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3 years ago

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

1 answer:

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