All categories

3 years ago

Select the correct answer.

1 answer:

6 0

<span>The correct answer is C:Waves transfer energy, but not matter. A wave does not move matter in the direction of its propagation. It only transfers energy just like the ocean wave traveling many miles away with the water just moving up and down.</span>

You might be interested in

A 1.2 kg hammer slams down on a nail at 5.0 m/s and bounces off at 1.0 m/s. If the impact lasts 1.0 ms, what average force is ex

Delvig [45]

Answer:

Explanation:

Impulse results in a change of momentum

FΔt = mΔV

F = mΔV/Δt

The impulse acting on the hammer will equal the impulse acting on the nail

If we assume upward is the positive direction

F = m(vf - vi)/t

F = 1.2(1.0 - (-1.5)) / 0.001

F = 3000 N

7 0

2 years ago

A ball is thrown up into the air with an initial velocity of 18 m/s. A) How high does the ball go? B) Calculate the time needed

kaheart [24]

Answer:

B) t = 1.83 [s]

A) y = 16.51 [m]

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f} =v_{o} -g*t$

where:

Vf = final velocity = 0

Vo = initial velocity = 18 [m/s]

g = gravity acceleration = 9.81 [m/s²]

t = time [s]

Note: the negative sign in the above equation means that the acceleration of gravity is acting in the opposite direction to the motion.

A) The maximum height is reached when the final velocity of the ball is zero.

0 = 18 - (9.81*t)

9.81*t = 18

t = 18/9.81

t = 1.83 [s], we found the answer for B.

Now using the following equation.

$y = y_{o} + v_{o}*t - 0.5*g*t^{2}\\$

where:

y = elevation [m]

Yo = initial elevation = 0

y = 18*(1.83) - 0.5*9.81*(1.83)²

y = 16.51 [m]

7 0

2 years ago

Two spheres, 1.00 kg each, whose centers are 2.00 m apart, would have what gravitational force between them? A. 3.14 X 10-17 N

Angelina_Jolie [31]

Answer: B

Explanation: the teacher just told us the answer

4 0

2 years ago

An object with a mass m slides down a rough 370 inclined plane where the coefficient of kinetic friction is 0.20. If the plane i

Svetllana [295]

Answer:

$v \approx 9.312\,\frac{m}{s}$

Explanation:

The Free Body Diagram of the system is presented in the image attached below. The final speed is determined by means of the Principle of Energy Conservation and the Work-Energy Theorem:

$K_{A} + U_{g,A} = K_{B} + U_{g,B} + W_{loss}$

$K_{B} = K_{A} + U_{g,A}-U_{g,B} - W_{loss}$

$\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot s\cdot \sin \theta - \mu_{k}\cdot m \cdot g \cdot s \cos \theta$

$\frac{1}{2}\cdot v^{2} = g\cdot s \cdot (\sin \theta - \mu_{k}\cdot \cos \theta)$

$v = \sqrt{2\cdot g \cdot s \cdot (\sin \theta - \mu_{k}\cdot \cos \theta)}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (10\,m)\cdot (\sin 37^{\textdegree} - 0.2\cdot \cos 37^{\textdegree})}$

$v \approx 9.312\,\frac{m}{s}$

3 0

3 years ago

Un atleta de 70 kg de masa que ha efectuado un salto de altura cae una vez que ha

Allushta [10]

Answer:

a) the elastic force of the pole directed upwards and the force of gravity with dissects downwards

Explanation:

The forces on the athlete are

a) at this moment the athlete presses the garrolla against the floor, therefore it acquires a lot of elastic energy, which is absorbed by the athlete to rise and gain potential energy,

therefore the forces are the elastic force of the pole directed upwards and the force of gravity with dissects downwards

b) when it falls, in this case the only force to act is batrachium by the planet, this is a projectile movement for very high angles

c) When it reaches the floor, it receives an impulse that opposes the movement created by the mat. The attractive force is the attraction of gravity.

3 0

2 years ago