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seropon [69]
3 years ago
13

A 10 N block is at the bottom of a frictionless incline as shown to the right. How much work must be done against gravity to mov

e it to the top of the incline?
600j

1000j

4800j

800j

Physics
1 answer:
Andreyy893 years ago
7 0

Explanation:

We have,

According to attached figure, the height of the inclined plane is 60 m and force acting on the block is 10 N. It is required to find the work must be done against gravity to move it to the top of the incline. The work done is given by :

W = mgh

or

W=F\times h\\\\W=10\times 60\\\\W=600\ J

Hence, the correct option is (A) "600 J".

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The correct answer is A.

A power station works on the principle of boiling water to create steam, which turns a turbine, generating a potential difference in a transformer with the magnets. The transformer is connected to a circuit, which hence induces a current, generating power.
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4 years ago
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An apple falls out of a tree from a height of 2.3m. what is the impact speed of the apple
Anuta_ua [19.1K]

Answer:6.71 m/s

Explanation:

Given

Apple fall from a height of h=2.3 m  

We need to find the impact speed of apple which can be given by using

v^2-u^2=2gh  

where v=final velocity

u=initial velocity

h=Displacement

Assuming initial velocity to be zero

substituting the value we get

v^2-0=2\times 9.8\times 2.3  

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5 0
3 years ago
Two positive charges of 20 micro coulomb and 100 micro coulomb and the distance between them is 150cm.What will be the electrica
Roman55 [17]

Answer:

0.8 N

Explanation:

From coulomb's law,

Formula:

F = kqq'/r²........................ Equation 1

Where F = Force of repulsion, k = coulomb's constant, q = first positive charge, q' = second positive charge, r = distance between the charge.

Given: q = 20 μC = 20×10⁻⁶ C, q' = 100 μC = 100×10⁻⁶ C, r = 150 cm = 1.5 m.

Constant:  k = 9×10⁹ Nm²/C²

Substitute these values into equation 1

F = (20×10⁻⁶ )( 100×10⁻⁶)(9×10⁹)/1.5²

F = 1800×10⁻³/2.25

F = 1.8/2.25

F = 0.8 N

3 0
3 years ago
A shell is fired at an angle of 35° above the horizontal at a velocity of 40 m/s. (a) What is it's speed at the highest point of
Fudgin [204]

Answer:

Part a)

v_f = v_x = 32.77 m/s

Part b)

T = 4.68 s

Explanation:

Part a)

Shell is fired at speed of 40 m/s at angle of 35 degree

so here we have

v_x = 40 cos35 = 32.77 m/s

v_y = 40 sin35 = 22.94 m/s

since gravity act opposite to vertical speed of the shell so at the highest point of its trajectory the vertical component of the speed will become zero

so at the highest point the speed is given

v_f = 32.77 m/s

Part b)

After completing the motion we know that the displacement of the object will be zero in Y direction

so we have

\Delta y = 0

0 = v_y t - \frac{1}{2}gt^2

T = \frac{2v_y}{g}

T = \frac{2(22.94)}{9.81} = 4.68 s

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A wave has a wavelength of 20 mm and a frequency of 5 Hz what is the speed?
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The answer is 100mm/s. I hope this helps :)

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