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katrin [286]
3 years ago
8

a busis moving with the initial velocity 10m/s . after 4 seconds, the velocity becomes 30m/s . find the acceleration produce by

bus,......please I need help​
Physics
1 answer:
Oksana_A [137]3 years ago
7 0

Answer:

5 m/s²

Explanation:

Use the acceleration formula: a=\frac{v_f-v_i}{t}

  • a = acceleration (m/s²)
  • vf = final velocity (m/s)
  • vi = initial velocity (m/s)
  • t = time (s)

Based on the information given to us by the prompt, we know:

  • vi = 10 m/s
  • vf = 30 m/s
  • t = 4 s

Substitute these values for the variables to calculate the acceleration:

a=\frac{30-10}{4}\\\\a=\frac{20}{4}\\\\a=5

Therefore, the acceleration of the bus is 5 m/s².

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Answer:

The separation distance between the parallel planes of an atom is hc/2sinθ(EK - EL)

Explanation:

The relationship between energy and wavelength is expressed below:

E = hc/λ

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Considering the condition of Bragg's law:

2dsinθ = mλ

For the first order Bragg's law of reflection:

2dsinθ = (1)λ

2dsinθ = hc/EK - EL

d = hc/2sinθ(EK - EL)

Where 'd' is the separation distance between the parallel planes of an atom, 'h' is the Planck's constant, 'c' is the velocity of light, θ is the angle of reflection, 'EK' is the energy of the K shell and 'EL' is the energy of the K shell.

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5 0
3 years ago
If the force that propels the cannonball forward is 500N, how much force will move the cannon backward?
Licemer1 [7]

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Explanation:

6 0
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A plane is flying east when it drops some supplies to a designated target below. The supplies land after falling for 10 seconds.
Ivanshal [37]

Part 1)

here we know that supply took 10 s to reach the ground

so here we will have

y = \frac{1}{2}gt^2

y = \frac{1}{2}\times 9.8 \times 10^2

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y = 490 m

Part 2)

Here all the supply covered horizontal distance of 650 m in 10 s interval of time

so here we can say

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Explanation :

It is given that,

Mass of the car, m = 1000 kg              

Force applied by the motor, F_A=1000\ N

The static and dynamic friction coefficient is, \mu=0.5

Let a is the acceleration of the car. Since, the car is in motion, the coefficient of sliding friction can be used. At equilibrium,

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\dfrac{F_A-\mu mg}{m}=a

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