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zheka24 [161]
3 years ago
7

A cannonball is shot up in the air with a vertical speed of 24 miles what is the cannon balls vertical speed just before it hits

the ground?
Physics
1 answer:
meriva3 years ago
8 0

Answer:

I think it will back towards the earth because earth gravitional field will attract to Wards

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A cooling fan is turned off when it is running at 850 rev/min. It turns 1500 revolutions before it comes to a stop. (a) What was
8_murik_8 [283]

Answer

given,

cooling fan revolution = 850 rev/min

fan turns before revolution = 1500 revolutions

\omega = 850 \dfrac{2\pi}{60}

\omega = 89\ rad/s

θ = 1500 revolution

θ = 1500 x 2 x π

θ = 9424.78 rad

a) using equation of rotation

ω² = ω₀² + 2 α θ

ω = 0 because body comes to rest

0 = 89² + 2 x α x 9424.78

α = -0.42 rad/s²

b) time take for the fan to stop

ω = ω₀ + α t

0 = 89 - 0.42 t

t = \dfrac{89}{0.42}

t = 211.9 s

5 0
3 years ago
Which of the following statements is correct?
Mumz [18]

Answer: c. Generally, metals are ductile.

Explanation:

From the options given in the question, the correct statement is that"Generally, metals are ductile.

Ductility of a metal simply means that a metal can be plastically deform before it is then fractured. It implies that metals can be drawn to thin wires. The only exception we have in this case is mercury.

7 0
3 years ago
*<br> Check all of the things that can affect power<br><br> Time<br> Work<br> Distance<br> Force
gladu [14]

Answer:

fhddmvxmvydlyghclhchc

3 0
3 years ago
A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and
alexira [117]

Answer:

A) \omega_f=17.503\ rad.s^{-1}

B) t=55.6822\ s

C) \theta=1312\ rad

Explanation:

Given:

  • mass of flywheel, m=40\ kg
  • diameter of flywheel, d=0.72\ m
  • rotational speed of flywheel, N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}
  • duration for which the power is off, t_0=35\ s
  • no. of revolutions made during the power is off, \theta=180\times 2\pi=360\pi\ rad

<u>Using equation of motion:</u>

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2

\alpha=-0.8463\ rad.s^{-2}

Negative sign denotes deceleration.

A)

Now using the equation:

\omega_f=\omega_i+\alpha.t

\omega_f=15\pi-0.8463\times 35

\omega_f=17.503\ rad.s^{-1} is the angular velocity of the flywheel when the power comes back.

B)

Here:

\omega_f=0\ rad.s^{-1}

Now using the equation:

\omega_f=\omega_i+\alpha.t

0=15\pi-0.8463\times t

t=55.6822\ s is the time after which the flywheel stops.

C)

Using the equation of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2

\theta=1312\ rad revolutions are made before stopping.

3 0
3 years ago
A wave is described by y(x,t) = 0.1 sin(3x + 10t), where x is in meters, y is in centimeters and t is in seconds. The angular wa
Mila [183]

Answer: 3 radians/meter.

Explanation:

The general sinusoidal function will be something like:

y = A*sin(k*x - ω*t) + C

Where:

A is the amplitude.

k is the wave number.

x is the spatial variable

ω is the angular frequency

t is the time variable.

C is the mid-value.

The rule that we can use to solve this problem, is that the argument of the sin( ) function must be in radians (or in degrees)

Then if x is in meters, the wave-number must be in radians/meters, so when these numbers multiply the "meters" part is canceled.

Then for the case of the function:

y(x,t) = 0.1 sin(3x + 10t)

Where x is in meters, the units of the wave number (the 3) must be in radians/meters. Then the angular wave number is 3 radians/meter.

5 0
3 years ago
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