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ollegr [7]
3 years ago
8

A child pulls a wagon at a constant velocity along a level sidewalk. The child does this by applying a 22 newton force to the wa

gon handle, which is inclined at 35° to the sidewalk as shown below. What is the magnitude of the fore of friction on the wagon?
Physics
1 answer:
Jlenok [28]3 years ago
5 0

Answer:

The answer is 18 N.

Explanation:

A force can be divided into components x and y components. The component along the x-axis is called the horizontal component and along the y-axis is called the vertical component. In this case, as the force is in a horizontal direction and is also known as x-component of force. The x- component of force is  

Fx = Fcosθ

Fx = 22(cos 35°)

Fx = 22 x 0.819

Fx =  18 N

Child's horizontal pull forces are equal to that of frictional resistance force on the wagon.

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At what angle does the sun hit the temperate zone?​
Nostrana [21]

Answer:

at the 90 degree angle... i think

Explanation:

7 0
3 years ago
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You're in your room blasting music with door shut, your mom opens your door. Now music is heard through out your home. this is e
Sliva [168]

Answer:

A

Explanation:

Diffraction, as the waves spread out (specifically spread to the whole house) by passing the door.

5 0
2 years ago
On a hot day, the deck of a small ship reaches a temperature of 48
AlekseyPX

The final temperature of the seawater-deck system is 990°C.

<h3>What is heat?</h3>

The increment in temperature adds up the thermal energy into the object. This energy is Heat energy.

The deck of a small ship reaches a temperature Ti= 48.17°C seawater on the deck to cool it down. During the cooling, heat Q =3,710,000 J are transferred to the seawater from the deck. Specific heat of seawater= 3,930 J/kg°C.

Suppose for 1 kg of sea water, the heat transferred from the system is given by

3,710,000 = 1 x 3,930 x (T - 48.17)

T = 990°C  to the nearest tenth.

The final temperature of the seawater-deck system is 990°C.

Learn more about heat.

brainly.com/question/13860901

#SPJ1

6 0
1 year ago
(a) Calculate the magnitude of the gravitational force exerted by Mars on a 80 kg human standing on the surface of Mars. (The ma
andrew11 [14]

Answer:

a) F=1.044\times 10^9\ N

b)F'=1.044\times 10^9\ N

c) F_p=1.0672\times10^{-7}\ N

d) Treat the humans as though they were points or uniform-density spheres.

Explanation:

Given:

  • mass of Mars, M=6.4\times 10^{23}\ kg
  • radius of the Mars, r=3.4\times 10^{6}\ m
  • mass of human, m=80\ kg

a)

Gravitation force exerted by the Mars on the human body:

F=G.\frac{M.m}{r^2}

where:

G=6.67 \times 10^{-11}\ m^3.kg^{-1}.s^{-2} = gravitational constant

F=6.67\times10^{-11}\times \frac{6.4\times 10^{23}\times 80}{(3.4\times 10^{6})^2}

F=1.044\times 10^9\ N

b)

The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.

F'=F

F'=1.044\times 10^9\ N

c)

When a similar person of the same mass is standing at a distance of 4 meters:

F_p=6.67\times10^{-11}\times \frac{80\times 80}{4}

F_p=1.0672\times10^{-7}\ N

d)

The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.

  • Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.
  • Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.
4 0
3 years ago
If a system has 225 kcal of work done to it, and releases 5.00 × 102 kj of heat into its surroundings, what is the change in int
vovikov84 [41]

We can solve the problem by using the first law of thermodynamics:

\Delta U = Q-W

where

\Delta U is the change in internal energy of the system

Q is the heat absorbed by the system

W is the work done by the system on the surrounding


In this problem, the work done by the system is

W=-225 kcal=-941.4 kJ

with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is

Q=-5 \cdot 10^2 kJ=-500 kJ

with a negative sign as well because it is released by the system.


Therefore, by using the initial equation, we find

\Delta U=Q-W=-500 kJ+941.4 kJ=441.4 kJ

8 0
3 years ago
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