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VashaNatasha [74]
3 years ago
8

Answer this for me please and thank you

Physics
1 answer:
Zinaida [17]3 years ago
7 0
Pretty sure its the 2nd one hope this helps :)
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You can send your picture<br>and type the answer to the questions
Snowcat [4.5K]
The answer is B because I took the test and had the question
3 0
3 years ago
A car initially going 61 ft/sec brakes at a constant rate (constant negative acceleration), coming to a stop in 7 seconds.
Bond [772]

Answer:

See the attachment below for the graphics in part (a)

The initial velocity for this time interval is u = 61ft/sec and the final velocity is 0m/s because the car comes to a stop.

This a constant acceleration motion considering the given time interview over which the brakes are applied. So the equals for constant acceleration motion apply here.

Explanation:

The full solution can be found in the attachment below.

Thank you for reading. I hope this post is helpful to you.

4 0
3 years ago
3. A man hits a 0.2 kg golf ball with a force of 4
pogonyaev

Answer:

Force(f)= mass x acceleration

Acceleration (a) is the rate of change in velocity.

F=4N

M=0.2kg

a=F/M

a=4/0.2

a=20m/s^2

Explanation:

8 0
2 years ago
1. What is the intensity of a sound when an observer is 10.0 m from the speaker
Nana76 [90]

Answer:

About 133 db.

Explanation:

Sound Intensity Level in db (SIL db) is equal to 10log (base 10) times the ratio of the sound intensity at 200 watts (I) relative to the sound intensity of  the reference sound intensity (I sub 0), which by default is equal to 10⁻¹² W/m² or 0 dB.

I = 200 w / 10 m^2 = 20 w per square meter  

I sub 0 = 10^-12 w per square meter

SIL = 10log ( I / I sub o) = 20 / 10^-12 = 10log ( 20^12) = 10 ( 13.3 ) = 133 db

Hope I typed this part correctly. Hard to get it in without being able to do exponents, etc. :D

8 0
3 years ago
the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

d_x(t)=vo*cos(A)*t\\

R=Range

R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

7 0
3 years ago
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