A golf ball is hit of the ground and undergoes projectile motion. It is given an initial

velocity vector with a horizontal velocity component (parallel to the surface of the earth) of

speed of 30 m/s and a vertical velocity component (perpendicular to the surface of the earth)

of speed 40 m/s

(a) What is the initial speed of the golf ball? (HINT: Use the Theorem of Pythagoras for a triangle

as discussed in class.)

A. 7.0 m/s

B. 50 m/s

C. 1,200 m/s

(b) How much time does it take to reach its maximum height?

A. 4.1

B. 82

C. 0.22

(c) What is the maximum height of its path?

A. 403

B. 82

C. 1.2

(d) What is the horizontal distance the golf ball travelled when it finally hits the ground?

A. 246

B. 1,230

C. 34

A orbiting piece of debris with a mass of 140 kg is orbiting the Earth at a height of 51,000 km above the surface.

The Earth's mass and radius are 5.98×10^24 kg and 6,385 km. Resolution At a distance of 51000 km from the surface of the Earth, the gravitational field strength of the debris is obtained using the formula;$$g=\frac{GM}{r^2}$$

Where; G is the universal gravitational constant M is the mass of the Earth r is the radius of orbit of the debris

We know that G=6.6743×10^-11 m^3 kg^-1 s^-2,

M=5.98×10^24 kg,

and r= (6385 km + 51000 km)

= 57385 km

= 5.7385 ×10^7m.

Substitute these values into the formula,

$$g=\frac{6.6743×10^{-11} × 5.98×10^{24}}{(5.7385×10^7)^2}

= 0.22 N/kg$$

Thus, the gravitational field strength at this height is 0.22 N/kg.

Weight of the satellite: We can find the weight of the satellite using the formula;$$F=mg$$

Where; F is the weight of the satellite g is the gravitational field strength m is the mass of the satellite We know that m=140 kg and g=0.22 N/kg.

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The maximum current that can be drawn from the battery without the terminal voltage dropping below 8.82 V is approximately 5 Amperes.

To find the maximum current that can be drawn from the battery without the terminal voltage dropping below 8.82 V, we can use Ohm's Law and consider the voltage drop across the internal resistance.

The terminal voltage (Vt) is given by:

Vt = emf - (internal resistance) * (current)

We want to find the maximum current (I) that can be drawn while keeping Vt above 8.82 V.

8.82 V = 9.00 V - (0.036 Ω) * I

Rearranging the equation:

(0.036 Ω) * I = 9.00 V - 8.82 V

(0.036 Ω) * I = 0.18 V

Dividing both sides by 0.036 Ω:

I = 0.18 V / 0.036 Ω

I ≈ 5 A

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In conclusion, based on the given information, we cannot find the values of (a), (b), and (c).(a) To find the magnitude of the current density J by free electrons, we need to use the formula J = σE, where σ is the conductivity and E is the electric field.

However, the given question only provides information about the magnetic flux density and relative permeability, which are not directly related to current density.

Therefore, we cannot determine the magnitude of the current density J by free electrons with the given information.

(b) Similarly, to find the size of the current density Jb by a magnetic dipole, we would need information about the dipole moment and the area of the dipole.

Since the question only provides information about the magnetic flux density and relative permeability, we cannot determine the size of the current density Jb.

(c) The size of the x-axis, y-axis, and z-axis of magnetization M cannot be determined with the given information. The question does not provide any information about the dimensions or characteristics of the magnetic body.

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The weight of the object on the moon is 16.33 N.

Given,

The weight of the object on earth is 98 N.

The gravitational acceleration on the moon is 1/6 of that on Earth.

Let the weight of the object on the moon be w.

The weight of the object can be found using the formula;

w = mg

where,

m = mass of the object

g = acceleration due to gravity

Now, Acceleration due to gravity on the Earth,

g_Earth = 9.8 m/s²

Acceleration due to gravity on the Moon,

g_Moon = 1/6

g_Earth = 9.8 / 6 m/s²

             = 1.633 m/s²

The mass of the object can be found using the formula;

w_Earth = mg_Earth

             => 98

             = m x 9.8

             => m

             = 98 / 9.8

            = 10 kg

The weight of the object on the moon,

w = m x g_Moon

   = 10 x 1.633

   = 16.33 N

Therefore, the weight of the object on the moon is 16.33 N.

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In a straight part of a stream, the water moves fastest d) in the center of the stream, away from the bottom and the banks.

This is due to the principle of laminar flow, where water molecules in the center experience less friction compared to those near the bottom or the banks. The flow of water in a stream is influenced by factors such as gravity, channel shape, and friction. As the stream flows, the water near the bottom and the banks experiences more frictional resistance, causing it to slow down. In contrast, the water in the center of the stream has less interaction with the stream's boundaries, allowing it to move more quickly. This creates a faster flow in the center of the stream compared to the other areas.

