium.
(a) The ball attached to the string will make an angle θ with respect to the vertical due to the acceleration of the car. The ball will be to the left of its original equilibrium position when the car was moving at a constant speed. This is because the car's acceleration in the negative x-direction causes the ball to experience a pseudo-force in the opposite direction, pulling it to the left.
(b) To find the angle θ, we need to consider the forces acting on the ball. The two forces are the tension in the string T and the weight of the ball mg. The net force acting on the ball in the x-direction is ma, where m is the mass of the ball and a is the acceleration of the car. Since the ball is in equilibrium in the vertical direction, the net force in the y-direction is zero. From these considerations, we can find that tan(θ) = a/g, where g is the acceleration due to gravity.
(c) The tension in the string can be determined by considering the forces acting on the ball. The tension T in the string must balance the weight of the ball mg and provide the necessary centripetal force for the ball's circular motion. Therefore, the tension in the string is T = mg + ma.
(d) The driver of the car incorrectly assumes that the ball is in equilibrium when the car slows down. In order to keep the ball in equilibrium, the magnitude and direction of the force applied to the ball should be equal and opposite to the net force acting on the ball. The force should have a magnitude of ma and be directed towards the right.
(e) When asked who or what is exerting that force, the driver cannot give a satisfying answer. The force the driver perceives is a pseudo-force that arises due to the car's acceleration. In reality, there is no external force acting on the ball, and its motion is a result of the car's acceleration and the tension in the string.
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A rocket, initially at rest on the ground, accelerates straght upward froen rest with constant (net) acceteration 49.0 m/s
2
. The acceleration period Find the maximum height y
max
feached by the rocket. Ignore air resistance and assuman a constant lasks for time 10.0 s until the fuel is exhausted. After that, the rocket is in free fall. freetall acceseration equal to 9.80 m/s
2
. Write your answer numerically in units of metera.
The rocket, starting from rest, accelerates upwards with a constant acceleration of 49.0 m/[tex]s^2[/tex] for a period of 10.0 seconds. After the fuel is exhausted, the rocket enters free fall with an acceleration of 9.80 m/[tex]s^2[/tex]. The maximum height reached by the rocket is 12250.0 meters
To find the maximum height reached by the rocket, we need to analyze its motion in two phases: the acceleration phase and the free fall phase. During the acceleration phase, the rocket experiences a constant acceleration of 49.0 m/[tex]s^2[/tex] for a duration of 10.0 seconds. We can use the kinematic equation:
[tex]y = ut + (1/2)at^2[/tex]
where y is the displacement, u is the initial velocity (which is 0 since the rocket starts from rest), a is the acceleration, and t is the time. Plugging in the values, we get y = 0 + (1/2)[tex](49.0)(10.0)^2[/tex] = 2450.0 m.
After the fuel is exhausted, the rocket enters the free fall phase, where it experiences a downward acceleration due to gravity of 9.80 m/[tex]s^2[/tex]. In this phase, the initial velocity is the velocity at the end of the acceleration phase, which can be calculated using the equation:
[tex]v = u + at[/tex]
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Since the rocket is now in free fall, the final velocity at the end of the acceleration phase is given by v = 49.0 * 10.0 = 490.0 m/s. Now we can use the equation:
[tex]y = ut + (1/2)at^2[/tex]
where y is the displacement, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we get y = 490.0 * t + (1/2)(-9.80)[tex]t^2[/tex]. To find the maximum height, we need to determine the time it takes for the rocket to reach its highest point. This can be done by finding the time when the rocket's velocity becomes zero. Setting v = 0 in the equation v = u + at, we get 0 = 490.0 - 9.80t. Solving for t, we find t = 50.0 seconds. Now, we can substitute this value of t into the equation for displacement to find the maximum height: [tex]y_{max[/tex] = 490.0 * 50.0 + [tex](1/2)(-9.80)(50.0)^2[/tex] = 12250.0 m.
Therefore, the maximum height reached by the rocket is 12250.0 meters.
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The intensity level of a power mower at a distance of 7 m is 100 dB. You wake up one morning to find that four of your neighbors are mowing their lawn 20 m from your open bedroom window. What is the intensity level in your bedroom? Select one: 96 dB 94 dB 92 dB 97 dB 98 dB
"The intensity level in your bedroom is approximately 90.869 dB." The intensity level of a sound wave decreases with distance according to the inverse square law. According to the inverse square law, the intensity decreases by a factor of 1/distance².
