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3 years ago

15

A train car has a mass of 10,000 kg and is moving at +3.0 m/s. It strikes an identical train car that is at rest. The train cars

combine. What is the magnitude of the final velocity of the combined cars? 0 m/s 1.5 m/s 3.0 m/s 6.0 m/s

1 answer:

jolli1 [7]3 years ago

7 0

Answer:1.5

Explanation:

Edg

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Photoelectrons with a maximum speed of 7.00 · 105 m/s are ejected from a surface in the presence of light with a frequency of 8.

aliya0001 [1]

Answer:

2.2 x 10-19

Explanation:

Kinetic Energy = 1/2 m v ^2

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3 years ago

M=3000km v=25m/s what’s the momentum

valina [46]

<h3>Answer:</h3>

Momentum of the given body will be : 75000 Kg m/s

<h3>Explanation:</h3>

According to Newton's first law of motion, all bodies continue to be in the state of rest or motion unless an external force is applied on the body. We can use this in the case of momentum also

The formula of momentum is given by :

$:\implies \sf\quad \sf \: P = mv$

Here, we are given the mass of the body ( m ) as 3000kg and the velocity of the body ( v ) as 25 m/s. On putting the values in the formula:

$\begin{aligned}&:\implies \sf\quad \sf \: P = mv \\& :\implies \sf\quad \sf \: P = 3000 \times 25 \\ & :\implies \sf\quad \sf \: \boxed{ \sf \: P = 75000kgm {s}^{ - 1} } \end{aligned}$

Momentum is associated with the mass of the moving body and can be defined as the quantity of motion measured as a product of mass and velocity.

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2 years ago

What is the accletation of a 1500kg with a net force of 7500 N

matrenka [14]

Acceleration formulae is:
a=Fnet/mass
According to the question
a=7500N/1500kg
a=5m/s sq.

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3 years ago

Which type of nuclear reaction is used in commercial power generation?

Alisiya [41]

B. Fission

Should be correct

4 0

3 years ago

Read 2 more answers

Consider the speed of light in another universe to be only 100 m/s. Two cars are traveling along a highway in opposite direction

olganol [36]

Answer:

-62.45m/s and +62.45m/s

Explanation:

The formula for relativistic speed

This is the speed of A with respect to B

$V_{AB}=\frac{V_{A}-V_{B}}{1-\frac{V_{A}V_{B}}{C^2} }$

where

$V_{A}$ will be the velocity of person 1: 39m/s

$V_{B}$ will be the velocity of person 2: -31m/s (negative because is travelling in opposite direction)

and $C$ the velocity of light: 100m/s

The velocity of person 1 measured by person 2 is:

$V_{AB}=\frac{39m/s-(-31m/s)}{1-\frac{(39m/s)(-31m/s)}{(100m/s)^2}}=62.45m/s$

The velocity of person 2 measured by person 1 is:

$V_{BA}=\frac{V_{B}-V_{A}}{1-\frac{V_{B}V_{A}}{C^2} }$

$V_{BA}=\frac{-31m/s-39m/s}{1-\frac{(-31m/s)(39m/s)}{(100)^2} }=-62.45m/s$

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3 years ago