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If a rocket initially at rest accelerates at a rate of 49.5 m/s/s for 6 seconds, its speed will be 297.0 m/s.

calculate the final speed of the rocket, we can use the equation of motion:

v = u + at

v is the final velocity

u is the initial velocity (in this case, 0 m/s)

a is the acceleration

t is the time

u = 0 m/s

a = 49.5 m/s²

t = 6 s

Substituting the values into the equation:

v = 0 + (49.5 m/s²) * (6 s)

v = 297 m/s

Rounding the answer to one decimal place:

v ≈ 297.0 m/s

The speed of the rocket after 6 seconds of acceleration is 297.0 m/s.

The final speed of the rocket can be calculated using the equation v = u + at, where v is the final velocity, u is the initial velocity (which is 0 m/s since the rocket starts at rest), a is the acceleration, and t is the time.

Substituting the given values of a = 49.5 m/s² and t = 6 s into the equation, we find that v = 297 m/s. Rounding the answer to one decimal place, the speed of the rocket after 6 seconds of acceleration is  297.0 m/s.

This means that the rocket will be moving at a speed of 297.0 meters per second after accelerating at a rate of 49.5 meters per second squared for 6 seconds.

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The magnetic domains in a non-magnetized piece of iron are oriented randomly.

In a non-magnetized piece of iron, the magnetic domains are not aligned in any specific direction. The magnetic domains consist of small regions within the iron where the atomic magnetic moments are aligned. However, the orientations of these domains are random, resulting in a net magnetic field of zero. When an external magnetic field is applied to the iron, these domains start to align in the direction of the external field, leading to magnetization of the material. So, in their initial state, the magnetic domains in non-magnetized iron are randomly oriented, lacking any specific alignment or orientation.

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The length of the SLAC accelerator as measured in the reference frame of the electrons is 3,2 km. Here's an explanation of the SLAC accelerator and its length.

The SLAC accelerator is a linear particle accelerator located in Menlo Park, California. It was created in 1962 by the Stanford Linear Accelerator Center (SLAC), which is now known as the SLAC National Accelerator Laboratory. The SLAC accelerator is a 3.2 km-long linear accelerator that uses microwaves to accelerate electrons to extremely high speeds.

The length of the SLAC accelerator as measured in the reference frame of the electrons is 3,2 km because the electrons are moving very quickly relative to the laboratory reference frame, causing them to contract in length. This phenomenon is known as length contraction and is a consequence of Einstein's theory of special relativity.

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a) The average speed between t = 3.20 s and t = 4.50 s is 5.06 m/s.

b) The instantaneous speed at t = 3.20 s is 20.8 m/s and at t = 4.50 s is 30.8 m/s.

c) The average acceleration between t = 3.20 s and t = 4.50 s is 10 m/s².

d) The instantaneous acceleration at t = 3.20 s is 7.4 m/s².

e) The object is at rest at 0.27 s.

(a) Average speed is given by;

`V = (x2 - x1) / (t2 - t1)`

Substituting values;

`V = (3.7(4.5)² - 2(4.5) + 3) - (3.7(3.2)² - 2(3.2) + 3) / (4.5 - 3.2)

`V = 5.06 m/s

Therefore, the average speed between t = 3.20 s and t = 4.50 s is 5.06 m/s.

(b) To find instantaneous speed, differentiate the given equation of displacement with respect to time.

`v = 7.4t - 2`

Substituting values;

`v = 7.4(3.2) - 2 = 20.8 m/s`

Therefore, the instantaneous speed at t = 3.20 s is 20.8 m/s.

To find instantaneous speed, differentiate the given equation of displacement with respect to time.

`v = 7.4t - 2`

Substituting values;

`v = 7.4(4.5) - 2 = 30.8 m/s`

Therefore, the instantaneous speed at t = 4.50 s is 30.8 m/s

(c) Average acceleration is given by;

`a = (v2 - v1) / (t2 - t1)`

Substituting values;

`a = (30.8 - 20.8) / (4.5 - 3.2)`

a = 10 m/s²

Therefore, the average acceleration between t = 3.20 s and t = 4.50 s is 10 m/s²

(d) To find instantaneous acceleration, differentiate the equation of velocity with respect to time.

`a = 7.4`

Therefore, the instantaneous acceleration at t = 3.20 s is 7.4 m/s².

To find instantaneous acceleration, differentiate the equation of velocity with rest.

(e) For the object to be at rest, its velocity should be zero.

`v = 7.4t - 2 = 0``7.4t = 2``t = 0.27 s`

Therefore, the object is at rest at 0.27 s.

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To solve the problem, first let's find the weight of all seven books. The weight of a single book is 5 N, therefore the weight of all seven books is:5 N x 7 = 35 N Let's now calculate the maximum static frictional force that must be overcome to get the top six books sliding off the bottom one. The coefficient of static friction between the books is 0.19.

The formula for calculating the force of friction is:F = μN where F is the force of friction, μ is the coefficient of friction, and N is the normal force acting perpendicular to the surface between the two objects.So, the maximum force of static friction between the books can be calculated as:F = 0.19 x 35 N = 6.65 N Now, to start sliding the top six books off the bottom one, a force greater than 6.65 N must be applied horizontally.

Therefore, the minimum horizontal force that must be applied to get the top six books sliding is:6.65 N x 2 = 13.3 N (since the force will need to overcome the static friction between the bottom book and the ground and between the bottom book and the second book)Therefore, the horizontal force needed to start sliding the top six books off the bottom one is 13.3 N.

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The magnitude of the resultant displacement is approximately 30.83 cm, and the direction is approximately 132° with respect to due west. To find the resultant displacement of the grasshopper, we can add up the individual displacement vectors using vector addition.

To find the resultant displacement of the grasshopper, we can add up the individual displacement vectors using vector addition.

Given:

Vector 1: Magnitude = 31.0 cm, Direction = due west (0°)

Vector 2: Magnitude = 33.0 cm, Direction = 28.0° south of west (-28.0°)

Vector 3: Magnitude = 24.0 cm, Direction = 58.0° south of east (-58.0°)

Vector 4: Magnitude = 16.0 cm, Direction = 52.0° north of east (52.0°)

(a) To find the magnitude of the resultant displacement, we can use the Pythagorean theorem:

Resultant Displacement = √((Δx)^2 + (Δy)^2)

where Δx and Δy are the horizontal and vertical components of the displacement.

Horizontal component:

Δx = -31.0 cm + (33.0 cm)cos(-28.0°) + (24.0 cm)cos(-58.0°) + (16.0 cm)cos(52.0°)

Δx = -31.0 cm + 29.283 cm + 12.500 cm + 9.757 cm

Δx = 20.540 cm

Vertical component:

Δy = (33.0 cm)sin(-28.0°) + (24.0 cm)sin(-58.0°) + (16.0 cm)sin(52.0°)

Δy = -15.051 cm - 20.060 cm + 12.075 cm

Δy = -23.036 cm

Resultant Displacement = √((20.540 cm)^2 + (-23.036 cm)^2)

Resultant Displacement = √(421.3136 cm^2 + 529.0365 cm^2)

Resultant Displacement = √950.35 cm^2

Resultant Displacement ≈ 30.83 cm

(b) To find the direction of the resultant displacement, we can use trigonometry:

θ = tan^(-1)(Δy/Δx)

θ = tan^(-1)(-23.036 cm/20.540 cm)

θ ≈ -48.0°

Since the question asks for the direction as a positive angle with respect to due west, we add 180° to get the positive angle:

Direction = -48.0° + 180°

Direction ≈ 132°

Therefore, the magnitude of the resultant displacement is approximately 30.83 cm, and the direction is approximately 132° with respect to due west.

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Critical angle is the minimum angle of incidence at which total internal reflection occurs. It happens when the angle of incidence in an optically dense medium is more than the angle of refraction in an optically less dense medium.

When a ray of light moves from a denser medium to a rarer medium, it bends away from the normal. If the angle of incidence is increased continuously, at one point, the angle of refraction will become 90°. This is called the critical angle.

Explanation:

The formula for critical angle is sin c = n2 / n1,

where

c is the critical angle,

n1 is the refractive index of the denser medium,

and n2 is the refractive index of the rarer medium.

The condition for the critical angle to exist is that the angle of incidence must be greater than the critical angle. If the angle of incidence is equal to or less than the critical angle, then the light will refract into the rarer medium.

If the angle of incidence is greater than the critical angle, then total internal reflection will occur.

In summary, critical angle is the minimum angle of incidence at which total internal reflection occurs. The condition for the critical angle to exist is that the angle of incidence must be greater than the critical angle, which is calculated using the formula sin c = n2 / n1, where c is the critical angle, n1 is the refractive index of the denser medium, and n2 is the refractive index of the rarer medium.

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The answers are that the horizontal distance travelled by the artillery shell is 285.375 m; the coordinates where the artillery shell explodes are (285.375 m, 232.03 m). using the information in the question we can solve the problem as follows.

1) Calculation of horizontal distance (x):
We know that the horizontal component of the initial velocity remains constant throughout the projectile motion. Therefore, u=300 m/s, θ=51.5°
u_x = u cos θ = 300 × cos 51.5° = 190.25 m/s
Let x be the distance traveled by the artillery shell. Now using the formula of motion under constant acceleration we have,
x = u_x × t + (1/2)axt²
where t is the time taken by the artillery shell to explode.
x = (190.25 × 1.5) + (1/2) (0) (1.5)²
x = 285.375 m
Therefore, the horizontal distance travelled by the artillery shell is 285.375 m.

2) Calculation of vertical distance (y):
We know that the vertical component of the initial velocity changes uniformly under gravity. Therefore,
u = 300 m/s
θ = 51.5°
u_y = u sin θ = 300 × sin 51.5° = 234.98 m/s
The vertical distance traveled by the artillery shell in 1.5 s can be calculated as follows,
y = u_yt + (1/2)ayt²
where ay is the acceleration due to gravity which is -9.8 m/s² (negative as it acts in the downward direction)
y = (234.98 × 1.5) + (1/2) (-9.8) (1.5)²
y = 232.03 m
Therefore, the vertical distance travelled by the artillery shell is 232.03 m.
Hence, the coordinates where the artillery shell explodes are (285.375 m, 232.03 m).

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To explain the relationship between the given variables, let's consider two cars: a Toyota Camry and a Hummer, traveling at different speeds.

The force required to maintain a constant speed can be expressed as the sum of the drag force and the force of friction (rolling resistance). The drag force is proportional to the square of the speed and is influenced by the drag coefficient and the density of air. A higher drag coefficient indicates greater resistance to motion.

Comparing the two cars, the Toyota Camry has a drag coefficient of 0.28, while the Hummer has a higher coefficient of 0.64. This implies that the Hummer experiences higher air resistance, leading to a larger drag force at the same speed compared to the Camry.

When the speed increases from 76 km/h to 98 km/h, both cars experience an increase in drag force due to the squared relationship with speed. However, since the Hummer has a higher drag coefficient, it will experience a larger increase in drag force compared to the Camry.

Therefore, to maintain a constant speed, the engine of the Hummer needs to generate a higher force to overcome the increased drag force, while the Camry requires a relatively smaller increase in force due to its lower drag coefficient.

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The friction force acting on sled 3 is 196 N. The tension force in rope A is 476 N. The applied force Fpull is 672 N.

A. The following is the free-body diagram for sled 3:

Draw a diagram representing sled 3 as a rectangle or a box.

Label the downward direction as the positive y-axis.

Identify and draw arrows for the following forces:

a. Friction force (FFr): Draw an arrow to the left.

b. Tension force in rope B (FTB): Draw an arrow to the right.

c. Weight (Fg): Draw an arrow downward.

Label each force with its respective symbol (e.g., FFr, FTB, Fg).

Here's a list of forces acting on the sled: Friction force FFr, Tension force FTB, and Weight Fg.FFr is directed to the left, FTB is directed to the right, and Fg is directed downwards using our coordinate system's positive y-axis

.Note that the net force Fnet of sled 3 is equal to the product of mass m3 and acceleration a: Fnet = m3a.Using the formula above, we can calculate the friction force acting on sled 3 as follows: Fnet = Fg - FFr - FTB => FFr = Fg - FTB - Fnet => FFr = m3g - FTB - m3a  Substituting the given values: FFr = 20 kg x 9.8 m/s2 - 280 N - 20 kg x 10 m/s2 FFr = 196 N  

Therefore, the friction force acting on sled 3 is 196 N.

B. The following is the free-body diagram for sled 2:

Draw a diagram representing sled 2 as a rectangle or a box.

Label the downward direction as the positive y-axis.

Identify and draw arrows for the following forces:

a. Friction force (FFr): Draw an arrow to the left.

b. Tension force in rope A (FTA): Draw an arrow to the left.

c. Tension force in rope B (FTB): Draw an arrow to the right.

d. Weight (Fg): Draw an arrow downward.

Label each force with its respective symbol (e.g., FFr, FTA, FTB, Fg).

Here's a list of forces acting on the sled: Friction force FFr, Tension force FTA, Tension force FTB, and Weight Fg.FFr is directed to the left, FTA is directed to the left, FTB is directed to the right, and Fg is directed downwards using our coordinate system's positive y-axis.

Note that the net force Fnet of sled 2 is equal to the product of mass m2 and acceleration a: Fnet = m2a.Using the formula above, we can calculate the tension force in rope A as follows: Fnet = FTA - FFr - FTB => FTA = FFr + FTB + Fnet => FTA = m2a + FFr + FTB Substituting the given values: FTA = 15 kg x 10 m/s2 + 196 N + 280 N FTA = 476 N  

Therefore, the tension force in rope A is 476 N.

C. The following is the free-body diagram for sled 1:

Draw a diagram representing sled 1 as a rectangle or a box.

Label the downward direction as the positive y-axis.

Identify and draw arrows for the following forces:

a. Friction force (FFr): Draw an arrow to the left.

b. Applied force (Fpull): Draw an arrow to the right.

c. Tension force in rope A (FTA): Draw an arrow to the left.

d. Weight (Fg): Draw an arrow downward.

Label each force with its respective symbol (e.g., FFr, Fpull, FTA, Fg).

Here's a list of forces acting on the sled: Friction force FFr, Applied force Fpull, Tension force FTA, and Weight Fg.

FFr is directed to the left, Fpull is directed to the right, FTA is directed to the left, and Fg is directed downwards using our coordinate system's positive y-axis.

Note that the net force Fnet of sled 1 is equal to the product of mass m1 and acceleration a: Fnet = m1a.

Using the formula above, we can calculate the applied force Fpull as follows: Fnet = Fpull - FFr - FTA => Fpull = FFr + FTA + Fnet => Fpull = m1a + FFr + FTA Substituting the given values: Fpull = 10 kg x 10 m/s2 + 196 N + 476 N Fpull = 672 N Therefore, the applied force Fpull is 672 N.

A. The friction force acting on sled 3 is 196 N. B. The tension force in rope A is 476 N. C. The applied force Fpull is 672 N.

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The weight of an object is determined by the gravitational force acting on it. If the man is transported to the planet X, his weight would be 80 N.

