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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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Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant
Reika [66]

Answer:

v = 21 m / s

Explanation:

We can solve this exercise with the kinematics equations, let's start by finding the acceleration of the train with the initial data

            v = v₀ + a t

the initial speed is the speed within the city 6 m / s, the final speed is v = 11 m / s and the time is t = 8 s

             a = (v-v₀) / t

             a = (11 - 6) / 8

             a = 0.625 m / s²

when it leaves the city with speed vo = 11 m / s it accelerates for t = 16 s

            v = v₀ + a t

            v = 11 + 0.625 16

             v = 21 m / s

4 0
4 years ago
When you rub a balloon on a sweater or your hair, the what is transferred to the balloon giving it a what charge. The balloon wi
Elis [28]

Answer:

When you rub a balloon on a sweater or your hair, the <u>electrons</u> are transferred to the balloon giving it a <u>negative</u> charge. The balloon will then <u>repel</u> any object that has a different charge.

hope this helps :)

5 0
3 years ago
Pls help i’ll give brainliest if you give a correct answer!!
Anit [1.1K]

Answer:

south

im not sure with this answer

3 0
3 years ago
A screen is placed 1.60 m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40 c
Lunna [17]

Answer:

distance between the two second-order minima is 2.8 cm

Explanation:

Given data

distance = 1.60 m

central maximum = 1.40 cm

first-order diffraction minima = 1.40 cm

to find out

distance between the two second-order minima

solution

we know that fringe width = first-order diffraction minima /2

fringe width = 1.40 /2 = 0.7 cm

and

we know fringe width of first order we calculate slit d

β1 = m1λD/d

d = m1λD/β1

and

fringe width of second order

β2 = m2λD/d

β2 = m2λD / ( m1λD/β1 )

β2 = ( m2 / m1 ) β1

we know the two first-order diffraction minima are separated by 1.40 cm

so

y = 2β2 = 2 ( m2 / m1 ) β1

put here value

y = 2 ( 2 / 1 ) 0.7

y = 2.8 cm

so distance between the two second-order minima is 2.8 cm

6 0
3 years ago
The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\ti
deff fn [24]

Answer:

Explanation:

Given

side of square shape a=5\ cm

Electric flux \phi =3\times 10^{-5}\ N.m^2/C

Permittivity of free space \epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}

Flux is given by

\phi =EA\cos \theta

where E=electric field strength

A=area

\theta=Angle between Electric field and area vector

E=\frac{\phi }{A\cos (0)}

E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}

E=0.012\ N/C

and Electric field  by a uniformly charged sheet is given by

E=\frac{\sigma }{2\epsilon_0}

where \sigma=charge density

=\frac{\sigma }{\epsilon_0}

\sigma =0.012\times 8.85\times 10^{-12}

\sigma =2.12\times 10^{-13}\ C/m^2    

5 0
3 years ago
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