All categories

3 years ago

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

You might be interested in

If 6.24×10¹⁸ electron pass through a wire in 1s how many pass through it during a time interval of 2hr, 47min and 10s

levacccp [35]

Answer = 6.24x10^18 x ((2 x 3600) + (47 x 60) + 10)

3 0

3 years ago

A variable that should not change throughout the experiment is a(n):

Natalka [10]

The answer is control variable

7 0

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

F = 200 N

So, the resultant force on the mass is 200 N.

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

Read 2 more answers