The weight of an object is determined by the gravitational force acting on it. The formula to calculate weight is W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity. On the surface of Earth, man's weight is 640 N. Since the mass of man remains the same, the change in weight is due to the change in gravitational acceleration on planet X.

The gravitational acceleration on planet X can be determined using the formula for the acceleration due to gravity: g' = (GM)/(r'^2), where G is the gravitational constant, M is the mass of the planet, and r' is the radius of the planet. Since planet X has the same mass as Earth but a radius that is eight times larger, the gravitational acceleration on planet X is (1/64) times the gravitational acceleration on Earth.

Therefore, the weight of the man on planet X is (1/64) times his weight on Earth, which is 80 N.

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the index of refraction for the unknown material is approximately 1.525.

A ray of unpolarized light that travels through ethanol, having an index of refraction (n) of 1.361, hits an unknown material at an angle of 22 ∘ with respect to the surface and refracts at an angle of 63.658 ∘ with respect to the surface.

To find the index of refraction for the unknown material, we use Snell's law, which states that the ratio of the sines of the angles of incidence (i) and refraction (r) is equal to the ratio of the indices of refraction (n1 and n2) of the two media.n1sin(i) = n2sin(r)We are given:n1 = 1.361i = 22 ∘r = 63.658 ∘

We can plug these values into the equation and solve for n2:n2 = (n1sin(i))/sin(r)n2 = (1.361sin(22))/sin(63.658)  = 1.525 (rounded to three significant figures)

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The approximate contact time between the stick and puck is 0.8 s.

The given situation can be analyzed by applying the conservation of linear momentum principle which is given as:

m1v1i + m2v2i = m1v1f + m2v2f

where m, v and subscripts i and f represent mass, velocity, initial and final values respectively.

Let m1 be the mass of the puck and m2 be the mass of the stick. Before the collision, the stick is at rest, therefore its initial velocity is zero. After the collision, the stick breaks, therefore its final velocity is zero. Thus, the above equation becomes:

m1v1i = m1v1f + m2v2f.

From the problem statement, we can infer that the collision is one-dimensional and that only the y-component of the velocity of the puck changes. Let us represent the y-component of the velocity of the puck before and after the collision as v1iy and v1fy respectively.

Let t be the time of collision and F be the force exerted on the puck by the stick. According to Newton's second law of motion,

F = m1(v1fy − v1iy)/t.

Further, we can represent the change in the y-component of the velocity of the puck as: ∆v1y = v1fy − v1iy.

Substituting the values in the above equation, we get:

F = m1∆v1y/t.

Given that the stick breaks under a force of 270 N, we can equate the above equation to 270 N and solve for t to get the time of collision between the stick and the puck.

The solution will be as follows:

Given data: Initial velocity of the puck, v1i = 20 m/s

Position of the puck at the time of collision, r = 5i + 3j m

Position of the puck after the collision, r = 13i + 21j m

Breaking force of the stick, F = 270 N

Mass of the stick, m2 = ?

Using the distance formula, we can find the displacement of the puck during the collision as follows:

s = r2 − r1 = (13i + 21j) − (5i + 3j) = 8i + 18j m

Using the time formula, we can find the time of collision as follows:

t = s/∆v1y = (8i + 18j)/∆v1y

Let us now find ∆v1y by using the conservation of linear momentum principle which is as follows:

m1v1i = m1v1f + m2v2f20 m/s = v1fy + m2 × 0

Since the x-component of the velocity of the puck remains unchanged, we can say that:

v1fx = v1ix = 20 m/s

Thus, the y-component of the velocity of the puck before and after the collision is given by:

v1iy = 0 m/sv1fy = (m1 − m2) × (20/ m1)

Note that the magnitude of the velocity of the puck remains the same before and after the collision.

Therefore, we can say that ∆v1y = |v1fy − v1iy| = v1fy.

Substituting the values in the above equations, we obtain:

m1 × 20 = (m1 − m2) × (20/ m1) + m2 × 0m2

= m1/2∆v1y = |v1fy − v1iy|

= v1fy = (m1 − m2) × (20/ m1) = 10 m/s

Hence, t = s/∆v1y

= (8i + 18j) / (10 j/s) = 0.8 s

Approximately, the contact time between the stick and the puck is 0.8 s.

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The magnitude of the electric field at the particle's position is approximately 1.4 × 10^6 N/C.

To determine the magnitude of the electric field at the particle's position, we can use the equation that relates the electric force (F) experienced by a charged particle to the electric field (E) and the charge (q) of the particle:

F = q * E

Given that the charged particle experiences an electric force of 0.021 N and has a charge of 15 nC (nanocoulombs), we can rearrange the equation to solve for the electric field (E):

E = F / q

Substituting the given values:

E = 0.021 N / 15 nC

To simplify the units, we need to convert nanocoulombs (nC) to coulombs (C):

1 nC = 1 × 10^(-9) C

E = 0.021 N / (15 × 10^(-9) C)

E ≈ 1.4 × 10^6 N/C

Therefore, the magnitude of the electric field at the particle's position is approximately 1.4 × 10^6 N/C.

