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Artist 52 [7]
3 years ago
11

a 2kg marble is moving at 3 m/s when it strikes another 4kg marble moving in the opposite direction at -3 m/s. What will be the

velocity of the 4kg marble after the collision of the 2kg marble's new velocity is -4 m/s. Also state the direction of the 2kg marble after they hit.
Physics
1 answer:
Elan Coil [88]3 years ago
4 0
Momentum is conserved after the collision 

Momentum of 2 Kg before collision = 2 * 3 = 6
Momentum of 4 kg  before collision = 4 * -3 = -12

so 6 + -12 = 2 * -4 + 4 *x   where x is velocity of 4kg marble.

4x - 8 = -6
4x = 2
x = 0.5 

Velocity of 4 kg marble is 0.5 m/s after collision

The 2 kg marble will move in the opposite direction to which it was moving before the collision.
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Which of the following properties is the same for all electromagnetic radiation in a vacuum?
Crazy boy [7]
It is wavelength i think so!!!!
3 0
3 years ago
A converging lens has a focal length of 20 cm. An object 1 cm tall is placed 10 cm from the center of the lens. What is the heig
SCORPION-xisa [38]

Answer: 2 cm

Explanation:

Given , for a converging lens

Focal length : f=20\ cm

Height of object : h=1\ cm

Object distabce from lens : u=-10\ cm

Using lens formula: \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}, we get

\dfrac{1}{20}=\dfrac{1}{v}+\dfrac{1}{10}, where v = image distance from the lens.

On solving aboive equation , we get

\dfrac{1}{v}=\dfrac{1}{20}-\dfrac{1}{10}=\dfrac{1-2}{20}=\dfrac{-1}{20}\Rightarrow\ v=-20\ cm

Formula of Magnification : m=\dfrac{v}{u}=\dfrac{h'}{h} , where h' is the height of image.

Put value of u, v and h in it , we get

\dfrac{-20}{-10}=\dfrac{h'}{1}\\\\\Rightarrow\ h'=2\ cm

Hence, the height of the image is 2 cm.

3 0
3 years ago
A certain car can accelerate from 0 to 60 mph in 7.9 s. What is the car's average acceleration in mph/s?
Anna007 [38]

Answer:

<em>a = 7.6\ mph/s</em>

Explanation:

<u>Motion With Constant Acceleration </u>

It's a type of motion in which the velocity of an object changes uniformly in time.

The equation that describes the change of velocities is:

v_f=v_o+at

Where:

a   = acceleration

vo = initial speed

vf  = final speed

t    = time

Solving the equation [for a:

\displaystyle a=\frac{v_f-v_o}{t}

The car accelerates from vo=0 to vf=60 mph in t=7.9 s, thus the acceleration is:

\displaystyle a=\frac{60 \ mph-0}{7.9}

a = 7.6\ mph/s

7 0
3 years ago
A proposed communication satellite would revolve around the earth in a circular orbit in the equatorial plane at a height of 358
aleksandr82 [10.1K]

Answer:

Period is 86811.5 seconds.

Explanation:

{ \boxed{ \bf{T {}^{2} =  (\frac{4 {\pi}^{2} }{GM}) {r}^{3}   }}}

{ \tt{T {}^{2}  =  \frac{4 {(3.14)}^{2} }{(6.6 \times  {10}^{ - 11} ) \times (5.98 \times  {10}^{24} )} \times  {((35880\times  {10}^{3}) } + (6370 \times  {10}^{3} )) {}^{3}   }} \\  \\ { \tt{T {}^{2}  =  7.54 \times {10}^{9} }} \\ { \tt{T =  \sqrt{7.54 \times  {10}^{9} } }} \\ { \tt{T = 86811.5 \: seconds}}

6 0
3 years ago
1500 kg Peugeot car is traveling at 16.67 m s ⁻¹ and accelerates to 30.56 m s ⁻¹ for 2 minutes. Calculate the impulse of the car
Elis [28]

Answer:

2084 kg*m/s

Explanation:

Impulse is change in momentum

Mathematically;

Impulse, J= F*t=mΔv   where

F= ma = 1500 * { 30.56 - 16.67}/2*60

F= 1500 *0.11575

F=  174 N

J=F*t

J= 174*120*0.1

J= 2084 kg*m/s

5 0
3 years ago
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