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2 years ago

A car has a mass of 1.00 × 103 kilograms, and it has an acceleration of 4.5 meters/second2. What is the net force on the car

Physics

1 answer:

Novay_Z [31]2 years ago

0 0

Answer:

$4.5 *10^{3}\ kgm/s^{2}$

Explanation:

Given:

$Mass= 1.00 * 10^{3} \ kilograms,\\Acceleration =4.5\ meters/second^{2}$

As we know that

Force=Mass * Acceleration

Putting the value of mass and Acceleration we get ,

$Force\ = \ 1.00 * 10^{3} * 4.5\ m/s^{2}$

$Force\ =4.5 * \ 10^{3} \ kgm/s^{2}$

Therefore Net force is : $4.5 *10^{3}\ kgm/s^{2}$

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Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is

Ronch [10]

Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is $c = 28853.78 \ m^2 /s^2$

Explanation:

From the question we are told that

The acceleration is $a = \frac{c}{s}\ m/s^2$

The initial position of the projectile is s= 1.5m

The final position of the projectile is $s_f = 3 \ m$

The velocity is $v = 200 \ m/s$

Generally $time = \frac{ds}{dv}$

and acceleration is $a = \frac{v}{time }$

$a = v \frac{dv}{ds}$

=> $vdv = a ds$

$vdv = \frac{c}{s} ds$

integrating both sides

$\int\limits^a_b vdv = \int\limits^c_d \frac{c}{s} ds$

Now for the limit

a = 200 m/s

b = 0 m/s

c = s= 3 m

d = $s_f$ = 1.5 m

So we have

$\int\limits^{200}_{0} vdv = \int\limits^{3}_{1.5} \frac{c}{s} ds$

$[\frac{v^2}{2} ] \left | 200} \atop {0}} \right. = c [ln s]\left | 3} \atop {1.5}} \right.$

$\frac{200^2}{2} = c ln[\frac{3}{1.5} ]$

=> $c = \frac{20000}{0.69315}$

$c = 28853.78 \ m^2 /s^2$

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3 years ago

describe the physics modeling process in moving a basketball and say what effects you should overlook

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Explanation:

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2 years ago

Which of the following terms is given to a pair of stars that appear to change position in the sky, indicating that they are orb

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Answer:

option A

Explanation:

The correct answer is option A

The binary star system is the system in which two stars are continuously orbiting each other,

In the eclipsing binary system, two stars revolve about there center of mass and in this system one one-star eclipse another star.

Spectroscopic binary stars are found from the observation of radial velocity and the brighter member of such binary can be seen to have continuously changed the wavelength and periodic velocity.

When the pairs of stars appear to change position in the sky then it is known as visual binary.

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2 years ago

A 75.0kg bicyclist (including the bicycle) is pedaling to the right, causing her speed to increase at a rate of 2.20m/s^2, despi

malfutka [58]

1) 4 forces

2) 165 N

3) 225 N

Explanation:

There are in total 4 forces acting on the bicylist:

- The gravitational force on the byciclist, acting vertically downward, of magnitude $mg$ , where m is the mass of the bicyclist and g is the acceleration due to gravity

- The normal force exerted by the floor on the bicyclist and the bike, N, vertically upward, and of same magnitude as the gravitational force

- The force of push F, acting horizontally forward, given by the push exerted by the bicylist on the pedals

- The air drag, R, of magnitude R = 60.0 N, acting horizontally backward, in the direction opposite to the motion of the bicyclist

The magnitude of the net force on the bicyclist can be calculated by considering separately the two directions.

- Along the vertical direction, we have the gravitational force (downward) and the normal force (upward); these two forces are equal in magnitude, since the acceleration of the bicyclist along this direction is zero, therefore the net force in this direction is zero.

- Along the horizontal direction, the two forces (forward force of push and air drag) are balanced, since the acceleration is non-zero, so we can use Newton's second law of motion to find the net force on the bicylist:

$F_{net}=ma$

where

$F_{net}$ is the net force

m = 75.0 kg is the mass of the bicyclist

$a=2.20 m/s^2$ is its acceleration

Solving, we find the net force:

$F_{net}=(75.0)(2.20)=165 N$

In this part, we basically want to find the forward force of push, F.

We can rewrite the net force acting on the bicyclist as

$F_{net}=F-R$

where:

F is the forward force of push

R is the air drag

We know that:

$F_{net}=165 N$ is the net force on the bicyclist

R = 60.0 N is the magnitude of the air drag

Therefore, by re-arranging the equation, we can find the force generated by the bicylicst by pedaling:

$F=F_{net}+R=165+60=225 N$

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2 years ago

The number of significant figures in 0.060900 is

Lady bird [3.3K]

Answer:

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Explanation:

You must start conting your sig figs until you continue to hit zeros at the end. Those zeroes at the end are disregarded. So 0.0609 is where you get your <em>sig figs</em> from.

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3 years ago