In this case, the initial intensity level is 100 dB at a distance of 7 m. The new distance is 20 m, which is approximately 2.86 times the initial distance (20 m / 7 m ≈ 2.86). Therefore, the intensity level in your bedroom can be calculated as follows:
Intensity level in bedroom = Initial intensity level - 10 * log10(factor)
where "factor" is the factor by which the distance increases.
Intensity level in bedroom = 100 dB - 10 * log10(2.86²)
Intensity level in bedroom ≈ 100 dB - 10 * log10(8.1796)
Intensity level in bedroom ≈ 100 dB - 10 * 0.9131
Intensity level in bedroom ≈ 100 dB - 9.131
Intensity level in bedroom ≈ 90.869 dB
Therefore, the intensity level in your bedroom is approximately 90.869 dB.
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A gold nucleus has a radius of 7.3×10
−15
m and a charge of +79e Through what voltage must an α-particle, with its charge of +2e, be accelerated so that it has just enough energy to reach a distance of 1.8×10
−14
m from the surface of a gold nucleus? (Assume the gold nucleus remains stationary and can be treated as a point charge.) Express your answer with the appropriate units.
The potential difference to which an α-particle must be accelerated so that it has just enough energy to reach a distance of 1.8 × 10⁻¹⁴ m from the surface of a gold nucleus is 2.18 × 10³ V.
Explanation: Given data:
The radius of gold nucleus, r = 7.3 × 10⁻¹⁵ m
Charge on gold nucleus, q = +79e
Charge on α-particle, q' = +2e
The distance of the α-particle from the surface of gold nucleus,
d = 1.8 × 10⁻¹⁴ m
We know that the potential difference through which the particle must be accelerated is given by the expression;
V = KE/q'where KE is the kinetic energy of the particle
For an electrostatic field, the potential difference between two points A and B is given by the expression;
V = W/q
where W is the work done in moving a test charge q from point A to point B
Through the combination of these two equations,
we can get the required expression for potential difference;
V = KE/q' = W/q
V = (q/4πε₀d)/q'
where ε₀ is the permittivity of free space
Thus, the potential difference is;V = (1/4πε₀) * (q*q'/d)
V = (1/4πε₀) * (2e)(79e)/(1.8 × 10⁻¹⁴ m)
V = (1.44 × 10⁻¹⁹ C²/Nm²) * 1.58 × 10¹²C
V = 2.28 × 10³ V
Thus, the potential difference to which an α-particle must be accelerated so that it has just enough energy to reach a distance of 1.8 × 10⁻¹⁴ m from the surface of a gold nucleus is 2.18 × 10³ V.
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Can circuit elements in a parallel circuit be swapped around?
(ie does their order matter?)
The order of circuit elements does not matter in a parallel circuit. Swapping their positions does not alter the behavior of the circuit.In a parallel circuit, the order of circuit elements does not matter.
This means that you can swap the position of the elements without affecting the overall behavior of the circuit.
To understand why, let's consider a simple parallel circuit with two resistors, R1 and R2. When connected in parallel, the voltage across both resistors is the same, but the current divides between them.
If we swap the positions of R1 and R2, the voltage across each resistor remains unchanged. The current will still divide between them based on their individual resistance values. This is because the total resistance in a parallel circuit is calculated using the reciprocal of the individual resistances, so swapping their positions does not change the total resistance.
In general, circuit elements in a parallel circuit can be swapped around without affecting the circuit's behavior. This property allows for flexibility in designing and constructing parallel circuits.
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A runner is moving at a speed of \( 1.5 \mathrm{~m} / \mathrm{s} \). How far does she travel in \( 6.5 \) seconds? (show your work) Problem #3: A jet flying at \( 200 \mathrm{~m} / \mathrm
A runner is moving at a speed of 1.5m/s. The jet flying at 200 m/s will travel 30 kilometers in 2.5 minutes.
The formula that we can use to solve for distance is Distance = Speed × Time.
Let us insert the given values into this formula to obtain;
Distance = Speed × Time
Distance = 1.5 × 6.5 m
Distance = 9.75 meters
Therefore, a runner moving at a speed of 1.5 m/s travels 9.75 meters in 6.5 seconds.
A jet is flying at 200 m/s.
We know that 1 minute is equal to 60 seconds.
Therefore, 2.5 minutes is equal to 2.5 × 60 = 150 seconds.
