Answer:Current across 10ohm, I =0.59 A
Explanation:
Total Resistance n parallel s given as
1/Rt = 1/R1 + 1/R2 + 1/R3
1/Rt = 1/4 + 1/6 + 1/10
1/Rt = 0.51666
Rt = 1/0.51666
Rt = 1.935 ohms
which is now connected In series with a ohm resstor
Rnet in series = R1+ R2
Rnet = 1.935 + 2.0 = 3.935 ohms
Since a voltage of 12 V s connected with the resistors in series, we have that the net current flowing across the 2ohm and 1.935ohm will be
Current, I = Voltage, V/ Resistance, R net
I= I2 / 3.935= 3.049 A
the Voltage across the 1.935 ohm resistance
V= 1R
3.049 X 1.935 = 5.899 Volts
remember that 1.935 ohm s the resultant resistance connected in parallel, therefore The voltage across the 1.935 ohm is same for the 10 ohm, 4ohm and 6ohm resistances
Therefore, the current passing through the 10ohm resistance will be
Current across 10ohm, I = V/R = 5.899/10 = 0.5899 Ampere ≈0.59 A
Question 2: If a runner travels 50 m in 5 s, his acceleration is
1 point
250 m/s/s
50 m/s/s
10 m/s/s
2 m/s/s
An athlete swings a 6.50-kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.900 m at an angular speed of 0.700 rev/s. (a) What is the tangential speed of the ball
Answer:
v = 3.951 m/s
Explanation:
Given that,
Mass of a ball, m = 6.5 kg
Radius of the circle, r = 0.9 m
Angular speed of the ball, [tex]\omega=0.7\ rev/s=4.39\ rad/s[/tex]
Let v is the tangential speed of the ball. It is given in terms of angular speed is follows :
[tex]v=r\omega\\\\v=0.9\times 4.39\\\\v=3.951\ m/s[/tex]
So, the tangential speed of the ball is 3.951 m/s.
A dentist’s drill starts from rest. After 7.28 s of constant angular acceleration it turns at a rate of 16740 rev/min. Find the drill’s angular acceleration. Answer in units of rad/s 2 .
Answer:
[tex]\alpha =240.79\ rad/s^2[/tex]
Explanation:
Given that,
Initial angular velocity, [tex]\omega_i=0[/tex]
Final angular velocity, [tex]\omega_f=16740\ rev/min = 1753\ rad/s[/tex]
We need to find the angular acceleration of the drill. It is given by the formula as follows :
[tex]\alpha =\dfrac{\omega_f-\omega_i}{t}\\\\\alpha =\dfrac{1753-0}{7.28}\\\\\alpha =240.79\ rad/s^2[/tex]
So, the angular acceleration of the drill is [tex]240.79\ rad/s^2[/tex].
A particle starts from rest and is acted on by a net force that does work at a rate that is proportional to the time t. The speed of the particle is proportional to:_____
a. √t
b. t
c. t^2
d.1/√t
e. 1/t
Answer:
The speed of the particle is proportional to:
c. t²
Explanation:
Given;
initial velocity of the particle, u = 0
let the net force on the object = F
Work done on the particle is given by;
W = F x d
W ∝ t
[tex]Fd \ \alpha \ t\\\\\frac{mv}{t}d \ \ \alpha \ t\\\\mvd \ \ \alpha \ t^2\\\\v \ \ \alpha \ t^2[/tex]
Therefore, the speed of the particle is proportional to t²
The speed of the particle is proportional to t.
It should be noted that work is a change in kinetic energy. Therefore, power will be calculated as:
Power = Work / Time = force × velocity
Power is proportional to t. Since force is a constant, then the velocity will be proportional to t. Therefore, the speed of the particle is proportional to t.
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A 54.0-kg ice skater is moving at 3.94 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.900 m around the pole. (a) Determine the magnitude of the force exerted by the horizontal rope on her arms. .931416 Correct: Your answer is correct. kN (b) Compare this force with her weight. Frope W
Answer:
Explanation:
a ) magnitude of force = centripetal force = m v² / R , m is mass , v is velocity of skater , R is radius of circular path .
= 54 x 3.94² / .9
= 931.41 N
= .93141 kN.
b ) Her weight = mg = 54 x 9.8 = 529.2 N
= .5292 kN .
