a race car is traveling at a speed of 80.0m/s on a circular race track of radius 450m what is the centripetal acceleration

The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit

Margaret [11]

Let us consider two bodies having masses m and m' respectively.

Let they are separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as - $F = G\frac{mm'}{r^{2} }$ where G is the gravitational force constant.

From the above we see that F ∝ mm' and $F\alpha \frac{1}{r^{2} }$

Let the orbital radius of planet A is $r_{1}$ = r and mass of planet is $m_{1}$ .

Let the mass of central star is m .

Hence the gravitational force for planet A is $f_{1} =G \frac{m_{1}*m }{r^{2} }$

For planet B the orbital radius $r_{2} =2r_{1}$ and mass $m_{2} = 3 m_{1}$

Hence the gravitational force $f_{2} =G\frac{m m_{2} }{r^{2} }$

$f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }$

$= \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }$

Hence the ratio is $\frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2} }{Gmm_{1}/r_{1} ^2 }$

$=\frac{3}{4}$ [ ans]

3 0

4 years ago

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A homeowner uses a snowblower to clear his driveway. knowing that the snow is discharged at an average angle of 40° with the hor

Nesterboy [21]

given that snow is projected at an angle of 40 degree

It range is given as a = 19 ft

$a = 19 * 0.3048 = 5.8 m$

now we can use the formula of horizontal range

$R = \frac{v_o^2 sin2\theta}{g}$

$5.8 =\frac{ v_o^2 sin(2*40)}{9.8}$

$5.8 = \frac{v_o^2 * sin80}{9.8}$

$v_o^2 = 57.7$

$v_o = 7.6 m/s$

<u>so its initial speed must be 7.6 m/s</u>

8 0

4 years ago

A 250 g object is hung onto a spring. It stretches 18 cm. Find the spring's spring constant

Juliette [100K]

<h2>Answer: 13.61 N/m</h2>

Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force $F$ applied to it, <u>as long as the spring is not permanently deformed</u>:

$F=k (x-x_{o})$ (1)

Where:

$k$ is the elastic constant of the spring. The higher its value, the more work it will cost to stretch the spring.

$x_{o}$ is the length of the spring without applying force.

$x$ is the length of the spring with the force applied.

According to this, we have a spring where only the force due gravity is applied.

In other words, the force applied is the weigth $W$ of the block:

$W=m.g$ (2)

Where $m=250g=0.25kg$ is the mass of the block and $g=9.8\frac{m}{s^{2}}$ is the gravity acceleration.

$W=(0.25kg)(9.8\frac{m}{s^{2}})$ (3)

$W=2.45N$ (4)

Knowing the force applied $W$ and $x=18cm=0.18m$ and $x_{o}=0$ , we can substitute the values in equation (1) and find $k$ :

$W=k (x-x_{o})$ (5)

$2.45N=k (0.18m-0m)$ (6)

<u>Finally:</u>

$k=13.61\frac{N}{m}$

4 0

3 years ago

If the force of a golf club on a golf ball is 200 N forward, what will the force of the ball on the club be? A. 200 N forward B.

Korolek [52]

The ball should put 200 N of force towards the golfer.

Newton's Third Law is every action has an equal and opposite reaction.

It's the ball exerting 200 N of force towards the club as well, but the opposite reaction is that it flies away.

8 0

4 years ago

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Dfine ' weight of moist air '

grandymaker [24]

Answer:

First of all, “moist air” is air with a high water vapor content. Water vapor, the invisible, gaseous form of water, occurs in highly variable amounts in the atmosphere. Water is composed of a hydrogen atom and two oxygen atoms (H2O) and has a molecular weight of 18 grams per mole.

6 0

3 years ago