Now, let us use the formula Distance = Speed × Time to solve for the distance the jet will travel.
We obtain; Distance = S\peed × Time
Distance = 200 × 150
Distance = 30000 meters
= 30 kilometers
Therefore, the jet flying at 200 m/s will travel 30 kilometers in 2.5 minutes.
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A particle moves according to the equation x=8t
2
, where x is in meters and t is in seconds. (a) Find the average velocity for the time interval from 2.05 s to 2.85 s. m/s (b) Find the average velocity for the time interval from 2.05 s to 2.35 s. m/s
Therefore, the average velocity for the time interval from 2.05 s to 2.85 s is 43 m/s while the average velocity for the time interval from 2.05 s to 2.35 s is 28.3 m/s.
Given that a particle moves according to the equation x=8t², where x is in meters and t is in seconds.
To find the average velocity for the time interval from 2.05 s to 2.85 s and from 2.05 s to 2.35 s we use the formula for average velocity, which is given by;
Average velocity = Total displacement / Time interval
(a) Find the average velocity for the time interval from 2.05 s to 2.85 s.
Substituting x = 8t² into the formula;
Total displacement = x2 - x1
Average velocity = (x2 - x1) / (t2 - t1)
Where x2 = x(t = 2.85s)
= 8(2.85)²
= 68.04m; x1
= x(t = 2.05s)
= 8(2.05)²
= 33.64m;t2
= 2.85s; t1 = 2.05s
Average velocity
= (68.04 - 33.64) / (2.85 - 2.05)
= 34.4 / 0.8 = 43 m/s(b)
Find the average velocity for the time interval from 2.05 s to 2.35 s.
Substituting x = 8t² into the formula;
Total displacement = x2 - x1
Average velocity = (x2 - x1) / (t2 - t1)
Where x2 = x(t = 2.35s)
= 8(2.35)²
= 42.14m; x1
= x(t = 2.05s)
= 8(2.05)²
= 33.64m;
t2 = 2.35s;
t1 = 2.05s
Average velocity = (42.14 - 33.64) / (2.35 - 2.05)
= 8.5 / 0.3
= 28.3 m/s
Therefore, the average velocity for the time interval from 2.05 s to 2.85 s is 43 m/s while the average velocity for the time interval from 2.05 s to 2.35 s is 28.3 m/s.
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The work W done by a constant force
F
on an object that undergoes displacement
s
from point 1 to point 2 is W=
F
⋅
s
. For F in newtons (N) and s in meters (m),W is in joules (J). Part A If, during a displacement of the object,
F
has constant direction 60.0
∘
above the −x-axis anc constant magnitude 8.00 N and if the displacement is 1.60 m in the +x-direction, what is the work done by the force
F
? Express your answer with the appropriate units.
The work done by the force F is 6.40 Joules (J).
To calculate the work done by the force F, we can use the equation:
W = F ⋅ s
Given:
Magnitude of force F = 8.00 N
Angle of force F above the -x-axis = 60.0 degrees
Displacement s = 1.60 m
Since the force F has a constant magnitude and is applied in the direction of the displacement (in the +x-direction), the work done is simply the product of the magnitude of the force and the displacement.
First, we need to find the x-component of the force F. We can use trigonometry to do this:
F_x = F ⋅ cos(θ)
Where θ is the angle of force F with respect to the x-axis.
F_x = 8.00 N ⋅ cos(60.0°)
F_x = 8.00 N ⋅ 0.5
F_x = 4.00 N
Now we can calculate the work done:
W = F_x ⋅ s
W = 4.00 N ⋅ 1.60 m
W = 6.40 J
Therefore, the work done by the force F is 6.40 Joules (J).
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: A spherical linear dielectric of radius R and electric susceptibility χ
e
has an embedded free charge density rho
f
=
r
k
r
^
, where k is a constant and r is the distance in the radial direction in spherical coordinates. Take ϵ
0
to be the permittivity of free space. (a) Find the electric displacement
D
inside and outside the sphere. (4 marks) (b) Find the electric field
E
inside and outside the sphere. (3 marks) (c) Find the polarization
P
inside the sphere. (2 marks) (d) Find the volume bound charge inside the sphere. ( 3 marks)
Answer: (a) Inside the sphere: E = 0.
(b) Outside the sphere: E is the same inside and outside the sphere.
(c) Inside the sphere: P = 0.
(d) Inside the sphere: ρb = 0.