Ratio = .93141 / .5292 = 1.76
What constant acceleration is required to bring a rocket to an altitude of 650 m [up] from rest in 7.2 s? What is the rocket’s velocity at that point?
Answer:
a = 15.68 [m/s²]; v = 112.92 [m/s]
Explanation:
To solve this problem we must use the combination of two interesting topics of physics, the principle of energy conservation and then the application of kinematics.
The principle of Energy Conservation tells us that kinetic energy is transformed into potential energy or vice versa. For this particular problem, we must imagine the rocket at a height above 650 [m], suddenly this rocket falls to the ground. We will propose the reference point of potential energy at ground level, at this point the potential energy is equal to zero.
[tex]E_{pot}=E_{kin}[/tex]
[tex]E_{pot}=m*g*h\\E_{kin}=\frac{1}{2}*m*v^{2}[/tex]
where:
m = mass of the rocket [kg]
g = gravity acceleration = 9.81[m/s²]
v = velocity of the rocket [m/s]
m*g*h = 0.5*m*v²
9.81*650 = 0.5*v²
v = √((6376.5)/0.5)
v = 112.92 [m/s]
Now using the following kinematic equation we have:
[tex]v_{f} = v_{o} + (a*t)[/tex]
where:
Vf = final velocity = 112.92 [m/s]
Vo = initial velocity = 0 (at the begining the rocket is at rest)
a = rocket acceleration [m/s²]
t = time = 7.2 [s]
Note: in the kinematics equation the positive sign of acceleration means that the rocket accelerates in the direction of motion, i.e. activates its thrusters to descend and descend to 650 [m] in 7.2 [s].
112.92 = 0 + (a*7.2)
a = 15.68 [m/s²]
What is conserved in physical changes?
O shape
O energy
O mass
Odensity
A 3.00-kg block starts from rest at the top of a 35.5° incline and slides 2.00 m down the incline in 1.45 s. (a) Find the acceleration of the block.
Answer:
0.344 m/s^2
Explanation:
The first step is to calculate V
V= distance /time
= 2/1.45
= 1.379 m/s
Therefore the acceleration of the block can be calculated as follows
a= V - U/2d
= 1.379 - 0/2×2
= 1.379/4
= 0.344 m/s^2
A 80-kg person stands at rest on a scale while pulling vertically downwards on a rope that is above them. Use g = 9.80 m/s2.
With what magnitude force must the tension in the rope be pulling on the person so that the scale reads 25% of the persons weight?
Answer:
The tension in the rope is 588 N
Explanation:
I have attached a free body diagram of the problem.
First we calculate the force applied downwards by the person's weight:
[tex]F_{weight} =m*a = 80 kg * 9.8 \frac{m}{s^{2}} = 784 N[/tex]
If the scale reads a 25% of the person' weight:
[tex]F_{Scale} = F_{weight}*0.25=784N*0.25=196N[/tex]
which is the same value (but opposite) to the effective weight of the persons ([tex]F_{Person}=-F_{Scale}[/tex])
The other 75% of the total weight is the force of the rope pulling the person upwards.
[tex]F_{Pull}= F_{weight}*0.75=784N*0.75=588N[/tex]
To verify, we can use 1st Newton's law
∑F = 0
[tex]F_{Rope}-F_{Pull}+F_{Person}-F_{Scale}=0 \\588N-588N+196N-196N=0[/tex]
PLEASE HELP!!!!
An object is launched horizontally from a cliff. The cliff is 80 m high and the object has an initial launch velocity of 50 m/s.
What is the initial horizontal velocity?
What is the initial vertical velocity?
What is the final horizontal velocity?
How much time did it take for the object to hit the ground?
What is the final vertical velocity?
What is the final resultant speed?
How far from the base of the cliff will the projectile land?
Answer:
a. 50 m/s b. 0 m/s c. 50 m/s d. 4.04 s e. -39.6 m/s f. 63.78 m/s g. 202 m
Explanation:
a. What is the initial horizontal velocity?
Since the object is launched horizontally, it initial horizontal velocity is 50 m/s
b. What is the initial vertical velocity?
Since the object is launched horizontally, it has no initial vertical component. So, its initial vertical velocity is 0 m/s
c. What is the final horizontal velocity?
