What is the kinetic energy in joules of a 0.05. kg bullet traveling 310 m/s

(V) What are the objectives of the space mission?

Svetach [21]

Explanation:

I hope this helps, I got this from another person

3 0

3 years ago

What distance will be traveled if you are going 120km/hr for 30 min? steps please

ArbitrLikvidat [17]

D= Speed x Time
Then you convert 120km/hr to m/s depends in wich unifs you want them
120/hr to m/s it's 33.3 m/s and then 30min to seconds it's 1800
D= 33.3m/s(1800s)
D= 59940m

8 0

3 years ago

Object A has 25 J of kinetic energy. Object B has one-quarter the mass of object A.By what factor does the speed of each object

Natalka [10]

Answer:

The speed of object B is 2 times of speed of object A.

Explanation:

Given that,

Kinetic energy of object A = 25 J

Mass of object B $m_{B}=\dfrac{1}{4}m_{A}$

Work done = -20 J

We need to calculate the factor of the speed of the object

Suppose the final kinetic energy is same for both object.

$K.E_{f_{A}}=K.E_{f_{B}}$

$\dfrac{1}{2}m_{A}v_{A}^2=\dfrac{1}{2}m_{B}v_{B}^2$

Put the value into the formula

$\dfrac{1}{2}m_{A}\times v_{A}^2=\dfrac{1}{2}\times\dfrac{1}{4}m_{A}\times v_{B}^2$

$\dfrac{v_{A}}{v_{B}}=\dfrac{1}{2}$

$v_{A}=\dfrac{1}{2}v_{B}$

$v_{B}=2v_{A}$

Hence, The speed of object B is 2 times of speed of object A.

4 0

3 years ago

Read 2 more answers

12*3a car is stopped at a traffic light. it then travels along a straight road so that its distance from the light is given by w

Digiron [165]

Missing parts in the text of the exercise:
- The distance from the traffic light is given by $x(t) = bt^2 -ct^3$ , where $b=2.4 m/s^2$ and $c=0.13 m/s^3$

Solution:

part a) calculate the average velocity of the car for the time interval t=0 to t=8.0 s
- The average velocity is given by the ratio between the distance covered in the time interval:
 $v_{ave} = \frac{x(8.0 s)-x(0 s)}{8.0 s-0 s}$
x(0 s), the distance covered after t=0 s, is zero, while the distance after t=8.0 s is
$x(8.0 s)=b(8 s)^2-c(8 s)^3 = (2.4 m/s^2)(8 s)^2 -(0.13 m/s^3)(8 s)^3=$
$=87.04 m$
Therefore, the average velocity is
$v_{ave}= \frac{x(8.0 s)-x(0)}{8.0s-0}= \frac{87.04 m}{8.0s}= 10.88 m/s$

part b) calculate the instantaneous velocity of the car at t=0, t=4.0 s and t=8.0 s
- The instantaneous velocity can be found by performing the derivation of x(t):
 $v(t) = \frac{dx(t)}{dt}=2bt-3ct^2$
So now we just have to substitute t=0, t=4 s and t=8 s:
- t=0: v(0)=0
- t=4 s: $v(4.0 s)=2(2.4 m/s^2)(4.0 s)-3(0.13 m/s^2)(4.0s)^2=12.96 m/s$
- t=8 s: $v(8.0 s)=2(2.4 m/s^2)(8.0 s)-3(0.13 m/s^2)(8.0s)^2=13.44 m/s$

part c) how long after starting from rest is the car again at rest?
- To solve this part we must find the value of t for which v(t)=0, so:
 $2bt-3ct^2=0$
$t(2b-3ct)=0$
The first solution is t=0 s, which corresponds to the beginning of the motion, so we are not interested in this value. The second solution is
 $t= \frac{2b}{3c}= \frac{2\cdot 2.4 m/s^2}{3 \cdot 0.13 m/s^3}=12.31 s$
and this is the time at which the car is at rest again.

8 0

3 years ago

"Which of the following is most like the rotation" of stars in the disk of the Milky Way? cars moving at a constant speed on a c

xxTIMURxx [149]

Answer:

cars moving at a constant speed on a circular race track.

Explanation:

The rotation of stars in the disk of milky way is more like that of cars moving at a constant speed on a circular race track.

That is all stars are rotating at fixed point in a fixed orbit , such that no two stars are crossing each other. Speed of all stars are different from each other but all stars are rotating at a constant speed.

4 0

3 years ago