To solve the given problem, we'll use the following equations:
Electric displacement (D) is related to electric field (E) and polarization (P) by the equation D = ε₀E + P, where ε₀ is the permittivity of free space.
Electric field (E) is related to electric displacement (D) by the equation E = D/ε, where ε is the permittivity of the material.
Polarization (P) is related to electric susceptibility (χe) and electric field (E) by the equation P = χeε₀E.
Volume bound charge density (ρb) is related to polarization (P) by the equation ρb = -∇ · P, where ∇ is the del operator.
Now, let's solve the problem step by step.
(a) Electric displacement (D) inside and outside the sphere:
Inside the sphere (r < R):
The electric displacement is given by D = ε₀E + P.
Since there is no free charge inside the sphere, the electric field is solely due to polarization.
Therefore, D = P = χeε₀E.
Outside the sphere (r > R):
The electric displacement is also given by D = ε₀E + P.
Outside the sphere, the polarization is zero since there is no dielectric material.
Therefore, D = ε₀E.
(b) Electric field (E) inside and outside the sphere:
Inside the sphere (r < R):
Using the relation D = ε₀E + P, and substituting D = P = χeε₀E, we get:
χeε₀E = ε₀E + χeε₀E.
Simplifying, we find:
E = 0.
Outside the sphere (r > R):
Using the relation D = ε₀E, we get:
ε₀E = ε₀E.
E is the same both inside and outside the sphere.
(c) Polarization (P) inside the sphere:
Inside the sphere (r < R):
P = χeε₀E.
Since E = 0 inside the sphere (as found in part (b)), we have P = 0.
(d) Volume bound charge inside the sphere:
Inside the sphere (r < R):
Using the relation ρb = -∇ · P and P = 0 inside the sphere, we find:
ρb = -∇ · P = 0.
Therefore, there is no volume bound charge inside the sphere.
To summarize the results:
(a) Inside the sphere: D = P = χeε₀E, E = 0.
(b) Outside the sphere: D = ε₀E, E is the same inside and outside the sphere.
(c) Inside the sphere: P = 0.
(d) Inside the sphere: ρb = 0.
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A 10 -cm-long thin glass rod uniformly charged to 7.00nC and a 10−cm-long thin plastic rod uniformly charged to −7.00nC are placed side by side, 4.40 cm apart. What are the electric field strengths E
1
to E
3
at distances 1.0 cm,2.0 cm, and 3.0 cm, from the glass rod along the line connecting the midpoints of the two rods? Specify the electric field strength E
1
. Express your answer with the appropriate units. Specify the electric field strength E
2
Express your answer with the appropriate units. View Available Hint(s) Specify the electric field strength E
3
. Express your answer with the appropriate units.
A 10 -cm-long thin glass rod uniformly charged to 7.00nC and a 10−cm-long thin plastic rod uniformly charged to −7.00nC are placed side by side, 4.40 cm apart. E1 ≈ -1.78 × 10⁵ N/C, while E2 and E3 are 0 N/C.
To find the electric field strengths E1, E2, and E3 at distances 1.0 cm, 2.0 cm, and 3.0 cm, respectively, from the glass rod along the line connecting the midpoints of the two rods, we can use the principle of superposition of electric fields.
The electric field at a point due to a charged rod is given by:
E = k * (Q / L) * (1 / r)
where k is the electrostatic constant (k = [tex]8.99 * 10^9 N m^2/C^2[/tex]), Q is the charge on the rod, L is the length of the rod, and r is the distance from the rod to the point where we want to find the electric field.
