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3 years ago

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Car A of mass 1200kg traveling at 10m/s , collided in too the back of the car B, which is stationary. Following the collision,.

The two cars stick together and travel at 6m/s. What is the mass of a car B?

2 answers:

Nina [5.8K]3 years ago

4 0

Answer:

Car B has a mass of 800 kg.

General Formulas and Concepts:

<u>Momentum</u>

Law of Conservation of Momentum: $\displaystyle m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_{f}$

Explanation:

<u>Step 1: Define</u>

<em>Identify variables</em>

[Given] m₁ = 1200 kg

[Given] v₁i = 10 m/s

[Solve] m₂

[Given] v₂i = 0 m/s

[Given] vf = 6 m/s

<u>Step 2: Solve for m₂</u>

Substitute in variables [Law of Conservation of Momentum]: (1200 kg)(10 m/s) + m₂(0 m/s) = (1200 kg + m₂)(6 m/s)
Multiply: 12000 kg · m/s = (1200 kg + m₂)(6 m/s)
Isolate m₂ term: 2000 kg = 1200 kg + m₂
Isolate m₂: 800 kg = m₂

rewona [7]3 years ago

3 0

Answer:

Explanation:

Caraofmass1200kg

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You exert a force of 25 newtons while you move a rock 15 meters. How much work did you perform?

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3 years ago

Two 51 g blocks are held 30 cm above a table. As shown in the figure, one of them is just touching a 30-long spring. The blocks

vivado [14]

The concept of this question can be well understood by listing out the parameters given.

The mass of the block = 51 g = 51 × 10⁻³ kg
The distance of the block from the table = 30 cm
Length of the spring = 30 cm

The purpose is to determine the spring constant.

Let us assume that the two blocks are Block A and Block B.

At point A on block A, the initial velocity on the block is zero

i.e. u = 0

We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.

$\mathsf{S = ut + \dfrac{1}{2}gt^2}$

From the above formula,

The distance (S) = 30 cm; we need to convert the unit to meter (m).

Since 1 cm = 0.01 m
Then, 30cm = 0.3 m

The acceleration (g) due to gravity = 9.8 m/s²

∴

inputting the values into the equation above, we have;

$\mathsf{0.3 = (0)t + \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 = \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 =4.9*(t^2)}$

By dividing both sides by 4.9, we have:

$\mathsf{t^2 = \dfrac{0.3}{4.9}}$

$\mathsf{t^2 = 0.0612}$

$\mathsf{t = \sqrt{0.0612}}$

$\mathsf{t =0.247 \ seconds}$

However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.

By applying the equation of the time period of a simple harmonic motion.

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$

where the relation between time (t) and period (T) is:

$\mathsf{t = \dfrac{T}{2}}$

T = 2t

T = 2(0.247)

T = 0.494 seconds

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$

By making the spring constant k the subject of the formula:

$\mathsf{\dfrac{T}{2 \pi } = \sqrt{ \dfrac{m}{k}}}$

$\Big(\dfrac{T}{2 \pi }\Big)^2 = { \dfrac{m}{k}$

$\dfrac{T^2}{(2 \pi)^2 }= { \dfrac{m}{k}$

$\mathsf{ T^2 *k = 2 \pi^2*m} \\ \\ \mathsf{ k = \dfrac{2 \pi^2*m}{T^2}}$

$\mathsf{ k =\Big( \dfrac{(2 \pi)^2*(51 \times 10^{-3})}{(0.494)^2} \Big) N/m}$

$\mathbf{ k =8.25 \ N/m}$

Therefore, we conclude that the spring constant as a result of instantaneous rest caused by the compression of the spring is 8.25 N/m.

Learn more about simple harmonic motion here:

brainly.com/question/17315536?referrer=searchResults

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3 years ago

calculate the frequency and time period of sound wave of 35 m wave length propagating at a speed of 3500m/s

zheka24 [161]

Answer:

Frequency = 100 Hz

Time period = 0.01 sec

Explanation:

f= 3500/35 = 100

T = 1/f = .01

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2 years ago

Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o

kobusy [5.1K]

Answer:

$\Delta \lambda=14.3\ nm$

Explanation:

It is given that,

The number of lines per unit length, N = 900 slits per cm

Distance between the formed pattern and the grating, l = 2.3 m

n the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm, $\Delta Y=2.98\ mm = 0.00298\ m$

Let d is the slit width of the grating,

$d=\dfrac{1}{N}$

$d=\dfrac{1}{900\ cm}$

$d=1.11\times 10^{-5}\ m$

For the first wavelength, the position of maxima is given by :

$y_1=\dfrac{L\lambda_1}{d}$

For the other wavelength, the position of maxima is given by :

$y_2=\dfrac{L\lambda_2}{d}$

So,

$\Delta \lambda=\dfrac{\Delta y d}{l}$

$\Delta \lambda=\dfrac{0.00298\times 1.11\times 10^{-5}}{2.3}$

$\Delta \lambda=1.43\times 10^{-8}\ m$

or

$\Delta \lambda=14.3\ nm$

So, the difference between these wavelengths is 14.3 nm. Hence, this is the required solution.

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3 years ago