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Lina20 [59]
3 years ago
9

Which of the following are true for acceleration?

Physics
1 answer:
Fittoniya [83]3 years ago
6 0

The SI unit for acceleration is m/s2 ( D)

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What is the potential energy of a 3 kilogram-ball that is on the ground?
Aleks [24]
The answer is 0
potential energy= mass• gravity (9.8m/s)• height.
Your mass 3kg. I believe your need to convert this to grams (3000g) and then multiply by 9.8. Since your height is 0 because the ball is on the ground then your potential energy is 0
3 0
3 years ago
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A motorcycle that is slowing down uniformly covers 2.0 successive km in 80 s and 120 s, respectively. Calculate (a) the accelera
8090 [49]

Answer:

acceleration = -0.042 m/s²

velocity at beginning =  14.167 m/s

velocity at  end = 5.7183 m/s

Explanation:

given data

distance d1 = 1 km

distance d2 = 2 km

time  t1 = 80 s

time t2 = 120 s + 80s = 200 s

to find out

acceleration and velocity at beginning and end

solution

we apply here law of motion that is

d = vt + 1/2×at²    

put value

1000 = v(80) + 1/2×a(80)²           ........................1

and

2000 = v(200) + 1/2×a(200)²      ........................2

so from equation 1 and 2 we get a and v

a = -0.042 m/s² and

v = 14.167 m/s

so by kinematic final velocity will be

V² = v² + 2ad

V² = (14.167)² + 2×(-0.042)×(2000)

V²  = 32.70

V = 5.7183 m/s

so

acceleration = -0.042 m/s²

velocity at beginning =  14.167 m/s

velocity at  end = 5.7183 m/s

8 0
3 years ago
To solve a problem using the equation for keplerâs third law, enrico must convert the average distance of mars from the sun from
Natali5045456 [20]
1 astronomical unit = 149597870700m
Enrico should divide distance in meters with this number.
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Witch is the property of matter in witch a substance can transfer heat or electricity
saw5 [17]

Answer:

B. Conductivity.

Explanation:

Conductivity is the quantity of heat passing per second through a slab of unit cross-sectional area when the temperature gradient between the two faces is unity when put in heat.

4 0
4 years ago
Light of wavelength O is passed through a diffraction grating with N lines/meter and then lands on a screen a distance L from th
rodikova [14]

Condition for diffraction

dsin\theta = m\lambda

Where

a = Distance between slits

m = Order of the fringes

\lambda = Wavelength

\theta = At the angle between the ray of light and the projected distance perpendicular between the two objects

For small angles

sin\theta = \approx tan\theta

Where

tan\theta = \frac{Y}{L}

Where L is the distance between the slits and Y the length of the light.

Replacing we have

d\frac{Y}{L} = \lambda m

Y = \frac{m\lambda L}{d}

The distance between slits d can be expressed also as d= \frac{L}{N} Where N is the number of the fringes, then

Y_n = mN\lambda L

Similarly when there is added a new Fringe we have the change of the distance would be :

Y_{n+1} = (m+1)N\lambda L

Linear distance between fringes is

\Delta Y = \Delta Y_{m+1}-Y_m

\Delta Y = (m+1)N\lambda L - mN\lambda L

Therefore the answer is

\Delta Y = N\lambda L

8 0
3 years ago
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