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3 years ago

What is true about an object in free fall?

1 answer:

5 0

The only force being acted upon it is gravity. A ball that was thrown downward off a building isn't in freefall, because it had initial velocity. A piece of paper can't necessarily be in free fall because it is affected by air resistance.

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A 53-N force is needed to keep a 50.0-kg box sliding across a flat surface at a constant velocity. What is the coefficient of ki

earnstyle [38]

The weight of the box is (mass) x (gravity) = (50 kg) x (9.8m/s²) = 490 newtons.

If the box is sliding at constant speed, and not speeding up or slowing down,
that means that the horizontal forces on it add up to zero.

Since you're pushing on it with 53N in that direction, friction must be pulling
on it with 53N in the other direction.

The 53N of friction is (the weight) x (the coefficient of kinetic friction).

53N = (490N) x (coefficient).

Divide each side by 490N : Coefficient = (53N) / (490N) = 0.1082 .

Rounded to the nearest hundredth, that's 0.11 . (choice 'd')

5 0

3 years ago

Read 2 more answers

You observe a plane approaching overhead and assume that its speed is 600 miles per hour. The angle of elevation of the plane is

scZoUnD [109]

Answer:4.34 miles

Explanation:

first Elevation = $19^{\circ}$

After 1 minute Elevation changes to $59^{\circ}$

Ditsance travelled in 1 minute = $600\times \frac{1}{60}$ =10 mile

Now

tan59= $\frac{H}{x}$

H=xtan59

tan19= $\frac{H}{10+x}$

H= $\left ( 10+x\right )tan19$

Equating H

we get

1.319x=10tan19

x=2.61 miles

H= $2.61\times tan59$ =4.34 miles

4 0

3 years ago

What are (a) the energy of a photon corresponding to wavelength 6.0 nm, (b) the kinetic energy of an electron with de Broglie wa

dlinn [17]

Explanation:

Given that,

Wavelength = 6.0 nm

de Broglie wavelength = 6.0 nm

(a). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}$

$E=3.315\times10^{-17}\ J$

(b). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}$

$E=6.709\times10^{-21}\ J$

(c). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}$

$E=3.315\times10^{-11}\ J$

(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}$

$E=6.709\times10^{-9}\ J$

Hence, This is the required solution.

6 0

3 years ago

A bowling ball is traveling at 3 m/s. It starts to roll up a ramp. How high above the ground will the ball be when it stops roll

Zepler [3.9K]

Answer:

FREE FREE FREE FREEE FREEEEEEEEEEE FREEEEEEEEEEEEEEEE FREEEEEEEEEEEEEEEEEE

Explanation:

FREE FREE FREE FREEE FREEEEEEEEEEE FREEEEEEEEEEEEEEEE FREEEEEEEEEEEEEEEEEE

8 0

3 years ago

Read 2 more answers

What gravitational force does the earth exert on a person

lara [203]

The gravitational force exerted by the earth on a person standing on the earth's surface is 602.74 N.

<h3>What is the gravitational force of the earth on the person?</h3>

The gravitational force exerted by the earth on a person standing on the earth's surface is given below as follows:

$F = \frac{Gm^{1}m^{2}}{r^{2}}$

where

G = 6.67 * 10⁻¹¹

m¹ = 62 kg

m² = 5.97 * 10²⁷ kg

r = 6.4 * 10⁶ m

$F = \frac{5.97*10^{24}*62*6.67*10^{-11}}{(6.4*10^{6}){2}} = 602.74\:N$

Therefore, the gravitational force exerted by the earth on a person standing on the earth's surface is 602.74 N.

Learn more about gravitational force at: brainly.com/question/940770

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7 0

2 years ago