PHYSICS COLLEGE

Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 1979.(a) He flew for 169 min at an average velocity of 3.53 m/s in a direction 45° south of east. What was his total displacement?

(b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion relative to the Earth. What was his average velocity relative to the air?

(c) What was his total displacement relative to the air mass?

Answers

Answer 1

Answer:

Answer:

a) $s=35794.2\ m$

b) $v_w=3.53\ m.s^(-1)$

c) $s_w=56074.2\ m$

Explanation:

Given:

duration of flight, $t=169* 60=10140\ s$

velocity of flight, $v=3.53\ m.s^(-1)$

direction of flight, $45^(\circ)$ to the south of east

Now the total displacement:

$s=v.t$

$s=3.53* 10140$

$s=35794.2\ m$

Velocity of air, $v_a=2\ m.s^(-1)$

When the aircraft encounters a headwind in the opposite direction to the velocity of motion then the speed of the aircraft is lowered with respect to the ground.

But when the speed is observed with respect to the wind the reduced velocity of the aircraft is observed from an opposite moving wind having a magnitude equal to the difference in velocity of the aircraft. This results in no change in the apparent velocity of the aircraft.

Mathematically:

Velocity of the aircraft with respect to the ground:

$v_(g)=v-v_a$

$v_(g)=3.53-2$

$v_g=1.53\ m.s^(-1)$

Now the velocity of the aircraft with respect to the wind:

$v_w=v_g+v_a$

$v_w=1.53+2$

$v_w=3.53\ m.s^(-1)$

Now the total displacement with respect to the wind:

$s_w=v_w.t+v_a.t$

$s_w=3.53* 10140+2* 10140$

$s_w=56074.2\ m$

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What is the density of the paint if the mass of a tin containing 5000 cm3 paint is 7 kg. If the mass of the empty tin, including the lid is 0.5 kg.

Answers

We are given:

Mass of the Paint bucket (with paint) = 7000 grams

Mass of the paint bucket (without paint) = 500 grams

Volume of Paint in the Bucket = 5000 cm³

Mass of Paint in the Bucket:

To get the mass of the paint in the bucket, we will subtract the mass of the bucket from the mass of the paint bucket (with paint)

Mass of Paint = Mass of Paint bucket (with paint) - Mass of the paint Bucket (without paint)

Mass of Paint = 7000 - 500

Mass of Paint = 6500 grams

Density of the Paint:

We know that density = Mass / Volume

Density of Paint = Mass of Paint / Volume occupied by Paint

Density of Paint = 6500/5000

Density of Paint = 1.3 grams / cm³

As a delivery truck travels along a level stretch of road with constant speed, most of the power developed by the engine is used to compensate for the energy transformations due to friction forces exerted on the delivery truck by the air and the road. If the power developed by the engine is 2.43 hp, calculate the total friction force acting on the delivery truck (in N) when it is moving at a speed of 24 m/s. One horsepower equals 746 W.

Answers

Answer:

F=75.53N

Explanation:

To calculate the power we define the equation,

$P=Fv$

Where,

F= Force

V= Velocity,

Here we have that 2.43hp is equal to 1812.78W,

clearing F,

$F=(P)/(v) = (1812)/(24)\nF=75.53N$

An interstellar space probe is launched from Earth. After a brief period of acceleration it moves with a constant velocity, 70.0% of the speed of light. Its nuclear-powered batteries supply the energy to keep its data transmitter active continuously. The batteries have a lifetime of 15.9 years as measured in a rest frame. (a) How long do the batteries on the space probe last as measured by mission control on Earth? yr
(b) How far is the probe from Earth when its batteries fail, as measured by mission control?
ly
(c) How far is the probe from Earth, as measured by its built-in trip odometer; when its batteries fail?
ly

Answers

Answer:

22.26 years

, 15.585 light years , 11.13 light years

Explanation:

$t' = t/(√(1-(v/(c*v)/c))$

= $15.9/√((1-0.7*0.7))$

= 22.26 years

0.7*c*22.26 years

=15.585 light years

0.7*c*15.9

=11.13 light years

Two children, each with a mass of 25.4 kg, are at fixed locations on a merry-go-round (a disk that spins about an axis perpendicular to the disk and through its center). One child is 0.78 m from the center of the merry-go-round, and the other is near the outer edge, 3.14 m from the center. With the merry-go-round rotating at a constant angular speed, the child near the edge is moving with translational speed of 11.5 m/s.a. What is the angular speed of each child?
b. Through what angular distance does each child move in 5.0 s?
c. Through what distance in meters does each child move in 5.0 s?
d. What is the centripetal force experienced by each child as he or she holds on?
e. Which child has a more difficult time holding on?

