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3 years ago

A ball with an initial velocity of zero is dropped from a table onto the ground. After 2 s, the

Physics

2 answers:

WINSTONCH [101]3 years ago

7 0

Answer:

Explanation:

I don't know 200000000000000000000

GenaCL600 [577]3 years ago

5 0

Initial velocity=u=0m/s
Acceleration=a=10m/s^2
Time=t=2s

Final velocity=v

$\\ \sf\longmapsto v=u+at$

$\\ \sf\longmapsto v=0+10(2)$

$\\ \sf\longmapsto v=20m/s$

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A 1. 18 kg gold cube hangs at the end of a 4. 00 m long string. Rhogold = 19. 3 × 103 kg/m3; rhomercury = 13. 6 × 103 kg/m3. Whe

VashaNatasha [74]

When the gold cube is immersed in mercury, the tension in the string in Newtons is 3.142N.

<h3>What is tension?</h3>

Tension is the force acting on the linear object like string, chain or rope due to pulling.

Volume of gold V = mass / density

V = 1.18 /19.3x 10³ =61.1 x 10⁻⁶ m³

Tension in the string after immersing will be

T = [ρ(Gold) -ρ(Hg)] g. V

T =[ 19.3x 10³ - 13.6 x 10³] x 9.81 x 61.1 x 10⁻⁶

T =3.416 N
Thus, the tension in the string is 3.42 N.

Learn more about tension.

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6 0

3 years ago

2. A marble is rolling at a velocity of 1.5 m/s with a momentum of 3.0 kg×m/s. What is its mass?

Nikolay [14]

Answer:

m= 2[kg]

Explanation:

Linear momentum can be calculated as the product of mass by Velocity.

$P=m*v$

where:

P = linear momentum = 3 [kg*m/s]

v = velocity = 1.5 [m/s]

m = mass [kg]

$m=P/v\\m=3/1.5\\m=2[kg]$

5 0

3 years ago

(a) What is the best coefficient of performance for a refrigerator that cools an environment at −30.0ºC and has heat transfer to

kari74 [83]

Answer:

a) 3.242

b) 1291.178 KJ

c) 3.59 cents

Explanation:

a) First all the temperature units should be in Kelvin. -30°C + 273.15 = 243.15K and 45°C + 273.15 = 318.15K. Coefficient of performance for refrigerator is COP = TL/(TH-TL) where TL is Lower Temperature and TH is the Higher Temperature. COP = 243.15/(318.15-243.15) = 3.242

b) COP = Cooling Effect/Work Input. Cooling Effect (or Desired Output) is 4186 kJ.

So Work Input = Cooling Effect/COP = 4186KJ/3.242= 1291.178 KJ

c) 3.6 MJ (3600kJ) is for 10 cents, using ratio for cost: energy used

$\frac{10}{3600 kJ} =\frac{x}{1291.178kJ}$