Ask question

Ask question

All categories

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?

1 answer:

meriva3 years ago

8 0

Answer:

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

You might be interested in

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