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2 years ago

HELLPPP NOWWA!!!!!!!!!!!!

Physics

2 answers:

elena55 [62]2 years ago

8 0

Answer:

it would be a

Explanation:

i remeber taking the test im pretty sure its a

prisoha [69]2 years ago

7 0

Answer:

Explanation:

the gravitational pull also helps with that but

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A ball is thrown up with a speed of 15m/s. How high will it go before it begins to fall? ( g = 10m/s2 )

LekaFEV [45]

Answer:

Height is 11.25m

Explanation:

Given the following data;

Initial velocity, u = 0

Final velocity, v = 15m/s

Acceleration due to gravity, g = 10m/s²

To find the height, we would use the third equation of motion;

$V^{2} = U^{2} + 2aS$

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

S represents the displacement (height) measured in meters.

$V^{2} = U^{2} + 2aS$

Making S the subject, we have;

$S = \frac {V^{2} - U^{2}}{2a}$

But a = g = 10m/s²

Substituting into the equation, we have;

$S = \frac {15^{2} - 0^{2}}{2*10}$

$S = \frac {225 - 0}{20}$

$S = \frac {225}{20}$

S = 11.25m

Therefore, the ball will reach a height of 11.25m before it begins to fall.

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1 year ago

Waves that can travel only through a medium are called ______________ waves. Sound waves are ______________ waves. (Hint: same w

Nana76 [90]

First blank is radiator

Second blank is radiator

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2 years ago

An athlete runs 200 m by running 100 m in a straight line and then turning around and running 100m back to the start point.

siniylev [52]

Answer:

Explanation:

The athlete ran a total distance of zero because they ran 100m forward then turned around so they went back to their starting position

6 0

2 years ago

How can we teach our kitty to behave properly

Dmitry_Shevchenko [17]

Answer:

I have no idea.

Explanation:

I just wanted to tell you I absolutely read this wrong.

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2 years ago

The angular velocity of a process control motor is (13−12t2) rad/s, where t is in seconds. Part A At what time does the motor re

mihalych1998 [28]

Answer:

Explanation:

Given

$\omega =13-\frac{1}{2}\cdot t^2$

Motor reverse its direction when \omega =0

$13-0.5t^2=0$

$26=t^2$

$t=\sqrt{26}=5.099\approx 5.1 s$

(b)

$\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega$

$\int d\theta =\int_{0}^{5.1}\omega dt$

$\int d\theta =\int_{0}^{t}(13-.05t^2)dt$

$\theta =(13t-0.1667\times t^3)_0^{5.1}$

$\theta =44.192^{\circ}$

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2 years ago