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IrinaK [193]
3 years ago
6

The drag coefficient for a newly designed hybrid car is predicted to be 0.21. the cross-sectional area of the car is 30 ft2. det

ermine the aerodynamic drag on the car when it is driven through still air at 55 mph.
Physics
1 answer:
Nadya [2.5K]3 years ago
3 0
Drag, D = 1/2*Drag coefficient*density of air*velocity^2*Frontal area

Where,
Density of air ≈ 1.225 kg/m^3
Velocity = 55 mph = 24.5872 m/s
Drag coefficient = 0.21
Frontal area = 30 ft^2 = 2.78709 m^2

Substituting;
D = 0.5*0.21*1.225*24.5872^2*2.78709 = 216.72 N
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The speed of light in a transparent medium is 0.6 times that of its speed in vacuum. Find the refractive index of the medium.
grin007 [14]
<h2>Answer:</h2>

The refractive index is 1.66

<h2>Explanation:</h2>

The speed of light in a transparent medium is 0.6 times that of its speed in vacuum .

Refractive index of medium = speed of light in vacuum / speed of light in medium  

So

RI = 1/0.6 = 5/3 or 1.66

3 0
3 years ago
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A cyclist racing in the keiran is riding at the top of the track at 5m/s. Then he sprints downhill in the sprinting to the finis
kumpel [21]
The cyclist is moving by uniformly accelerated motion, with an initial velocity of v_i=5~m/s and an acceleration of a=9~m/s^2. 
The acceleration is given by
a= \frac{v_f-v_i}{\Delta t}
where v_f is the final velocity and \Delta t is the time between the end and the beginning of the motion, and in our case it is 1.75 s. Therefore, from this relationship we can find the final velocity:
v_f=v_i + a \Delta t=5~m/s+9~m/s^2 \cdot 1.75~s=20.75~m/s
6 0
4 years ago
The bowling ball is whizzing down the bowling lane at 4 m/s. If the mass of the bowling ball is 7 kg, what is its kinetic energy
Lisa [10]
Kinetic Energy = 1/2xmassx(velocity)^2
Input values;
K.E=1/2x7kgx(4m/s)^2
K.E.=56J
3 0
4 years ago
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At 20°c, the resistance of a sample of nickel is 525 ohms. what is the resistance when the sample is heated to 70°C?​
Ipatiy [6.2K]

The resistance of the sample is 682.5\Omega

Explanation:

The relationship between resistance of a material and temperature is given by the equation

R(T)=R_0(1+\alpha (T-T_0))

where

R_0 is the resistance at the temperature T_0

\alpha is the temperature coefficient of resistance

For the sample of nickel in this problem, we have:

R_0 = 525 \Omega when the temperature is T_0 = 20^{\circ}C

While the temperature coefficient of resistance of nickel is

\alpha = 0.006/^{\circ}C

Therefore, the resistance of the sample when its temperature is

T=70^{\circ}C

is

R=(525)(1+0.006(70-20))=682.5 \Omega

Learn more about resistance:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

#LearnwithBrainly

3 0
3 years ago
How much work should be done to lift a 5kg brick to the height of 12m
Goshia [24]

Answer: 588 joules

Explanation:

Work is done when force is applied on an object over a distance ( whether vertical or horizontal). It is measured in joules.

Thus, Workdone = Force X distance

- Vertical distance to be moved by the brick = 12 metres

- Mass of box = 5kg

- Acceleration due to gravity when box was lifted represented by g is a constant with value of 9.8m/s^2

Now, recall that Force = Mass x acceleration due to gravity

i.e Force = 5kg x 9.8m/s^2

Force = 49 Newton

So, Workdone = Force X Distance

Workdone = 49 Newton X 12 metres

Workdone = 588 joules

Thus, 588 joules of work was done.

8 0
3 years ago
Read 2 more answers
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