Problem 1: You go outside on a hot, sunny summer day and you feel the warm sun on your skin. What causes this warmth?A. The sun heats the air directly around your skin.

B. Waves of light carry energy directly to your skin.

C. Light causes surface waves on your skin.

D. Water vapor in the air warms your skin.

PRoblem 2: When a water wave runs into you at the beach, what causes you to get knocked down?

A. air molecules in front of the wave

B. the wave's energy

C. periodic motion

D. friction

Answers

Answer 1

Answer: 1) B
The sun transfers energy to the earth via radiation. This radiation is a form of electromagnetic radiation so it heats your skin direclty
2) B
The waves' energy is transferred to you and you lose your footing and fall

Answer 2

Answer: B.waves of light carry energy directly to your skin.

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a warrior swings a slingshot in a horizontal circle above his head at a constant speed. The sling is 1.5 m long, and the stone has a mass of 50 g. The tension in the string is 3.3 N. When he releases the sling, what will the stone's speed be?

In the easiest definition what is inertia

Answers

Answer:

Aproperty of matter by which it remains at rest or in motion in the same straight line unless acted upon by some external force

When you hit a nail into a board using a hammer the head of the nail gets warm. In terms of kinetic and thermal energy describe why you think this happens?

Answers

The kinetic energy is transferred to thermal energy through friction

When we hit the nail, the particulate objects are set into kinetic energy and they begin to move. However, due to the rigid structure, the particulate objects cannot move from their equilibrium position, and therefore, it is dissipated through heat.

A soccer ball is traveling at a velocity of 50 m/s. The kinetic energy of the ball is 500 J. What is the mass of the soccer ball?

Answers

A soccer ball is traveling at a velocity of 50 m/s. The kinetic energy of the ball is 500 J.The mass of the soccer ball is 0.4 kgs. This answer is derived from the formula K=1/2 MV^2.So velocity and kinetic energy are given from that mass of the ball is calculated.By substituting the values 500=1/2*M*50*50 which gives M=0.4 Kgs.

Answer:

C.0.4 kg

This is your answer!

A car with a mass of 1500 kg is traveling at a speed of 30 m/s. Calculate its kinetic energy.step by step pls

Answers

Explanation:

KE = 1/2 mv²,

where KE is the kinetic energy,

m is the mass,

and v is the velocity.

Given m = 1500 kg and v = 30 m/s:

KE = 1/2 (1500 kg) (30 m/s)²

KE = 675,000 J

KE = 675 kJ

A 111 ‑turn circular coil of radius 2.11 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicular to the plane of the coil. The coil is connected to a 14.1 Ω resistor to create a closed circuit. During a time interval of 0.125 s, the magnetic field strength decreases uniformly from 0.669 T to zero. Find the energy, in millijoules, that is dissipated in the resistor during this time interval.

Answers

Answer:

0.0061 J

Explanation:

Parameters given:

Number of turns, N = 111

Radius of turn, r = 2.11 cm = 0.0211 m

Resistance, R = 14.1 ohms

Time taken, t = 0.125 s

Initial magnetic field, Bin = 0.669 T

Final magnetic field, Bfin = 0 T

The energy dissipated in the resistor is given as:

E = P * t

Where P = Power dissipated in the resistor

Power, P, is given as:

P = V² / R

Hence, energy will be:

E = (V² * t) / R

To find the induced voltage (EMF), V:

EMF = [-(Bfin - Bin) * N * A] / t

A is Area of coil

EMF = [-(0 - 0.669) * 111 * pi * 0.0211²] / 0.125

EMF = 0.83 V

Hence, the energy dissipated will be:

E = (0.83² * 0.125) / 14.1

E = 0.0061 J

A student releases a block of mass m at the top of a slide of height h1. the block moves down the slide and off the end of the table of height h2, landing on the floor a horizontal distance d from the edge of the table. Friction and air resistance are negligible. The overall height H of the setup is determined by the height of the room. Therefore, if h1 is increased, h2 must decrease by the same amount so that the sum h1+h2 remains equal to H. The student wants to adjust h1 and h2 to make d as large as possible.A) 1) Without using equations, explain why making h1 very small would cause d to be small, even though

h2 would be very large?
2) Without using equations, explain why making h2 very small would cause d to be small, even though
h1 would be large

B) Derive an equation for d in terms of h1, h2, m, and physical constants as appropriate.

Answers

(A)

(1) The reason for making $h_(1)$ very small is due to smaller value of horizontal component of launch velocity.

(2) The reason for making $h_(2)$ very small is due to smaller value of time of flight.

(B) The distance (d) covered by the block is $2\sqrt{h_(1)h_(2)}$ .

Given data:

The mass of block is, m.

The height of table from the top of slide is, $h_(1)$ .

The height of table at the end of slide is, $h_(2)$ .

The height of room is, H.

(A)

(1)

If the launch velocity of the block is v, then its horizontal component is very small, due to which adjusting the height $h_(1)$ to be very small will cause the d to be small.

(2)

The height $h_(2)$ is dependent on the time of flight, and since the time of flight taken by the block to get to the floor is very less, therefore the block will not get sufficient time to accomplish its horizontal motion. That is why making $h_(2)$ very small will cause d to be smaller.

(B)

The expression for the distance covered by the block is,

$v=(d)/(t)\nd = v * t$ ..............................(1)

Here, v is the launch speed of block and t is the time of flight.

The launch speed is,

$v^(2)=u^2+2gh_(1)\nv=\sqrt{u^2+2gh_(1)}\nv=\sqrt{0^2+2gh_(1)}\nv=\sqrt{2gh_(1)}$

And the time of flight is,

$h_(2)=ut+(1)/(2)gt^(2)\nh_(2)=0 * t+(1)/(2)gt^(2)\nh_(2)=0+(1)/(2)gt^(2)\nt=\sqrt(2h_(2))/(g)$

Substituting the values in equation (1) as,

$d = v * t\nd = \sqrt{2gh_(1)}* \sqrt(2 h_(2))/(g)}\nd=2\sqrt{h_(1)h_(2)}$

Thus, the distance (d) covered by the block is $2\sqrt{h_(1)h_(2)}$ .

Learn more about the time of flight here:

brainly.com/question/17054420?referrer=searchResults

A1) The reason why making h₁ very small would cause d to be small is; Because the horizontal component of the launch velocity would be very small.

A2) The reason why making h₂very small would cause d to be small is;

Because the time of flight it will take the object to get to the floor would be very small and as a result, the object would not possess enough time to move horizontally before the vertical motion.

B) The equation for d in terms of h₁ and h₂ is;

d = 2√(h₁ × h₂)

A) 1) The reason why making h₁ very small would cause d to be small is because the horizontal component of the launch velocity would be very small.

A) 2) The reason why making h₂ very small would cause d to be small is because the time of flight it will take the object to get to the floor would be very small and as a result, the object would not possess enough time to move horizontally before the vertical motion.

B) Formula for Launch Velocity is;

V = √(2gh₁)

h₁ was used because the top of the slide from where the student released the block has a height of h₁.

Also, the time it takes to fall which is time of flight is given by the formula;

t = √(2h₂/g)

h₂ was used because the height of the table the object is on before falling is h₂.

Now, we know that d is distance from edge of the table and formula for distance with respect to speed and time is;

distance = speed × time

Thus;

d = √(2gh₁) × √(2h₂/g)

g will cancel out and this simplifies to give;

d = 2√(h₁ × h₂)

Read more at; brainly.com/question/20427663

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