All categories

3 years ago

21. An object is moving at 5 m/s and a height of 4 m. If the object has a mass of 5 kg what is

Physics

1 answer:

Stels [109]3 years ago

6 0

Answer: 258.7 J

Explanation:

Velocity of object V=5m/s

Mass ; 5kg

Height; h=4m

Mechanical energy is equal to sum of kinetic and potential energy

E=K+Ep

Kinetic energy is calculeted using; K=m*V²/2

Potential energy is calculated using : Ep=m*G*h

-------------------------------------------------------------------------

V=5m/s

m=5kg

h=4m

E=K+Ep

E=m*V²/2+m*G*h

E=5kg*(5m/s)²/2+5kg*9.81m/s²*4m

E=62.5J+196.2J

E=258.7 J

You might be interested in

A bookcase has a base of 1 m long and 0.5 m wide. It has a mass of 300kg. Find the pressure it exerts on the floor in kPa.

Snezhnost [94]

$1m*0.5m=0.5m^2\\
g=10\frac{m}{s^2}\\
300kg*10\frac{m}{s^2}=3000N\\
3000N--0.5m^2\\
\ \ x\ \ --\ 1\\
x=6000Pa=60kPa$

4 0

2 years ago

Newton’s second law of motion addresses the relationship between what two variables that influence the force on a body?

Helen [10]

Answer:

A. Mass and acceleration

Explanation:

According to Newton's second law of motion, the resultant force is directly proportional to the rate of change in momentum
Therefore; F = ma , where F is the resultant force, m is the mass, and a is the acceleration of the body.
<u>Resultant force depends on the acceleration and the mass of a body in motion, an increase in acceleration causes a corresponding increase in resultant force.</u> A body with higher mass will have a larger strong force if the acceleration is kept constant.

3 0

2 years ago

Read 2 more answers

A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re

Lady_Fox [76]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

$P_1$ = Pressure at reservoir = 10 atm

$T_1$ = Temperature at reservoir = 300 K

$P_2$ = Pressure at exit = 1 atm

$T_2$ = Temperature at exit

$R_s$ = Mass-specific gas constant = 287 J/kgK

$\gamma$ = Specific heat ratio = 1.4 for air

For isentropic flow

$\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K$