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valina [46]
3 years ago
11

The 1350-kg car has a velocity of 24 km/h up the 8-percent grade when the driver applies more power for 18 s to bring the car up

to a speed of 65 km/h. Calculate the time average F of the total force tangent to the road exerted on the tires during the 18 s. Treat the car as a particle and neglect air resistance.
Physics
1 answer:
Brrunno [24]3 years ago
4 0

Answer:

Explanation:

We shall apply concept of impulse to solve the problem  .

Impulse = force x time

impulse = change in momentum

force x time = change in momentum

initial speed u = 24 km/h = 6.67 m /s

final speed v = 65 km/h = 18.05 m /s

change in momentum = m v - mu

= m ( v-u )

= 1350 ( 18.05 - 6.67 )

= 15363 kg m/s

F x 18 = 15363

F = 853.5 N .

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Over 4 seconds, a car's momentum decreases by 1000 kg m/s how much force did it take to make this happen?
kotykmax [81]

Answer:

250N

Explanation:

Given parameters:

Time  = 4s

Momentum  = 1000kgm/s

Unknown:

Force  = ?

Solution:

To solve this problem, we use Newton's second law of motion;

      Ft  = Momentum

F is the force

t is the time

So;

          F x 4 = 1000kgm/s

          F  = 250N

6 0
3 years ago
If a current is two amps and the resistance is 3 ohms, how much voltage was needed?
Hunter-Best [27]

Answer:

6 V

Explanation:

We can solve the problem by using Ohm's law:

V=RI

where

V is the voltage in the circuit

R is the resistance

I is the current

In this problem, we know the current, I=2 A, and the resistance, R=3 \Omega, therefore we can find the voltage in the circuit:

V=RI=(3 \Omega )(2 A)=6 V

7 0
3 years ago
A spring on Earth has a 0.500 kg mass suspended from one end and the mass is displaced by 0.3 m. What will the displacement of t
julia-pushkina [17]

To solve this problem we will apply the concepts related to the Force of gravity given by Newton's second law (which defines the weight of an object) and at the same time we will apply the Hooke relation that talks about the strength of a body in a system with spring.

The extension of the spring due to the weight of the object on Earth is 0.3m, then

F_k = F_{W,E}

kx_1 = mg

The extension of the spring due to the weight of the object on Moon is a value of x_2, then

kx_2 = mg_m

Recall that gravity on the moon is a sixth of Earth's gravity.

kx_2 = m\frac{g}{6}

kx_2 = \frac{1}{6} mg

kx_2 = \frac{1}{6} kx_1

x_2 = \frac{1}{6} x_1

We have that the displacement at the earth was x_1 = 0.3m, then

x_2 = \frac{1}{6} 0.3

x_2 = 0.05m

Therefore the displacement of the mass on the spring on Moon is 0.05m

6 0
3 years ago
What is the momentum of a 1550kg car that is traveling 38.0 m/s?
konstantin123 [22]

Answer:

p = 58,900 kg m/s

Explanation:

p = m × v

p = 1,550 × 38.0

p = 58,900 kg m/s

5 0
2 years ago
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