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SVEN [57.7K]
3 years ago
9

A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 20 m/s by a 5620 N braking force actin

g opposite the car's motion. How far does the car move during 2.5 s? *
Physics
1 answer:
Dennis_Churaev [7]3 years ago
5 0

Answer:

the distance traveled by the car is 42.98 m.

Explanation:

Given;

mass of the car, m = 2500 kg

initial velocity of the car, u = 20 m/s

the braking force applied to the car, f = 5620 N

time of motion of the car, t = 2.5 s

The decelaration of the car is calculated as follows;

-F = ma

a = -F/m

a = -5620 / 2500

a = -2.248 m/s²

The distance traveled by the car is calculated as follows;

s = ut + ¹/₂at²

s = (20 x 2.5) + 0.5(-2.248)(2.5²)

s = 50 - 7.025

s = 42.98 m

Therefore, the distance traveled by the car is 42.98 m.

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Answer:

486nm

Explanation:

in order for an electron to transit from one level to another, the wavelength emitted is given by Rydberg Equation which states that

\frac{1}{wavelength}=R.[\frac{1}{n_{f}^{2} } -\frac{1}{n_{i}^{2} }] \\n_{f}=2\\n_{i}=4\\R=Rydberg constant =1.097*10^{7}m^{-1}\\subtitiute \\\frac{1}{wavelength}=1.097*10^{7}[\frac{1}{2^{2} } -\frac{1}{4^{2}}]\\\frac{1}{wavelength}= 1.097*10^{7}*0.1875\\\frac{1}{wavelength}= 2.06*10^{6}\\wavelength=4.86*10{-7}m\\wavelength= 486nm\\

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4 0
3 years ago
Several large firecrackers are inserted into the holes of a bowling ball, and the 6.3 kg ball is then launched into the air with
Anni [7]

Answer

given,

mass of the ball = 6.3 kg

speed of the ball = 10.4 m/s

angle made with horizontal = 43°

m_a = 1.8 kg               v_a = 2.2 m/s

m_b = 1.6 kg               v_b = 1.8 m/s

mass of third particle = 6.3 - 1.8 - 1.6

                                   = 2.9 kg

u cos θ = 10.4 x cos 43° = 7.61 m/s

by using conservation momentum along x-axis

6.3 x 7.61 = 1.8 × (-2.2) + 0 + 1.6 × V₃ₓ

V₃ₓ = 32.44 m/s (toward right)

by using conservation momentum along y-axis

0 = 0 + 1.6 x 1.8 + 1.6 × V₃y

V₃y = -1.8 m/s (indicate downward)

velocity of the third particle

v = \sqrt{32.44^2 + (-1.8)^2}

v = 32.49 m/s

tan \theta = \dfrac{-1.8}{32.44}

\theta = tan^{-1}(\dfrac{-1.8}{32.44})

θ = 3.176° (downward with horizontal)

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3 years ago
What is the relationship between mass and weight
sergiy2304 [10]
Mass is how heavy is it, weight is the size both are the same
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3 years ago
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lora16 [44]

d)through a railroad track

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3 years ago
Consider the displacement vectors Ā=(i +6j)m, B = (3i– 7j)m,
andrezito [222]

Answer:

Dx = -0.5

Dy = -0.25

Explanation:

Two vectors are given in rectangular components form as follows:

A = i + 6j

B = 3i - 7j

It is also given that:

A - B - 4D = 0

so, we solve this to find D vector:

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D = - (2/4)i - (1/4)j

D = - (1/2)i - (1/4)j

<u>D = - 0.5i - 0.25j</u>

Therefore,

<u>Dx = -0.5</u>

<u>Dy = -0.25</u>

8 0
3 years ago
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