The electric field represents the force experienced by a charged particle per unit charge. In this case, the particle experiences a force of 0.021 N, which is divided by its charge of 15 nC to find the electric field strength. The resulting value indicates that at the particle's position, the electric field has a magnitude of 1.4 × 10^6 N/C. This means that if another charged particle with a charge of 1 C were placed in this electric field, it would experience a force of 1.4 × 10^6 N. The electric field provides important information about the interactions between charged particles and helps understand their behavior in electromagnetic systems.

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(a)The impedance of the circuit (Z) is 132.91 ohms. (b)The rms current flowing in the circuit is approximately 1.204 A. (c) It is not possible to determine the phase angle or the degree value of θ.

The magnitude of the current in the circuit at the instant the voltage across the generator is at its maximum value is 1.204 A. This value is obtained by calculating the rms current flowing in the circuit using the given values of impedance, which is 132.91 ohms.

The phase angle between the current and the voltage supplied by the AC source is not provided. To find the magnitude of the current in the circuit at the instant the voltage across the generator is at its maximum value, we need to calculate the rms current flowing in the circuit. The rms current can be determined using the formula:

I_rms = V_rms / Z

where I_rms is the rms current, V_rms is the rms voltage of the AC source, and Z is the impedance of the circuit. From the given information, the rms voltage of the AC source is 160 V, and the impedance of the circuit (Z) is 132.91 ohms.

Substituting the values into the formula, we get:

I_rms = 160 V / 132.91 ohms

I_rms ≈ 1.204 A

Therefore, the rms current flowing in the circuit is approximately 1.204 A.

The phase angle between the current in the circuit and the voltage supplied by the AC source (θ) is not provided in the given information. Hence, it is not possible to determine the phase angle or the degree value of θ.

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In a conical pendulum: (a) The tension in the string is given by T = mg / cosθ, (b) The centripetal acceleration is ac = g tanθ, (d) The radius of the circular path is r = L sinθ, and (e) The speed of the mass is v = √(rgtanθ).

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Therefore, the average speed over the entire trip is 3.61 m/s.(b) To find the average velocity, we need to calculate the displacement and the time taken. Since the person returns to the initial position, the displacement is zero. Therefore, the average velocity is also zero. Hence, the answer is 0 m/s.

(a) To find the average speed, we can use the formula:Average speed = Total distance / Total timeLet's consider the distance from point A to point B as 'd'.

The distance from B to A is also 'd' since she is walking back along the same path.

Let's also consider the time taken to go from A to B as 't1' and the time taken to come back from B to A as 't2'.To calculate the total time taken for the entire trip, we can add the time taken to go from A to B and the time taken to come back from B to A.

Therefore,Total time taken = t1 + t2

Now, speed = distance / time

Using this formula, we can calculate the time taken to go from A to B as:t1 = d / 5.10

And, the time taken to come back from B to A as:t2 = d / 2.80

Therefore, the total time taken for the entire trip is:

t1 + t2 = d / 5.10 + d / 2.80

= (2.80d + 5.10d) / (5.10 × 2.80)

= 7.90d / 14.28 = 0.554d

Now, the total distance covered in the entire trip is:

Total distance = distance from A to B + distance from B to A

= d + d = 2d

Therefore, the average speed is:

Average speed = Total distance / Total time

= 2d / 0.554d

= 3.61 m/s

Therefore, the average speed over the entire trip is 3.61 m/s.(b) To find the average velocity, we need to calculate the displacement and the time taken. Since the person returns to the initial position, the displacement is zero. Therefore, the average velocity is also zero. Hence, the answer is 0 m/s.

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(a) The average speed over the entire trip is 3.57 m/s.
(b) The average velocity over the entire trip is 0 m/s.

(a) To find the average speed over the entire trip, we can use the formula:

average speed = Total distance / Total time

Since the person travels from point (A) to point (B) and then back to point (A), the total distance covered is twice the distance between (A) and (B). Let's assume this distance is represented by d.

Total distance = 2d

To find the total time, we need to calculate the time taken for each leg of the trip.

Time taken from (A) to (B) = Distance / Speed = d / 5.10 m/s

Time taken from (B) to (A) = Distance / Speed = d / 2.80 m/s

Total time = Time from (A) to (B) + Time from (B) to (A) = (d / 5.10) + (d / 2.80)

Now we can calculate the average speed:

Average speed = Total distance / Total time = 2d / [(d / 5.10) + (d / 2.80)]

Average speed = 2 * (1 / [(1 / 5.10) + (1 / 2.80)]) = 3.57 m/s

Therefore, the average speed over the entire trip is 3.57 m/s.

(b) To find the average velocity over the entire trip, we need to consider both the magnitude and direction of the displacement. Since the person starts and ends at the same point, the total displacement is zero.

Average velocity = Total displacement / Total time

Since the total displacement is zero, the average velocity over the entire trip is also zero.

Therefore, the average velocity over the entire trip is 0 m/s.

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 Charge of corner A: -qDistance between A and center of the square: `a/sqrt(2)`Electric potential at center of the square when a charge of -q is placed at corner A: -6.50 × 10^-2 V The net electric potential at the center of the square is the algebraic sum of the potential difference between the center and the other charges present at the corners of the square.

