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3 years ago

What is the acceleration of a 5 kg mass pushed by a 10 n force?

Physics

2 answers:

pav-90 [236]3 years ago

6 0

2Kg50m/
s2.5m/
s22m/
s2

lorasvet [3.4K]3 years ago

3 0

∑F=ma, newton's second law. m=5kg, ∑F= 10N, solve for a. 10N/5kg= 2m/s^2.

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B. have eight electrons in their outer shell

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3 years ago

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Everyone experiences a wide range of emotions, but when could they indicate a mental disorder?

Vikentia [17]

Answer:

Excessive paranoia, worry, or anxiety.

Long-lasting sadness or irritability.

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Explanation:

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3 years ago

A baseball travels 200 metes in 6 seconds, what is the baseball’s velocity?

goldenfox [79]

Answer:

33.33 m/sec

Explanation:

A baseball travels 200 metes in 6 seconds,

what is the baseball’s velocity?

use the formula: velocity = distance over time

where (d) distance = 200 m

and (t) time = 6 sec.

plugin values into the formula:

v = d / t

= 200 m / 6 sec

= 33.33 m/sec.

therefore, the baseball's velocity is 33.33 m/sec

8 0

3 years ago

A child’s toy that is made to shoot ping pong balls consists of a tube, a spring (k = 18 N/m) and a catch for the spring that ca

UkoKoshka [18]

Answer:

The height is 3.1m

Explanation:

Here we have a conservation of energy problem, we have a conversion form eslastic potencial energy to gravitational potencial energy, so:

$E_e=\frac{1}{2}K*x^2\\E_e=\frac{1}{2}18N/m*(9.5*10^{-2}m)^2\\E_e=0.081J$

then we have only gravitational potencial energy when the ball is at its maximun height.

$E_g=m*g*h$

because all the energy was transformed Eg=Ee

$h=\frac{0.081J}{9.8m/s^2*m}$

searching the web, the mass of a ping pong ball is 2.7 gr in average. so:

$h=\frac{0.081J}{9.8m/s^2*(2.7*10^{-3}kg)}\\h=3.1m$

6 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, $\theta=180\times 2\pi=360\pi\ rad$

Using equation of motion:

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

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

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

Negative sign denotes deceleration.

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.

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.

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