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mrs_skeptik [129]

3 years ago

12

16. At a particular point, a satellite in an elliptical orbit has a gravitational potential energy of 7000 MJ with respect to Ea

rth’s surface and a kinetic energy of 4000 MJ. At another point in its orbit, the satellite’s potential energy is 2000 MJ. What is its kinetic energy at that point? Show all calculations leading to your answer.

1 answer:

WARRIOR [948]3 years ago

8 0

Assuming conservation of energy:

Energy before = 7000 + 4000 = 11 000 MJ

Energy after = 2000 + Kinetic = 11 000

Therefore Kinetic = 11 000 - 2 000

Kinetic = 9 000 MJ

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sing a rope that will snap if the tension in it exceeds 361 N, you need to lower a bundle of old roofing material weighing 455 N

yaroslaw [1]

Answer:

a)-2m/s^2

b)27.2m/s

Explanation:

Hello! The first step to solve this problem is to find the mass of the block remembering that the definition of weight force is mass by gravity (g=9.8m / s ^ 2)

W=455N=weight

W=mg

W=455N=weight

$m=\frac{W}{g} =\frac{455}{9.81}=46.38kg$

The second step is to draw the free body diagram of the body (see attached image) and use Newton's second law that states that the sum of the forces is equal to mass by acceleration

$a=\frac{T-W}{m} =\frac{361-455}{46.38kg} =-2m/s^2$

for point b we use the equations of motion with constant acceleration to find the velocity

$Vf=\sqrt{X(2)(a)+Vo^2}$

Where

Vf = final speed

Vo = Initial speed =0

A = acceleration =2m/s

X = displacement =6.8m

Solving

$Vf=\sqrt{(6.8)(2)(2)+0^2}=27.2m/s$

4 0

3 years ago

A vector has a magnitude of 46.0 m and points in a direction 20.0° below the positive x-axis. A second vector, , has a magnitude

irina1246 [14]

Answer with Step-by -step explanation:

We are given that

b. $\mid A\mid=46 m$

$\theta=20^{\circ}$ below the positive x-axis

Therefore, the angle made by vector A in counter clockwise direction when measure from positive x-axis= $x=360-20=340^{\circ}$

x-component of vector A= $A_x=\mid A\mid cosx=46cos 340=46\times 0.94=43.24$

y-Component of vector A= $A_y=\mid A\mid sinx=46sin340=46(-0.34)=-15.64$

Magnitude of vector B=86 m

The vector B makes angle with positive x- axis= $x'=42^{\circ}$

x-component of vector B= $B_x=86cos42=63.64$

y-Component of vector B= $B_y=86sin42=57.62$

Vector A= $A_xi+A_yj=43.24i-15.64j$

Vector B= $B_xi+B_yj=63.64i+57.62j$

Vector C=A+B

Substitute the values

$C=43.24i-15.64j+63.64i+57.62j$

$C=106.88i+41.98j$

c.Direction= $\theta=tan^{-1}(\frac{y}{x})=tan^{-1}(\frac{41.98}{106.88})=21.5^{\circ}$

The direction of the vector C=21.5 degree

6 0

3 years ago

In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between st

bekas [8.4K]

Answer:

a) t = 746 s

b) t = 666 s

Explanation:

a)

Total time will be the sum of the partial times between stations plus the time stopped at the stations.
Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

$v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s (1)$

Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

$t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)$

Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

$x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m (3)$

In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

$t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)$

We can find the distance traveled while the train was decelerating as follows:

$x_{3} = (v_{1} * t_{3}) + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m (5)$

Finally, we need to know the time traveled at constant speed.
So, we need to find first the distance traveled at the constant speed of 26.4m/s.
This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)

The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

$t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s (7)$

Total time between two stations is simply the sum of the three times we have just found:
t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
tm = t*5 = 131.6 * 5 = 658.2 s (9)
Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

b)

Using all the same premises that for a) we know that the only difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
We can find the distance traveled at constant speed, rewriting (6) as follows:

x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)

The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

$t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s (12)$

Total time between two stations is simply the sum of the three times we have just found:
t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
tm = t*3 = 207.4 * 3 = 622.2 s (14)
Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)

7 0

2 years ago

What process is occurring in the picture?

Alex777 [14]

Two fat black arrows are swimming along together, when they see a single skinny black arrow coming toward them. They are afraid of strangers, and they know that the skinny one must be mean and tough if it's not afraid to travel alone. So they turn to the side and get out of the skinny arrow's way.

3 0

3 years ago

A 1 900-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.00 m before coming into contact

Leno4ka [110]

Answer:

471392.4 N

Explanation:

From the question,

Just before contact with the beam,

mgh = Fd.................... Equation 1

Where m = mass of the beam, g = acceleration due to gravity, h = height. F = average Force on the beam, d = distance.

make f the subject of the equation

F = mgh/d................ Equation 2

Given: m = 1900 kg, h = 4 m, d = 15.8 = 0.158 m

Constant: g = 9.8 m/s²

Substitute into equation 2

F = 1900(4)(9.8)/0.158

F = 471392.4 N

6 0

3 years ago