In the given problem, charges of -q are placed at the corners A, C, and D and a charge of +q is placed at the corner B. Thus, the net electric potential at the center of the square can be given as,

Vnet = VAB + VBC + VCD + VDANow, electric potential due to a charge q at a distance r is given as:

V=kq/r

VBC=kq/aVDV is the electric potential difference between charges at corner D and V:

VCD=kq/(a√2)Therefore,

Vnet = VAB + VBC + VCD + VDANow substituting the values,

Vnet=kq/(a√2) - kq/a + kq/(a√2) - kq/(a√2)

=kq/(a√2) - 2kq/aPutting the value of

k = 9.0 × 10^9 N m^2 C^-2 and

q = -q, we get, Hence, the electric potential at the center of the square when charges are placed at each corner of the square is 9.0 × 10^9 N m^2 C^-2 × q [2/a - 1/a√2].

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For the circuit shown in the figure, the ideal battery has an emf ε = 60 V. four resistors have varying resistances. The rate at which heat is being generated in the resistor R4 is 180.47 Watts.

To calculate the rate at which heat is being generated in resistor R4, we can use the formula:

Power = (Current)²* Resistance

First, we need to find the current flowing through resistor R4. To do that, we'll calculate the equivalent resistance of the circuit using series and parallel combination rules.

Resistors R1 and R2 are in parallel, so their equivalent resistance is given by:

1/Requiv = 1/R1 + 1/R2

1/Requiv = 1/14 Ω + 1/21 Ω

1/Requiv = 3/42 Ω + 2/42 Ω

1/Requiv = 5/42 Ω

Requiv = 42/5 Ω

Resistors R3 and R4 are also in parallel, so their equivalent resistance is given by:

1/Requiv' = 1/R3 + 1/R4

1/Requiv' = 1/21 Ω + 1/14 Ω

1/Requiv' = 2/42 Ω + 3/42 Ω

1/Requiv' = 5/42 Ω

Requiv' = 42/5 Ω

Now, the equivalent resistance of R1, R2, R3, and R4 in series is:

Req = Requiv + Requiv'

Req = 42/5 Ω + 42/5 Ω

Req = 84/5 Ω

Now, we can calculate the current (I) flowing through the circuit using Ohm's Law:

I = ε / Req

I = 60 V / (84/5 Ω)

I = 60 V * (5/84 Ω)

I = 300/84 A

I ≈ 3.57 A

Finally, we can calculate the rate at which heat is being generated in resistor R4:

Power = I² * R4

Power = (3.57 A)² * 14 Ω

Power ≈ 180.47 W

Therefore, the rate at which heat is being generated in resistor R4 is approximately 180.47 Watts.

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The magnitude of the electric field at 5.0 cm from the wire is 5,000 N/C. The correct option is c) 5,000 N/C.

Given that the charges on any 2.0 cm long segment of the wire is measured, it shows 0.56nC,

To find the electric field at a distance of 5.0 cm from the wire, we have to use the formula

Electric field E= kQ/r

where, r=5.0cm=0.05m

Q=charge=0.56n

C=5.6 x 10⁻¹⁰C

k= Coulomb's constant= 9 × 10⁹ N.m²/C²

Now we can find E, the magnitude of electric field

E= kQ/r

=9 × 10⁹ N.m²/C² × 5.6 × 10⁻¹⁰ C/0.05 m

= 1.008 × 10⁻⁴ N/C

= 0.0001008 N/C≈ 5,000 N/C

So, the magnitude of the electric field at 5.0 cm from the wire is 5,000 N/C.

Therefore, the correct option is c) 5,000 N/C.

Thus, we have found the magnitude of the electric field at 5.0 cm from the wire using the formula E= kQ/r, where Q=charge, r=distance and k=Coulomb's constant.

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A force of 4.7 x 10^6 N is applied to the top face of an aluminum cube, causing it to move 0.3083 m to the right.

The distance the top face of the cube moves relative to the stationary bottom face is given by:

x = \frac{F \cdot d}{G \cdot A}

where:

F is the force applied to the top face of the cube

d is the side length of the cube

G is the shear modulus of aluminum

A is the area of the top face of the cube

The shear modulus of aluminum is 7×10 10 N/m^2, the side length of the cube is 4.400×10 −1 m, and the area of the top face of the cube is 4.400×10 −1 ×4.400×10 −1

=1.936×10 −2 m^2.

Plugging in these values, we get:

x = \frac{4.7 \times 10^{6} \cdot 4.400 \times 10^{-1}}{7 \times 10^{10} \cdot 1.936 \times 10^{-2}}

x = 0.3083 m

Therefore, the top face of the cube moves 0.3083 meters to the right.

