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jeyben [28]
3 years ago
8

A car with two people traveling down the road has a mass of 100 kg and a velocity of 5 m/s. The car pulls over and picks up two

people. The mass of the car is now 150 kg. What is the new momentum if the velocity stays the same?
Physics
1 answer:
Andrej [43]3 years ago
3 0

Answer:

New_momentum = 750 [kg.m/s]

Explanation:

Momentum is defined as

M  = mass* velocity

Original momentum

M_old = 100 Kg * 5 m/s = 500 [kg.m/s]

If the car pulls over and picks up two people. The mass of the car is now 150 kg, but the velocity stays the same.

New momentum

M_new = 150 Kg * 5 m/s = 750 [kg.m/s]

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A 5 meter long ladder leans against a wall. The bottom of the ladder slides away from the wall at the constant rate of 1 3 m/s.
Oksana_A [137]

Answer:9.75 m/s

Explanation:

Given

Length of ladder (L)=5 m

Foot the ladder is moving away with speed of \frac{\mathrm{d} x}{\mathrm{d} t}=13 m/s

From diagram

x^2+y^2=L^2------1

at x=3

y^2=25-9=16

y=4 m

Now differentiating equation 1 w.r.t time

2x\frac{\mathrm{d} x}{\mathrm{d} t}+2y\frac{\mathrm{d} y}{\mathrm{d} t}=0

x\frac{\mathrm{d} x}{\mathrm{d} t}=-y\frac{\mathrm{d} y}{\mathrm{d} t}

3\times 13=-4\times \frac{\mathrm{d} y}{\mathrm{d} t}

\frac{\mathrm{d} y}{\mathrm{d} t}=-\frac{3\times 13}{4}=-9.75 m/s

negative indicates distance is decreasing with time

5 0
3 years ago
Which of these best describes the relationship between the incident ray, the reflected ray, and the normal for a curved mirror?(
lions [1.4K]

For a curved mirror, all points have the same normal and the angle of incidence is also equal to the angle of reflection.

According to the laws of reflection, the incident ray, reflected ray and normal all lie on the same plane. For a curved mirror, the normal remains the same at all points along the curved mirror.

Again, the angle made between the incident ray and the normal is the same as the angle made between the reflected ray and the normal. Therefore, the angle of reflection is equal to the angle of incidence.

Learn more: brainly.com/question/17638582

8 0
2 years ago
Car A hits car B (initially at rest and of equal mass) from behind while going 15 m/s Immediately after the collision, car B mov
Mamont248 [21]

Given :

Initial speed of car A is 15 m/s and initial speed of car B is zero.

Final speed of car A is zero and final speed of car B is 10 m/s.

To Find :

What fraction of the initial kinetic energy is lost in the collision.

Solution :

Initial kinetic energy is :

K.E_i = \dfrac{15^2m}{2} + 0\\\\K.E_i = \dfrac{225 m}{2}

Final kinetic energy is :

K.E_f = \dfrac{10^2m}{2} + 0\\\\K.E_f = \dfrac{100m}{2}

Now, fraction of initial kinetic energy loss is :

Loss = \dfrac{\dfrac{225m}{2}-\dfrac{100m}{2}}{\dfrac{100m}{2}}\\\\Loss = \dfrac{125}{100}\\\\Loss = 1.25

Therefore, fraction of initial kinetic energy loss in the collision is 1.25 .

6 0
3 years ago
1. What is the momentum of a 1550 kg car that is traveling leftward at a velocity of 15 m/s?
Alik [6]

Answer:

Momentum, p = 23250 kg m/s

Explanation:

Given that

Mass of a car, m = 1550 kg

Speed pf car, v = 15 m/s

We need to find the momentum of the car. The formula for the momentum of an object is given by :

p = mv

Substituting all the values in the above formula

p = 1550 kg × 15 m/s

p = 23250 kg m/s

So, the momentum of the car is 23250 kg m/s.

3 0
3 years ago
From a set of graphed data the slope of the best fit line is found to be 1.35 m/s and the slope of the worst fit line is 1.29m/s
Svetradugi [14.3K]

Solution:

Let the slope of the best fit line be represented by 'm_{best}'

and the slope of the worst fit line be represented by 'm_{worst}'

Given that:

m_{best} = 1.35 m/s

m_{worst} = 1.29 m/s

Then the uncertainity in the slope of the line is given by the formula:

\Delta m = \frac{m_{best}-m_{worst}}{2}               (1)

Substituting values in eqn (1), we get

\Delta m = \frac{1.35 - 1.29}{2} = 0.03 m/s

8 0
3 years ago
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