All categories

3 years ago

Identify the motion when a car is moving with constant speed along a curve

Physics

2 answers:

zavuch27 [327]3 years ago

5 0

Answer:

Uniform circular motion

Explanation:

When a car is moving with constant speed along a curve, it will said to have uniform circular motion. The speed of the car remains constant in this case. But its velocity keeps on changing. The velocity can be calculated by drawing the tangent on the circle. Hence, the motion of the car is uniform circular motion.

nasty-shy [4]3 years ago

4 0

Answer:

Uniform motion

Explanation:

The motion when a car is moving with constant speed along a curve is an example of uniform motion.

A car moving at constant speed covers equal distances in equal intervals of time. The average speed remains the same in uniform motion.

The curve defines the path of the motion and the speed remains constant.

You might be interested in

A gas has a volume of 8.21 mL and exerts a pressure of 2.9 atm. What new volume will the gas have if the pressure is changed to

labwork [276]

Answer:

5.95 ml

Explanation:

Given info

P1=2.9 atm

V1=8.21 ml

P2=4 atm

From Boyle's law we know that p1v1=p2v2 where p and v are pressure and volume respectively. This is at a constant temperature. Making v2 the subject of formula then

V2=p1v1/p2

V2= 2.9*8.21/4=5.95 ml

3 0

3 years ago

A 74 kg man holding a 13 kg box rides on a skateboard at a speed of 11m/s. He throws the box behind him,giving it velocity of 6

Ivahew [28]

Answer:

After throwing the object the, the velocity of the man is 13.98 m/s

Explanation:

Given:

Let,

mass of man, m1 = 74 kg

mass of box, m2 = 13 kg

Initial velocity u = 11 m/s (initially both are together hence initial velocity will be same for both)

Final velocity of man = v1

Final velocity of box = v2 = -6 m/s (the velocity is recoil velocity therefore it is negative)

To Find:

Final velocity of man,after throwing the object = v1 = ?

Solution:

Recoil velocity:

It is the backward velocity experienced.

Here recoil velocity is the backward velocity experience while throwing the box behind.Hence the velocity of the box is negative 6 m/s.

The recoil velocity is the result of conservation of linear momentum of the system. Therefore we will follow the law of conservation of momentum.

Law of conservation of momentum :

Total momentum of an isolated system before collision is always equal to total momentum after collision

$\textrm{total momentum before collision}=\textrm{total momentum after collision}\\m_{1}\times u+ m_{2}\times u=m_{1}\times v_{1}+m_{2}\times v_{2}$

substituting the values which are given above we get

$74\times 11 + 13\times 11 = 74\times v_{1} +13\times -6\\ 957 = 74\times v_{1} -78\\v_{1}=\frac{1035}{74} \\v_{1}=13.98\ m/s$

Therefore, After throwing the object the, the velocity of the man is 13.98 m/s

3 0

3 years ago

Two round rods, one steel andthe other copper, are joined end to end. Each rod is 0.750 mlong and 1.50 cmin diameter. The combin

nikitadnepr [17]

Answer: a) Strain on Steel rod = 0.0001078

b) elongation on the steel rod = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain on Copper Rod = 0.000189

Explanation: a) To obtain the strain of the steel rod, we invoke Hooke's law which states that, provided the elastic limit of A material isn't exceeded, the stress it undergoes is directly proportional to its strain.

(Stress, σ) ∝ (Strain, ε)

The constant of proportionality is called Young's modulus, E.

σ = Eε

For steel, Younger Modulus as obtained from literature = 210GPa.

Strain = Stress/Young's Modulus

Stress = (Force or Load applied)/ Cross sectional Area.

Force applied For the steel = 4000N

Cross sectional Area = (π(D^2))/4

D = 1.50cm = 0.015m

A = 0.0001767 m2

σ = 4000/0.0001767 = 22637238.257 N/m2

ε = σ/E = 22637238.257/(210 × (10^9)) = 0.0001078.

b) To solve for elongation.

Strain, ε = (elongation, dl)/(original length, lo)

Elongation, dl = strain × original length

dl = 0.0001078 × 0.75 = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain in Copper

ε = σ/E; σ = 22637238.257 N/m2

Young's modulus of Copper, from literature, = 120GPa

ε = 22637238.257/(120 × (10^9)) = 0.000189

4 0

3 years ago

How do lines of latitude affect how direct or indirect the Sun’s rays are on the Earth?

Sonja [21]

Answer:

Explanation: We have seasons because Earth's axis – the imaginary line that goes through the Earth and around which the Earth spins — is tilted. It's tilted about 23.5 degrees relative to our plane of orbit (the ecliptic) around the Sun. As we orbit our Sun, our axis always points to the same fixed location in space. Our northern axis points almost directly toward Polaris, the North Star.

6 0

4 years ago

Point charge μC is located at x =, y = , point charge is located at x = 0m. What are (a)the magnitude and (b)direction of the to

pishuonlain [190]

Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0

Explanation:

a) First of all find the distance between the two charges;

x = 0, y = 0.30 and x = 0.40 m, y = 0

r = √( 0.4² + 0.3²)

= 0.5m

hence, the force F = 2Kq1q2cosθ /r²...............equation 1

but cosθ = y/r = 0.3/0.5

cosθ = 0.6

plugging back to equation 1;

F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2

F = 540 x 10^-3

Magnitude of Force = 0.54N

b) Direction is at angle 90

6 0

3 years ago