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SVEN [57.7K]
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
Physics
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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You stand on a bathroom scale in a moving elevator. what happens to the scale reading if the cable holding the elevator suddenly
Viefleur [7K]

A bathroom scales works due to gravity. Under normal conditions, a reading can be obtained when your body is pushing some force on the scale. However in this case, since you and the scale are both moving downwards, so your body is no longer pushing on the scale. Therefore the answer is:

<span>The reading will drop to 0 instantly</span>

7 0
2 years ago
Read 2 more answers
A 0.600-mm diameter wire stretches 0.500% of its length when it is stretched with a tension of 20.0 n. what is the young's modul
Rashid [163]
The Young modulus is given by:
E= \frac{F /A}{\Delta L / L_0}
where
F is the force applied
L_0 is the initial length of the wire
A is the cross-sectional area of the wire
\Delta L is the stretch of the wire

The wire in the problem stretches by 0.5% of its length, this means 
\frac{\Delta L}{L_0}  = 0.005

We can also calculate the area of the wire; its radius is in fact half the diameter:
r= \frac{d}{2}= \frac{0.600 mm}{2}=0.300 mm=0.3 \cdot 10^{-3} m
and so the area is
A=\pi r^2 = \pi (0.3 \cdot 10^{-3} m)^2 = 2.83 \cdot 10^{-7} m^2

We know the force applied to the wire, F=20 N, so now we have everything to calculate the Young modulus:

E=  \frac{F/A}{\Delta L / L_0} = \frac{20 N/(2.83 \cdot 10^{-7} m^2)}{0.005}=1.42 \cdot 10^{10} N/m^2
3 0
2 years ago
Solve A and B using energy considerations.
Alisiya [41]
 <span>Use the kinematic equation vf^2 = vi^2 + 2ad where; 
vf = ? 
vi = 0 m/s 
a = 9.8 m/s^2 
d1 = 10 m 
d2 = 25 m 

final velocity at the ground (d1): vf = sqrt(2)(9.8)(10) = 14 m/s 
final velocity to the bottom of the cliff (d2): vf = sqrt(2)(9.8)(25) = 22.14 m/s 
</span>
7 0
2 years ago
Suppose that now you want to make a scale model of the solar system using the same ball bearing to represent the sun. How far fr
Dahasolnce [82]

Answer:

d = 0.645 m <em>(assuming a radius of the ball bearing of 3 mm)</em>

Explanation:

<u>The given information is:</u>

  • <em>The distance from the center of the sun to the center of the earth is 1.496x10¹¹m = d_{e}</em>
  • <em>The radius of the sun is 6.96x10⁸m = r_{s}</em>

<u>We need to assume a radius for the ball bearing, so suppose that the radius is 3 mm = r_{b}</u>.  

First, we need to find how many times the radius of the sun is bigger respect to the radius of the ball bearing, which is given by the following equation:

\frac{r_{s}}{r_{b}} = \frac{6.96\cdot 10^{8}m}{3\cdot 10^{-3}m} = 2.32\cdot 10^{11}

Now, we can calculate the distance from the center of the sun to the center of the sphere representing the earth, d_{s}:  

[tex] d_{s} = \frac{d_{e}}{r_{s}/r_{b}} = \frac{1.496 \cdot 10^{11} m}{2.32\cdot 10^{11}} = 0.645 m

I hope it helps you!

7 0
2 years ago
it possible that the net kinetic energy for two objects be zero while the net momentum is zero? Explain.
svp [43]
Of course. That's what you have when both objects are at rest. I'm guessing that you left a word out of the question, and it actually says that the net kinetic energy is NOT zero. In that case, the answer is still 'yes', but you have to think about it for a second.
4 0
2 years ago
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