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3 years ago

An airplane travels east at an air speed of 400.0 m/s into a head wind of 35.0 m/s. What is the airplane's ground speed?

Physics

1 answer:

mezya [45]3 years ago

7 0

Velocity of the plane relative to the air = +400.0 m/s

Velocity of the air relative to the ground = -35.0 m/s (negative because it's moving opposite to the eastward direction)

Velocity of the plane relative to the ground = (velocity of the plane relative to the air) + (velocity of the air relative to the ground) = 365.0 m/s to the east

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3 years ago

Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and

djverab [1.8K]

The magnitude of the average friction force exerted on the collar (F)= 8.641 N

<h3>How can we calculate the magnitude of the average friction force exerted on the collar?</h3>

To calculate the magnitude of the average friction force exerted on the collar we are using the formula,

$\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c} = mgy$

Here we are given,

k = The spring has a spring constant.

= 25.5 N/m.

$x_f$ = Final length of the spring .

= $\sqrt{1.25^2+1.8^2} -0.60$

= 1.591 m

$x_i$ = The initial length of the spring.

= 1.25−0.60

=0.65 m

y=The collar then travels downward a distance.

= 1.80 m.

m= The mass of the collar.

=3.55 kg

$v_c$ = the velocity of the collar.

= 3.39 m/s.

g = The acceleration due to gravity.

= 9.81 m/s²

We have to calculate the magnitude of the average friction force exerted on the collar = F

Now we put the known values in the above equation, we get;

$\frac{1}{2} k(x^2_f - x^2_i ) + F\times y + \frac{1}{2} m v^{2} _{c} = mgy$

Or, $\frac{1}{2} \times 25.5 \times((1.591)^2 - (0.65)^2 ) + F\times 1.80 + \frac{1}{2}\times 3.55\times (3.39)^{2} = 3.55\times 9.81\times 1.80$

Or, F= 8.641 N

From the above calculation we can conclude that,

The magnitude of the average friction force exerted on the collar (F)= 8.641 N

Learn more about friction:

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Disclaimer: This question is incomplete in the portal. Here is the complete question.

Question:

The 3.55 kg collar shown below is attached to a spring and released from rest at A. The collar then travels downward a distance of y = 1.80 m. The spring has a spring constant of k = 25.5 N/m. The distance a is given as 1.25 m. The datum for gravitational potential energy is set at the horizontal line through A and B. Determine the magnitude of the average friction force exerted on the collar when the velocity of the collar at c is 3.39 m/s and the spring has an unstretched length of 0.60 m .

6 0

2 years ago

A 1900kg car starts from rest and drives around a flat 65-m-diameter circular track. The forward force provided by the car's dri

s344n2d4d5 [400]

Answer:

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Explanation:

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3 years ago

Sean climbs a tower that is 71.3 m high to make a jump with a parachute. The mass of Sean plus the parachute is 81.4 kg. If U =

myrzilka [38]

Answer:

U = 56877.4 J

Explanation:

The potential energy of a body is that which it possesses because it is located at a certain height above the surface of the earth and can be calculated using the following formula:

U = mgh Formula (1)

Where:

U is the potential energy in Joules (J)

m is the mass of the body in kilograms (kg)

g is the acceleration due to gravity (m/s²)

h is the height at which the body is found from the surface of the earth in meters (m)

Data

m= 81.4 kg

g= 9.8 m/s²

h = 71.3 m

Potential energy of Sean and the parachute at the top of the tower

We replace data in the formula (1)

U = m*g*h

U = (81.4 kg)*(9.8 m/s²)*(71.3 m)

U = 56877.4 N*m

U = 56877.4 J

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4 years ago