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2 years ago

How much force is needed to accelerate a 66 kg skier at 2 m/sec^2?

Physics

1 answer:

Ymorist [56]2 years ago

5 0

The equation of force is M(mass)•A(acceleration)=132N

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A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a

Elza [17]

Answer:

$\dfrac{dz}{dt}=0.65\ ft/s$

Explanation:

Given that

x= 150 ft

$\dfrac{dy}{dt}= 7\ ft/s$

y= 14 ft

From the diagram

$z^2=x^2+y^2$

When ,x= 150 ft and y= 14 ft

$z^2=150^2+14^2$

$z=\sqrt{150^2+15^2}$

z=150.74 ft

$z^2=x^2+y^2$

By differentiating with respect to time t

$2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}$

$z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}$

Here x is constant that is why

$\dfrac{dx}{dt}=0$

$z\dfrac{dz}{dt}= y\dfrac{dy}{dt}$

Now by putting the values in the above equation we get

$150.74\times \dfrac{dz}{dt}=14\times 7$

$\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s$

$\dfrac{dz}{dt}=0.65\ ft/s$

Therefore the distance between balloon and observer increasing with 0.65 ft/s.

5 0

3 years ago

A 900 kg car speeds up from 30 M/S to 80 m/s how much momentum did the car gain?

AlexFokin [52]

The change in momentum of the car is 45,000 kg m/s

Explanation:

The change in momentum of an object is given by:

$\Delta p = m(v-u)$

where

m is the mass of the object

u is its initial velocity

v is its final velocity

For the car in this problem, we have

m = 900 kg

u = 30 m/s

v = 80 m/s

Therefore, the change in momentum is:

$\Delta p = (900)(80-30)=45,000 kg m/s$

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

#LearnwithBrainly

3 0

3 years ago

A ball is released from a tower at a height of 100 meters toward the roof of another tower that is 25 meters high. The horizonta

I am Lyosha [343]

-- During the time the ball is flying from the high roof to the low roof,
it's going to fall (100-25) = 75 meters.
How long does it take an object dropped from rest to fall 75 meters ?

Distance = (1/2) · (gravity) · (time)²

75 m = (4.9 m/s²) · (time)²

Time² = (75 m) / (4.9 m/s²)
Time² = 15.31 sec²
Time = √(15.31 sec²) = 3.91 seconds

So the ball has to cover the horizontal distance of 20 meters
in 3.91 seconds.

Distance = (speed) · (time)

20 m = (speed) · (3.91 sec)

Speed = (20 m) / (3.91 sec)

Speed = 5.11 m/s

7 0

3 years ago

Peak hydraulic systems demands may be met by the use of what ?

nikklg [1K]

By the use of a Accumulators

8 0

2 years ago

Read 2 more answers

A testable question is a question that _____. A. might be asked on a science test B. can be answered with either a yes or no C.

Cloud [144]

Answer:

D. can be answered by gathering observations.