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2 years ago

9

A car is traveling in a race.The car went from initial velocity of 35m/s to the final velocity of 65m/s in 5 seconds what was th

e acceleration

2 answers:

I am Lyosha [343]2 years ago

8 0

Acceleration is the change in velocity divided by time. The change in velocity is -30m/s and time is 5s. If you divide -30m/s by 5s, you get -6m/s<span>².</span>

OverLord2011 [107]2 years ago

6 0

Answer:

Acceleration, $a=6\ m/s^2$

Explanation:

Given that,

Initial speed of the car, u = 35 m/s

Final speed of the car, v = 65 m/s

Time taken, t = 5 s

We need to find the acceleration of the car. It is given by :

$a=\dfrac{v-u}{t}$

$a=\dfrac{65-35}{5}$

$a=6\ m/s^2$

So, the acceleration of the car is $6\ m/s^2$ . Hence, this is the required solution.

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A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s

Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8 $^{-3}$

Explanation:

$m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg$

$v_{bullet}=341\frac{m}{s}$

Momentum of the motion the first part of the motion have a momentum that is:

$P_{1}=m_{bullet}*v_{bullet}$

$P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529$

The final momentum is the motion before the action so:

a).

$P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}$

$P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}$

$P_{1}=P_{2}$

$2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}$

b).

kinetic energy

$K=\frac{1}{2}*m*(v)^{2}$

Kinetic energy after

$Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J$

Kinetic energy before

$Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J$

Ratio = $\frac{Ka}{Kb}$

$R=\frac{1.14}{406.4}\\R=2.8x10^{-3}$

3 0

3 years ago

The collision between a hammer and a nail can be considered to be approximately elastic. estimate the kinetic energy acquired by

Setler [38]

Here we can use momentum conservation as in this type of collision there is no external force on it

$m_1v_{1i} + m_2v_{2i} = m_1 v_{1f} + m_2v_{2f}$

now here we can say

$m_1 = 10 g$

$v_{1i} = 0$

$m_2 = 550 g$

$v_{2i} = 3.5 m/s$

now here we can say

$10*0 + 550 * 3.5 = 10 v_{1f} + 550 v_{2f}$

$192.5 = v_{1f} + 55 v_{2f}$

now by coefficient of restitution

for elastic collision we know that e = 1

$v_{2f} - v_{1f} = e(v_{1i} - v_{2i})$

$v_{2f} - v_{1f} = 0 - 3.5$

now by solving the two equation

$56v_{2f} = 189$

$v_{2f} = 3.375 m/s$

also we know that

$v_{1f} = v_{2f} + 3.5 = 3.375 + 3.5 = 6.875 m/s$

so final speed of the nail is 6.875 m/s

6 0

2 years ago

Read 2 more answers

You adjust the temperature so that a sound wave travels more quickly through the air. You increase the temperature from 30 degre

katen-ka-za [31]

To calculate the velocity of the sound wave, we use this formula:
V = 331 + [0.6*T],
Where V is the velocity and T represents temperature.
When the temperature is 36 degree Celsius, we have
V = 331 + [0.6 * 36]
V = 331 + 21.6 = 352.6
Therefore, V = 352.6 m/s.

7 0

2 years ago

Read 2 more answers

What is the difference between kinetic and potential energy and how do they work?

Iteru [2.4K]

To explain, I will use the equations for kinetic and potential energy:

$PE = mgh\\KE = \frac{1}{2}mv^{2}$

<h3>Potential energy </h3>

Potential energy is the potential an object has to move due to gravity. An object can only have potential energy if 1) <u>gravity is present</u> and 2) <u>it is above the ground at height h</u>. If gravity = 0 or height = 0, there is no potential energy. Example:

An object of 5 kg is sitting on a table 5 meters above the ground on earth (g = 9.8 m/s^2). What is the object's gravitational potential energy? <u>(answer: 5*5*9.8 = 245 J</u>)

(gravitational potential energy is potential energy)

<h3>Kinetic energy</h3>

Kinetic energy is the energy of an object has while in motion. An object can only have kinetic energy if the object has a non-zero velocity (it is moving and not stationary). An example:

An object of 5 kg is moving at 5 m/s. What is the object's kinetic energy? (<u>answer: 5*5 = 25 J</u>)

<h3>Kinetic and Potential Energy</h3>

Sometimes, an object can have both kinetic and potential energy. If an object is moving (kinetic energy) and is above the ground (potential), it will have both. To find the total (mechanical) energy, you can add the kinetic and potential energies together. An example:

An object of 5 kg is moving on a 5 meter table at 10 m/s. What is the objects mechanical (total) energy? (<u>answer: KE = .5(5)(10^2) = 250 J; PE = (5)(9.8)(5) = 245 J; total: 245 + 250 = 495 J</u>)

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2 years ago

brick, dry soil, paper, and water. If all four substances were exposed to sunlight for the same amount of time, which substance

denis-greek [22]

Water because the light is able to enter the water and allow heat to enter faster.

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3 years ago