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boyakko [2]
3 years ago
11

A 20.0 kg mass moving at a velocity of + 3.0 m/s is stopped by a constant force of 15.0 n. how many seconds must the force act o

n the mass to stop it? html editor keyboard shortcuts
Physics
1 answer:
slamgirl [31]3 years ago
4 0
The mass is 20 kg, initial speed is 30 m/s and the final speed is 0 m/s (rest), 
while the force is 15 N
Using second newton's law of motion which states that the rate of change in momentum is directly proportional to the resultant force. Then
 Ft = MV-MU
15 t = 20 (0-3)
15t = -60
  t = -4 sec ( negative because the force is acting on opposite direction to the motion)
Therefore, the time taken is 4 seconds
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. A 79 g sample of water at 21oC is heated until it becomes steam with a temperature of 143oC. Find the change in heat content o
mote1985 [20]

Answer:

40479.6 J

Explanation:

Applying,

q = cm(t₂-t₁).................... Equation 1

Where q = change in heat content of the system, c = specific heat capacity of the system, m = mass of the system, t₁ = initial temperature, t₂ = final temperature.

From the question,

Given: m = 79 g = 0.079 kg, t₁ = 21°C, t₂ = 143°C

Constant: c = 4200 J/kg.°C

Substitute these values into equation 1

q = 4200(0.079)(143-21)

q = 331.8(122)

q = 40479.6 J

4 0
3 years ago
When is a zero not significant?
maria [59]

Answer:

Not between significant digits.

Explanation:

A zero not significant when it's not between significant digits.

8 0
3 years ago
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______ can occur when water-saturated soil turns from a solid to a liquid as a result of an earthquake.
jek_recluse [69]
The answer is liquefaction
6 0
2 years ago
1. The Moon's mass is 7.34 x 1022 kg, and it is 3.8 x 105 km away from Earth. Earth's
sineoko [7]

Answer:

2.03 x 10²⁴N

Explanation:

Given parameters:

Mass of moon = 7.34 x 10²²kg

Mass of the earth  = 5.97 x 10²⁴kg

Distance  = 3.8 x 10⁵km

Unknown:

Gravitational force of attraction  = ?

Solution:

To find the gravitational force of attraction between the masses, we use the expression below;

   F = \frac{Gm_{1} m_{2}  }{r^{2} }

G is the universal gravitation constant

m is the mass

1 and 2 represents moon and earth

r is  the distance

  F = \frac{6.67 x 10^{-11}  x 7.34 x 10^{22} x 5.97 x 10^{24}  }{(3.8 x 10^{5})^{2}  }

 F = \frac{2.92 x 10^{35} }{1.44 x 10^{11} }  = 2.03 x 10²⁴N

8 0
3 years ago
0.5-lbm of a saturated vapor is converted to asaturated liquid by being cooled in a weighted piston-cylinder device maintained a
klio [65]

Answer:

The boiling point temperature of this substance when its pressure is 60 psia is  480.275 R

Explanation:

Given the data in the question;

Using the Clapeyron equation

(\frac{dP}{dT} )_{sat } = \frac{h_{fg}}{Tv_{fg}}

(\frac{dP}{dT} )_{sat } = \frac{\frac{H_{fg}}{m} }{T\frac{V_{fg}}{m} }

where h_{fg is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu

T is the temperature ( 15 + 460 )R

m is the mass of water ( 0.5 Ibm )

V_{fg is specific volume ( 1.5 ft³ )

we substitute

(\frac{dP}{dT} )_{sat } =( \frac{250Btu\frac{778Ibf-ft}{Btu} }{0.5}) / ( (15+460)\frac{1.5}{0.5})  

(\frac{dP}{dT} )_{sat } = 272.98 Ibf-ft²/R

Now,

(\frac{dP}{dT} )_{sat } = (\frac{P_2 - P_1}{T_2 - T_1})_{sat

where P₁ is the initial pressure ( 50 psia )

P₂ is the final pressure ( 60 psia )

T₁ is the initial temperature ( 15 + 460 )R

T₂ is the final temperature = ?

we substitute;

T_2 = ( 15 + 460 ) + \frac{(60-50)psia(\frac{144in^2}{ft^2}) }{272.98}

T_2 = 475 + 5.2751\\

T_2 = 480.275 R

Therefore, boiling point temperature of this substance when its pressure is 60 psia is  480.275 R

3 0
3 years ago
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