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2 years ago

What is the net force acting on a car cruising at a constant velocity of 70 km/h (a) on a level road, (b) on an uphill road?

Physics

1 answer:

xeze [42]2 years ago

6 0

The net force on the car for either case will be zero, since the car is moving at a constant velocity.

<h3>Net force on the car</h3>

The net force on the car is the sum of all the forces acting on the car.

∑F = ma

<h3>When the car is on a level road</h3>

When the car is on a level road, the only two forces acting on the car is the applied force and frictional force of the road.

F - Ff = 0

The net force will be zero, since the car is moving at a constant velocity.

<h3>When the car is on uphill</h3>

The forces acting on the car at the uphill is the applied force, weight of the car acting downwards and the frictional force acting opposite direction.

F - Wsinθ - μFₙcosθ = 0

The net force will be zero, since the car is moving at a constant velocity.

Learn more about net force at constant velocity here: brainly.com/question/14392124

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A gas contained within a piston-cylinder assembly, initially at a volume of 0.1 m3 , undergoes a constant-pressure expansion at

Gnom [1K]

Answer:

Work: 4.0 kJ, heat: 4.25 kJ

Explanation:

For a gas transformation at constant pressure, the work done by the gas is given by

$W=p(V_f -V_i)$

where in this case we have:

$p = 2 bar = 2\cdot 10^5 Pa$ is the pressure

$V_i = 0.1 m^3$ is the initial volume

$V_f = 0.12 m^3$ is the final volume

Substituting,

$W=(2\cdot 10^5)(0.12-0.10)=4000 J = 4.0 kJ$

The 1st law of thermodynamics also states that

$\Delta U = Q-W$

where

$\Delta U$ is the change in internal energy of the gas

Q is the heat absorbed by the gas

Here we know that

$\Delta U = +0.25 kJ$

Therefore we can re-arrange the equation to find the heat absorbed by the gas:

$Q=\Delta U + W = 0.25 kJ + 4.0 kJ = 4.25 kJ$

7 0

4 years ago

Please help me guys never mind the calculations

vlada-n [284]

The shape is connected in parallel so;

5.1) Ans;

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} \\ \frac{1}{R} = \frac{1}{2} + \frac{1}{3} \\ \frac{1}{R} = \frac{3 + 2}{6} = \frac{5}{6} \\ R = \frac{6}{5} = 1.2 \: \: ohm$

5.2) Ans;

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} \\ \frac{1}{R} = \frac{1}{8} + \frac{1}{10} \\ \frac{1}{R} = \frac{5 + 4}{40} = \frac{9}{40} \\ R = \frac{40}{9} = 4.4 \: \: ohm$

I hope I helped you^_^

7 0

3 years ago

The space shuttle travels at a speed of about 7.6times10^3 m/s. The blink of an astronaut's eye lasts about 110 ms. How many foo

sveta [45]

Answer:

It covers distance of 9.15 football fields in the said time.

Explanation:

We know that

$Distance=Speed\times Time$

Thus distance covered in blinking of eye =

$Distance=7.6\times 10^{3}m/s\times 110\times 10^{-3}s\\\\Distance=836 meters$

Thus no of football fields= $\frac{936}{91.4}=9.15Fields$

7 0

3 years ago

Consider a transformer. used to recharge rechargeable flashlight batteries, that has 500 turns in its primary coil, 3 turns in i

Rashid [163]

Answer:

<em>c) 2.304 Watts</em>

Explanation:

A transformer is used to step-up or step-down voltage and current. It uses the principle of electromagnetic induction. When the primary coil is greater than the secondary coil, the it is a step-down transformer, and when the primary coil is less than the secondary coil, the it is a step-up transformer.

number of primary turns = $N_{p}$ = 500 turns