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Sav [38]
3 years ago
10

Two balls of equal mass at the bottom of an incline are rolled upward without slipping at the same initial velocity. one ball is

solid and the other is a thin-walled hollow ball. which rolls higher up the incline before coming to a stop?
Physics
1 answer:
Andrei [34K]3 years ago
6 0
The appropriate response would be the hollow ball.The purpose for it has more kinetic energy at the last (2/3 as opposed to 2/5) so it will have more potential energy (more height) at the best. Pondering it, more inertia implies its less impervious to change meaning the hollow ball with more inertia wont need to stop as fast.
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Question 4 of 10
kiruha [24]
B hey what do u know i took that test to
5 0
3 years ago
a skinny girl devours food that is equivalent to 600 kcal, as she is anorexic she repents and wants to lose twice as many calori
blondinia [14]

Answer:

Explanation:

Work done in lifting the weight once = mgh

= 20 x 9.8 x (1.9+1.7)

= 705.6 J  

= 705.6 / 4.2 calorie

= 168 cals

Total energy to be spent = 600 x 10³ cals

No of times weight is required to be lifted

= 600 x 10³ / 168

= 3.57  x 10³ times

Total time to be taken = 2 x 3.57 x 10³

= 7.14 x 10³ s

=119 minutes .

4 0
3 years ago
You are driving at 50 miles per hour. If you decrease the time it takes you to travel 1 mile by 8 seconds, what is your new spee
Dmitriy789 [7]

Answer:

The new speed is 56.25 miles/hour.

Explanation:

Since speed = distance/time;

time = distance/speed.

While driving at 50 miles/hour, time taken for one to complete 1 mile is (1/50) hour

(1/50) hour = (1/50) × 3600s = 72 seconds.

So, if this time to complete 1 mile (72 seconds) is reduced by 8 seconds,

New time to complete 1 mile will be = 72 - 8 = 64 seconds = (64/3600) hour = 0.0178 hour

New speed would be = (1 mile/64 seconds) = (1 mile/0.0178 hour) = 56.25 miles/hour.

Hope this Helps!!!

7 0
4 years ago
A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by υ⃗ =[5.00m/
Reptile [31]
It would possibly be [7.00m/s(0.5613m/s)
8 0
3 years ago
A 0.5 kg mass on a spring undergoes simple harmonic motion with a total mechanical energy of 12 J. If the oscillation amplitude
Darya [45]

Answer:

The frequency of the oscillation is 2.45 Hz.

Explanation:

Given;

mass of the spring, m = 0.5 kg

total mechanical energy of the spring, E = 12 J

Determine the spring constant, k as follows;

E = ¹/₂kA²

kA² = 2E

k = (2E) / (A²)

k = (2 x 12) / (0.45²)

k = 118.519 N/m

Determine the angular frequency, ω;

\omega = \sqrt{\frac{k}{m} } \\\\\omega =  \sqrt{\frac{118.519}{0.5} } \\\\\omega = 15.396 \ rad/s

Determine the frequency of the oscillation;

ω = 2πf

f = (ω) / (2π)

f = (15.396) / (2π)

f = 2.45 Hz

Therefore, the frequency of the oscillation is 2.45 Hz.

8 0
3 years ago
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