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aniked [119]
3 years ago
11

Roger pushes a box on a 30° incline. If he applies a force of 60 newtons parallel to the incline and displaces the box 10 meters

along the incline, how much work will he do on the box?
Physics
2 answers:
Delvig [45]3 years ago
7 0

Answer:

(60 newtons) x (10 meters)  =  600 joules.

Explanation:

melomori [17]3 years ago
4 0

Pretty sneaky, I must say !  

                   Work = (force) x (distance)

That formula says nothing about the directions of the force OR
the distance.  As long as they're both in the same direction, they
can both be horizontal (sliding a crate on the floor), vertical (lifting
something), or on any slant in between.

Roger's force and distance are both in the same direction ... the
direction of the incline ... so

                   Work = (60 newtons) x (10 meters)  =  600 joules.

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A rifle fires a 2.90 x 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the
iren2701 [21]

Answer: 586.60N/m

Explanation:

In this scenario, the elastic potential energy of the spring is converted into potential energy.

0.5*K*x^2 = mgh

Thus K = 2mgh/x^2

=(2*2.90*10^-2*9.8*7.23)/(8.37*10^-2)^2

=586.599

Therefore K = 586.60N/m

7 0
4 years ago
Read 2 more answers
Difference between static and kinetic friction in physics
KIM [24]
The force of static friction keeps a stationary object at rest. Once the force of static friction is overcome, the force of kinetic friction is what slows down a moving object.
4 0
3 years ago
If a wave has amplitude of 2 meters, a wavelength of 2 meters, and a frequency of 10 Hz, and a period of 1 second, then at what
Serhud [2]

Answer:

20 m/s

Explanation:

The speed of a wave is given by:

v=\lambda f

where

\lambda is the wavelength

f is the frequency

v is the speed

For the wave in this problem,

f = 10 Hz is the frequency

\lambda=2 m is the wavelength

So the speed is

v=(10 Hz)(2 m)=20 m/s

6 0
3 years ago
A 2kg hockey puck is sliding across the ice skating rink at 2 m/s. A player hits the puck so it's velocity increases to 10 m/s.
konstantin123 [22]

The work done on the puck is 96 J

Explanation:

According to the work-energy theorem, the work done on the hockey puck is equal to the change in kinetic energy of the puck.

Mathematically:

W=K_f -K_i= \frac{1}{2}mv^2-\frac{1}{2}mu^2

where

K_f = \frac{1}{2}mv^2 is the final kinetic energy of the puck, with

m = 2 kg being the mass of the puck

v = 10 m/s is the final speed

K_i = \frac{1}{2}mu^2 is the initial kinetic energy of the puck, with

u = 2 m/s being the initial speed of the puck

Substituting numbers into the equation, we find the work done by the player on the puck:

W=\frac{1}{2}(2)(10)^2 - \frac{1}{2}(2)(2)^2=96 J

Learn more about work and kinetic energy:

brainly.com/question/6763771  

brainly.com/question/6443626  

brainly.com/question/6536722

#LearnwithBrainly

6 0
3 years ago
Giving brainliest!! if you answer correctly :) (30pts)
myrzilka [38]

Answer:

mass = 4kg

Explanation:

Kinetic Energy = 1/2 x m x v²

where m = mass and v = velocity

So,

KE = 50

1/2 × m × 5² = 50

1/2 × m × 25 = 50

m = (50 x 2)/25

m = 100/25

m = 4 kg

8 0
3 years ago
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