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Ierofanga [76]
2 years ago
6

A car has a mass of 1600 kg. It is stuck in the snow and is being pulled out by a cable that applies a force of 7560 N due north

. The resistance of the snow and mud also applies a force to the car, which has a magnitude of 7340 N and points due south. What is the acceleration of the car?
Physics
1 answer:
sergiy2304 [10]2 years ago
7 0

Answer:

Acceleration of the car will be a=0.1375m/sec^2

Explanation:

We have given mass of the ball m = 1600 kg

Force in north direction F= 7560 N

Resistance force which opposes the movement of car F_R=7340N

So net force on the car F_{net}=F-F_R=7560-7340=220N

According to second law of motion we know that F=ma

So 220=1600\times a

a=0.1375m/sec^2

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A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the
yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE          

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0     (initial KE = 0 because the ball starts from rest)  (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

3 0
2 years ago
A certain satellite travels in an approximately circular orbit of radius 2.0 × 106 m with a period of 7 h 11 min. Calculate the
kap26 [50]

Answer: Mass of the planet, M= 8.53 x 10^8kg

Explanation:

Given Radius = 2.0 x 106m

Period T = 7h 11m

Using the third law of kepler's equation which states that the square of the orbital period of any planet is proportional to the cube of the semi-major axis of its orbit.

This is represented by the equation

T^2 = ( 4π^2/GM) R^3

Where T is the period in seconds

T = (7h x 60m + 11m)(60 sec)

= 25860 sec

G represents the gravitational constant

= 6.6 x 10^-11 N.m^2/kg^2 and M is the mass of the planet

Making M the subject of the formula,

M = (4π^2/G)*R^3/T^2

M = (4π^2/ 6.6 x10^-11)*(2×106m)^3(25860s)^2

Therefore Mass of the planet, M= 8.53 x 10^8kg

5 0
2 years ago
calculate the workdone to stretch an elastic string by 40cm if a force of 10N produces an extension of 4cm in it
Lera25 [3.4K]
The force of F=10 N produces an extension of
x=4 cm=0.04 m
on the string, so the spring constant is equal to
k= \frac{F}{x}= \frac{10 N}{0.04 m}=250 N/m

Then the string is stretched by \Delta x=40 cm=0.40 m. The work done to stretch the string by this distance is equal to the variation of elastic potential energy of the string with respect to its equilibrium position:
W= \Delta U= \frac{1}{2}k(\Delta x)^2  = \frac{1}{2}(250 N/m)(0.40 m)^2=20 J
5 0
2 years ago
Suppose you first walk 12.0 m in a direction 20 owest of north and then 20.0 m in a direction 40.0osouth of west. How far are yo
alexgriva [62]

Answer:

R=19.5m

\theta = 4.65° S of W

Explanation:

Refer the attached fig.

displacement  of the x and y components

x-component displacement is (R_{x}) = A_{x}+B_{x}

= A \sin(20°) + B \cos(40°)

= -12.0\sin(20°) + 20.0\cos(40°)

= -19.425m

x-component displacement is (R_{y}) = A_{y}+ B_{y}

=  A \cos(20°) - B \sin(40°)

= 12.0\cos(20°) - 20.0\sin(40°)

= -1.579

resultant displacement

∴

R = \sqrt{R_{x}^{2} +R_{y}^{2} }  }

=\sqrt{(-19.425)^{2}+(-1.579)^{2}  }

=19.5m

\theta = \tan^{-1}\left | \frac{R_{x}}{R_{y}} \right |

\theta = \tan^{-1}\left | \frac{1.579}{19.425} \right |

\theta = 4.65° S of W

6 0
2 years ago
When a perfume bottle is opened, some liquid changes to gas and the fragrance spreads around the room. Which sentence explains t
monitta
A. 

since they do not have a definite volume, this causes it to spread out in the air 
7 0
2 years ago
Read 2 more answers
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