Answers

Answer:

a) ω₁ = ω₂ = 3.7 rad/sec

b) Δθ₁ = Δθ₂ = 18.5 rad

c) d₁ = 14.5 m d₂ = 57.5 m

d) Fc1 = 273.9 N Fc2 = 1069.8 N

e) The boy near the outer edge.

Explanation:

Since the merry-go-round is a rigid body, any point on it rotates at the same angular speed.
However, linear speeds of points at different distances from the center, are different.
Applying the definition of angular velocity, and the definition of angle, we can write the following relationship between the angular and linear speeds:

$v = \omega*r (1)$

Since we know the value of v for the child near the outer edge, and the value of r for this point, we can find the value of the angular speed, as follows:

$\omega = (v_(out) )/(r_(out) ) = (11.5m/s)/(3.14m) = 3.7 rad/sec (2)$

As we have already said, ωout = ωin = 3.7 rad/sec

Since the angular speed is the same for both childs, the angle rotated in the same time, will be the same for both also.
Applying the definition of angular speed, as the rate of change of the angle rotated with respect to time, we can find the angle rotated (in radians) as follows:
$\Delta \theta = \omega * t = 3.7 rad/sec* 5.0 sec = 18.5 rad (3)$

⇒ Δθ₁ = Δθ₂ = 18.5 rad.

The linear distance traveled by each child, will be related with the linear speed of them.
Knowing the value of the angular speed, and the distance from each boy to the center, we can apply (1) in order to get the linear speeds, as follows:

$v_(inn) = \omega * r_(inn) = 3.7 rad/sec * 0.78 m = 2.9 m/s (4)$

vout is a given of the problem ⇒ vout = 11. 5 m/s

Applying the definition of linear velocity, we can find the distance traveled by each child, as follows:

$d_(inn) = v_(inn) * t = 2.9m/s* 5.0 s = 14.5 m (5)$

$d_(out) = v_(out) * t = 11.5 m/s* 5.0 s = 57.5 m (6)$

The centripetal force experienced by each child is the force that keeps them on a circular movement, and can be written as follows:

$F_(c) = m*(v^(2))/(r) (7)$

Replacing by the values of vin and rin, since m is a given, we can find Fcin (the force on the boy closer to the center) as follows:

$F_(cin) = m*(v_(in)^(2))/(r_(in)) = 25.4 kg* ((2.9m/s)^(2) )/(0.78m) = 273.9 N (8)$

In the same way, we get Fcout (the force on the boy near the outer edge):

$F_(cout) = m*(v_(out)^(2))/(r_(out)) = 25.4 kg* ((11.5m/s)^(2) )/(3.14m) = 1069.8 N (9)$

The centripetal force that keeps the boys in a circular movement, is not a different type of force, and in this case, is given by the static friction force.
The maximum friction force is given by the product of the coefficient of static friction times the normal force.
Since the boys are not accelerated in the vertical direction, the normal force is equal and opposite to the force due to gravity, which is the weight.
As both boys have the same mass, the normal force is also equal.
This means that for both childs, the maximum possible static friction force, is the same, and given by the following expression:
$F_(frs) = \mu_(s) * m* g (10)$

If this force is greater than the centripetal force, the boy will be able to hold on.
So, as the centripetal force is greater for the boy close to the outer edge, he will have a more difficult time holding on.

A spherical shell rolls without sliding along the floor. The ratio of its rotational kinetic energy (about an axis through its center of mass) to its translational kinetic energy is:

Answers

Answer:

The ratio is $(RE)/(TE) = (2)/(3)$

Explanation:

Generally the Moment of inertia of a spherical object (shell) is mathematically represented as

$I = (2)/(3) * m r^2$

Where m is the mass of the spherical object

and r is the radius

Now the the rotational kinetic energy can be mathematically represented as

$RE = (1)/(2)* I * w^2$

Where $w$ is the angular velocity which is mathematically represented as

$w = (v)/(r)$

=> $w^2 = [(v)/(r)] ^2$

$RE = (1)/(2)* [(2)/(3) *mr^2] * [(v)/(r) ]^2$

$RE = (1)/(3) * mv^2$

Generally the transnational kinetic energy of this motion is mathematically represented as

$TE = (1)/(2) mv^2$

$(RE)/(TE) = ((1)/(3) * mv^2)/((1)/(2) * m*v^2)$

$(RE)/(TE) = (2)/(3)$

If R = 12 cm, M = 430 g, and m = 60 g , find the speed of the block after it has descended 50 cm starting from rest. Solve the problem using energy conservation principles. (Treat the pulley as a uniform disk.)