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2 years ago

In the equation for the gravitational force between two objects, which quantity must be squared?

Physics

1 answer:

bezimeni [28]2 years ago

8 0

Answer:

Explanation:

The quantity that must be squared in the equation of gravitational force is distance d.

According to the universal gravitational law, the square of the distance between two objects is inversely proportional to the force of gravity.

Therefore, the quantity to be squared is d

The formula is given as:

Fg = $\frac{G m_{1} m_{2} }{d^{2} }$

So d is the quantity that must be squared

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Jeff is a landscaping contractor and lifts a rock weighing 600 pounds by wedging a board under the rock. Jeff weighs 150 pounds

jonny [76]

Answer: 4

The mechanical advantage is the ratio of the force exerted by the object to the force applied to do work on it.

Here, Jeff tried to lift a rock weighing 600 pounds by wedging board under the rock. Jeff who weighs 150 pounds uses all his weight to exert force on lever and lift rock.

Mechanical advantage, $M.A.=\frac{weight\hspace{1mm}of\hspace{1 mm}rock}{weight\hspace{1mm}of\hspace{1 mm}Jeff}=\frac{600 pounds}{150 pounds}=4.$

Therefore, the mechanical advantage that lever provided to Jeff in lifting rock is 4.

6 0

2 years ago

Read 2 more answers

A high-jump athlete leaves the ground, lifting her center of mass 1.8 m and crossing the bar with a horizontal velocity of 1.4 m

Romashka [77]

Answer:

The minimum speed when she leave the ground is 6.10 m/s.

Explanation:

Given that,

Horizontal velocity = 1.4 m/s

Height = 1.8 m

We need to calculate the minimum speed must she leave the ground

Using conservation of energy

$K.E+P.E=P.E+K.E$

$\dfrac{1}{2}mv_{1}^2+0=mgh+\dfrac{1}{2}mv_{2}^2$

$\dfrac{v_{1}^2}{2}=gh+\dfrac{v_{2}^2}{2}$

Put the value into the formula

$\dfrac{v_{1}^2}{2}=9.8\times1.8+\dfrac{(1.4)^2}{2}$

$\dfrac{v_{1}^2}{2}=18.62$

$v_{1}=\sqrt{2\times18.62}$

$v_{1}=6.10\ m/s$

Hence, The minimum speed when she leave the ground is 6.10 m/s.

6 0

3 years ago

How do i get a negative cube when i calculate density

fgiga [73]

374848282!,!( rnanxjcifjejzxj

8 0

3 years ago

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If the car’s speed decreases at a constant rate from 64 mi/h to 30 mi/h in 3.0 s, what is the magnitude of its acceleration, ass

mixas84 [53]

Answer: $3.874 m/s^2$

Explanation:

Given

Car speed decreases at a constant rate from 64 mi/h to 30 mi/h

in 3 sec

$60mi/h \approx 26.8224m/s$

$34mi/h \approx 15.1994 m/s$

we know acceleration is given by $=\frac{velocity}{Time}$

$a=\frac{15.1994-26.8224}{3}$

$a=-3.874 m/s^2$

negative indicates that it is stopping the car

Distance traveled

$v^2-u^2=2as$

$\left ( 15.1994\right )^2-\left ( 26.8224\right )^2=2\left ( -3.874\right )s$

$s=\frac{488.419}{2\times 3.874}$

s=63.038 m

7 0

2 years ago

1. The large wheels on the train have a mass of 0.5 kg, while the small wheels have a mass of 0.15 kg. The train body, and child

blsea [12.9K]

Answer: 60.975 Joules

KE = 1/2 mv^2

=> 1/2 * 40.65 * 3

=> 60.975 J

5 0

2 years ago