Table C: Average Speeds for Lower Racetrack

. A rope is being used to pull a mass of 10 kg vertically upward. Determine the tension on the rope, if starting from rest, the

telo118 [61]

$\text{Given that,}\\\\\text{Mass, m =10 kg}\\\\\text{Time, t = 8 sec}\\\\\text{Velocity, v = 4~m/s}\\\\\text{When a body is moving upwards,}\\\\\text{Tension,}~ T=mg +ma\\\\~~~~~~~~~~~~~~~=mg+m\left(\dfrac{v-u}t \right)\\\\~~~~~~~~~~~~~~~=10(10)+10\left(\dfrac{4-0}8\right)\\\\~~~~~~~~~~~~~~~=100+10\left(\dfrac 12\right)\\\\~~~~~~~~~~~~~~~=100+5\\\\~~~~~~~~~~~~~~~=105~N$

5 0

2 years ago

An airplane flies at 150 km/hr. (a) The airplane is towing a banner that is b = 0.8 m tall and l = 25 m long. If the drag coef-

maw [93]

Answer:

Power requirement P for the banner is found to be 30.62 W
Power requirement P for the solid flat plate is found to be 653.225 W
Answer for part(c) is explained below in the explanation section and can be summarized as: The main difference between the drags and power requirements of the two objects of same size was due to their significantly different drag-coefficients. The Cd for banner was given, whereas the Cd for a flat plate is generally found to be around 1.28 which is the value we used in our calculations that resulted in a huge increase of power to tow the flat plate
Power requirement P for the smooth spherical balloon was found to be 40.08 W

Explanation:

First of all we will establish variables and equations known that are known to us to solve this question. Since we are given the velocity of the airplane:

v = velocity of airplane i.e. 150 km/hr. To convert it into m/s we will divide it by 3.6 which gives us 41.66 m/s
The density of air at s.t.p (standard temperature pressure) is given as d = 1.225 kg / m^3
The power can be determined this equation: P = F . v, where F represents the drag-force that we will need to determine and v represents the velocity of the airplane
The equation to determine drag-force is: $F = 1/2 * d * C_d * A$

In the drag-force equation Cd represents the co-efficient of drag and A represents the frontal area of the banner/plate/balloon (the object being towed)

Frontal area A of the banner is : 25 x 0.8 = 20 m^2

Part a) We will plug in in the values of Cd, d, A in the drag-force equation i.e. Fd = 1/2 * 0.06* 1.225 * 20 = 0.735 N. Now to find the power P we will use P = F . v i.e. 0.735 * 41.66 = 30.62 W

Part b) For this part the only thing that has fundamentally changed is the drag-coefficient Cd since it's now of a solid flat plate and not a banner. The drag-coefficient of a flat plate is approximately given as : Cd_fp = 1.28

Now we will plug-in our values into the same equations as above to determine drag-force and then power. i.e. Fd = 1/2 * 1.28 * 1.225 * 20 = 15.68 N. Using Fd to determine power, P = 15.68 * 41.66 = 653.225 W

Part c) The main reason for such a huge power difference between two objects of same size was due to their differing drag-coefficients, as drag-coefficients are generally large for objects that are not of a streamlined shape and leave a large wake (a zone of low air pressure behind them). The flat plate being solid had a large Cd where as the banner had a considerably low Cd and therefore a much lower power consumption

Part d) The power of a smooth sphere can be calculated in the same manner as the above two. We just have to look up the Cd of a smooth sphere which is found to be around 0.5 i.e. Cd_s = 0.5. Area of sphere A is given as : pi* r^2 (r = d / 2). Now using the same method as above:

Fd = 1/2 * 0.5 * 3.14 * 1.225 = 0.962 N

P = 0.962 * 41.66 = 40.08 W

4 0

3 years ago

Two satellites, A and B are in different circular orbits

jek_recluse [69]

Answer:

The ratio of the orbital time periods of A and B is $\frac{1}{2}$

Solution:

As per the question:

The orbit of the two satellites is circular

Also,

Orbital speed of A is 2 times the orbital speed of B

$v_{oA} = 2v_{oB}$ (1)

Now, we know that the orbital speed of a satellite for circular orbits is given by:

$v_{o} = \farc{2\piR}{T}$

where

R = Radius of the orbit

Now,

For satellite A:

$v_{oA} = \farc{2\piR}{T_{a}}$

Using eqn (1):

$2v_{oB} = \farc{2\piR}{T_{a}}$ (2)

For satellite B:

$v_{oB} = \farc{2\piR}{T_{b}}$ (3)

Now, comparing eqn (2) and eqn (3):

$\frac{T_{a}}{T_{b}} = \farc{1}{2}$

6 0

2 years ago

What is the ratio of the intensities of an earthquake P wave passing through the Earth and detected at two points 14 km and 49 k

Molodets [167]

Answer:

$\dfrac{I_1}{I_2}=12.25$

Explanation:

$r_1$ = 14 km

$r_2$ = 49 km

Intensity of a wave is inversely proportional to distance

$I\propto \dfrac{1}{r^2}$

So,

$\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{49^2}{14^2}\\\Rightarrow \dfrac{I_1}{I_2}=12.25$

The ratio of the intensities is $\dfrac{I_1}{I_2}=12.25$

6 0

3 years ago

Properties of convex mirror

Ahat [919]

Answer:

The image result of an object reflected by a convex mirror is typically virtual, upright, and smaller. Discover how moving the object farther away from the mirror's surface affects the size of the virtual image formed behind the mirror

Explanation:

6 0

3 years ago