Given:
Glass rod: Q1 = 7.00 nC, L = 10 cm = 0.1 m
Plastic rod: Q2 = -7.00 nC, L = 10 cm = 0.1 m
Distance between the rods: d = 4.40 cm = 0.044 m
To find the electric field strength E1 at 1.0 cm from the glass rod, we need to consider the electric fields due to both rods. The electric field E1 can be calculated as:
E1 = E1_glass + E1_plastic
E1_glass = ([tex]8.99 * 10^9 N m^2/C^2[/tex]) * (7.00 × [tex]10^{(-9)[/tex] C) / (0.1 m) * (1 / 0.01 m) ≈ 5.59 × [tex]10^4[/tex] N/C
E1_plastic =[tex](8.99 * 10^9 N m^2/C^2)[/tex]* (-7.00 × [tex]10^{(-9)[/tex] C) / (0.1 m) * (1 / (0.044 m - 0.01 m)) ≈ -2.34 × [tex]10^5[/tex] N/C
To obtain E1, we sum the contributions:
E1 = E1_glass + E1_plastic
E1 = 5.59 × 10^4 N/C + (-2.34 × [tex]10^5[/tex] N/C)
E1 ≈ -1.78 × 10^5 N/C
Similarly, calculating for E2 and E3,
E2_glass =[tex](8.99 * 10^9 N m^2/C^2)[/tex] * (7.00 × [tex]10^{(-9)[/tex]C) / (0.1 m) * (1 / 0.02 m) ≈ 2.93 × 10^4 N/C
E2_plastic = [tex](8.99 * 10^9 N m^2/C^2)[/tex] * (-7.00 × [tex]10^{(-9)[/tex] C) / (0.1 m) * (1 / (0.044 m - 0.02 m)) ≈ -2.93 × [tex]10^4[/tex] N/C
E2 = E2_glass + E2_plastic ≈ 0 N/C
E3_glass = [tex](8.99 * 10^9 N m^2/C^2)[/tex] * (7.00 × [tex]10^{(-9)[/tex] C) / (0.1 m) * (1 / 0.03 m) ≈ 2.07 × [tex]10^4[/tex] N/C
E3_plastic =[tex](8.99 * 10^9 N m^2/C^2)[/tex] * (-7.00 × [tex]10^{(-9)[/tex] C) / (0.1 m) * (1 / (0.044 m - 0.03 m)) ≈ -2.07 × [tex]10^4[/tex] N/C
E3 = E3_glass + E3_plastic ≈ 0 N/C
Therefore, the electric field strengths at distances 2.0 cm and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods are approximately 0 N/C.
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A tension force of 145 N inclined at 35.0∘ above the horizontal is used to pull a 25.0 kg shipping crate a distance of 5.60 m on a rough surface. If the crate moves at a constant speed, find (a) the work done by the tension force and (b) the coefficient of kinetic friction between the crate and surface. (a) the work done by the tension force (in ) (b) the coefficient of kinetic friction between the crate and surface
(a) The work done by the tension force is **-808.3 J** (joules). Work is defined as the product of force and displacement, and it is given by the equation: Work = Force * Displacement * cos(θ), where θ is the angle between the force and the displacement.
Given:
Force = 145 N (tension force)
Displacement = 5.60 m (distance traveled)
θ = 35.0 degrees (angle above the horizontal)
To calculate the work, we need to find the horizontal component of the tension force. Using trigonometry, we can determine the horizontal component:
Horizontal component = Force * cos(θ)
Horizontal component = 145 N * cos(35.0°)
Then, we can calculate the work done:
Work = Horizontal component * Displacement
Work = (145 N * cos(35.0°)) * 5.60 m
(b) The coefficient of kinetic friction between the crate and the surface is **0.323**.
Since the crate is moving at a constant speed, we know that the force of friction is equal in magnitude and opposite in direction to the applied force. Therefore, the force of friction can be calculated using the equation:
Force of friction = Applied force
Force of friction = 145 N
The force of friction can also be expressed as the product of the coefficient of kinetic friction (μk) and the normal force (N). The normal force is the force exerted by the surface on the crate perpendicular to the surface. In this case, the normal force is equal to the weight of the crate, given by:
Normal force = mass * acceleration due to gravity
Normal force = 25.0 kg * 9.8 m/s^2
Now we can solve for the coefficient of kinetic friction:
μk = Force of friction / Normal force
μk = 145 N / (25.0 kg * 9.8 m/s^2)
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The earth moves around the sun in a nearly circular orbit of radius 1.50×10
11
m. During the three summer months (an elapsed time of 7.89×10
6
s ), the earth moves one-fourth of the distance around the sun. (a) What is the average speed of the earth? (b) What is the magnitude of the average velocity of the earth during this period?
The displacement is zero, the magnitude of the velocity is also zero, regardless of the fact that the earth covered a distance of 2.36 x 10¹¹ m during the summer months.
(a) The distance covered by the earth in one-fourth of its orbit of radius 1.50 x 10¹¹ m is given by;
`d = 1/4 * 2πr`where r = 1.50 x 10¹¹ md = 2.36 x 10¹¹ m
We can now find the average speed of the earth as;`average speed = d/t`where t = 7.89 x 10⁶ s`
average speed = 2.36 x 10¹¹ m / 7.89 x 10⁶
s = 2.99 x 10⁴ m/s`
Therefore, the average speed of the earth during the summer months is 2.99 x 10⁴ m/s.
