PHYSICS COLLEGE

If my weight on Earth is 140lbs, what is my mass?

Answers

Answer 1

Answer:

Answer:

63.57 kg

Explanation:

weight = 140 lbs

Let the mass is m.

1 lbs = 4.45 N

The weight of an object is defined as the force with which our earth attracts the body towards its centre.

Weight is the product of mass of the body and the acceleration due to gravity of that planet.

W = m x g

On earth surface g = 9.8 m/s^2

Now convert lbs in newton

So, 140 lbs = 140 x 4.45 = 623 N

So, m x 9.8 = 623

m = 63.57 kg

Thus, the mass is 63.57 kg.

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A river 500 ft wide flows with a speed of 8 ft/s with respect to the earth. A woman swims with a speed of 4 ft/s with respect to the water.1) If the woman heads directly across the river, how far downstream is she swept when she reaches the opposite bank?
2) If she wants to be swept a smaller distance downstream, she heads a bit upstream. Suppose she orients her body in the water at an angle of 37° upstream (where 0° means heading straight accross, how far downstream is she swept before reaching the opposite bank?
3) For the conditions, how long does it take for her to reach the opposite bank?

Answers

Answer:

1) $\Delta s=1000\ ft$

2) $\Delta s'=998.11\ ft.s^(-1)$

3) $t\approx125\ s$

$t'\approx463.733\ s$

Explanation:

Given:

width of river, $w=500\ ft$

speed of stream with respect to the ground, $v_s=8\ ft.s^(-1)$

speed of the swimmer with respect to water, $v=4\ ft.s^(-1)$

Now the resultant of the two velocities perpendicular to each other:

$v_r=√(v^2+v_s^2)$

$v_r=√(4^2+8^2)$

$v_r=8.9442\ ft.s^(-1)$

Now the angle of the resultant velocity form the vertical:

$\tan\beta=(v_s)/(v)$

$\tan\beta=(8)/(4)$

$\beta=63.43^(\circ)$

Now the distance swam by the swimmer in this direction be d.

so,

$d.\cos\beta=w$

$d* \cos\ 63.43=500$

$d=1118.034\ ft$

Now the distance swept downward:

$\Delta s=√(d^2-w^2)$

$\Delta s=√(1118.034^2-500^2)$

$\Delta s=1000\ ft$

On swimming 37° upstream:

The velocity component of stream cancelled by the swimmer:

$v'=v.\cos37$

$v'=4* \cos37$

$v'=3.1945\ ft.s^(-1)$

Now the net effective speed of stream sweeping the swimmer:

$v_n=v_s-v'$

$v_n=8-3.1945$

$v_n=4.8055\ ft.s^(-1)$

The component of swimmer's velocity heading directly towards the opposite bank:

$v'_r=v.\sin37$

$v'_r=4\sin37$

$v'_r=2.4073\ ft.s^(-1)$

Now the angle of the resultant velocity of the swimmer from the normal to the stream:

$\tan\phi=(v_n)/(v'_r)$

$\tan\phi=(4.8055)/(2.4073)$

$\phi=63.39^(\circ)$

Now let the distance swam in this direction be d'.

$d'* \cos\phi=w$

$d'=(500)/(\cos63.39)$

$d'=1116.344\ ft$

Now the distance swept downstream:

$\Delta s'=√(d'^2-w^2)$

$\Delta s'=√(1116.344^2-500^2)$

$\Delta s'=998.11\ ft.s^(-1)$

Time taken in crossing the rive in case 1:

$t=(d)/(v_r)$

$t=(1118.034)/(8.9442)$

$t\approx125\ s$

Time taken in crossing the rive in case 2:

$t'=(d')/(v'_r)$

$t'=(1116.344)/(2.4073)$

$t'\approx463.733\ s$

A rocket with a mass of 62,000 kg (including fuel) is burning fuel at the rate of 150 kg/s and the speed of the exhaust gases is 6,000 m/s. If the rocket is fired vertically upward from the surface of the Earth, determine its height after 744 kg of its total fuel load has been consumed. Since the mass of fuel consumed is small compared to the total mass of the rocket, you can consider the mass of the rocket to be constant for the time interval of interest.

Answers

Answer:

h≅ 58 m

Explanation:

GIVEN:

mass of rocket M= 62,000 kg

fuel consumption rate = 150 kg/s

velocity of exhaust gases v= 6000 m/s

Now thrust = rate of fuel consumption×velocity of exhaust gases

=6000 × 150 = 900000 N

now to need calculate time t = amount of fuel consumed÷ rate

= 744/150= 4.96 sec

applying newton's law

M×a= thrust - Mg

62000 a=900000- 62000×9.8

acceleration a= 4.71 m/s^2

its height after 744 kg of its total fuel load has been consumed

$h= (1)/(2)at^2$

$h= (1)/(2)4.71*4.96^2$

h= 58.012 m

h≅ 58 m

A crate of eggs is located in the middle of the flatbed of a pickup truck as the truck negotiates a curve in the flat road. The curve may be regarded as an arc of a circle of radius 35.0 m. If the coefficient of static friction between crate and truck is 0.600, how fast can the truck be moving without the crate sliding?

