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Serggg [28]
2 years ago
14

Please help me! Uniform acceleration problem sheet:

Physics
1 answer:
Brrunno [24]2 years ago
7 0

From the calculation, the value of the acceleration is 5.8 m/s^2.

<h3>What is uniform acceleration?</h3>

The term uniform acceleration refers to a situation in which the velocity increases by equal amounts in equal time intervals.

Given the fact that the car started from rest and reached a velocity of  780.34 mph or 348.84 m/s in 1 minute of 60 seconds;

v = u + at

a = v/t

a = 348.84 m/s/ 60 seconds

a = 5.8 m/s^2

Learn more about acceleration:brainly.com/question/12550364?

#SPJ1

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Rashid [163]
Hello!

The slope of the line given by graphing pressure vs 1/Volume at constant temp for one mole of gas will give you the value for nRT from equation PV=nRT

So set nRT=slope and take the constant number mole of gas and the constant temp and solve for R the universal gas constant. You arm for pressure and litters for volume to get R in units of L*atm/mol*k

Hope this helps you! Thanks!!
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3 years ago
How many moles are in 73.4 grams of Phosphorus?
mylen [45]
23.2


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3 years ago
2. A car accelerates down the road. What is the "reaction" to the tires pushing on the road?
Artist 52 [7]
The answer would be friction
7 0
3 years ago
An isolated charged point particle produces an electric field with magnitude E at a point 2 m away. At a point 1 m from the part
guajiro [1.7K]

Explanation:

The electric field at a distance r from the charged particle is given by :

E=\dfrac{kq}{r^2}

k is electrostatic constant

if r = 2 m, electric field is given by :

E_1=\dfrac{kq}{(2)^2}\\\\=\dfrac{kq}{4}\ .....(1)

If r = 1 m, electric field is given by :

E_2=\dfrac{kq}{r_2^2}\\\\=\dfrac{kq}{1}\ ....(2)

Dividing equation (1) and (2) we get :

\dfrac{E_1}{E_2}=\dfrac{\dfrac{kq}{4}}{kq}\\\\\dfrac{E_1}{E_2}=\dfrac{1}{4}\\\\E_2=4\times E_1

So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.

4 0
3 years ago
Two pianos each sound the same note simultaneously, but they are both out of tune. On a day when the speed of sound is 349 m/s,
SSSSS [86.1K]

Answer:

Time period between the successive beats will be 0.1703 sec

Explanation:

We have given speed of the sound v = 349 m/sec

Wavelength of piano A\lambda _A=0.766m

Wavelength of piano  B\lambda _B=0.776m

So frequency of piano A f_1=\frac{v}{\lambda _1}=\frac{349}{0.766}=455.61Hz

Frequency of piano B f_2=\frac{v}{\lambda _1}=\frac{349}{0.776}=449.74Hz

So beat frequency f = 455.61 - 449.74 = 5.87 Hz

So time period T=\frac{1}{f}=\frac{1}{5.87}=0.1703sec

So time period between the successive beats will be 0.1703 sec

4 0
3 years ago
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