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3 years ago

PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer these correctly!!!! (40pts)

Physics

2 answers:

Cerrena [4.2K]3 years ago

8 0

Answer:

B false

Explanation:

dont have any explanation

777dan777 [17]3 years ago

6 0

Answer:

Explanation:

you see when something is changing directions it is accelerating, not in a sence is speed is changing altho its volocity is (speed and direction)

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1. Give three examples, from the lab, where potential energy was converted to kinetic energy:

Lubov Fominskaja [6]

Answer:

A book on a table before it falls.

A yoyo before it is released.

A raised weight.

Explanation:

These are all examples of potential energy. So I hope you can find something that is comparable from the lab.

3 0

3 years ago

Pure water has a pH of 7. Pure water _______. A. Is a neutral substance B. Could be either an acid or a base C. Is a base D. Is

Hoochie [10]

A. is a neutral substance a pH of 7 describes something of neutral pH where anything less than 7 is an acid and higher is a base

5 0

3 years ago

Read 2 more answers

A transformer has 100 turns in its primary coil and 75 turns in its secondary coil. If the input voltage is 12.0 V, what is the

Sindrei [870]

Is 54 por que es igual

7 0

3 years ago

An assault rifle fires an eight-shot burst in 0.40 s. Each bullet has a mass of 7.5 g and a speed of 300 m/s as it leaves the gu

myrzilka [38]

Answer:

The average recoil force on the gun during that 0.40 s burst is 45 N.

Explanation:

Mass of each bullet, m = 7.5 g = 0.0075 kg

Speed of the bullet, v = 300 m/s

Time, t = 0.4 s

The change in momentum of an object is equal to impulse delivered. So,

$F\times t=mv\\\\F=\dfrac{mv}{t}$

For 8 shot burst, average recoil force on the gun is :

$F=\dfrac{8mv}{t}\\\\F=\dfrac{7.5}{1000}\cdot\dfrac{300}{0.4}\cdot8\\\\F=45\ N$

So, the average recoil force on the gun during that 0.40 s burst is 45 N.

5 0

3 years ago

Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father

Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity $\omega_o = 0$

angular acceleration $\alpha$ = 4.44 rad/s²

Using the formula:

$\omega = \omega_o+ \alpha t$

Making t the subject of the formula:

$t= \dfrac{\omega- \omega_o}{ \alpha }$

where;

$\omega = 1.53 \ rad/s^2$

∴

$t= \dfrac{1.53-0}{4.44 }$

t = 0.345 s

Using the formula:

$\omega ^2 = \omega _o^2 + 2 \alpha \theta$

here;

$\theta$ = angular displacement

∴

$\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }$

$\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }$

$\theta =0.264 \ rad$

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

$x = \dfrac{0.264 \times 1}{2 \pi}$

x = 0.042 revolutions

Here; force = 270 N

radius = 1.20 m

The torque = F * r

$\tau = 270 \times 1.20 \\ \\ \tau = 324 \ Nm$

However;

From the moment of inertia;

$Torque( \tau) = I \alpha \\ \\ Since( I \alpha) = 324 \ Nm. \\ \\ Then; \\ \\ \alpha= \dfrac{324}{I}$

given that;

I = 84.4 kg.m²

$\alpha= \dfrac{324}{84.4} \\ \\ \alpha=3.84 \ rad/s^2$

For re-tardation; $\alpha=-3.84 \ rad/s^2$

Using the equation

$t= \dfrac{\omega- \omega_o}{ \alpha }$

$t= \dfrac{0-1.53}{ -3.84 }$

$t= \dfrac{1.53}{ 3.84 }$

t = 0.398s

The required time it takes= 0.398s

5 0

3 years ago