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Shkiper50 [21]
3 years ago
13

The harder I push an object the _______ it will travel.

Physics
2 answers:
lions [1.4K]3 years ago
8 0

It's b, because the more force an object it is given the harder it will be for it to slow down.

STatiana [176]3 years ago
3 0
The answer is b because it’s b
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Answer = 6.24x10^18 x ((2 x 3600) + (47 x 60) + 10)
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A variable that should not change throughout the experiment is a(n):
Natalka [10]
The answer is control variable
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3 years ago
A person just supports a mass of 20kg suspended from a rope.
Georgia [21]

Answer:

F = 200 N

Explanation:

Given that,

The mass suspended from the rope, m = 20 kg

We need to find the resultant force acting on the rope. The resultant force on the rope is equal to its weight such that,

F = mg

Where

g is acceleration due to gravity

Put all the values,

F = 20 kg × 10 m/s²

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7 0
3 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the
Margaret [11]

Hi there!

\boxed{\omega = 0.38 rad/sec}

We can use the conservation of angular momentum to solve.

\large\boxed{L_i = L_f}

Recall the equation for angular momentum:

L = I\omega

We can begin by writing out the scenario as a conservation of angular momentum:

I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)

I_m = moment of inertia of the merry-go-round (kgm²)

\omega_m = angular velocity of merry go round (rad/sec)

\omega_f = final angular velocity of COMBINED objects (rad/sec)

I_b = moment of inertia of boy (kgm²)

\omega_b= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

I = MR^2

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

\omega = \frac{v}{r}

L_b = MR^2(\frac{v}{R}) = MRv

Plug in the given values:

L_b = (20)(3)(5) = 300 kgm^2/s

Now, we must solve for the boy's moment of inertia:

I = MR^2\\I = 20(3^2) = 180 kgm^2

Use the above equation for conservation of momentum:

600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}

8 0
3 years ago
The image shows street lights powered by solar panels. Which sequence shows the energy transformations taking place in these lig
yawa3891 [41]

the answer is C. for plato

8 0
3 years ago
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