Its final horizontal velocity is 50 m/s since no force acts on it in the horizontal direction to change its value.
d. How much time did it take for the object to hit the ground?
We use the equation s = ut - 1/gt² since the object is falling under gravity where u = initial vertical velocity = 0 m/s, s = height of cliff = 80 m, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the object to hit the ground.
s = ut - 1/2gt²
80 m = 0 × t - 1/2 × -9.8 m/s² × t²
80 m = 4.9 m/s² × t²
t² = 80 m ÷ 4.9 m/s²
t² = 16.33 s²
t = √(16.33 s²)
t = 4.04 s
e. What is the final vertical velocity?
Using v = u + at where u = initial vertical velocity = 0 m/s, v = final vertical velocity, a = acceleration = -g = -9.8 m/s²and t = time it takes object to reach the ground = 4.04 s.
Substituting these values into the equation, we have
v = u + at
v = 0 m/s + (-9.8 m/s²) × 4.04 s
v = -39.6 m/s
f. What is the final resultant speed?
The final resultant speed v' is the resultant of the final horizontal velocity and the final vertical velocity. Let u' = final vertical velocity = 50 m/s.
v' = √(u'² + v²)
v' = √((50 m/s)² + (-39.6 m/s)²)
v' = √(2500 m²/s² + 1568.16 m²/s²)
v' = √(4068.16 m²/s²)
v' = 63.78 m/s
g. How far from the base of the cliff will the projectile land?
The distance from the base of the cliff, d where the projectile lands is
d = u't where u' = horizontal velocity = 50 m/s and t = time it takes object to land = 4.04 s
d = 50 m/s × 4.04 s
d = 202 m
In 2 to 3 sentences explain the roles of chloroplast and mitochondria in the process of plant cells gaining energy.
Answer:
Explanation:
Mitochondria is the powerhouse of the cell. And Chroloplast helps plants capture energy
Car is going from 0 to 60 miles per hour in 6 seconds what is the advertised acceleration in mph squared
Answer:
10 m/h^2
Explanation:
a=v/t
a=60/6
a=10 m/h^2
A race car accelerates from rest with a displacement of 25.0 m. If the
acceleration is 1.2 m/s2, what is the car's final velocity?
Answer:
Vf = Vi + at
t = Vf - Vi = 0.0m/s - 1.75 m/s
--------- = ------------------------- = 8.8s
a -0.20 m/s(squared)
Explanation:
This is what I got I am pretty sure it's correct but correct me if I'm wrong : )
The earth travels at 68,000 miles/hour as it moved around the sun. How many miles does the earth travel in one trip around the sun
Answer:
595,680,000 miles
(= 5.96 x 10⁸ miles)
Explanation:
we are given the speed of travel as 68,000 mi/hr
if we assume one common year
= 365 days
= 365 days x 24 hours/day
= 8760 hours
Then the distance travelled
= speed x time
= 68,000 x 8760
= 595,680,000 miles
= 5.96 x 10⁸ miles (rounded to 3 sig. fig and expressed in scientific form)
Answer:
Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).
Hope this helps!
An Indy 500 race car's velocity increases from 2 mis to 40 mis over a 40 s time interval. What is its
acceleration ?
Answer:8
Explanation:
What is centripetal force?Define it with example.
Answer:
Centripetal force is defined as, "the force that is necessary to keep an object moving in a curved path and that is directed inward toward the center of rotation,"
Explanation:
so if you put a small ball por grape and put it on a table and you take a wine glass and spin it around the grape the grape will continue spinning even if you stop spinning the glass hope this helped
;)
Answer:
A force acting on a moving body at an angle to the direction of motion, tending to make the body follow a circular or curved path. The force of gravity acting on a satellite in orbit is an example of a centripetal force; the friction of the tires of a car making a turn similarly provides centripetal force on the car.
Explanation:
4. A 5.0 kg block moving to the right at 12.0 m/sec collides with a 4.0 kg block moving to the right at 2 m/sec. If the 4.0 kg moves to the right at 10 m/sec after the collision, what will be the speed of the 5.0 kg block? How much energy was lost during the collision?
Answer:
Energy lost = 55.28Joules
Explanation:
Using the law of momentum:
m1u1+m2u2 = m1v1+m2v2
m1 and m2 are the masses of the objects
u1 and u2 are the initial velocities.
v1 and v2 are final velocities
Given
m1 = 5kg
m2 = 4kg
u1 = 12m/s
u2 = 2m/s
v1 = 10m/s
v2 = ?
