All categories

3 years ago

???????????!??!!!!!!!!

Physics

1 answer:

Diano4ka-milaya [45]3 years ago

6 0

The answer would be B.

You might be interested in

Consider an eraser, weighing 20g, sitting 5cm from the center of a turntable (flat disk attached to a motor which can spin the d

Artist 52 [7]

Answer:

A. 42.3 revolution per minute

B. Centripetal Force

Explanation:

20 g = 0.02 kg

5 cm = 0.05 m

Let g = 9.8 m/s2

A.The friction force acting on the eraser is the product of normal force and its coefficient:

$F_f = N\mu = mg\mu = 0.02*9.8*0.1 = 0.0196 N$

For the eraser to begin slipping, its centripetal acceleration must win over the acceleration created by friction force, which is

$a_f = F_f / m = 0.0196 / 0.02 = 0.98 m/s^2 = a_c$

We can calculate the angular velocity from this:

$a_c = \omega^2 r$

where r = 0.05 is the radius of rotation, which is distance from the center to the eraser

$\omega^2 = a_c/r = 0.98 / 0.05 = 19.6$

$\omega = \sqrt{19.6} = 4.43 rad/s$

If it spins and angle of 4.43 rad per second then for every minute (60 seconds) it would spin an angle of 4.43 * 60 = 265.6 rad/min

Since 1 revolution is 2π rad, then in a minute it would spin 265.6 / 2π = 42.3 revolution/minute

B. The force responsible for making the eraser go faster while the turntable is spinning up would be the centripetal force.

7 0

3 years ago

A placekicker kicks a football upward at an angle of 30 and a speed of 5.0 m/s at what other angle will he have to kick a second

Flura [38]

Answer: $60^{\circ}$

Explanation:

Given

speed of ball $u=5 m/s$

launch angle $\theta =30^{\circ}$

Range of a Projectile $R=\frac{u^2\sin 2\theta }{g}$

Range will be common for two angles i.e. $\theta [tex] and [tex]90-\theta$

for $\theta R=\frac{5^2\sin 2\cdot 30}{g}$

$R=2.209 m$

For $90-\theta$

$90-30=60^{\circ}$

$R=\frac{5^2\sin 2\cdot 60}{g}$

$R=2.209 m$

6 0

3 years ago

Read 2 more answers

Joe decides to go bungee jumping. He secures a bungee to his ankles and leaps off a tall bridge. At one location, for one instan

slavikrds [6]

Gravitational energy i believe

6 0

3 years ago

Two 22.7 kg ice sleds initially at rest, are placed a short distance apart, one directly behind the other, as shown in Fig. 1. A

boyakko [2]

Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced

(a) The final speeds of the ice sleds is approximately 0.49 m/s each

(b) The impulse on the cat is 11.0715 kg·m/s

The reason for arriving at the above values is as follows:

The given parameters are;

The masses of the two ice sleds, m₁ = m₂ = 22.7 kg

The initial speed of the ice, v₁ = v₂ = 0

The mass of the cat, m₃ = 3.63 kg

The initial speed of the cat, v₃ = 0

The horizontal speed of the cat, v₃ = 3.05 m/s

(a) The required parameter:

The final speed of the two sleds

For the first jump to the right, we have;

By the law of conservation of momentum

Initial momentum = Final momentum

∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'

Where;

v₁' = The final velocity of the ice sled on the left

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05

∴ 22.7 × v₁' = -3.63 × 3.05

v₁' = -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

For the second jump to the left, we have;

By conservation of momentum law, m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'

Where;

v₂' = The final velocity of the ice sled on the right

v₃' = The final velocity of the cat

Plugging in the values gives;

22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05

∴ 22.7 × v₂' = -3.63 × 3.05

v₂' = -3.63 × 3.05/22.7 ≈ -0.49

The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)

The final speed ≈ 0.49 m/s

(b) The required parameter;

The impulse of the force

The impulse on the cat = Mass of the cat × Change in velocity

The change in velocity, Δv = Initial velocity - Final velocity

Where;

The initial velocity = The velocity of the cat before it lands = 3.05 m/s

The final velocity = The velocity of the cat after coming to rest =

∴ Δv = 3.05 m/s - 0 = 3.05 m/s

The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s

(c) The required information

The average velocity

Impulse = $F_{average}$ × Δt

Where;

Δt = The time of collision = The time it takes the cat to finish landing = 12 ms

12 ms = 12/1000 s = 0.012 s

We get;

$F_{average} = \mathbf{\dfrac{Impulse}{\Delta \ t}}$

∴ $F_{average} = \dfrac{11.0715 \ kg \cdot m/s}{0.012 \ s} = 922.625 \ kg\cdot m/s^2 = 922.625 \ N$

The average force on the right sled applied by the cat while landing, $\mathbf{F_{average}}$ = 922.625 N

Learn more about conservation of momentum here:

brainly.com/question/7538238

brainly.com/question/20568685

brainly.com/question/22257327

8 0

3 years ago

A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav

gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

Explanation:

(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

$K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2$ (1)