Which property of a wave is labeled "x" on the diagram?

A car is traveling at a constant speed on the highway. Its tires have a diameter of 68.0 cm and are rolling without sliding or s

Bad White [126]

Answer:

37.4m/s

Explanation:

since the car doesn't accelerate, we can use the formula v=ωr where v is linear speed, ω is angular speed (rads/second) and r is radius. Substitute values for equation:

v=55*0.68

v=37.40

5 0

2 years ago

The origins of Yoga can be traced back to which ancient scholar?

Mashcka [7]

Answer:

Rig Veda

Explanation:

5 0

3 years ago

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Bill forgets to put his car in park and it starts rolling west. When it is moving at a speed of 3.5 m/s, it collides with Tanya'

djverab [1.8K]

PART A)

momentum of Bill's car

mass = 900 kg

speed = 3.5 m/s towards west

so momentum of this car given as

$P_1 = 900(3.5) = 3150 kg m/s$

Similarly momentum of Tanya's car

mass = 1100 kg

speed = 2 m/s towards East

so momentum will be

$P_2 = 1100(2) = 2200 kg m/s$

Total momentum of both cars will be

$P = 3150 -2200 = 950 kg m/s$ Towards west

PART B)

Since there is no external force on the system of two cars

So here total momentum of both cars will remains conserved

So final momentum of two cars will be same as initial momentum

$P = 950 kg m/s$ towards west

Part C)

final momentum of Bills car

$P_1 = (900)(2) = 1800 kg m/s$ Towards East

now we have

$P_1 + P_2 = 950$

$1100 v - 1800 = 950$

$v = 2.5 m/s$ towards west

3 0

3 years ago

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A cannon fires a cannonball 500.0 m downrange when set at a 45.0o angle. At what velocity does the cannonball leave the cannon?

kobusy [5.1K]

Answer:

v = 70 m/s

Explanation:

Range on the cannon ball is given as

$d = 500.0 m$

here the angle of the projection of the ball is given as 45 degree

now we know that if the velocity of the ball is "v" then its two components will be given as

$v_x = vcos45$

$v_y = vsin45$

so here time of flight of the motion is given as

$T = \frac{2v_y}{g}$

$T = \frac{2vsin45}{g}$

also the range is given as

$R = v_x T$

$R = (vcos45)(\frac{2vsin45}{g})$

now plug in all data in this equation

$500.0 = \frac{v^2(2sin45cos45)}{g}$

$v = 70 m/s$

4 0

3 years ago

A spring has a force constant of 500.0 N/m. Show that the potential energy stored in the spring is as follows: a. 0.400 J when t

kogti [31]

Answer:

(a) Hence, the potential energy = 0.400 J.

(b) Hence the potential Energy = 0.225 J.

(c) Hence the potential energy = 0 J

Explanation:

Potential Energy; This is the energy of a body, by virtue of its position in the gravitational field. The unit of potential energy is Joules (J)

the potential energy stored in a spring is

Ep = 1/2ke²..................................... Equation 1

Where Ep = potential Energy, k = force constant, e = extension.

a.

When spring is stretched 4.00 cm, e = 4.00 cm = 4/100 = 0.04 m, and k = 500 N/m.

Substituting into equation 1

Ep = 1/2(500)(0.04)²

Ep = 0.4 J.

Hence, the potential energy = 0.400 J.

(b)

When the spring is compressed 3.00 cm, e = 3.00 cm = 3/100 = 0.03 m, and k = 500 N/m.

Substitute into equation 1,

Ep = 1/2(500)(0.03)²

Ep = 250(0.0009)

Ep = 0.225 J.

Hence the potential Energy = 0.225 J.

(c) When the spring is unstretched, e = 0 cm = 0 m and k = 500 N/m.

Substituting into equation 1

Ep = 1/2(500)(0)²

Ep = 250(0)

Ep = 0.

Hence the potential energy = 0 J

3 0

3 years ago