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3 years ago

Describe your acceleration as you skate down your neighborhood sidewalk.

Physics

1 answer:

Paraphin [41]3 years ago

6 0

Answer: when you increase or decrease your speed.

Explanation:

Moving a skate at rest, you need to apply force in order to cause acceleration.

F = ma Where

F = force applied

m = mass of the skate

a = acceleration

The initial velocity u will be equal to zero and the skate will acceleration to a certain velocity.

as you skate down your neighborhood sidewalk, you will accelerate when you increase your speed. Because

Acceleration is the rate of change of velocity. That is,

Acceleration = change in velocity/ time.

And also, you will decelerate when you reduce the speed or velocity down your neighborhood sidewalk.

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How much work is done on a satellite in a circular orbit about earth?

alexgriva [62]

Answer:

0 J

Explanation:

$W$ = Work done on the satellite in circular orbit about earth by earth

$F$ = Force on gravity on satellite by earth

$d$ = displacement of the satellite

$\theta$ = Angle between the force on gravity and displacement = 90

We know that, Work done is given as

$W = F d Cos\theta\\W = F d Cos90\\W = F d (0)\\W = 0 J$

3 0

3 years ago

if a ball with an original velocity of 0 is dropped from a tall structure and takes 7 Seconds to hit the ground what velocity do

krok68 [10]

$a_y=\dfrac{v_y-v_{0y}}t\implies-9.81\,\dfrac{\mathrm m}{\mathrm s^2}=\dfrac{v_y-0}{7\,\mathrm s}\implies v_y=-68.7\,\dfrac{\mathrm m}{\mathrm s}$

5 0

3 years ago

Two trains A and B of length 400 m each are moving on two parallel tracks with

Vinvika [58]

Answer:

The initial distance between the trains is 1450 m. 

Explanation:

In the question two trains are of equal length 400 m and moves at a uniform speed of 72 km/h. train A is moving ahead of train B. If the train B has to overtake train A it should accelerate.

Train B’s acceleration is $1m/s^2$ and it accelerated for 50 seconds.

 $a=1 m/s^2$ 

t=50 s 

initial speed u=72km/h 

we have to convert this speed into m/s 

 $u=72 \times 5/18=20 m/s$ 

Distance covered in accelerating phase $S=ut+1/2 at^2$ 

 $=20 \times 50+1/2 \times 1 \times 50^2$ 

 $=1000+1250=2250 m$ 

If a train is just behind another, the distance covered by the train located behind during overtaking phase will be equal to the sum of the lengths of the trains.

Here length of train A+length of train $B=400+400=800 m$ 