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AVprozaik [17]
3 years ago
9

How much work is done if a 15 newton suitcase is lifted 2 m ?​

Physics
1 answer:
nirvana33 [79]3 years ago
3 0

Answer:

196 J

Explanation:

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If I was a scientist and I wanted to measure the intensity of an earthquake I would use
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Answer:

A seismograph

Explanation:

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2 years ago
A gas was compressed to 30.0 mL at 1.5 atm from 65<br>mL. What was the original pressure?​
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A gas as 30.0 Ml because pile of 1.4 add 1 that equals. To 1.5 and remove 10 so thebpresssure is 1399.93
4 0
3 years ago
You throw a ball downward from a window at a speed of 2.5 m/s. how fast will it be moving when it hits the sidewalk 2.1 m below?
Otrada [13]
List out all the variables that you do know; 

acceleration=-9.8 ms⁻¹ (this remains constant on Earth)
Final velocity=?
Displacement (s)= -2.1 m 
Initial Velocity(u)=2.5 ms⁻¹

v²=u²+2as 
v²=(2.5)²+2(-9.8)(-2.1)
v²=47.41 
v=√47.41 
v=6.88549 ≈ 6.9 ms⁻¹ 

Hope I helped :) 
4 0
3 years ago
A 2.0-kilogram ball traveling north at 4.0 meters per second collides head on with a 1.0-kilogram ball traveling south at 8.0 me
enyata [817]

Answer:

We know the momentum after the collision MUST be equal to the momentum BEFORE the collision.  

Momentum is a VECTOR quantity having both magnitude and direction.  The first ball has momentum P =m*v = 2*4 = 8 at 90degrees.  The second ball has momentum P = 1*8 = 8 at -90 or 270 degrees.  They sum to zero when you perform vector addition.

Explanation:

6 0
2 years ago
(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W
Kitty [74]

Answer:

The minimum value of width for first minima is λ

The  minimum value of width for 50 minima is 50λ

The  minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

D\sin\theta=n\lambda

D=\dfrac{n\lambda}{\sin\theta}

Where, D = width of slit

\lambda = wavelength

Put the value into the formula

D=\dfrac{n\lambda}{\sin\theta}

Here, \sin\theta should be maximum.

So. maximum value of \sin\theta is 1

Put the value into the formula

D=\dfrac{1\times\lambda}{1}

D=\lambda

(b). If the minimum number  is 50

Then, the width is

D=\dfrac{50\times\lambda}{1}

D=50\lambda

(c). If the minimum number  is 1000

Then, the width is

D=\dfrac{1000\times\lambda}{1}

D=1000\lambda

Hence, The minimum value of width for first minima is λ

The  minimum value of width for 50 minima is 50λ

The  minimum value of width for 1000 minima is 1000λ

4 0
3 years ago
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