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Alja [10]
2 years ago
3

A 1400 kg car is traveling on a level road at a constant speed of 26.8 m/s (60 miles per hour.) The drag force exerted by air on

the car and the frictional force exerted by the road on the car’s tires add to give a net force of 680N pointing opposite to the direction of the car’s motion. What is the rate of work done (power) by the air and the road on the car
Physics
1 answer:
VLD [36.1K]2 years ago
5 0

Answer:

Power, P = 18224 watts

Explanation:

It is given that,

Mass of car, m = 1400 kg

Speed of the car, v = 26.8 m/s

The frictional force exerted by the road on the car’s tires, F = 680 N

We need to find the rate of doing work by the air and the road on the car. We know that rate of doing work is called power exerted. It is equal to the product of force and velocity. It is given by :

P=F\times v

P=680\ N\times 26.8\ m/s

P = 18224 watts

So, the power delivered by the air and the road on the car is 18224 watts. Hence, this is the required solution.

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What is the direction of the magnetic field b⃗ net at point a? Recall that the currents in the two wires have equal magnitudes.
andrew11 [14]

Answer:

Explanation:

The direction of a magnetic field indicates where the magnetic inluence on the electric charges are directed to.

From the given  question, we are to determine the direction of the magnetic field bnet at a point A.

Also, having the notion that  the currents in the two wires have equal magnitudes, Then:

\bar{B_{net}} = \bar{B_1} + \bar{B_2}

\bar{B_{net}} = \frac{\mu_oI}{2 \pi r } \bar {k}+ \frac{\mu_oI}{2 \pi r } \bar {k}

\bar{B_{net}} = \frac{2 \mu_oI}{2 \pi r } \bar {k} \ out

Thus; \bar{B_{net}} points out of the screen at A.

6 0
2 years ago
There are lots of examples of ideal gases in the universe, and they exist in many different conditions. In this problem we will
elena-14-01-66 [18.8K]

Answer:

P = ρRT/M

Explanation:

Ideal gas equation is given as follows generally:

PV = nRT (1)

P = pressure in the containing vessel

V = volume of the containing vessel

n = number of moles

R = gas constant

T = temperature in K

n = m/M

m = mass of the gas contained in the vessel in g

M = molar mass in g/mol

ρ = m/V

Density of the gas = ρ

Substituting for n in (1)

PV = mRT/M. (2)

Dividing equation (2) through by V

P = m/V ×RT/M

P = ρRT/M

5 0
2 years ago
A circular coil 17.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this locati
sergeinik [125]

Answer:

The torque in the coil is  4.9 × 10⁻⁵ N.m  

Explanation:

T = NIABsinθ

Where;

T is the  torque on the coil

N is the number of loops = 9

I is the current = 7.8 A

A is the area of the circular coil = ?

B is the Earth's magnetic field = 5.5 × 10⁻⁵ T

θ is the angle of inclination = 90 - 56 = 34°

Area of the circular coil is calculated as follows;

A = \frac{\pi d^2}{4} \\\\A = \frac{\pi 0.17^2}{4} =0.0227 m^2

T = 9 × 7.8 × 0.0227 × 5.5×10⁻⁵ × sin34°

T = 4.9 × 10⁻⁵ N.m

Therefore, the torque in the coil is  4.9 × 10⁻⁵ N.m

5 0
2 years ago
A blue puck has a velocity of 0i – 3j m/s and a mass of 4 kg. A gold puck has a velocity of 12i – 5j m/s and a mass of 6 kg. Wha
Mnenie [13.5K]
By definition, the kinetic energy is given by:
 K = (1/2) * m * v ^ 2
 where
 m = mass
 v = speed
 We must then find the speed of both objects:
 blue puck
 v = root ((0) ^ 2 + (- 3) ^ 2) = 3
 gold puck
 v = root ((12) ^ 2 + (- 5) ^ 2) = 13
 Then, the kinetic energy of the system will be:
 K = (1/2) * m1 * v1 ^ 2 + (1/2) * m2 * v2 ^ 2
 K = (1/2) * (4) * (3 ^ 2) + (1/2) * (6) * (13 ^ 2)
 K = <span> 525</span> J
 answer
 The kinetic energy of the system is<span> <span>525 </span></span>J
6 0
2 years ago
You are called as an expert witness to analyze the following auto accident: Car B, of mass 2100 kg, was stopped at a red light w
Artemon [7]

Answer:

Explanation:

Force of friction at car B ( break was applied by car B ) =μ mg = .65 x  2100 X 9.8  = 13377 N .

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If v be the common velocity of both the cars after collision

kinetic energy of both the cars = 1/2 ( 2100 + 1500 ) x v²

= 1800 v²

so , applying work - energy theory ,

1800 v² = 97652.1

v² = 54.25

v = 7.365 m /s

This is the common velocity of both the cars .

To know the speed of car A , we shall apply law of conservation of momentum  .Let the speed of car A before collision be v₁ .

So , momentum before collision = momentum after collision of both the cars

1500 x v₁ = ( 1500 + 2100 ) x 7.365

v₁ = 17.676 m /s

= 63.63 mph .

( b )

yes Car A was crossing speed limit by a difference of

63.63 - 35 = 28.63 mph.

7 0
2 years ago
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