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pishuonlain [190]
3 years ago
10

If a car travels 30 miles in 5 hours, how many miles per hour is the car traveling?

Physics
1 answer:
Ivenika [448]3 years ago
3 0

Answer:

6 mph

Explanation:

30 miles/5 hours = 6 mph

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TECHNOLOGY A Lamborghini Aventador accelerates from rest to 100 km / h in 2.9 s. The mass of the car is 1575 kg Calculate the po
DerKrebs [107]

At 100 km/hr, the car's kinetic energy is

KE = (1/2) (mass) (speed)²

KE = (1/2) (1575 kg) ( [100 km/hr] x [1000 m/km] x [1 hr/3600 sec] )²

KE = (787.5 kg) (27.78 m/s)²

KE = 607,639 Joules

In order to deliver this energy in 2.9 seconds, the engine must supply

(607,639 J / 2.9 sec) = 209,531 watts

<em>Power = 281 HP</em>

6 0
3 years ago
Both formal and informal
Tema [17]

Explanation:

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5 0
2 years ago
A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in
german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

\Sigma F = -f = m\cdot a (Eq. 1)

Where:

f - Kinetic friction force, measured in newtons.

m - Mass of the box, measured in kilograms.

a - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight (W = m\cdot g) and uniform accelerated motion (v = v_{o}+a\cdot t), we expand the previous expression:

-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)

And the magnitude of the friction force exerted on the box is calculated by this formula:

f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right) (Eq. 1b)

Where:

W - Weight, measured in newtons.

g - Gravitational acceleration, measured in meters per square second.

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

t - Time, measured in seconds.

If we know that W = 52.4\,N, g = 9.807\,\frac{m}{s^{2}}, v_{o} = 1.37\,\frac{m}{s}, v = 0\,\frac{m}{s} and t = 2.8\,s, the magnitud of the kinetic friction force exerted on the box is:

f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s}  }{2.8\,s} \right)

f = 2.614\,N

The magnitude of the friction force exerted on the box is 2.614 newtons.

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3 years ago
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The motorist travels (a) 58 km/h and (b) ~16.1 m/sec
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