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son4ous [18]
2 years ago
6

Question 2 (1 point)

Physics
1 answer:
sveticcg [70]2 years ago
7 0
according to the facts it should be a numerous amount of wrong answers so ignore dis
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A 1000.0 kg truck accelerates from 20.0 m/s to 25.0 m/s over a distance of 300.0 m. What is the average net force on the truck?
choli [55]

Answer:

The average net force on the truck is 375 Newtons.

Explanation:

Using Newton's 3rd equation of motion, we have :

v^{2} - u^{2} = 2×a×s

where, v = final velocity = 25 m/s

u = initial velocity = 20 m/s

a = acceleration

s = distance traveled = 300 m

Using these values in the above equation, we get acceleration = 0.375 m/s^{2}

Using Newton's second law, we have:

F=m×a

where m = mass = 1000 kg

a= acceleration = 0.375 m/s^{2}

Putting values we have F=375 N

3 0
3 years ago
How long does it took for 1000-watt microwave oven to do 20000 joules of work
klio [65]

Answer:

Work = power * time

time = 20000 joules / 1000 joules / sec = 20 sec

4 0
2 years ago
How much time would it take for the sound of thunder to travel 1500 meters if sound travels at a speed of 330 m/s
Elis [28]

Data given:

Δx=1500m

v=330m/s

t=?

Formula:

V=Δx/t

Solution:

t=1500m/330m/s

t=4.5s

7 0
3 years ago
Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ
Angelina_Jolie [31]

Answer:

a). \frac{\dot{W}}{m}= 311 kJ/kg

b). \frac{\dot{\sigma _{gen}}}{m}=0.9113 kJ/kg-K

Explanation:

a). The energy rate balance equation in the control volume is given by

\dot{Q} - \dot{W}+m(h_{1}-h_{2})=0

\frac{\dot{Q}}{m} = \frac{\dot{W}}{m}+m(h_{1}-h_{2})

\frac{\dot{W}}{m}= \frac{\dot{Q}}{m}+c_{p}(T_{1}-T_{2})

\frac{\dot{W}}{m}= -30+1.1(980-670)

\frac{\dot{W}}{m}= 311 kJ/kg

b). Entropy produced from the entropy balance equation in a control volume is given by

\frac{\dot{Q}}{T_{boundary}}+\dot{m}(s_{1}-s_{2})+\dot{\sigma _{gen}}=0

\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+(s_{2}-s_{1})

\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+c_{p}ln\frac{T_{2}}{T_{1}}-R.ln\frac{p_{2}}{p_{1}}

\frac{\dot{\sigma _{gen}}}{m}=\frac{-30}{315}+1.1ln\frac{670}{980}-0.287.ln\frac{100}{400}

\frac{\dot{\sigma _{gen}}}{m}=0.0952+0.4183+0.3978

\frac{\dot{\sigma _{gen}}}{m}=0.9113 kJ/kg-K

5 0
3 years ago
Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, t
stellarik [79]

Answer:

0.243 m/s

Explanation:

From law of conservation of motion,

mu+m'u' = V(m+m')................. Equation 1

Where m = mass of the first car, m' = mass of the second car, initial velocity of the first car, u' = initial velocity of the second car, V = Final velocity of both cars.

make V the subject of the equation

V = (mu+m'u')/(m+m')................. Equation 2

Given: m = 260000 kg, u = 0.32 m/s, m' = 52500 kg, u' = -0.14 m/s

Substitute into equation 2

V = (260000×0.32+52500×(-0.14))/(260000+52500)

V = (83200-7350)/312500

V = 75850/312500

V = 0.243 m/s

8 0
3 years ago
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