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kodGreya [7K]
2 years ago
14

3. The space shuttle aims for an orbit about 250 km above the surface of the earth. In orbit, the mass

Physics
1 answer:
Contact [7]2 years ago
7 0

The orbital speed of the space shuttle is 2.12 x 10⁴ m/s.

<h3>Orbital speed of the space shuttle</h3>

The orbital speed of the space shuttle is calculated as follows;

V = \sqrt{\frac{GM}{r} }

where;

  • G is gravitational constant
  • M is mass of the Earth
  • r is the distance from the center point of the planet

V = \sqrt{\frac{6.67 \times 10^{-11} 5.98 \times 10^{24}}{(250 + 638) \times 10^3} }\\\\V = 2.12 \times 10^4 \ m/s

Thus, the orbital speed of the space shuttle is 2.12 x 10⁴ m/s.

Learn more about orbital speed here: brainly.com/question/22247460

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Masteriza [31]
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8 0
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A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass,
Citrus2011 [14]

Answer:

v = 7934.2 m/s

Explanation:

Here the total energy of the Asteroid and the Earth system will remains conserved

So we will have

-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2

now we know that

v_0 = 660 m/s

M = 5.98 \times 10^{24} kg

m = 5 \times 10^9 kg

r = 4 \times 10^9 m

R = 6.37 \times 10^6 m

now from above formula

GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2

now we have

2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2

now plug in all data

2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2

v = 7934.2 m/s

5 0
3 years ago
Find the mass and center of mass of the solid E with the given density function ρ. E lies under the plane z = 3 + x + y and abov
makvit [3.9K]

Answer:

The mass of the solid is 16 units.

The center of mass of the solid lies at (0.6875, 0.3542, 2.021)

Work:

Density function: ρ(x, y, z) = 8

x-bounds: [0, 1], y-bounds: [0, x], z-bounds: [0, x+y+3]

The mass M of the solid is given by:

M = ∫∫∫ρ(dV) = ∫∫∫ρ(dx)(dy)(dz) = ∫∫∫8(dx)(dy)(dz)

First integrate with respect to z:

∫∫8z(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8x+8y+24](dx)(dy)

Then integrate with respect to y:

∫[8xy+4y²+24y]dx, evaluate y from 0 to x

= ∫[8x²+4x²+24x]dx

Finally integrate with respect to x:

[8x³/3+4x³/3+12x²], evaluate x from 0 to 1

= 8/3+4/3+12

= 16

The mass of the solid is 16 units.

Now we have to find the center of mass of the solid which requires calculating the center of mass in the x, y, and z dimensions.

The z-coordinate of the center of mass Z is given by:

Z = (1/M)∫∫∫ρz(dV) = (1/16)∫∫∫8z(dx)(dy)(dz)

<em>Calculate the integral then divide the result by 16.</em>

First integrate with respect to z:

∫∫4z²(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[4(x+y+3)²](dx)(dy)

= ∫∫[4x²+24x+8xy+4y²+24y+36](dx)(dy)

Then integrate with respect to y:

∫[4x²y+24xy+4xy²+4y³/3+12y²+36y]dx, evaluate y from 0 to x

= ∫[28x³/3+36x²+36x]dx

Finally integrate with respect to x:

[7x⁴/3+12x³+18x²], evaluate x from 0 to 1

= 7/3+12+18

Z = (7/3+12+18)/16 = <u>2.021</u>

The y-coordinate of the center of mass Y is given by:

Y = (1/M)∫∫∫ρy(dV) = (1/16)∫∫∫8y(dx)(dy)(dz)

<em>Calculate the integral then divide the result by 16.</em>

First integrate with respect to z:

∫∫8yz(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8xy+8y²+24y](dx)(dy)

Then integrate with respect to y:

∫[4xy²+8y³/3+12y²]dx, evaluate y from 0 to x

= ∫[20x³/3+12x²]dx

Finally integrate with respect to x:

[5x⁴/3+4x³], evaluate x from 0 to 1

= 5/3+4

Y = (5/3+4)/16 = <u>0.3542</u>

<u />

The x-coordinate of the center of mass X is given by:

X = (1/M)∫∫∫ρx(dV) = (1/16)∫∫∫8x(dx)(dy)(dz)

<em>Calculate the integral then divide the result by 16.</em>

First integrate with respect to z:

∫∫8xz(dx)(dy), evaluate z from 0 to x+y+3

= ∫∫[8x²+8xy+24x](dx)(dy)

Then integrate with respect to y:

∫[8x²y+4xy²+24xy]dx, evaluate y from 0 to x

= ∫[12x³+24x²]dx

Finally integrate with respect to x:

[3x⁴+8x³], evaluate x from 0 to 1

= 3+8 = 11

X = 11/16 = <u>0.6875</u>

<u />

The center of mass of the solid lies at (0.6875, 0.3542, 2.021)

4 0
3 years ago
The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t
AVprozaik [17]

Answer:

v=115 m/s

or

v=414 km/h

Explanation:

Given data

A_{area}=0.140m^{2}\\  p_{air}=1.21 kg/m^{3}\\  m_{mass}=80kg

To find

Terminal velocity (in meters per second and kilometers per hour)

Solution

At terminal speed the weight equal the drag force

mg=1/2*C*p_{air}*v^{2}*A_{area}\\   v=\sqrt{\frac{2*m*g}{C**p_{air}*A_{area}} }\\ Where C=0.7\\v=\sqrt{\frac{2*9.8*80}{1.21*0.14*0.7} }\\ v=115m/s

For speed in km/h(kilometers per hour)

To convert m/s to km/h you need to multiply the speed value by 3.6

v=(115*3.6)km/h\\v=414km/h

5 0
3 years ago
To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm throug
Romashka [77]

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

\tau = Fr

Where,

F = Force

r = Radius

Replacing we have that,

\tau = Fr

\tau = 21cm (\frac{1m}{100cm})* 550N

\tau = 11.55Nm

The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore

I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2

I = 0.394kg\cdot m^2

Finally, angular acceleration is a result of the expression of torque by inertia, therefore

\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}

\alpha = \frac{11.55}{0.394}

\alpha = 29.3 rad/s^2

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians (\pi / 3), therefore

W = \tau \theta

W = 11.5* \frac{\pi}{3}

W = 12.09J

4 0
3 years ago
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