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In-s [12.5K]
2 years ago
8

5. Which statement about the acceleration of an object is corr

Physics
1 answer:
zavuch27 [327]2 years ago
6 0

Answer:

a

Explanation:

you will find out that an object that has more weight requires more energy to move than the one with less weight

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Is there a serve zone?
Lera25 [3.4K]

Answer:

When it comes to serving, the court is divided into six zones. Right back is zone one, right front is zone two, middle front is zone three, left front is zone four, left back is zone five and middle back is zone six.

Explanation:

7 0
3 years ago
Read 2 more answers
What is the minimum force require to move a 5kg wooden crate on a wooden floor?
kolbaska11 [484]

You need to know the coefficient of static friction between a wooden object and a wooden surface. I'll denote it with <em>µ</em>. If you're given a specific value you should obviously use that.

By Newton's second law, the horizontal and vertical net forces are

• net horizontal:

∑ <em>F</em> = <em>p</em> - <em>f</em> = 0

• net vertical:

∑ <em>F</em> = <em>n</em> - <em>w</em> = 0

where

<em>p</em> = magnitude of the <u>p</u>ushing force

<em>f</em> = mag. of <u>f</u>riction

<em>n</em> = mag. of the <u>n</u>ormal force

<em>w</em> = <u>w</u>eight of the crate

The second equation gives

<em>n</em> = <em>w</em> = (5 kg) (9.8 m/s²) = 49 N

Friction is proportional to the normal force by a factor of <em>µ</em>, so

<em>f</em> = <em>µ</em> (49 N) = 49<em>µ</em> N

To overcome static friction, the push has to exceed this in magnitude, so that

<em>p</em> > 49<em>µ</em> N

For instance, if <em>p</em> = 0.25, then <em>p</em> would need to greater than 12.25 N. (This example isn't particularly helpful, though, since both possibly correct options are larger than 12.25 N...)

7 0
3 years ago
How do you calculate energy lost due to friction in an experiment?
Snowcat [4.5K]

Answer:

A treadmill get it? but its   Ff * d cos theta

Explanation:

6 0
3 years ago
During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau
Viktor [21]

Answer:

6.75 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 16 m/s²

g = Acceleration due to gravity = 9.81 m/s²

Let y be the distance the rocket is accelerating

960-y is the distance traveled in free fall

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 16\times y+0^2}\\\Rightarrow v^2=32y\ m/s

In free fall

v^2-u^2=2g(960-y)\\\Rightarrow 0-32y=2g(960-y)\\\Rightarrow -32y=2\times -9.81(960-y)\\\Rightarrow 960-y=\dfrac{-32}{2\times -9.81}y\\\Rightarrow 960-y=1.63098878695y\\\Rightarrow 960=2.63098878695y\\\Rightarrow y=\dfrac{960}{2.63098878695}\\\Rightarrow y=364.881828749\ m

The distance the rocket will keep accelerating is 364.881828749 m

After which it will travel 960-364.881828749 = 595.118171251 m in free fall

s=ut+\frac{1}{2}at^2\\\Rightarrow 364.881828749=0t+\frac{1}{2}\times 16\times t^2\\\Rightarrow t=\sqrt{\frac{364.881828749\times 2}{16}}\\\Rightarrow t=6.75353452598\ s

The time the rocket is accelerating is 6.75 seconds

5 0
3 years ago
An AC adapter for a telephone answering machine uses a transformer to reduce the line voltage of 120 V to a voltage of 5.00 V. T
Maksim231197 [3]

Answer:

35

Explanation:

We are given that

Initial voltage,V_1=120 V

Final voltage, V_2=5 V

Number of tuns in primary coil of the transformer, N_p=840

Rms current, I_{rms}=580mA=580\times 10^{-3} A

1 mA=10^{-3} A

We have to find the number of turns  are there on the secondary coil.

We know that

\frac{N_s}{N_p}=\frac{V_2}{V_1}

Using the formula

\frac{N_s}{840}=\frac{5}{120}

N_s=\frac{5}{120}\times 840=35

Hence, there are  number of turns on the secondary coil=35

8 0
3 years ago
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