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NeX [460]
3 years ago
13

You wad up a piece of paper and throw it into the wastebasket. How far will

Physics
2 answers:
vitfil [10]3 years ago
6 0

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, g=9.8 m/s^2)

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:

d=\frac{u^2 sin 2\theta}{g}

where

u is the initial velocity

\theta is the angle of projection

g is the acceleration of gravity

For the piece of paper in this problem,

u = 4.3 m/s

\theta=65^{\circ}

Substituting,

d=\frac{(4.3)^2 sin(2\cdot 65^{\circ})}{9.8}=1.45 m \sim 1.4 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

aksik [14]3 years ago
4 0

Answer:

like the other person said, the answer is 1.4 m

Explanation:

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An 8 kilogram bowling ball is rolling in a straight line toward you. if its momentum is 16 kg* m/s, how fast is it traveling ?
morpeh [17]
From the information given, The mass of the bowling ball is 8 Kilograms and the momentum with which it is moving is 16 kg m/s. We use the formula p = m × v Where p is the momentum, m is the mass and v is the velocity. We need velocity so we rewrite the equation thus: P = mv, therefore p/m = v or v = p/m In our case p = 16 and m = 8 v = p/m v = 16/8 v = 2 Therefore the bowling ball is travelling at 2m/s
6 0
3 years ago
If you apply a force of 100 N to the level, how much force is applied to lift the crate?
Kaylis [27]
I believe that the answer is B. 133 N
4 0
3 years ago
The force between two charges, q, and 92, is F. If the distance between the
Sholpan [36]

Answer:

4F

Explanation:

F = kQ₁Q₂/d²

F' = k2Q₁2Q₂/d²

F' = 4(kQ₁Q₂/d²)

F' = 4F

4 0
3 years ago
A 71 kW radio station broadcasts its signal uniformly in all directions. - What is the average intensity of its signal at a dist
marshall27 [118]

Answer:

Explanation:

Energy of signal being radiated per second on all sides = 71 x 10³ J .

At a distance of 220 m it is spread over an area of 4 π x (220)² because it is spreading uniformly on all sides.

So energy crossing per unit area

= \frac{71\times10^3}{4 \times \pi\times(220)^2}

= 11.67 x 10⁻² Wm⁻²s⁻¹.

This is the intensity of the signal.

At 2200 m this intensity will further reduce by 100 times

So there it becomes equal to

11.67 x 10⁻⁴ Wm⁻² s⁻¹.

3 0
3 years ago
A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an ang
adell [148]

Answer:

A boxed 14.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an angle of 36.9 ∘ above the horizontal. The monitor's speed is a constant 2.30 cm/s.

how much work is done on the monitor by (a) friction, (b) gravity

work(friction) = 453.5J

work(gravity) = -453.5J

Explanation:

Given that,

mass = 14kg

displacement length = 5.50m

displacement angle = 36.9°

velocity = 2.30cm/s

F = ma

work(friction) = mgsinθ .displacement

                      = (14) (9.81) (5.5sin36.9°)

                       = 453.5J

work(gravity)

= the influence of gravity oppose the motion of the box and can be pushing down, on the box from and angle of (36.9° + 90°)

= 126.9°

work(gravity) = (14) (9.81) (5.5cos126.9°)

                      = -453.5J

8 0
3 years ago
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