Ask question

Ask question

All categories

3 years ago

12

A 1500-kg car traveling due north with a speed of 25 m/s collides head-on with a 4500-kg truck traveling due south with a speed

of 15 m/s. The two vehicles stick together after the collision. What is the total momentum of the system prior to the collision

1 answer:

Goryan [66]3 years ago

8 0

Answer:

OMG IM ON THE SAME QUESTION

Explanation:

You might be interested in

what is a point of view of an object used to determine another obejects motion i nedd help asap plsss

Igoryamba

A believe that’s called a reference point.

9 0

3 years ago

In a compression wave, particles in the medium move

nikdorinn [45]

in the same direction as the wave

Explanation:

In a compression wave, the particles in the medium moves in the same direction as the wave source.

A wave is generally defined as a disturbance that transmits energy.

There are two types of waves based on the direction through which they are propagated.
Transverse waves are directed perpendicularly in the direction of propagation.
Examples are electromagnetic waves.
Longitudinal waves are parallel to their source. Examples are sound waves, p-waves.
They are made up of series of rarefaction and compression.

learn more:

Waves brainly.com/question/3183125

#learnwithBrainly

8 0

2 years ago

Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B

andreyandreev [35.5K]

Explanation:

Let $\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}$ and $\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}$

The sum of the two vectors is

$\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}$

$= 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}$

The difference between the two vectors can be written as

$\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}$

$= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}$

8 0

3 years ago

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. Complete the following statemen

Eva8 [605]

Answer:

Keeping the speed fixed and decreasing the radius by a factor of 4

Explanation:

A ball is whirled on the end of a string in a horizontal circle of radius R at constant speed v. The centripetal acceleration is given by :

$a=\dfrac{v^2}{R}$

We need to find how the "centripetal acceleration of the ball can be increased by a factor of 4"

It can be done by keeping the speed fixed and decreasing the radius by a factor of 4 such that,

R' = R/4

New centripetal acceleration will be,

$a'=\dfrac{v^2}{R'}$

$a'=\dfrac{v^2}{R/4}$

$a'=4\times \dfrac{v^2}{R}$

$a'=4\times a$

So, the centripetal acceleration of the ball can be increased by a factor of 4.

7 0

3 years ago

The doctor writes a prescription for na heparin 20,000 units in 500 ml n.s. infuse over 8 hours. what is the flow rate in ml/hr?

Ne4ueva [31]

I think this type of equation could be conducted in simple division equation since it does not involve drop rate.

we know that there is 500 ml of substance and should be infused within 8 hours period.

So the flow rate in ml/hr would be:

500/8 = 62.5 ml/hr

8 0

2 years ago