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WINSTONCH [101]
3 years ago
10

A 1000 kg red car traveling at 40 m/s rear ends a 3000 kg blue car traveling at 35 m/s. the cars bounce off each other and the b

lue car is accelerated to 37.5 m/s. what is the velocity of the red car after the crash?
Physics
1 answer:
klemol [59]3 years ago
8 0

Answer:

32.5 ms⁻¹

Explanation:

<em><u>Theory </u></em>

<u> The law of conservation of linear momentum </u>

The sum of linear momentum of a system under no external force remains constant.

In this scenario although different forces act on two vehicles by each other when we  consider the total system of two cars no external unbalance force act on them. So we can apply this law.

As law says,

The sum of linear momentum before collision = The sum of linear momentum after collision.

Momentum = mass × velocity

1000×40+3000×35 = 1000×v + 3000×37.5

where v = velocity of the red car after the cash

v = 32.5 ms⁻¹

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Which of the following statements is a consequence of the equation E = mc2?
lara31 [8.8K]

Answer;

B. Mass and energy are equivalent.

Explanation;

-Each of the letters of E = mc2 stands for a particular physical quantity.

E is the energy, m is the mass and c is the speed of light (m/s)

 Energy = mass x the speed of light squared

-The equation E=mc2 explains nuclear fusion, how matter can be destroyed and converted to energy and energy can be converted back to mass. It explains the atomic energy produced by nuclear power plants and the atomic energy released by atomic bombs.

-The equation tells us that mass and energy are related, and, in those rare instances where mass is converted totally into energy, how much energy that will be.

7 0
3 years ago
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Two 10-cm-diameter charged rings face each other, 25.0 cm apart. Both rings are charged too + 50.0 nC . What is the electric fie
Kamila [148]

Answer:6773.54 V/m

Explanation:

Given data

Distance between two rings(x)=25cm

Charge on each ring(Q)=50\times 10^{-9} C

diameter of ring(d)=10 cm

Now Electric field strength at distance x from ring=\frac{kxQ}{\left ( x^{2}+r^{2}\right )^{1.5}}

Where K=coulomb's constant=\frac{1}{4\pi \epsilon_0}=8.98\times 10^{9]

Now electric field strength at midpoint between two rings is zero

because one is pointing towards positive x direction and other in negative z direction.

but electric field strength at center of ring  due to other ring is given by

\left ( E\right )_{ring}=\frac{kxQ}{\left ( x^2+r^{2}\right )^{1.5}}

\left ( E\right )_{ring}=\frac{8.98\times 10^{9}\times 0.25\times 50\times 10{-9}}\left ( 0.25^2+0.05^{2}\right )}^{1.5}}

\left ( E\right )_{ring}=6773.54 V/m

4 0
3 years ago
1. A car drives at a velocity of 40 [m/s].
Llana [10]

Answer:

V=40 m/s d=150m t=? if V=d/t t=d/V t=150m/40m/s

t=3.75s

Explanation:

i am not good at explaining but i hope this will help you

6 0
3 years ago
you have two solid substances that look the same. What measurements would you take and which tests would you preform to determin
musickatia [10]
To see how much they weigh
7 0
3 years ago
A box of mass m = 1.80 kg is dropped from rest onto a massless, vertical spring with spring constant k = 2.00 ✕ 102 N/m that is
Rudik [331]

Answer:

spring compressed is 0.724 m

Explanation:

given data

mass = 1.80 kg

spring constant k = 2 × 10²  N/m

initial height = 2.25 m

solution

we know from conservation of energy is

mg(h+x)  = 0.5 × k × x²       ...................1

here x is compression in spring

so put here value in equation 1 we get

1.8 × 9.8 × (2.25+x)  = 0.5 × 2× 10² × x²

solve it we get

x = 0.724344

so spring compressed is 0.724 m

3 0
4 years ago
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