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2 years ago

Science Net Forces. Could somebody help me?

Physics

1 answer:

tensa zangetsu [6.8K]2 years ago

7 0

2) Unbalanced. Mike will push the box with a force of 20 N. The forces would be balanced if the box responded with 30 N.

3) Balanced. Both boys are pulling with the same force. Neither is winning.

4) Unbalanced. The rope will move with 10 N to the west. The teachers are winning.

5) Unbalanced. The kids are pulling 220 N to the east. The kids are winning.

6) Balanced. You and the dog are pulling with the same force.

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At the top of a looped section of roller coaster track, the car and rider are completely upside down. Engineers calculated that

Pie

Answer:

m v^2 / R = m g where gravitational force provides centripetal force

R = v^2 / g = 14.3^2 m/s / 9.8 m/s^2 = 20.9 m

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3 years ago

Two separate masses on two separate springs undergo simple harmonic motion indefinitely (the surface is frictionless). In CASE 1

Artemon [7]

Answer:

$\frac{F_1}{F_2} =\frac{1}{2}$

Explanation:

Assuming the we have to find ratio maximum forces on the mass in each case

we know that in a spring mass system

F= Kx

K= spring constant

x= spring displacement

Case 1:

$F_1=2k\times d$

case 2:

$F_2=2k\times 2d$

therefore, $\frac{F_1}{F_2} = \frac{2K\times d}{2K\times 2d}$

$\frac{F_1}{F_2} =\frac{1}{2}$

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3 years ago

What happens to thermal energy when Kinetic energy is Increased/Decreased?

Elis [28]

Answer:If kinetic energy increases, so does the thermal energy, and vice versa.

Please brainliest!

3 0

3 years ago

A hill that has a 28.1% grade is one that rises 28.1 m vertically for every 100.0 ml of distance in the horizontal direction. At

Serhud [2]

Answer:

$\theta=15.70^\circ$

Explanation:

A right triangle is formed, in which the vertical elevation is the opposite cathetus and the horizontal distance is the adjacent cathetus, since we know these two values, we can calculate the angle of inclination using the definition of tangent:

$tan\theta=\frac{opp}{adj}\\\theta=arctan(\frac{opp}{adj})\\\theta=arctan(\frac{28.1m}{100m})\\\theta=15.70^\circ$

6 0

3 years ago

An unbanked (flat) curve of radius 150 m is rated for a maximum speed of 32.5 m/s. At what maximum speed, in m/s, should a flat

Rasek [7]

Answer:

The maximum speed is 21.39 m/s.

Explanation:

Given;

radius of the flat curve, r₁ = 150 m

maximum speed, $v_{max}$ = 32.5 m/s

The maximum acceleration on the unbanked curve is calculated as;

$a_c_{max} = \frac{V_{max}^2}{r} \\\\a_c_{max} = \frac{32.5^2}{150} \\\\a_c_{max} = 7.04 \ m/s^2$

the radius of the second flat curve, r₂ = 65.0 m

the maximum speed this unbanked curve should be rated is calculated as;

$a_c_{max} = \frac{V_{max}^2}{r_2} \\\\V_{max}^2 = a_c_{max} \ \times \ r_2\\\\V_{max} = \sqrt{a_c_{max} \ \times \ r_2} \\\\V_{max} =\sqrt{7.04 \ \times \ 65} \\\\V_{max} = 21.39 \ m/s$

Therefore, the maximum speed is 21.39 m/s.

3 0

3 years ago