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3 years ago

. Two people are pushing a car of mass 2000 kg.

Physics

1 answer:

Viktor [21]3 years ago

5 0

Answer:

two people who are not going to be able to make it to class today because of the day and then I will be there at the house and then we can go

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Attracting means pulling toward you and repelling means pushing away

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2 years ago

Swinging a golf club toward a golf ball and hitting it off the tee.

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2 years ago

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Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° rela

NikAS [45]

The acceleration of the wagon is found by applying Newton's Second Law of motion.

1. The responses for question 1 are;

x-component of the tension in the rope on the right is approximately 91.93 N

y-component of the tension in the rope on the right is approximately 71.135 N

x-component of the tension in the rope on the left is -80.0 N

y-component of the tension in the rope on the left is 0

2. The net force in the x-direction is approximately 11.93 N

3. The net acceleration of the wagon in the horizontal direction is approximately 0.1193 m/s².

Reasons:

The given parameters are;

Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

Tension in the rope on the left = 80.0 N

1. The x and y component of the tension in the rope on the right are;

x-component = 120.0 N × cos(40.0°) ≈ 91.93 N

y-component = 120.0 N × sin(40.0°) ≈ 77.135 N

The x and y component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = -80.0 N

y-component = 80.0 N × sin(180°) = 0.0 N

2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = 11.93 N

3. The net acceleration of the block is given as follows;

According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

$\displaystyle a_x = \frac{F_x}{m} \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}$

The acceleration of the wagon in the horizontal direction, aₓ ≈ 0.1193 m/s².

Learn more here:

brainly.com/question/20357188

8 0

2 years ago

Two identical asteroids travel side by side while touching one another. If the asteroids are composed of homogeneous pure iron a

victus00 [196]

Answer:

diameter = 21.81 ft

Explanation:

The gravitational force equation is:

$F=\frac{GMm}{R^{2} }$

Where:

F => Gravitational force or force of attraction between two masses
M => Mass of asteroid 1
m => Mass of asteroid 2
R => Distance between asteroids 1 and 2 (from center of gravity)

We also know that the asteroids are identical so their masses are identical:

Since R is the distance between centers of the two asteroids and their diameters are identical (see attachment), we can conclude that:

R=d=2r

We don´t know the mass of the asteroids but we know they are composed of pure iron, so we can relate their masses to their density:

m=ρV

This is going to be helpful because the volume of a sphere is:

$\frac{4}{3}\pi r^{3}$

And know we can write our original force of gravity equation in terms of the radius of the asteroids:

$F=\frac{GMm}{R^{2} } =\frac{Gmm}{(2r)^{2} } =\frac{Gm^{2} }{4r^{2} }$
$F=\frac{G ( \frac{4}{3}\pi r^{3}ρ)^{2} }{4r^{2} }$
$F= \frac{G(16)\pi ^{2} r^{6} ρ^{2}}{(9)(4)r^{2} } =\frac{G(16)\pi ^{2} r^{4}ρ^{2} }{36}$

Now let´s plug in the values we know:

$F = 1 lb$ mutual gravitational attraction force
$G = 6.67(10)^{-11}$ gravitational constant
$ρ_{iron} =491.5 \frac{lb}{ft^{3} }$

$1= \frac{6.67(10)^{-11} \pi ^{2} r^{4} (491.5)^{2}}{36}$

Solve for r and multiply by 2 because 2r = diameter

$d=2\sqrt[4]{\frac{1}{7.07(10)^{-5} } }$

Result is d = 21.81 Feet

6 0

2 years ago