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Karolina [17]
3 years ago
11

A 1,450 kg car drives toward a 60 kg shopping cart that has a velocity of -1.2 m/s toward the car. The two objects collide, givi

ng the car a final velocity of 5.13 m/s, and the shopping cart a velocity of 11.75 m/s. What was the initial velocity of the car?
A. 5.67 m/s
B. 5.36 m/s
C. -5.36 m/s
D. -5.67 m/s
Physics
2 answers:
Margarita [4]3 years ago
8 0

Answer:

5.67 m/s

Just did it.

Y_Kistochka [10]3 years ago
4 0

Answer:

A) v₁ = 5.66 [m/s]

Explanation:

To solve this problem we must use the definition of linear momentum conservation, which tells us that momentum is conservation before and after a collision.

The linear momentum is equal to the mass by the product of the Velocity.

P = m*v

where:

P = lineal momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

Now, to the right side of the equal sign will take the linear momentum before the collision and to the left side of the equal sign as after the collision.

Pbefore = Pafter

(m₁*v₁) - (m₂*v₂) = (m₁*v₃) + (m₂*v₄)

where:

m₁ = mass of the car = 1450 [kg]

v₁ = velocity of the car before the collision [m/s]

m₂ = mass of the shopping cart = 60 [kg]

v₂ = velocity of the shopping cart before the collision = -1.2 [m/s]

v₃ = velocity of the car after the collision = 5.13 [m/s]

v₄ = velocity of the shopping cart after the collision = 11.75 [m/s]

Now replacing:

(1450*v₁) - (60*1.2) = (1450*5.13) + (60*11.75)

1450*v₁ - 72 = 7438.5 + 705

1450*v₁  = 7438.5 + 705 + 72

1450*v₁ = 8215.5

v₁ = 5.66 [m/s]

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A beam of light in air is incident at an angle of 30º to the surface of a rectangular block of clear plastic (n = 1.46). The lig
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Answer:

θ = 30°

Explanation:

Firts, the angle when the beam of light passes through the block cam be calculated using Snell Law:

n_{1}sin(\theta_{1}) = n_{2}sin(\theeta_{2})

<u>Where</u>:

n₁: is the index of refraction of the incident medium (air) = 1

θ₁: is the incident angle = 30°

n₂: is the medium 2 (plastic) = 1.46

θ₂: is the transmission angle

Hence, θ₂ is:

sin(\theta_{2}) = \frac{n_{1}*sin(\theta_{1})}{n_{2}} = \frac{1*sin(30)}{1.46} = 0.34 \rightarrow \theta_{2} = 20.03 ^{\circ}

Now, when the beam of light re-emerges from the opposite side, we have:

n₁: is the index of refraction of the incident medium (plastic) = 1.46  

θ₁: is the incident angle = 20.03°                

n₂: is the medium 2 (air) = 1

θ₂: is the transmission angle

Hence, the angle to the normal to that surface (θ₂) is:

sin(\theta_{2}) = \frac{n_{1}*sin(\theta_{1})}{n_{2}} = \frac{1.46*sin(20.03)}{1} = 0.50 \rightarrow \theta_{2} = 30 ^{\circ}  

 

Therefore, we have that the beam of light will come out at the same angle of when it went in, since, it goes from air and enters to a plastic medium and then enters again in this medium to go out to air again. This was proved using the Snell Law.    

I hope it helps you!                

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Light is reflected from a crystal of table salt with an index of refraction of 1.544. An analyser is placed to intercept the ref
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Explanation:

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Your mother asks you to carry a load of wood to the fireplace and lower it to the hearth. In doing this you have used your arms
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Answer: The correct answer is the weight of the wood.

Explanation:

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A motorcyclist travelling at 30m/s starts to apply his brake when he is 50m from the traffic light that has just turned red .if
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Answer:

9m/s²

Explanation:

Given parameters:

Initial velocity  = 30m/s

Final velocity  = 0m/s

Distance traveled  = 50m

Unknown:

Deceleration  = ?

Solution:

To solve this problem, we apply the proper motion equation;

          V²   = U²   +  2aS

V is the final velocity

U is the initial velocity

a is the acceleration

S is the distance

  Insert the parameters and solve;

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              a = -9m/s²

The negative value indicates deceleration.

The motorcycle decelerates at a rate of 9m/s²

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What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0
3 years ago
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