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the percent error is 226.90%

The graph between velocity (v) and √h is linear with a slope of √2g and an intercept of zero for Torricelli’s law. Here is how to calculate the slope of the linear graph and the experimental value for g using the data table provided:

Data Table:Height, h ± 1 (cm)Speed v ±0.2 (m/s)1004.21205.01405.21605.61806.02006.42206.82407.0Plot the graph with velocity (v) on the y-axis and the square root of height (√h) on the x-axis. Calculate the slope of the linear graph using the formula:

slope = Δv/Δ(√h)Where Δv is the change in velocity and Δ(√h) is the change in the square root of height. To calculate Δv and Δ(√h), use the following formulae:Δv = v2 - v1Δ(√h) = (√h2 - √h1)

Slope: Using the data table, let's calculate the values of Δv and Δ(√h) for the first two rows:Δv = v2 - v1= 5.0 - 4.2= 0.8Δ(√h) = (√h2 - √h1)= (√120 - √100)= 2.9154759 - 2.8284271= 0.0870488

Now calculate the slope:slope = Δv/Δ(√h)= 0.8/0.0870488= 9.1932936 g experimental: Now use the slope to calculate an experimental value for g:√2g = slopeg = (slope/√2)²= (9.1932936/√2)²= 32.0862589 m/s²Percent error: The experimental value for g is 32.0862589 m/s², while gtheor is 9.81 m/s². T

he percent error is calculated as follows:percent error = (|gtheor - gexperimental|/gtheor) × 100%= (|9.81 - 32.0862589|/9.81) × 100%= 226.90%Therefore, the percent error is 226.90%. .

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The angle that the blade has turned is approximately 598.8 radians.

(a) To find the angular acceleration, formula used is:

Angular acceleration (α) = Change in angular velocity / Time

Given:

Initial angular velocity (ωi) = 0 rad/s (since the blade starts from rest)

Final angular velocity (ωf) = 200 rad/s

Time (t) = 6.00 s

Angular acceleration (α) = (ωf - ωi) / t

α = (200 rad/s - 0 rad/s) / 6.00 s

α = 200 rad/s / 6.00 s

α ≈ 33.3 rad/s²

Therefore, the angular acceleration of the circular saw blade is approximately 33.3 rad/s².

(b) To find the angle that the blade has turned, formula used is:

θ = ωi * t + (1/2) * α * t²

Given:

Initial angular velocity (ωi) = 0 rad/s

Time (t) = 6.00 s

Angular acceleration (α) = 33.3 rad/s²

[tex]θ = 0 rad/s * 6.00 s + (1/2) * 33.3 rad/s² * (6.00 s)²θ = 0 rad + (1/2) * 33.3 rad/s² * 36.00 s²θ = 0 rad + 0.5 * 33.3 rad * 36.00θ = 0 + 0.5 * 33.3 * 36.00 radθ ≈ 598.8 rad[/tex]

Therefore, the angle bearing the blade has turned is approximately 598.8 radians.

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To write a laboratory report to prove conservation of momentum, one should use the simulation to conduct the experiment and collect data for this section.

Conservation of momentum can be proven using a simulation in a laboratory report. To carry out this experiment, you should use a simulation and collect data for this section. The simulation will enable you to gather data that will be used to determine the velocity, momentum, and collision of objects. When conducting this experiment, one should take note of the masses of the objects involved in the collision as it affects the momentum of the system.

The conservation of momentum states that in a closed system, the total momentum of the system is conserved before and after the collision. It means that the total momentum of the system is constant even after the collision. To prove conservation of momentum, you should compare the total momentum before and after the collision. If they are equal, then the conservation of momentum holds true. A laboratory report should be written detailing the experiment's purpose, materials, procedures, data, observations, and conclusions.

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(a) Calculation of resistance: Resistance = Power/ (Voltage)²= 1000/ (230)²= 0.019 Ω (b) Calculation of power: Given: V = 210VResistance = 0.019 Ω Using Ohm's law, Current I = V/R= 210/0.019= 11053.3 A Using P = VI= 210 x 11053.3= 2.33 x 10⁶W= 2.33 MW (approx).  Therefore, the drift velocity is 4.18 x 10⁻⁵ m/s.

(Note: the resistance of the electric fire does not change because the change in voltage is not too drastic).Part 2: Calculation of the mean time between collisions: Given: Resistivity ρ = 1.7 x 10⁻⁸ Ωm Electron density n = 8.5 x 10²⁸ m⁻³Electric field E = 0.5 V/m Mass of electron m = 9.11 x 10⁻³¹ kg, The equation for mean time between collisions isτ = m/ (ne²ρ)where e is the electron charge. Substituting the given values in the above equation,τ = (9.11 x 10⁻³¹)/(8.5 x 10²⁸ x e² x 1.7 x 10⁻⁸)Since e = 1.6 x 10⁻¹⁹ C,τ = 4.74 x 10⁻¹⁴ s

Part 3: Calculation of the drift velocity: Given: Electric field E = 0.5 V/m Mass of electron m = 9.11 x 10⁻³¹ kg Using the equation, The drift velocity is vd = (eEτ)/m where e is the electron charge. Substituting the given values in the above equation, vd = (1.6 x 10⁻¹⁹ x 0.5 x 4.74 x 10⁻¹⁴)/9.11 x 10⁻³¹vd = 4.18 x 10⁻⁵ m/s Therefore, the drift velocity is 4.18 x 10⁻⁵ m/s.

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