An object that falls and accelerates solely as a result of gravity is said to be in

It takes 40 J of energy to push a large box 4m across a floor. Assuming the push is the same direction as the movement, what is

Jobisdone [24]

Answer:

10 N

Explanation:

Work done is the dot product of the force magnitude with distance and cosine of the angle betweenthem.

Equation to use: W=|F|*|d|*cosθ

Your unknown is the force F; You have one equation, oneunknown.

40j=(F)*(4m)*cos(0)

F=40j/4m

F=10N

3 0

3 years ago

Read 2 more answers

Find the speed vfinal of the joined cars after the collision. mastering physics

Tanya [424]

<span>Px = 0 Py = 2mV second, Px = mVcosĎ† Py = â€“mVsinĎ† add the components Rx = mVcosĎ† Ry = 2mV â€“ mVsinĎ† Magnitude of R = âš(RxÂ˛ + RyÂ˛) = âš((mVcosĎ†)Â˛ + (2mV â€“ mVsinĎ†)Â˛) and speed is R/3m = (1/3m)âš((mVcosĎ†)Â˛ + (2mV â€“ mVsinĎ†)Â˛) simplifying Vf = (1/3m)âš((mVcosĎ†)Â˛ + (2mV â€“ mVsinĎ†)Â˛) Vf = (1/3)âš((VcosĎ†)Â˛ + (2V â€“ VsinĎ†)Â˛) Vf = (V/3)âš((cosĎ†)Â˛ + (2 â€“ sinĎ†)Â˛) Vf = (V/3)âš((cosÂ˛Ď†) + (4 â€“ 2sinĎ† + sinÂ˛Ď†)) Vf = (V/3)âš(cosÂ˛Ď†) + (4 â€“ 2sinĎ† + sinÂ˛Ď†)) using the identity sinÂ˛(Ď)+cosÂ˛(Ď) = 1 Vf = (V/3)âš1 + 4 â€“ 2sinĎ†) Vf = (V/3)âš(5 â€“ 2sinĎ†)</span>

6 0

4 years ago

Describe a scenario where a car's speed could stay the same, but the acceleration changes.

k0ka [10]

Answer:

An object's acceleration is the rate its velocity (speed and direction) changes. Therefore, an object can accelerate even if its speed is constant - if its direction changes.

Explanation:

8 0

3 years ago

A 62.0-kg skier is moving at 6.30 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.90

Klio2033 [76]

Here are the missing questions:
(a) How fast is the skier moving when she gets to the bottom of the hill?
(b) How much internal energy was generated in crossing the rough patch?
Part A
The initial kinetic energy of the skier is:
$E_{k0}=m\frac{v_0^2}{2}$
Part of this energy is then used to do work against the force of friction. Force of friction on the horizontal surface can be calculated using following formula:
$F_f=mg\mu$
The work is simply the force times the length:
$W_f=F_f\cdot L=mg\mu L$
So when the skier passes over the rough patch its energy is:
$E=E_{k0}-W_f$
When the skier is going down the skill gravitational potential energy is transformed into the kinetic energy:
$E_p=E_{k1}\\ mgh=E_{k1}$
So the final energy of the skier is:
$E_f=E_{k0}-W_f+E_{k1}\\ E_f=m\frac{v_0^2}{2}-mg\mu L+mgh=1856.86$J$
This energy is the kinetic energy of the skier:
$E_f=m\frac{v_f^2}{2}\\ v_f=\sqrt{\frac{2E_f}{m}}=7.74\frac{m}{s}$
Part B
We know that skier lost some of its kinetic energy when crossing over the rough patch. This energy is equal to the work done by the skier against the force of friction.
$E_{int}=W_f\\
E_{int}=mg\mu L=894.1$J$

4 0

4 years ago

Which statement about electromagnetic waves is true?

zhuklara [117]

Electromagnetic waves don't require a medium to travel.

5 0

3 years ago

Read 2 more answers