Substitute
5(12)+4(2)=4(10)+5v2
60+8 = 40+5v2
68-40 = 5v2
28 = 5v2
v2 = 28/5
v2 = 5.6m/sec
Hence the speed of the 5kg block is 5.6m/s
Energy lost = Kinetic energy after collision - kinetic energy before collision
KE after collision = 1/2m1v1²+1/2m2v2²
KE after collision = 1/2(5)10²+1/2(4)(5.6)²
= 250+62.72
= 312.72Joules
KE before collision = 1/2m1u1²+1/2m2u2²
KE before collision = 1/2(5)12²+1/2(4)(2)²
= 360+8
= 368Joules
Energy lost = 368-312.72
Energy lost = 55.28Joules
A jogger runs at a speed of 4 m/s for 500]s, slows to 2.5 m/s for the next
300 s and then travels the final 900 s at a speed of 3.5 m/s. What is the
average speed of the jogger in m/s?
O 10 m/s
O 3.47 m/s
O 5 m/s
O 3.33 m/s
Answer:
3.33 m/s
Explanation:
Just add up all of the speeds and divide it by 3 (since there are 3 different speeds)
(4 m/s + 2.5 m/s + 3.5 m/s) / 3 = 3.33 m/s
1. a. Describe the type of energy that is observed when a musician's finger plucks a guitar string. does it give your ear kenetic or potential energy .
Answer:
kinetic
Explanation:
Answer:
kinetic energy.
Explanation:
Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.30 mm high. Assume it starts from rest and rolls without slipping.
Answer:
9.77 m/s
Explanation:
Since the cylinder is rolling, it will have both translational and rotational kinetic energy.
From conservation of energy, its initial mechanical energy equals its final mechanical energy.
So, K' + U' = K + U where K and U are the initial kinetic and potential energies respectively and K' and U' are the final kinetic and potential energies respectively.
So, Since the cylinder starts from rest, its initial kinetic energy K = 0, its initial potential energy = mgh where m = mass of cylinder, g = acceleration due to gravity = 9.8 m/s² and h = height of incline = 7.30 mm = 7.3 × 10⁻³ m . Its final kinetic energy K' = 1/2Iω² + 1/2mv² where I = moment of inertia of cylinder = 1/2mr² where r = radius of cylinder and v = translational velocity of cylinder. Its final potential energy U' = 0. Substituting these into the equation, we have
K' + U' = K + U
1/2Iω² + 1/2mv² + 0 = 0 + mgh
1/2(1/2mr²)ω² + 1/2mv² = mgh
1/4mr²ω² + 1/2mv² = mgh
1/4m(rω)² + 1/2mv² = mgh v = rω
1/4mv² + 1/2mv² = mgh
3/4mv² = mgh
v² = 4gh/3
v = √(4gh/3)
v = √(4 × 9.8 m/s² × 7.3 × 10⁻³ m/3)
v = (√286.16/3) m/s
v = (√95.39) m/s
v = 9.77 m/s
A dog with a mass of 57.0 kg slides down a wet slope with negligible friction. The dog starts from rest and has a speed of 2.10 m/s at the bottom. What is the height of the slope (in m)?
The dog starts from rest and has a speed of 2.10 m/s at the bottom. The height of the slope will be 0.225 m
What is friction?Friction is a type of force that resists or prevents the relative motion of two physical objects when their surfaces come in contact.
The friction force prevents any two surfaces of objects from easily sliding over each other or slipping across one another. It depends upon the force applied to the object.
As there is almost zero friction between the total potential energy height is equal to the kinetic energy
m*g*h = 1/2* m*v²
Where m is the mass of the dog
g is the acceleration due to gravity,h is the height of the slope
v is the velocity of the sliding dog
g*h = 1/2* v²
9.81*h = 0.5* 2.1²
=0.225 m
The height of the slope would be 0.225 m.
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What type of energy is possessed by a pear falling from a tree,
just before it touches the ground?
zero energy
kinetic energy
potential energy
kinetic and potential energy
Answer:
Kinetic Energy
Explanation:
I can't really explain it, but that's the correct answer.
