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2 years ago

What acceleration would you expect a 100 kg individual to experience from a 1000N force?

Physics

1 answer:

german2 years ago

7 0

Answer: 10 m/2^2

F=ma

Explanation:

Force = 1000N

1 N is 1 (kg*m/s^2)

Mass = 100kg

F=ma

a = F/m

a = (1000kg*m/s^20/(100kg)

a = 10 m/s^2

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A 55 kilogram person jumps off a cliff and hits the water 5.8 seconds later, how high is the cliff above the water?

ElenaW [278]

R=ut+gt^2/2
r- displacement (height to find)
u - initial speed (zero)
t - time taken

r=0*5.8 + 10*5.8^2 /2 = 168.2 meters

8 0

2 years ago

Air conditioners operate on the same principle as refrigerators. Consider an air conditioner that has 7.00 kg of refrigerant flo

kozerog [31]

Answer:

5.15J

Explanation:

First. 54% of the 7kg refrigerant is liquid

So we find mass of vapour at inlet generator

M1 = ( 1-0.54)*7= 3.2kg

At compressor mass of vapour will be

M2= 0.95*7= 6.7kg

So the Mass of vapour at exit generator is

M2-M1= 3.5kg

So to find heat absorbed by refrigerant in evaporation

Its using

Q= mh

°= 3.5x 1.50×10^5 J/kg

=5.15J

4 0

3 years ago

The velocity of a car is 65 m/s and it’s mass is 2515 kg. What is it’s KE?

Aloiza [94]

<span>Example Problems. Kinetic Energy (KE = ½ m v2). 1) The velocity of a car is 65 m/s and its mass is 2515 kg. What is its KE? 2) If a 30 kg child were running at a rate of 9.9 m/s, what is his KE? Practice Problems. IN THIS ORDER…. Page 2: #s 6, 7, 8, 5. Potential Energy. An object can store energy as the result of its position.</span><span>
</span>

4 0

2 years ago

Read 2 more answers

(a) If two sound waves, one in a gas medium and one in a liquid medium, are equal in intensity, what is the ratio of the pressur

GarryVolchara [31]

Answer:

(a) The ratio of the pressure amplitude of the waves is 43.21

(b) The ratio of the intensities of the waves is 0.000535

Explanation:

Given;

density of gas, $\rho _g$ = 2.27 kg/m³

density of liquid, $\rho _l$ = 972 kg/m³

speed of sound in gas, $C_g$ = 376 m/s

speed of sound in liquid, $C_l$ = 1640 m/s

The of the sound wave is given by;

$I = \frac{P_o^2}{2 \rho C} \\\\P_o^2 = 2 \rho C I\\\\p_o = \sqrt{2 \rho CI}$

Where;

$P_o$ is the pressure amplitude

$P_o_g= \sqrt{2 \rho _g C_gI} -------(1)\\\\P_o_l= \sqrt{2 \rho _l C_lI}---------(2)\\\\\frac{P_o_l}{P_o_g} = \frac{\sqrt{2 \rho _l C_lI}}{\sqrt{2 \rho _g C_gI}} \\\\\frac{P_o_l}{P_o_g} = \sqrt{\frac{2 \rho _l C_lI}{2 \rho _g C_gI} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ \rho _l C_l}{ \rho _g C_g} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ (972)( 1640)}{ (2.27)( 376)} }\\\\\frac{P_o_l}{P_o_g} = 43.21$

(b) when the pressure amplitudes are equal, the ratio of the intensities is given as;

$I = \frac{P_o^2}{2 \rho C}\\\\I_g = \frac{P_o^2}{2 \rho _g C_g}-------(1)\\\\I_l = \frac{P_o^2}{2 \rho _l C_l}-------(2)\\\\\frac{I_l}{I_g} = (\frac{P_o^2}{2 \rho _l C_l})*(\frac{2\rho_gC_g}{P_o^2} )\\\\\frac{I_l}{I_g} = \frac{\rho _gC_g}{\rho_lC_l} \\\\\frac{I_l}{I_g} = \frac{(2.27)(376)}{(972)(1640)}\\\\ \frac{I_l}{I_g} = 0.000535$

3 0

3 years ago

A parcel of land is in the shape of an isosceles triangle. The base has a length of 425 ft.; the other sides, which are of equal

kogti [31]

Answer:

The answer to your question is 636.6 ft

Explanation:

Data

base = 425 ft

angle = 39°

See the picture below

1.- Divide the triangle to get two right triangles.

Now the superior angle will measure 19.5° and the opposite side will measure 212.5 ft

2.- Use the trigonometric function sine to find the hypotenuse

sin 19.5 = 212.5/hyp

solve for hyp

hyp = 212.5 / sin 19.5

Result

hyp = 212.5/ 0.333

hyp = 636.6 ft

4 0

2 years ago