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3 years ago

A satellite’s velocity is 30000m/s. After 60 secs, it’s velocity shows to 15000m/a. What is the satellite’s acceleration?

Physics

1 answer:

Orlov [11]3 years ago

4 0

Answer:

Acceleration = 9 × 10^5 m/s^2 ( deceleration )

Explanation:

From the first equation of motion:

V = u + at

15000 = 30000 + 60a

a = ( 15000-30000)/60

a = 9 × 10^5 m/s^2

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At the circus, the Human Cannonball is

svlad2 [7]

The correct answer is 1.4285714.

In physics, velocity is characterised as a vector measurement of the motion's direction and speed. To be more precise, the rate of change in an object's position relative to a frame of reference and time is another way to describe velocity. The definition of velocity simply states the rate of motion of an object in a specific direction. It determines how quickly or slowly something is going.

Velocity = distance/ time

Thus time = distance/velocity

Here velocity = 350m/s

diatnce = 500 m

time = 500/350

time = 1.42857142857

t= 200m /350m/s = 1.4285714

To learn more about velocity refer the link:

brainly.com/question/18084516

#SPJ9

6 0

1 year ago

With 51 gallons of fuel in its tank, the airplane has a weight of 2390.7 pounds. What is the weight of the plane with 81 gallons

Shtirlitz [24]

Answer: 2561.7 pounds

Explanation:

If we assume the total weight of an airplane (in pounds units) as a <u>linear function</u> of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:

$m=\frac{Y-Y_{1}}{X-X_{1}}$ (1)

Where:

$m=5.7$ is the slope of the line

$Y_{1}=2390.7pounds$ is the airplane weight with 51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)

$X_{1}=51gallons$ is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)

This means we already have one point of the graph, which coordinate is:

$(X_{1},Y_{1})=(51,2390.7)$

Rewritting (1):

$Y=m(X-X_{1})+Y_{1}$ (2)

As Y is a function of X:

$Y=f_{(X)}=m(X-X_{1})+Y_{1}$ (3)

Substituting the known values:

$f_{(X)}=5.7(X-51)+2390.7$ (4)

$f_{(X)}=5.7X-290.7+2390.7$ (5)

$f_{(X)}=5.7X+2100$ (6)

Now, evaluating this function when X=81 (talking about the 81 gallons of fuel in the tank):

$f_{(81)}=5.7(81)+2100$ (7)

$f_{(81)}=2561.7$ (8) This means the weight of the plane when it has 81 gallons of fuel in its tank is 2561.7 pounds.

3 0

3 years ago

Variations to the forehand stroke are not allowed. a. true b. false

timofeeve [1]

This is b. false

There are numerous ways of adjusting the forehand stroke and variations are allowed.

3 0

3 years ago

Read 2 more answers

2. Two cars A and B are heading forward. The velocity of A and B are 30 m/s and 20 m/s

KengaRu [80]

Answer:

A.) 27000 kgm/s

18000 kgm/s

B.) Va = 22 m/s

C.) 19800 kgm/s

25200 kgm/s

Explanation: Given that the velocity of A and B are 30 m/s and 20 m/s. And of the same mass M = 9 × 10^5g

M = 9×10^5/1000 = 900 kg

A.) Initial momentum of A

Mu = 900 × 30 = 27000 kgm/s

Initial momentum of B

Mu = 900 × 20 = 18000 kgm/s

B.) if they have an accident and then the velocity of the B is 28 m/s, find out velocity of A.

Momentum before impact = momentum after impact

Given that Vb = 28 m/s

27000 + 18000 = 900Va + 900 × 28

45000 = 900Va + 25200

900Va = 45000 - 25200

900Va = 19800

Va = 19800/900

Va = 22 m/s

C.) Momentum of A after impact

MV = 900 × 22 = 19800 kgm/s

Momentum of B after impact

MV = 900 × 28 = 25200 kgm/s

8 0

3 years ago

In medieval warfare, one of the greatest technological advancement was the trebuchet. The trebuchet was used to sling rocks into

NNADVOKAT [17]

Answer:

2) a_y= -g 3) vₓ=constant v_y = v_{oy} - g t, 4) vₓ = v₀ₓ - ax t

5) changes the horizontal speed, should change range

7) changes the vertical speed change the maximum height

Explanation:

1) After reading your long writing, we are going to solve the exercise, in the attachment you can see the different vectors.

2) The acceleration vectors are vertical and directed downwards due to the attraction of the Earth (gravity force) this force is constant, on the x axis there is no acceleration

3) the velocity vectors on the x-axis are constant because there are no relationships and the y-axis changes value according to the expression

v_y = v_{oy} - gt

at the point of maximum height, vy = 0 is equal to the maximum height

4) For someone to change the horizontal acceleration we must assume a friction with the air, in this case they relate it would be in the opposite direction to the horizontal speed

In the graph it would be directed to the left, therefore the velocity would be

vₓ = v₀ₓ - ax t

5 and 6) If someone changes the horizontal speed, they should change the range of the shot for greater horizontal speed, the rock goes further.

the equations of motion are

x = v₀ₓ t

y = v_{oy} t - ½ g t²

7) If someone changes the vertical speed change the maximum height, but not the scope of the shot, for higher speed higher maximum height,

the equations of motion are the same.

4 0

3 years ago