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Goshia [24]
3 years ago
11

Consider a keen little boy who is having a wagon race with a friend. He starts from rest and

Physics
1 answer:
MAXImum [283]3 years ago
7 0
A) His wagon will accelerate more.
B) His wagon will accelerate less. Both parts are answered by F=ma. Mass is inversely proportional to acceleration, and force is directly proportional to acceleration.
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What force does a trampoline have to apply to a 45.0-kg gymnast to accelerate her straight up at 7.50 m/s^2? (a) 104N (b) 338 N
Brilliant_brown [7]

Answer:

b) 338 N

Explanation: let m be the mass of the gymnast and a be the acceleration of the gymnast.

the force required to accelerate the gymnast is given by:

F = m×a

  = (45.0)×(7.50)

  = 337.5 N

Therefore, the force a trampoline has to apply is 138 N.

6 0
3 years ago
What is an example of a stable system
oksian1 [2.3K]
Some examples of stable system are:

1) functions of sine

2) DC

3) signum

4) unit step 

5) cosine.

Happy Studying! ^^


6 0
3 years ago
Read 2 more answers
If a 5 N force pushes a 20 kg mass on a frictionless surface, how fast is the mass going in 5 seconds.
stira [4]

Explanation:

Let me know if you have questions

3 0
3 years ago
If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is µs = 0.800, how fast c
Cloud [144]

Answer:

Before start of slide velocity will be 14.81 m/sec

Explanation:

We have given coefficient of static friction \mu =0.8

Angle of inclination is equal to \Theta =tan^{-1}\mu

\Theta =tan^{-1}0.8=38.65^{\circ}

tan{38.65^{\circ}}=0.8

Radius is given r = 28 m

Acceleration due to gravity g=9.8m/sec^2

We know that tan\Theta =\frac{v^2}{rg}

0.8=\frac{v^2}{28\times 9.8}

v^2=219.52

v=14.816m/sec

So before start of slide velocity will be 14.81 m/sec

3 0
3 years ago
A 1.53-kg bucket hangs on a rope wrapped around a pulley of mass 7.07 kg and radius 66 cm. This pulley is frictionless in its ax
levacccp [35]

Answer:

\alpha = 6.431\,\frac{rad}{s^{2}}

Explanation:

The pulley is modelled by the Newton's Laws, whose equation of equilibrium is:

\Sigma M = T \cdot R = \frac{1}{2}\cdot M \cdot R^{2}\cdot \alpha

Given that tension is equal to the weight of the bucket, the angular acceleration experimented by the pulley is:

T = \frac{1}{2}\cdot M \cdot R \cdot \alpha

m_{b}\cdot g = \frac{1}{2}\cdot M \cdot R \cdot \alpha

\alpha = \frac{2\cdot m_{b}\cdot g}{M\cdot R}

\alpha = \frac{2\cdot (1.53\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{(7.07\,kg)\cdot (0.66\,m)}

\alpha = 6.431\,\frac{rad}{s^{2}}

7 0
3 years ago
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