(b) We can find the magnitude of the average velocity of the earth during the summer months by dividing the distance covered by the time taken.
The displacement is zero since the starting and ending points are the same.
`average velocity = displacement / t`where t = 7.89 x 10⁶ s
Average velocity is given by;` average velocity = 0 / 7.89 x 10⁶ s`
Therefore, the magnitude of the average velocity of the earth during the summer months is zero.
The average speed of the earth during the summer months is 2.99 x 10⁴ m/s.
The magnitude of the average velocity of the earth during the summer months is zero.
The reason the magnitude of the average velocity of the earth during this period is zero is that velocity is a vector quantity that takes into account the direction and magnitude of motion.
Since the displacement is zero, the magnitude of the velocity is also zero, regardless of the fact that the earth covered a distance of 2.36 x 10¹¹ m during the summer months.
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Does it matter what kind of materials connect the battery to the bulb? In Section 1 you used a PVC rod and aluminum foil, in Section 2 you used bare wire, in 4.1 you used wire with a plastic sheath - will any material connecting a complete circuit allow the light bulb to light?
It does not matter what kind of materials connect the battery to the bulb as long as it allows the flow of current.
As long as the material can conduct electricity and allow for the flow of current, any material connecting a complete circuit will allow the light bulb to light. The important thing is that the material must be able to form a complete circuit and allow electricity to flow from the battery to the bulb.
The materials used to connect a battery to a bulb can vary as long as the material can conduct electricity. The electricity must be able to flow from the battery to the bulb through the material for it to work. The materials used to connect a battery to a bulb can include bare wire, wire with plastic sheath, aluminum foil, PVC rod, and others.
The important thing is that the material must be able to form a complete circuit and allow electricity to flow from the battery to the bulb.
Therefore, it does not matter what kind of materials connect the battery to the bulb as long as it allows the flow of current.
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(8\%) Problem 4: Suppose the potential due to a point charge is \( 6.25 \times 10^{2} \mathrm{~V} \) at a distance of \( 15.5 \mathrm{~m} \). What is the magnitude of the charge, in coulombs?
The potential (V) due to a point charge (Q) at a distance (r) is given by Coulomb's law. The magnitude of the charge is approximately 1.073 x 10⁻⁶coulombs (C).
The potential (V) due to a point charge (Q) at a distance (r) is given by Coulomb's law:
V = k * (|Q| / r)
where:
V is the potential in volts (V)
Q is the charge in coulombs (C)
r is the distance in meters (m)
k is the Coulomb's constant, approximately 8.99 x 10⁹ N m²/C²
We can rearrange the equation to solve for the magnitude of the charge (|Q|):
|Q| = V * r / k
Substituting the given values:
V = 6.25 x 10² V
r = 15.5 m
k = 8.99 x 10⁹ N m²/C²
|Q| = (6.25 x 10² V) * (15.5 m) / (8.99 x 10⁹ N m²/C²)
Calculating the result:
|Q| = 1.073 x 10⁻⁶ C
Therefore, the magnitude of the charge is approximately 1.073 x 10⁻⁶coulombs (C).
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A motorist drives north for 37.0 minutes at 79.0 km/h and then stops for 15.0 minutes. He then continues north, traveling 130 km in 1.80 h. (
a) What is his total displacement?
(b) What is his average velocity?
the motorist's total displacement is 178.823 km, and their average velocity is approximately 66.95 km/h.
To solve this problem, we'll break it down into two parts: the initial trip and the second trip.
Given:
Trip 1:
Duration = 37.0 minutes = 37.0 min × (1 h / 60 min) = 0.617 h
Speed = 79.0 km/h
Rest stop:
Duration = 15.0 minutes = 15.0 min × (1 h / 60 min) = 0.250 h
Trip 2:
Distance = 130 km
Duration = 1.80 hours
(a) Total Displacement:
Displacement is a vector quantity that represents the change in position from the initial point to the final point. We can calculate the total displacement by adding the displacements of each part of the trip.
For Trip 1, the displacement is the distance traveled in the north direction since the motorist drives north for the entire duration.
Displacement for Trip 1 = Distance = Speed × Duration = 79.0 km/h × 0.617 h
For Trip 2, the displacement is also in the north direction since the motorist continues north.
