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Paul [167]
3 years ago
14

"a force sets an object in motion. when the force is multiplied by the time of its application, we call the quantity impulse, an

d an impulse changes the momentum of that object. what do we call the quantity force multiplied by distance?"
Physics
1 answer:
dsp733 years ago
4 0
<span>The quantity force multiplied by distance is work.Work is done when a force pushes something and the object moves some distance in the direction it’s being pushed.work = force x distance where distance traveled in the direction the force is pushing counts.The unit for rate of working also called as Power.</span>
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A 30 kg child rides a 20 kg bicycle. Together, the child and the bicycle have a momentum of 110 kg-m/s. What is the velocity of
love history [14]

Answer:

The velocity of the boy and the bicycle is 2.2 m/s.

Explanation:

We have,

Mass of child is 30 kg and the mass of bicycle is 20 kg. The combined momentum of the child and the bicycle is 110 kg-m/s.

It is required to find the velocity of the boy and the bicycle. The momentum of an object is given in terms of mass and its velocity. So,

p=Mv

M is combined mass of child and bicycle

v=\dfrac{p}{M}\\\\v=\dfrac{110}{30+20}\\\\v=2.2\ m/s

So, the velocity of the boy and the bicycle is 2.2 m/s.

8 0
3 years ago
The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at posit
SpyIntel [72]

Answer:

Magnitude of the Frictional force = (mv₀²)/2x₁

Explanation:

For the frictional force to stop the box, it has to produce the deceleration of the box; thereby being the opposing force to the box's motion.

According to Newton's first law of motion

Frictional force = (mass of the box) × (deceleration experienced by the box)

Let the mass of the box be m

Then,

Frictional force = ma

Then we can obtain the deceleration using the equations of motion

v² = u² + 2ax

u = Initial velocity = v₀ m/s

v = Final velocity = 0 m/s (since the box comes to rest at the end)

x = horizontal distance covered = (x₁ - x₀) = x₁ (since x₀ = 0)

a = ?

v² = u² + 2ax

0 = (v₀)² + 2ax₁

2ax₁ = - v₀²

a = - (v₀²)/(2x₁) (minus sign, because it's a deceleration)

Magnitude of the Frictional force = ma = (mv₀²)/2x₁

4 0
3 years ago
Melissa uses bags of mulch that way 40 pounds each. She needs to carry 10 bags to her truck. She usually carries one bag at a ti
arsen [322]
No. She would be doing the same amount of work that way. Work is defined to be equal to the force multiplied by the distance. Carrying two bags at a time would cause her to exert twice the effort, so the total amount of work done in the end would still be the same.
4 0
3 years ago
PLS HELP!!! A 1200-kg whale swims horizontally to the right at a speed of 6.0 m/s. It suddenly collides directly with a stationa
Fittoniya [83]

Answer:

7200 kg.m/s

Explanation:

According the law of conservation of linear momentum, the sum of momentum before and after collision are equal.

Using this principle, the sum of initial momentum will be given as p=mv where p is momentum, m is mass and v is velocity

Initial momentum

Mass of whale*initial velocity of whale + mass of seal*initial seal velocity

Since the seal is initially stationary, its velocity is zero. By substitution and taking right direction as positive

Initial momentum will be

1200*6+(280*0)=7200 kg.m/s

Since both initial and final momentum should be equal, hence the final momentum will also be 7200 kg.m/s

7 0
3 years ago
A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from
Ulleksa [173]

Answer:

1270.64\ \text{J}

Explanation:

m = Mass of object = \dfrac{mg}{g}

mg = Weight of object = 20 N

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

\theta = Angle of slope = 30^{\circ}

f = Force of friction

fd = Work done against friction

The force balance of the system is

\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}

The work done against friction is 1270.64\ \text{J}.

8 0
3 years ago
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