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3 years ago

14

Kim holds a pinwheel in the air and says it can be used to model a source of energy. Identify the type of energy Kim's pinwheel

models. Explain the type of energy source used.

1 answer:

Natali5045456 [20]3 years ago

7 0

Answer:

it is kenetic

Explanation: its in motion:D

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A playground slide is inclined 40°. If a boy with a mass of 32 kg slides down for -3meters. How much work is done by gravity on

telo118 [61]

Answer:

the work done by gravity on the boy is 604.62 J

Explanation:

Given;

distance the boy slides, d = 3 m

angle of inclination of the playground, θ = 40⁰

mass of the boy, m = 32 kg

The vertical height, h, above the ground through which the boy falls represents the height of the triangle which is the opposite side.

The distance through which the boy slides, d, represents the hypotenuse side of the right triangle.

$sin \theta = \frac{opposite }{hypotenuse} \\\\sin \theta = \frac{h}{d} \\\\h = dsin\theta\\\\h = 3 \times sin(40^0)\\\\h = 1.928 \ m$

The work done by gravity on the boy is calculated as;

W = P.E = mgh

= 32kg x 9.8m/s² x 1.928m

= 604.62 J

Therefore, the work done by gravity on the boy is 604.62 J

8 0

3 years ago

If one of two interacting charges is doubled, the force between the charges will _____________.

malfutka [58]

If one of two interacting charges is doubled, the force between the charges will double.

Explanation:

The force between two charges is given by Coulomb's law

$F=\frac{k q1 q2}{r^{2}}$

K=constant= 9 x 10⁹ N m²/C²

q1= charge on first particle

q2= charge on second particle

r= distance between the two charges

Now if the first charge is doubled,

we get $F'=\frac{k (2q1) q2}{r^{2}}$

F'= 2 F

Thus the force gets doubled.

4 0

3 years ago

A 1.0 kg football is given an initial velocity at ground level of 20.0 m/s [37° above horizontal]. It gets blocked just after re

brilliants [131]

1) Data:

Vo = 20 m/s
α = 37°
Yo = 0
Y = 3m

2) Questions: V at Y = 3m and X at Y = 3 m

3) Calculate components of the initial velocity

Vox = Vo * cos(37°) = 15.97 m/s

Voy = Vo * sin(37°) = 12.04 m/s

4) Formulas

Vx = constant = 15.97 m/s

X = Vx * t

Vy = Voy - g*t

Y = Yo + Voy * t - g (t^2) / 2

5) Calculate t when Y = 3m (first time)

Use g ≈ 9.8 m/s^2

3 = 12.04 * t - 4.9 t^2

=> 4.9 t^2 - 12.04t + 3 = 0

Use the quadratic equation to solve the equation

=> t = 0.28 s and t = 2.18s

First time => t = 0.28 s.

6) Calculate Vy when t = 0.28 s

Vy = 12.04 m/s - 9.8 * 0.28s = 9.3 m/s

7) Calculate V:

V = √ [ (Vx)^2 + (Vy)^2 ] = √[ (15.97m/s)^2 + (9.30 m/s)^2 ] = 18.48 m/s

tan(β) = Vy/Vx = 9.30 / 15.97 ≈ 0.582 => β ≈ arctan(0.582) ≈ 30°

Answer: V ≈ 18.5 m/s, with angle ≈ 30°

8) Calculate X at t = 0.28s

X = Vx * t = 15.97 m/s * 0.28s = 4,47m ≈ 4,5m

Answer: X ≈ 4,5 m

4 0

3 years ago

How to do number 19?

ivann1987 [24]

The answer is 8/3. You have to simplify each side and isolate the variable.

5 0

3 years ago

Read 2 more answers

Puck 1 is moving 10 m/s to the left and puck 2 is moving 8 m/s to the right. They have the same mass, m.

Julli [10]

Answer:

(a) the total momentum of the system before the collision = -2m kg.m/s.

(b) the total momentum of the system after the collision = -2m kg.m/s.

(c) puck 1's velocity after the collision in component form = (5.44 i, 2.54 j)

Explanation:

Given;

mass of Puck 1 , = m

mass of Puck 2, = m (since they have the same mass m)

initial velocity of Puck 1, u₁ = 10 m/s to the left

initial velocity of Puck 2, u₂ = 8 m/s to the right

Let the rightward direction be positive direction

Let the leftward direction be negative direction

(a) the total momentum of the system before the collision;

P₁ = (initial momentum of Pluck 1) + (initial momentum of Pluck 2)

P₁ = (-mu₁) + mu₂

P₁ = mu₂ - mu₁

P₁ = m(u₂ - u₁)

P₁ = m(8 - 10)

P₁ = -2m kg.m/s

(b) the total momentum of the system after the collision;

Based on the principle of conservation of linear momentum, the total momentum before collision is equal to the total momentum after collision.

Thus, the total momentum of the system after the collision is -2m kg.m/s.

(c) puck 1's velocity after the collision in component form

$v = (v_x, v_y)\\\\v = (vcos \theta , vsin \theta)\\\\v = (6cos 25^0 , 6sin25^0)\\\\v = (5.44i, 2.54j)m/s$

8 0

3 years ago