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3 years ago

What is the net force on a bathroom scale when a 150-pound person stands on it

Physics

1 answer:

andreev551 [17]3 years ago

7 0

Answer:

Zero, the support force and gravitational force together equal a zero net force

Explanation:

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A book is sitting on a desk. What best describes the normal force acting on the book?

Marat540 [252]

Explanation:

in my opinion I'd say it's gravity because it's making the book stay in it's one and only position pulling the book at the center of the desk

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3 years ago

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Can kintic friction affect mechanical energy

Nesterboy [21]

Yes. Kinetic energy is a form of mechanical energy and friction will turn that kinetic energy into heat.

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3 years ago

In the 2016 Olympics in Rio, after the 50 mm freestyle competition, a problem with the pool was found. In lane 1 there was a gen

gogolik [260]

Answer:

Explanation:

still water speed is 50 m / 25.0 s = 2.00 m/s or 200 cm/s

In lane 1 the effective speed would be 201.2 cm/s

5000 cm / 201.2 cm/s = 24.85 s

The change is 25.00 - 24.85 = 0.15 s decrease in time

In lane 8, the effective speed would be 198.8 cm/s

5000 cm / 198.8 cm/s = 25.15 s

The change is 25.00 - 25.15 = 0.15 s increase in time

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3 years ago

A skydiver of 75 kg mass has a terminal velocity of 60 m/s. At what speed is the resistive force on the skydiver half that when

ankoles [38]

Answer:

The speed of the resistive force is 42.426 m/s

Explanation:

Given;

mass of skydiver, m = 75 kg

terminal velocity, $V_T = 60 \ m/s$

The resistive force on the skydiver is known as drag force.

Drag force is directly proportional to square of terminal velocity.

$F_D = kV_T^2$

Where;

k is a constant

$k = \frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2}$

When the new drag force is half of the original drag force;

$F_D_2 = \frac{F_D_1}{2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_1}{2V_{T2}^2} \\\\\frac{1}{V_{T1}^2} = \frac{1}{2V_{T2}^2}\\\\2V_{T2}^2 = V_{T1}^2\\\\V_{T2}^2= \frac{V_{T1}^2}{2} \\\\V_{T2}= \sqrt{\frac{V_{T1}^2}{2} } \\\\V_{T2}= \frac{V_{T1}}{\sqrt{2} } \\\\V_{T2}= 0.7071(V_{T1})\\\\V_{T2}= 0.7071(60 \ m/s)\\\\V_{T2}= 42.426 \ m/s$

Therefore, the speed of the resistive force is 42.426 m/s

8 0

3 years ago

A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo

Rasek [7]

Answer:

$t=12.25\ seconds$

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

$x=v_ocos\theta t$

$\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}$

Where vo is the magnitude of the initial velocity, $\theta$ is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

$\displaystyle t=\frac{v_osin\theta}{g}$

There are two times where the value of y is $y_o$ when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making $y=y_o$