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Travka [436]
3 years ago
15

Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an

y other forces, what is the maximum acceleration a this engine can produce in 1650-kg car?
Physics
1 answer:
joja [24]3 years ago
8 0

Answer:

Approximately 9.7\; \rm m \cdot s^{-2}.

Explanation:

Assuming that there is no other force on this vehicle, the 16000\; \rm N force from the road would be the only force on this vehicle. The net force would then be equal to this 16000\; \rm N\! force. The size of the net force would be 16000\; \rm N\!\!.

Let m denote the mass of this vehicle and let \Sigma F denote the net force on this vehicle.  

By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, a, would be:

\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}.

For this vehicle, \Sigma F = 16000\; \rm N whereas m = 1650\; \rm kg. The acceleration of this vehicle would be:

\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}.

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A 2500-kg car is being pushed up a hill at an angle of 35 degrees. Determine the gravitational
velikii [3]

Answer:

The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N

Explanation:

For this exercise we will use Newton's second law, let's set a reference system where the x axis is parallel to the plane, in the adjoint we can see the forces in the system.

           sin  35 = Wₓ / W

            cos 35 = W_y / W

            Wₓ = W sin 35

            W_y = W cos 35

            Wₓ = 2500 9.8 sin 35

             Wₓ = 14052.6 N

let's write the equations for each axis

and

Y axis  

       N-W_y = 0

       N = W_y

X axis  

       F -Wₓ = m a

       F = Wₓ + m a = mg sin 35 + m a

       F = m (a + g sin 35)

   

let's calculate

       F = 2500 (5 + 9.8 sin 35)

       F = 26552.6 N

The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N

3 0
3 years ago
A force of 5N produces an acceleration of 8m/s2 on mass m1, and an acceleration of 24m/s2 on a mass m2. What acceleration would
kotykmax [81]

The acceleration that the same force will provide if both masses are tied together is; 6.0 m/s².

<h3>How to find the Acceleration?</h3>

We are given;

Force; F = 5 N

Acceleration of the first mass, a₁ = 8.0 m/s²

Acceleration of the second mass, a₂ = 24 m/s²

Formula for force is;

F = ma

Let us find both masses; m₁ and m₂.

m₁ = F/a₁

m₂ = F/a₂

Thus;

m₁ = 5/8 kg

m₂ = 5/24 kg

Total mass is; m = m₁ + m₂

m = 5/8 + 5/24

m = 15 + 5/24

m = 20/24 kg

Thus, acceleration if they are both tied together is;

a = F/m

a = 5/(20/24)

a = 6.0 m/s².

Read more about Acceleration at; brainly.com/question/605631

#SPJ1

4 0
2 years ago
I need help!!!!<br> Please and thank u!
alexandr1967 [171]

Answer:

False

True

False

True

False

True

Explanation:

5 0
3 years ago
A 25.0-g sample of copper at 363 K is placed in I 00.0 g of water at 293 K. The copper and water quickly come to the sa me tempe
Simora [160]

Answer:

Final temperature is 295K

Explanation:

Where the sample of copper is placed in the water, the heat transferred from the copper is equal that the heat absorbed by the water.

The heat transferred from the copper is:

C×\frac{1mol}{63,546g}×mass×ΔT

Where C is molar heat capacity of copper (24,5J/molK)

Mass is 25,0g

And ΔT is final temperature - initial temperature (X-363K)

Also, the heat absorbed by the water is:

-C×\frac{1mol}{18,02g}×mass×ΔT

Where C is molar heat capacity of water (75,2J/molK)

Mass is 100,0g

And ΔT is final temperature - initial temperature (X-293K)

As heat transferred is equal to heat absorbed:

24,5J/molK×\frac{1mol}{63,546g}×25,0g×(X-363K) = -75,2J/molK×\frac{1mol}{18,02g}×100,0g× (X-293K)

9,64X J/K - 3499J = - 417X J/K + 122273J

426,64X J/K = 125772 J

<em>X = 295K</em>

<em></em>

Final temperature is 295K

I hope it helps!

6 0
3 years ago
A e B são dos blocos de massas 3,0 kg e 2,0 kg, respectivamente, que se movimentam juntos sobre uma superficie horizontal e perf
makvit [3.9K]

F = m*a

30 N = (ma + mb) * a

30 = 5*a

a = 6 m/s ^2

F de B em A

30 - F de B,A = ma * a

30 - F de B em A = 3 * 6

30 - 18 = F de B em A

12 = F de B em A


Resposta: 6 m/s^2 e 12N

Bate com o gabarito, man? Ou eu tô viajando aqui?

Abç!

6 0
3 years ago
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