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DaniilM [7]
3 years ago
9

Two cars leave towns 680 kilometers apart at the same time and travel toward each other. one car's rate is 10 kilometers per hou

r less than the other's. if they meet in 4 hours, what is the rate of the slower car?
Physics
1 answer:
Evgen [1.6K]3 years ago
5 0

D = distance between the cars at the start of time = 680 km

v₁ = speed of one car

v₂ = speed of other car = v₁ - 10

t = time taken to meet = 4 h

distance traveled by one car in time "t" + distance traveled by other car in time "t" = D

v₁ t + v₂ t = D

(v₁ + v₂) t = D

inserting the values

(v₁ + v₁ - 10) (4) = 680

v₁ = 90 km/h

rate of slower car is given as

v₂ = v₁ - 10

v₂ = 90 - 10 = 80 km/h

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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
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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.

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Give 2 reasons for fitting heavy commercial vehicles with many tyres​
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A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

Substituting (5) in (1):

x=(23 m/s) cos(0\°) 0.63 s (6)

Finally:

x=14.49 m

5 0
2 years ago
If the net force acting on a moving object CAUSES NO CHANGE IN ITS VELOCITY, what happens to the object's momentum?
SOVA2 [1]

If the net force acting on a moving object causes no change in its velocity, the object's momentum will stay the same.

<h3>What is momentum?</h3>

Momentum of a body in motion refers to the tendency of a body to maintain its inertial motion.

The momentum is the product of its mass and velocity.

This suggests that if the net force acting on a moving object causes no change in its velocity, the momentum of the object will remain the same.

Therefore, if the net force acting on a moving object causes no change in its velocity, the object's momentum will stay the same.

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5 0
1 year ago
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