Ask question

Ask question

All categories

2 years ago

11

A truck is driving north at 35 miles per hour and passes a car going south at 40 miles per hour. What is the speed of the car fr

om the truck driver’s frame of reference?

2 answers:

inna [77]2 years ago

7 0

Answer:

70

Explanation:

right...

Citrus2011 [14]2 years ago

7 0

Answer:

120

………………..........................................

You might be interested in

Are temperature and thermal energy the same thing justify your answer

Sever21 [200]

Temperature is the measurement of the average energy of the particles in a solid, liquid or gas and thermal energy is the total energy in a set amount of solid, liquid or gas. Therefore, the temperature and thermal energy is not the same thing. They are both about the particle theory, which is a theory that all particles of solid, liquid or gas are always in motion. But the difference between the two is that temperature is the "measurement" of the particles in a solid, liquid or gas and the thermal energy is the total energy in a set amount of solid, liquid or gas.

6 0

3 years ago

A rock is being twirled in a circle on the end of a string. The string provides the centripetal force needed to keep the ball mo

KonstantinChe [14]

Answer:

No

Explanation:

The force of tension exerted by the string on the rock acts as centripetal force, so its direction is always towards the centre of the circle.

However, the direction of motion of the rock is always tangential to the circle: this means that the force is always perpendicular to the direction of motion of the rock.

As we know, the work done by a force on an object is

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the force and the displacement

In this situation, F and d are perpendicular, so $\theta=90^{\circ}$ , therefore $cos \theta = 0$ and the work done is zero:

$W=0$

4 0

3 years ago

A 12,000kg. railroad car is traveling at +2m/s when it

ivann1987 [24]

<u>Answer:</u>

The final velocity of the two railroad cars is 1.09 m/s

<u>Explanation:</u>

Since we are given that the two cars lock together it shows that the collision is inelastic in nature. The final velocity due to inelastic collision is given by

$\mathrm{V}=\frac{V 1 M 1+V 2 M 2}{M 1+M 2}$

where

V= Final velocity

M1= mass of the first object in kgs = 12000

M2= mas of the second object in kgs = 10000

V1= initial velocity of the first object in m/s = 2m/s

V2= initial velocity of the second object in m/s = 0 (given at rest)

Substituting the given values in the formula we get

V = 2×12000 + 0x100012000 + 10000= 2400022000= 1.09 m/s

$\mathrm{V}=\frac{2 \times 1200+0 \times 1000}{12000+10000}=\frac{24000}{22000}=1.09 \mathrm{m} / \mathrm{s}$

Which is the final velocity of the two railroad cars

8 0

3 years ago

Once potential energy is lost on a roller coaster it cannot be gained all the way back true or false

nignag [31]

Answer:

true

Explanation:

because the roller coaster can't work without energy

4 0

3 years ago

Starting from Newton’s law of universal gravitation, show how to find the speed of the moon in its orbit from the earth-moon dis

WARRIOR [948]

Answer: 1010.92 m/s

Explanation:

According to Newton's law of universal gravitation:

$F=G\frac{Mm}{r^{2}}$ (1)

Where:

$F$ is the gravitational force between Earth and Moon

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

$M=5.972(10)^{24} kg$ is the mass of the Earth

$m=7.349(10)^{22} kg$ is the mass of the Moon

$r=3.9(10)^{8} m$ is the distance between the Earth and Moon

Asuming the orbit of the Moon around the Earth is a circular orbit, the Earth exerts a centripetal force on the moon, which is equal to $F$ :

$F=m.a_{C}$ (2)

Where $a_{C}$ is the centripetal acceleration given by:

$a_{C}=\frac{V^{2}}{r}$ (3)

Being $V$ the orbital velocity of the moon

Making (1)=(2):

$m.a_{C}=G\frac{Mm}{r^{2}}$ (4)

Simplifying:

$a_{C}=G\frac{M}{r^{2}}$ (5)

Making (5)=(3):

$\frac{V^{2}}{r}=G\frac{M}{r^{2}}$ (6)

Finding $V$ :

$V=\sqrt{\frac{GM}{r}}$ (7)

$V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24} kg)}{3.9(10)^{8} m}}$ (8)

Finally:

$V=1010.92 m/s$

5 0

3 years ago