A tired worker pushes a heavy (100-kg) crate that is resting on a thick pile carpet. The coefficients of static and kinetic friction are 0.6 and 0.4, respectively. The worker pushes with a horizontal force of 490 N. The frictional force exerted by the surface is
Answer:
The friction force exerted by the surface is 490 newtons.
Explanation:
From Physics, we remember that static friction force ([tex]f_{s}[/tex]), measured in newtons, for a particle on a horizontal surface is represented by the following inequation:
[tex]f_{s} \leq \mu_{s}\cdot m \cdot g[/tex] (Eq. 1)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.
[tex]m[/tex] - Mass of the crate, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If [tex]f_{s} = \mu_{s}\cdot m \cdot g[/tex], then crate will experiment an imminent motion. The maximum static friction force is: ([tex]\mu_{s} = 0.6[/tex], [tex]m = 100\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex])
[tex]f_{s} = (0.6)\cdot (100\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]f_{s} = 588.42\,N[/tex]
From Newton's Laws we get that current force of friction as reaction to the pulling force done by the worker on the crate is:
[tex]\Sigma F = F-f = 0[/tex]
[tex]f = F[/tex] (Eq. 2)
Where:
[tex]F[/tex] - Horizontal force done by the worker, measured in newtons.
[tex]f[/tex] - Static friction force, measured in newtons.
If [tex]F = 490\,N[/tex], then the static friction force exerted by the surface is:
[tex]f = 490\,N[/tex]
Given that [tex]f < f_{s}[/tex], the crate does not change its state of motion. The friction force exerted by the surface is 490 newtons.
A roller coaster stops at the top of a hill. What force brings the roller coaster back down to the ground?
Answer:
Simple enough that it seems like a trick question: gravity.
An object thrown straight upward reaches a height of 10 meters. If the object is thrown upward with twice this speed, it will reach a height of ____ m.
Answer:
If the object is thrown upward with twice the initial speed, it will reach a height of 40 meters.
Explanation:
Based on the Principle of Energy Conservation we find that the square of the initial speed of the object ([tex]v[/tex]), measured in meters per second, is directly proportional to its change in height ([tex]\Delta h[/tex]), measured in meters. That is:
[tex]v^{2} \propto \Delta h[/tex]
[tex]v^{2} = k\cdot \Delta h[/tex] (Eq. 1)
Where [tex]k[/tex] is the proportionality constant, measured in meters per square second.
Such constant is eliminated by using the following relationship:
[tex]\left(\frac{v_{2}}{v_{1}} \right)^{2} = \frac{\Delta h_{2}}{\Delta h_{1}}[/tex]
[tex]\Delta h_{2} = \left(\frac{v_{2}}{v_{1}} \right)^{2}\cdot \Delta h_{1}[/tex] (Eq. 2)
Where:
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speed of the object, measured in meters per second.
[tex]\Delta h_{1}[/tex], [tex]\Delta h_{2}[/tex] - Initial and final changes in height, measured in meters.
If we know that [tex]\frac{v_{2}}{v_{1}} = 2[/tex] and [tex]\Delta h_{1} = 10\,m[/tex], then the change in height is:
[tex]\Delta h_{2} = 2^{2}\cdot (10\,m)[/tex]
[tex]\Delta h_{2} = 40\,m[/tex]
If the object is thrown upward with twice the initial speed, it will reach a height of 40 meters.
When the object is thrown with double the speed, it reaches a height of 40m.
Laws of motion:Given that the height attained by the object is h = 10m.
Let the initial speed of the object be u
and its final speed will be v = 0 at the highest point.
Then from the third equation of motion, we get,
[tex]v^2=u^2-2gh\\\\0=u^2-2gh\\\\h=\frac{u^2}{2g}[/tex]
if the object is thrown at double the speed, that is u' = 2u, then the height attained will be:
[tex]h'=\frac{u'^2}{2g}\\\\h'=\frac{(2u)^2}{2g}\\\\h'=\frac{4u^2}{2g}\\\\h'=4h[/tex]
h' = 4 × 10m
h' = 40m
So, the height attained will be 40 m.
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If a force of 12 N is applied to a 4.0 kg object, what acceleration is produced?
