A wave produced by a doorbell

car travels 80 meters due north in 12 seconds then the car turns around and travels 30 Mi do South in 4 seconds calculate the av

Zinaida [17]

Answer:

1) 3.1 m/s

2) 7 m/s

Explanation:

Distance due north = 80 m

Distance due south = 30 m

Distance between north and south = (80 - 30) m = 50 m

Total time = (12 + 4) sec = 16 sec

1) Average speed = 50/16 = 3.1 m/s

2) Average velocity = Total distance/total time = (80 + 30) m/16 s = 110/16 = 7 m/s

8 0

4 years ago

I need help with a physics worksheet about Newtons Laws.

Viktor [21]

The diagram of the object on the inclined plane is shown below

The mass of 7kg is exerted on the plane surface. It is acted upon by gravity. The force exerted on the surface is calculated by applying the formula,

F = mass x acceleration due to gravity

Assumme acceleration due to gravity is 9.8m/s^2, then

Force = 7 x 9.8 = 68.6N

The surface exerts an opposite force of the same magnitude with the force of the object but in opposite direction. Since it is inclined at an angle of 36.9 degrees,

Normal force = mgCosθ = 68.6Cos36.9 = 54.86N

Recall, frictional force = normal reaction x coefficient of friction

Given that coefficient of kinetic friction = 0.35,

Frictional force = 54.86 x 0.35 = 19.201N

This is the force that must be overcome to keep the object moving.

The force pulling the object upwards along the inclined plane is

mgSinθ = 7 x 9.8 x Sin36.9 = 41.19N

Since the velocity is constant, it means that there is no acceleration. The net force is zero. The force required to pull the mass and make it move at constant velocity must be equal to the sum of the exerted force and the the frictional force(Since it involves coeffricient of kinetic friction, it would cause the object to keep moving)

Thus,

Required force = 19.2 + 41.19 = 60.4N

Option E is correct

8 0

1 year ago

A green light is submerged 2.70 m beneath the surface of a liquid with an index of refraction 1.31. What is the radius of the ci

Anettt [7]

Answer:

The radius is $r = 3.1905 \ m$

Explanation:

From the question we are told that

The distance beneath the liquid is $d = 2.70 \ m$

The refractive index of the liquid is $n_i = 1.31$

Now the critical value is mathematically represented as

$\theta = sin ^{-1} [\frac{1}{n_i} ]$

substituting values

$\theta = sin ^{-1} [\frac{1}{131} ]$

$\theta = 49.76^o$

Using SOHCAHTOA rule we have that

$tan \theta = \frac{ r}{d}$

=> $r = d * tan \theta$

substituting values

$r = 2.7 * tan (49.76)$

$r = 3.1905 \ m$

5 0

3 years ago

Two positive charges of 20 micro coulomb and 100 micro coulomb and the distance between them is 150cm.What will be the electrica

Roman55 [17]

Answer:

0.8 N

Explanation:

From coulomb's law,

Formula:

F = kqq'/r²........................ Equation 1

Where F = Force of repulsion, k = coulomb's constant, q = first positive charge, q' = second positive charge, r = distance between the charge.

Given: q = 20 μC = 20×10⁻⁶ C, q' = 100 μC = 100×10⁻⁶ C, r = 150 cm = 1.5 m.

Constant: k = 9×10⁹ Nm²/C²

Substitute these values into equation 1

F = (20×10⁻⁶ )( 100×10⁻⁶)(9×10⁹)/1.5²

F = 1800×10⁻³/2.25

F = 1.8/2.25

F = 0.8 N

3 0

3 years ago

Tom is throwing an baseball at an aluminum can,

pishuonlain [190]

Answer:

The question relates to the conservation of energy principle, the conservation of the linear momentum, and Newton's Laws of motion

Part A

1) Tom throwing a baseball at a can

The initial velocity of the baseball = v₂

The initial kinetic energy of the baseball, K.E.₂ = (1/2)·m₂·v₂²

∴ The final kinetic energy of the baseball, K.E.₂' = (1/2)·m₂·v₂'² < (1/2)·m₂·v₂²

Therefore, the energy of the ball before the collision is lesser than the energy of the ball after the collision

2) The evidence that would likely support the claim is that the baseball's height above the ground reduces rapidly immediately after the collision which is due to the reduced velocity, and therefore, the reduced (kinetic) energy

The final velocity of the baseball v₂' < v₂

Part B

1) The argument

The initial velocity of the can = v₁ = 0 (The can is initially at rest)

The initial kinetic energy of the can, K.E.₁ = (1/2)·m₁·v₁² = 0

The final velocity of the can v₁' > v₁ = 0

∴ The final kinetic energy of the can, K.E.₁ = (1/2)·m₁·v₁² > 0

Given that the velocity of the can increases from zero to a positive value after collision with the baseball, the kinetic energy of the can is increased from zero before the collision to a positive value after the collision

2) An evidence in support of the argument is the motion of the can which was initially at rest which is an indication of increase in energy podded by the can

Explanation:

8 0

3 years ago