A racing car has a mass of 1525kg. What is its kinetic energy if it has a speed of 108km/h

The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the drag force?

timurjin [86]

Answer: The drag force goes up by a factor of 4

Explanation:

The Drag Force equation is:

$F_{D}=\frac{1}{2}C_{D}\rho A_{D}V^{2}$ (1)

Where:

$F_{D}$ is the Drag Force

$C_{D}$ is the Drag coefficient, which depends on the material

$\rho$ is the density of the fluid where the bicycle is moving (air in this case)

$A_{D}$ is the transversal area of the body or object

$V$ the bicycle's velocity

Now, if we assume $C_{D}$ , $\rho$ and $A_{D}$ do not change, we can rewrite (1) as:

$F_{D}=C.V^{2}$ (2)

Where $C$ groups all these coefficients.

So, if we have a new velocity $V_{n}$ , which is the double of the former velocity:

$V_{n}=2V$ (3)

Equation (2) is written as:

$F_{D}=C.V_{n}^{2}=C.(2V)^{2}$

$F_{D}=4CV^{2}$ (4)

Comparing (2) and (4) we can conclude the Drag force is four times greater when the speed is doubled.

7 0

3 years ago

During the experiment it is determined that, as the cart rolls between two points on the track, the work done on the cart by the

natita [175]

Answer:

Explanation:

If the work done on the cart is NET work

Then the work will result in an increase in kinetic energy

KE₀ + W = KE₁

½mv₀² + W = ½mv₁²

½(0.80)(0.61²) + 0.91 = ½(0.80)v₁²

v₁ = 1.626991...

v₁ = 1.6 m/s

4 0

2 years ago

n outer space, a constant net force with a magnitude of 140 N is exerted on a 32.5 kg probe initially at rest. A) What accelerat

musickatia [10]

Answer:

a) 4.31 m/s²

b) 215.5 m

Explanation:

a) According to Newton's first law of motion

The net force applied to particular mass produced acceleration, a, according to

F = ma

F = 140 N

m = 32.5 kg

a = ?

140 = 32.5 × a

a = 140/32.5 = 4.31 m/s²

b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s

u = initial velocity of the probe = 0 m/s (since it was initially at rest)

a = 4.31 m/s²

t = 10 s

s = distance travelled = ?

s = ut + at²/2

s = 0 + (4.31×10²)/2 = 215.5 m

7 0

3 years ago

If a player through a basketball to the target with an initial velocity of 17 m/s making an angle of 30 degrees with the horizon

Svetllana [295]

Answer:

The final position made with the vertical is 2.77 m.

Explanation:

Given;

initial velocity of the ball, V = 17 m/s

angle of projection, θ = 30⁰

time of motion, t = 1.3 s

The vertical component of the velocity is calculated as;

$V_y = Vsin \theta\\\\V_y = 17 \times sin(30)\\\\V_y = 8.5 \ m/s$

The final position made with the vertical (Yf) after 1.3 seconds is calculated as;

$Y_f = V_yt - \frac{1}{2}g t^2\\\\Y_f = (8.5 \times 1.3 ) - (\frac{1}{2} \times 9.8 \times 1.3^2)\\\\Y_f = 11.05 \ - \ 8.281\\\\Y_f = 2.77 \ m$

Therefore, the final position made with the vertical is 2.77 m.

3 0

2 years ago

How would you compare and contrast the 3 types of friction?

vlabodo [156]

Rolling friction . the force that slows down the movement of a rolling object
sliding friction.
Sliding friction : The opposing force that comes into play when one body is actually sliding over the surface of the other body is called sliding friction. e.g. A flat block is moving over a horizontal table.

Kinetic or dynamic friction: If the applied force is increased further and sets the body in motion, the friction opposing the motion is called kinetic friction

7 0

3 years ago

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