Answers

Answer:

v = 14.35 m/s

Explanation:

As we know that crate is placed on rough bed

so here when pickup will take a turn around a circle then in that case the friction force on the crate will provide the necessary centripetal force on the crate

So here we have

$\mu mg = (mv^2)/(R)$

here we have

$\mu g = (v^2)/(R)$

now we know that

$v = √(\mu Rg)$

here we have

$\mu = 0.600$

R = 35 m

g = 9.81 m/s/s

now plug in all values in above equation

$v = √((0.600)(35)(9.81))$

$v = 14.35 m/s$

A spring-loaded gun, fired vertically, shoots a marble 9.0 m straight up in the air. What is the marble's range if it is fired horizontally from 1.8 m above the ground?

Answers

Final answer:

The range of the marble when fired horizontally from 1.8m above the ground can be calculated using the equations of motion in physics. First, the time of flight is found using the vertical motion and then the range is calculated using the time of flight and the initial velocity determined from the vertical launch. The marble's range is approximately 8.4m.

Explanation:

To solve this problem, we need to make use of the concept of projectile motion in physics. The most crucial part in solving this type of problem is to break the motion into its horizontal and vertical components.

First, we find the time the projectile is in the air using the vertical motion. Ignoring air resistance, the time a projectile is in the air is determined by the initial vertical velocity and the height from which it drops. Here, the height is given as 1.8m and we can use the equation h = 0.5gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. After calculating, we find that the time the marble is in the air is about 0.6 seconds.

Now, we can use the time to find the horizontal distance traveled by the marble, a.k.a the range. The range is given by R = vt, where v is the horizontal velocity, which is the same as the initial vertical velocity. From the problem, we know the marble reached a height of 9.0m when shot vertically, which we can use to find the initial velocity using the equation v = sqrt(2gh), where g is the acceleration due to gravity (9.8 m/s^2) and h is the height. We find that the initial velocity is about 14 m/s.

So, the range R = vt = 14m/s * 0.6s = 8.4m. Therefore, the marble's range when fired horizontally from 1.8m above the ground is approximately 8.4m.

Learn more about Projectile Motion here:

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The area of the top side of a piece of sheet metal is given below. The sheet metal is submerged horizontally in 8 feet of water. Find the fluid force on the top side. (The weight-density of water is 62.4 pounds per cubic foot.) 3 square feet

Answers

Answer:

Explanation:

Given that,

Height at which the metal sheet is submerge

H= 8ft

Fluid Force at the top side with area sheet of 3ft², A = 3ft²

Weight density of water

w = ρg = 62.4 lb/ft³

The fluid pressure can be calculated using

P = ρgh

But we are already give the weight density ρg

Then,

P = w•h

P = 62.4 × 8

P = 499.2 lb/ft²

The force from top side?

We can calculate the force using the formula for pressure

Pressure = Force /area

P = F/A

F = P×A

F = 499.2 × 3

F = 1497.6 lbs

Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal energy dissipated from brakes in a 1600 kg car that descends a 15 ∘ hill. The car begins braking when its speed is 95 km/h and slows to a speed of 40 km/h in a distance of 0.34 km measured along the road.

Answers

Answer:

1838216 J

Explanation:

95 km/h = 26.39 m/s

40 km/h = 11.11 m/s

Initial kinetic energy

= .5 x 1600 x(26.39)²

= 557145.67 J

Final kinetic energy

= .5 x 1600 x ( 11.11)²

= 98745.68 J

Loss of kinetic energy

= 458400 J

Loss of potential energy

= mg x loss of height

= 1600 x 9.8 x 340 sin 15

= 1379816 J

Sum of Loss of potential energy and Loss of kinetic energy

= 1379816 + 458400

= 1838216 J

This is the work done by the friction . So this is heat generated.

Final answer:

To calculate the thermal energy dissipated from the brakes of a car, use the equation Q = Mgh/10, where Q is the energy transferred to the brakes, M is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill. The temperature change of the brakes can then be calculated using the equation Q = mc∆T, where m is the mass of the brakes and c is its specific heat capacity.

Explanation:

The thermal energy dissipated from the brakes of a car can be calculated by converting the gravitational potential energy lost by the car into internal energy of the brakes. By using the equation Q = Mgh/10, where Q is the energy transferred to the brakes, M is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill, we can calculate the thermal energy dissipated. From there, the temperature change of the brakes can be calculated using the equation Q = mc∆T, where m is the mass of the brakes and c is its specific heat capacity.

Learn more about Thermal energy dissipation here:

brainly.com/question/16043710

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