The box is accelerated from rest to 4 m/s in a matter of 2.5 s, so its acceleration <em>a</em> is such that
4 m/s = <em>a</em> (2.5 s) → <em>a</em> = (4 m/s) / (2.5 s) = 1.6 m/s²
Then the force applied to the box has a magnitude <em>F</em> such that
<em>F</em> = (10 kg) (1.6 m/s²) = 16 N
Answer:
Explanation:
Stefan's formula for emission of radiation is
E = e σ A ( T⁴ - T₀⁴ )
E is energy radiated , e is emissivity , σ is stefan's constant , T is temperature of object and T₀ is temperature of surrounding. A is area of surface .
E = .35 x 5.67 x 10⁻⁸ ( 298⁴ - 268⁴ ) x 4π x .25²
= 1.9845 x 10⁻⁸ ( 78.86 - 51.58 ) x 10⁸ x .0625
= 3.38 J /s
The range of the piece of paper is C) 1.4 m
Explanation:
The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:
- A uniform motion along the horizontal direction, with constant velocity
- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity,
)
From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

where
u is the initial velocity
is the angle of projection
g is the acceleration of gravity
For the piece of paper in this problem,
u = 4.3 m/s

Substituting,

Learn more about projectile motion:
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Answer:
If an object is moving with a constant velocity, then by definition it has zero acceleration. So there is no net force acting on the object. The total work done on the object is thus 0 (that's not to say that there isn't work done by individual forces on the object, but the sum is 0 ).
Explanation:
In the middle, when the object was changing position at a constant velocity, the acceleration was 0. This is because the object is no longer changing its velocity and is moving at a constant rate.
They both use small amouts of nuclar waves