You pick up 20-newton book off the floor and put it on a shelf 2 meters high. How much work did you do

Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resi

nirvana33 [79]

Answer:

a) $F = 882.63\,N$ , b) $\dot W= 4413.15\,W$ , c) $\eta = 15.216\,\%$ .

Explanation:

a) Let assume that car travel on a horizontal surface. The equations of equilibrium of the car are:

$\Sigma F_{x} = F - \mu_{r}\cdot N = 0$

$\Sigma F_{y} = N - m\cdot g = 0$

After some algebraic handling, the following expression for the propulsion force is constructed:

$F = \mu_{r}\cdot m \cdot g$

$F = (0.06)\cdot (1500\,kg)\cdot (9.807\,\frac{m}{s^{2}} )$

$F = 882.63\,N$

b) The power require to move the car at a speed of 5 meters per second is:

$\dot W = F\cdot v$

$\dot W = (882.63\,N)\cdot (5\,\frac{m}{s} )$

$\dot W= 4413.15\,W$

c) The efficiency of the car is:

$\eta = \frac{(882.63\,N)\cdot (15\,mi)\cdot (\frac{1609\,m}{1\,mi} )}{(1.4\times 10^{8}\,J)} \times 100\,\%$

$\eta = 15.216\,\%$

3 0

2 years ago

A mass M tied to a light string is moving in a vertical circle. The tension in the string at the top is TT and TB at the bottom

natita [175]

Answer:

Explanation:

Tension provides centripetal force in the circular motion . In circular motion work done by force = torque x angle

torque is zero as , centripetal force passes through axis of rotation that is center.

So work done by centripetal force = 0

So work done by tension on M = 0

5 0

2 years ago

Synchronous communications satellites are placed in a circular orbit that is 3.59 107 m above the surface of the earth. What is

suter [353]

Answer: $0.223 m/s^{2}$

Explanation:

We can solve this with the Law of Universal Gravitation and knowing the acceleration due gravity $g$ of an object above the surface of the planet decreases with the distance (height) of this object from the center of the planet.

Well, according to the law of universal gravitation:

$F=G\frac{m_{E}m}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G=6.67(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the gravitational constant

$m_{E}=5.98(10)^{24} kg$ is the mass of the Earth

$m$ are the mass of each communications satellite

$r=R_{E}+h$ is the distance between the center of the Earth and the satellite

$R_{E}=6.38(10)^{6} m$ is the radius of the Earth

$h=3.59(10)^{7} m$ is the height of the satellite, measured from the Earth's surface

On the other hand, we know according to Newton's 2nd law of motion:

$F=mg$ (2)

Combining (1) and (2):

$G\frac{m_{E}m}{r^2}=mg$ (3)

Isolating $g$ :

$g=\frac{G M_{E}}{r^2}$ (4)

Remembering $r=R_{E}+h$ :

$g=\frac{G M_{E}}{(R_{E}+h)^2}$ (5)

$g=\frac{(6.67(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24} kg)}{(6.38(10)^{6} m+3.59(10)^{7} m)^2}$

Finally:

$g=0.223 m/s^{2}$

5 0

2 years ago

Unit 12 Exam

lapo4ka [179]

Answer:

You increase the acceleration of the car.

6 0

2 years ago

A plane lands on a runway with a speed of 115 m/s, moving east, and it slows to a stop in 13.0 s. What is the magnitude (in m/s2

marin [14]

Answer:

 The planes average acceleration in magnitude and direction = 8.846 m/s² moving east

Explanation:

Acceleration: This can be defined as the rate of change of velocity. The S.I Unit of acceleration is m/s². Acceleration is a vector quantity because it can be represented both in magnitude and in direction.

Acceleration can be represented mathematically as

a = v/t.................................... Equation 1

Where a = acceleration, v = velocity, t= time.

Given: v = 115 m/s, t = 13.0 s

Substituting these values into equation 1

a = 115/13

a = 8.846 m/s² moving east

Thus the planes average acceleration in magnitude and direction = 8.846 m/s² moving east

4 0

2 years ago