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pishuonlain [190]
3 years ago
5

Exam

Physics
2 answers:
Sergeeva-Olga [200]3 years ago
5 0
Answer: D- output force

Explanation: for example, you exert force on the handle of a shovel when you use it to lift soil. You exert the improve force that moves the machine on a certain distance, called the input distance. The force the machine exerts on the object (DIRT) is called the output force.
Sedaia [141]3 years ago
3 0
Out put work
explanation:
You might be interested in
A 5 kg pineapple is hanging completely still in mid air on a string and suddenly explodes
11111nata11111 [884]

Answer:

Explanation:

Conservation of momentum

Initial momentum is zero

3(15) + 2(v) = 0

v = - 22.5 m/s

v = 22.5 m/s downward

3 0
3 years ago
Which of the following has the lowest U value?
jarptica [38.1K]

Answer:

c. expanded polyurethane

Explanation:

Thermal performance of a building fabric is measured in terms of heat loss and is expressed as U-value or R-value. U-value is the rate of heat transferred through a structure divided by the difference in temperature across the structure with a unit of measurement of W/m²K.You can calculate the U-value of a by getting the reciprocal of the sum of thermal resistances , R, making the building material.

If you have the value of R, then U=1/R

Material                         size            R                      U

plywood                          1"              1.25                0.8

Poured concrete            2"              0.99               1.010

Expanded polyurethane  1"            6.5                   0.1538

Asbestos shingles            1"             0.03                33.33

The material with lowest U-value is expanded polyurethane

4 0
3 years ago
Units called BEATS measure the loudness of sounds.<br> true or false
gogolik [260]

Answer:

False.

Explanation:

Decibels (dB) measure sound levels

7 0
3 years ago
Read 2 more answers
A satellite is in a circular orbit around Mars, which has a mass M = 6.40 × 1023 kg and radius R = 3.40 ×106 m.
Pepsi [2]

Answer:

a) The orbital speed of a satellite with a orbital radius R (in meters) will have an orbital speed of approximately \displaystyle \sqrt\frac{4.27 \times 10^{13}}{R}\; \rm m \cdot s^{-1}.

b) Again, if the orbital radius R is in meters, the orbital period of the satellite would be approximately \displaystyle 9.62 \times 10^{-7}\, R^{3/2}\; \rm s.

c) The orbital radius required would be approximately \rm 2.04 \times 10^7\; m.

d) The escape velocity from the surface of that planet would be approximately \rm 5.01\times 10^3\; m \cdot s^{-1}.

Explanation:

<h3>a)</h3>

Since the orbit of this satellite is circular, it is undergoing a centripetal motion. The planet's gravitational attraction on the satellite would supply this centripetal force.

The magnitude of gravity between two point or spherical mass is equal to:

\displaystyle \frac{G \cdot M \cdot m}{r^{2}},

where

  • G is the constant of universal gravitation.
  • M is the mass of the first mass. (In this case, let M be the mass of the planet.)
  • m is the mass of the second mass. (In this case, let m be the mass of the satellite.)  
  • r is the distance between the center of mass of these two objects.

On the other hand, the net force on an object in a centripetal motion should be:

\displaystyle \frac{m \cdot v^{2}}{r},

where

  • m is the mass of the object (in this case, that's the mass of the satellite.)
  • v is the orbital speed of the satellite.
  • r is the radius of the circular orbit.

Assume that gravitational force is the only force on the satellite. The net force should be equal to the planet's gravitational attraction on the satellite. Equate the two expressions and solve for v:

\displaystyle \frac{G \cdot M \cdot m}{r^{2}} = \frac{m \cdot v^{2}}{r}.

\displaystyle v^2 = \frac{G \cdot M}{r}.

\displaystyle v = \sqrt{\frac{G \cdot M}{r}}.

Take G \approx 6.67 \times \rm 10^{-11} \; m^3 \cdot kg^{-1} \cdot s^{-2},  Simplify the expression v:

\begin{aligned} v &= \sqrt{\frac{G \cdot M}{r}} \cr &= \sqrt{\frac{6.67 \times \rm 10^{-11} \times 6.40 \times 10^{23}}{r}} \cr &\approx \sqrt{\frac{4.27 \times 10^{13}}{r}} \; \rm m \cdot s^{-1} \end{aligned}.

<h3>b)</h3>

Since the orbit is a circle of radius R, the distance traveled in one period would be equal to the circumference of that circle, 2 \pi R.

Divide distance with speed to find the time required.

\begin{aligned} t &= \frac{s}{v} \cr &= 2 \pi R}\left/\sqrt{\frac{G \cdot M}{R}} \; \rm m \cdot s^{-1}\right. \cr &= \frac{2\pi R^{3/2}}{\sqrt{G \cdot M}} \cr &\approx  9.62 \times 10^{-7}\, R^{3/2}\; \rm s\end{aligned}.

<h3>c)</h3>

Convert 24.6\; \rm \text{hours} to seconds:

24.6 \times 3600 = 88560\; \rm s

Solve the equation for R:

9.62 \times 10^{-7}\, R^{3/2}= 88560.

R \approx 2.04 \times 10^7\; \rm m.

<h3>d)</h3>

If an object is at its escape speed, its kinetic energy (KE) plus its gravitational potential energy (GPE) should be equal to zero.

\displaystyle \text{GPE} = -\frac{G \cdot M \cdot m}{r} (Note the minus sign in front of the fraction. GPE should always be negative or zero.)

\displaystyle \text{KE} = \frac{1}{2} \, m \cdot v^{2}.

Solve for v. The value of m shouldn't matter, for it would be eliminated from both sides of the equation.

\displaystyle -\frac{G \cdot M \cdot m}{r} + \frac{1}{2} \, m \cdot v^{2}= 0.

\displaystyle v = \sqrt{\frac{2\, G \cdot M}{R}} \approx 5.01\times 10^{3}\; \rm m\cdot s^{-1}.

5 0
3 years ago
How much kinetic energy the truck still has
seropon [69]

Answer:

47.5kJ

Explanation:

Before climbing the cliff

E_t = E_k

E_k=\frac{1}{2} mv^2\\E_k=\frac{1}{2} 500*15^2\\\\E_k = 56250J\\

At 2.

E_t = E_k+E_p

8 0
3 years ago
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