Answer:
The answer is 3 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]a = \frac{f}{m} \\ [/tex]
where
f is the force
m is the mass
From the question we have
[tex]a = \frac{12}{4} \\ [/tex]
We have the final answer as
3 m/s²Hope this helps you
9. A 50 kg halfback is in the process of making a turn on a football field.
The halfback makes 1/4 of a turn with a radius of 15 meters in 2.1 seconds
before being tackled. What is the net force acting on the halfback before
he is tackled?
Answer:
Net force = 419.5N
Explanation:
Given the following data;
Mass = 50kg
Radius = 15m
Time = 2.1 secs
Turns = 1/4 = 0.25
In order to find the net force, we would first of all solve for the speed and acceleration of the halfback.
To find speed;
Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.
Mathematically, speed is given by the equation;
[tex]Speed = \frac{distance}{time}[/tex]
But the distance traveled is given by the circumference of a circle = [tex] 2\pi r[/tex]
Since he covered 1/4th of a turn;
[tex] Distance = 0.25 * 2 \pi *r[/tex]
Substituting into the equation;
[tex] Speed, v = \frac {(0.25*2*3.142 * 15)}{2.1}[/tex]
[tex] Speed, v = \frac {23.565}{2.1}[/tex]
Speed, v = 11.22m/s
To find acceleration;
[tex] Acceleration, a = \frac {v^{2}}{r}[/tex]
Where, v = 11.22m/s and r = 15m
Substituting into the equation, we have;
[tex] Acceleration, a = \frac {11.22^{2}}{15}[/tex]
[tex] Acceleration, a = \frac {125.8884}{15}[/tex]
Acceleration, a = 8.39m/s²
To find the net force;
Force is given by the multiplication of mass and acceleration.
Mathematically, Force is;
[tex] F = ma[/tex]
Where;
F represents force.m represents the mass of an object.a represents acceleration.Substituting into the equation, we have;
[tex] F = 50 * 8.39[/tex]
F = 419.5N
Therefore, the net force acting upon the halfback is 419.5 Newton.
A pair of equal-length vectors at right angles to each other have a resultant. If the angle between the vectors is less than 90°, their resultant is
Answer:
Larger in magnitude
Explanation:
As the vectors get closer reducing the 90 degree angle, based on the "parallelogram rule" for vector addition, the magnitude of the resultant vector gets larger than when they are forming a 90 degree angle.
If 2000 ft³ of air is crossing an evaporator coil and is cooled from 80°F to 60°F, what would be the volume of air in ft³ exiting the evaporator coil?
Given :
If 2000 ft³ of air is crossing an evaporator coil and is cooled from 80°F to 60°F.
To Find :
The volume of air in ft³ exiting the evaporator coil.
Solution :
We know, relation between initial and final temperature and volume is given by :
[tex]\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}[/tex]
Here, temperature is in Kelvin .
We know, Kelvin temperature is given by :
[tex]K=(F-32)\times \dfrac{5}{9} + 273[/tex]
Putting F = 60° F,
[tex]K=(60-32)\times \dfrac{5}{9} + 273\\\\K = 288.56\ K[/tex]
Putting F = 80° F,
[tex]K=(80-32)\times \dfrac{5}{9} + 273\\\\K = 299.67\ K[/tex]
Putting all values, we get :
[tex]\dfrac{V_1}{2000}=\dfrac{60}{80}\\\\V_1=\dfrac{60\times 2000}{80}\ ft^3\\\\V_1=1500\ ft^3[/tex]
Hence, this is the required solution.
A student throws a water bottle upward into the air. It takes 1.2 seconds for the bottle to reach it's
maximum height. How much time will it take for the water bottle to return to the student's hand
(from its maximum height)?
Answer:
1.2 seconds
Explanation:
Formula for total time of flight in projectile is; T = 2u/g
Now, for the time to reach maximum height, it is gotten from Newton's first equation of motion;
v = u + gt
t is time taken to reach maximum height.
But in this case, g will be negative since motion is against gravity. Final velocity will be 0 m/s at max height.
Thus;
0 = u - gt
u = gt
t = u/g
Putting t for u/g in the equation of total time of flight, we have;
T = 2t
We are given t = 1.2
Thus, T = 2 × 1.2 = 2.4
Since total time of flight is 2.4 s and time to get to maximum height is 1.2 s, then it means time to fall from maximum height back to the students hand is; 2.4 - 1.2 = 1.2 s