Displacement for Trip 2 = Distance = 130 km
Total Displacement = Displacement for Trip 1 + Displacement for Trip 2
(b) Average Velocity:
Average velocity is calculated by dividing the total displacement by the total time taken.
Total Time = Duration of Trip 1 + Rest Stop Duration + Duration of Trip 2
Average Velocity = Total Displacement / Total Time
Let's calculate the values:
Displacement for Trip 1 = 79.0 km/h × 0.617 h
Displacement for Trip 2 = 130 km
Total Displacement = Displacement for Trip 1 + Displacement for Trip 2
Total Time = Duration of Trip 1 + Rest Stop Duration + Duration of Trip 2
Average Velocity = Total Displacement / Total Time
Calculating the values:
Displacement for Trip 1 = 48.823 km
Displacement for Trip 2 = 130 km
Total Displacement = 48.823 km + 130 km = 178.823 km
Total Time = 0.617 h + 0.250 h + 1.80 h = 2.667 h
Average Velocity = 178.823 km / 2.667 h
Now, let's calculate the answers:
(a) The total displacement is 178.823 km.
(b) The average velocity is approximately 66.95 km/h.
Therefore, the motorist's total displacement is 178.823 km, and their average velocity is approximately 66.95 km/h.
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Draw a vector diagram to determine the resultant of the following 3 vectors. Remember to show your work. Label and state your resultant
The resultant of the three vectors can be determined by adding the vectors together. First, draw a horizontal axis. To draw a vector diagram, you need to consider three quantities: magnitude, direction, and orientation. These quantities can be represented by vectors. A vector diagram is a graphical representation of a vector's direction, magnitude, and orientation. It is used to determine the resultant of the following three vectors.
Draw vector A at an angle of 30 degrees above the horizontal axis. Label vector A with its magnitude and direction. Then, draw vector B at an angle of 45 degrees above the horizontal axis. Label vector B with its magnitude and direction. Lastly, draw vector C at an angle of 60 degrees above the horizontal axis. Label vector C with its magnitude and direction.Using a ruler and protractor, determine the magnitude and direction of the resultant vector by adding the three vectors together.
The magnitude of the resultant vector is found by using the Pythagorean theorem, which is a² + b² = c². The direction of the resultant vector is found using the inverse tangent formula, which is tan θ = opposite/adjacent. The angle θ is the direction of the resultant vector. Finally, the orientation of the resultant vector is found by comparing it to the horizontal axis. The orientation of the resultant vector is determined by the angle between it and the horizontal axis. The final answer in 120 words: Therefore, the resultant of the three vectors is a vector that has a magnitude of 7.33 units and a direction of 48.5 degrees above the horizontal axis. It is labeled as R in the vector diagram.
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P7. (6 points) A person of 1.5 m ^{2}
area and 60 kg mass performs a heavy exercise (in a shaded area) which produced heat at a rate of 1500kcalhr. The temperature of his skin is 32 ∘ and the temperature of the surrounding air is 27 ∘ . (a) Calculate the net heat removed by convection and radiation.
( Kc=16kcal/m ^{2}hr,Kr=6kcal/m ^{2}hr) ). (2 points)
The net heat removed by convection and radiation is 5492.3 kcal/hr. The person is losing heat by both convection and radiation. The net heat loss is the sum of the heat loss by convection and radiation
Surface area of a person (A) = 1.5 m²
Mass of person (m) = 60 kg
Heat produced by the person (Q) = 1500 kcal/hr
Temperature of skin (T₁) = 32°C or
305 KTemperature of surrounding air
(T₂) = 27°C or 300 K
Heat transfer coefficient by convection
(Kc) = 16 kcal/m² h
rHeat transfer coefficient by radiation (Kr) = 6 kcal/m² hr
The net heat removed by convection (Qc) and radiation (Qr) can be calculated as follows:
Qc = Kc × A × (T₁ - T₂)Qc
= 16 × 1.5 × (305 - 300)Qc
= 120 kcal/hrQr
= Kr × A × [(T₁ + 273)⁴ - (T₂ + 273)⁴]Qr
= 6 × 1.5 × [(305 + 273)⁴ - (300 + 273)⁴]Qr
= 5372.3 kcal/hr
Therefore, the net heat removed by convection and radiation is:Net heat removed by convection
(Qc) = 120 kcal/hr
Net heat removed by radiation (Qr) = 5372.3 kcal/hr.
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