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3 years ago

How long will it take an object to hit the ground if dropped from 100 meters

Physics

2 answers:

mash [69]3 years ago

8 0

Answer:

The answer depends on what object you are dropping. Are you dropping a balloon or a car? (I'm joking 'bout that one.) If the mass of the object is very little, then it might drop slower. If the mass is bigger, then it might drop faster.

Good luck!

Explanation:

My name is Ann [436]3 years ago

4 0

Answer:10.24

Explanation: well, the acceleration of gravity is 9.8m/s².So, gravity pulls objects down at this rate

so now we divide 100/9.8 and we get our answer which is 10.24 seconds(APPROX)

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A 2000 N net force will give a car with some amount of mass an acceleration of 4 m/s2

hammer [34]

Answer:

2m/s²

Explanation:

Given parameters:

Net force on the car = 2000N

Acceleration = 4m/s²

Unknown:

Acceleration of a car twice the mass = ?

Solution:

Let us first find the mass of the car;

Force = mass x acceleration

Mass = $\frac{Force }{acceleration}$

Mass = $\frac{2000}{4}$ = 500kg

Now,

whose mass is twice that of the car

Mass of the new car = 2 x 500 = 1000kg

So;

Acceleration = $\frac{Net force }{mass}$

Acceleration = $\frac{2000}{1000}$ = 2m/s²

3 0

3 years ago

A radio has 100J of energy transferred by electricity. 20J are transferred by heating and 10J are transferred by sound. How many

seraphim [82]

Answer:

70 Joules

Explanation:

3 0

3 years ago

determine the pressure exerted on the surface of a submarine cruising 175 ft below the free surface of the sea. assume that the

Alenkasestr [34]

Answer:

92.81 psia.

Explanation:

The density of water by multiplying its specific gravity by the density of sea water.

SG = density of sea water/density of water

ρ = SG x ρw

1 kg/m3 = 62.4 lbm/ft^3

= 1.03 * 62.4

= 64.27lbm/ft^3.

The absolute pressure at 175 ft below sea level as this is the location of the submarine.

P = Patm +ρgh

= 14.7 + 64.27 * 32.2 * 175

Converting to pound force square inch,

= 14.7 + 64.27 * (32.2ft/s^2) * (175ft) * (1lbf/32.2lbm⋅ft/s^2) * (1ft^2/144in^2 )

= 14.7 + 78.11 psia

= 92.81 psia.

8 0

3 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant