PHYSICS COLLEGE

What two statements are true about energy transformations

Answers

Answer 1

Answer:

First statement:

Energy can neither be created nor destroyed.

Second statement:

Energy can be converted from one form to another.

Explanation:

According to the law of conservation of energy:energy can neither be created nor destroyed but can be converted from one form to another

Related Questions

Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the bow back 50 cm and holds it in position with a force of 150N . If the mass of the arrow is 50g and the "spring" is massless, what is the speed of the arrow immediately after it leaves the bow?
A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance d and is used to launch a mass m along the frictionless surface. What compression of the spring would result in the mass attaining double the kinetic energy received in the above situation?
The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming alpha particles exceeded 32MeV32MeV. This can be explained by the fact that the predictions of the formula apply when the only force involved is the electromagnetic force and will break down if the incoming particles make contact with the nucleus. Use the fact that Rutherford's prediction ceases to be valid for alpha particles with an energy greater than 32MeV32MeV to estimate the radius rrr of the gold nucleus.
A rock is thrown vertically upward with a speed of 14.0 m/sm/s from the roof of a building that is 70.0 mm above the ground. Assume free fall.A) In how many seconds after being thrown does the rock strike the ground? B) What is the speed of the rock just before it strikes the ground?
Dave rows a boat across a river at 4.0 m/s. the river flows at 6.0 m/s and is 360 m across.a. in what direction, relative to the shore, does dave’s boat go? b. how long does it take dave to cross the river? c. how far downstream is dave’s landing point? d. how long would it take dave to cross the river if there were no current?

A swimmer heads directly across a river, swimming at 1.00 m/s relative to still water. He arrives at a point 41.0 m downstream from the point directly across the river, which is 73.0 m wide. What is the speed of the river current?

Answers

Answer:

velocity of the river is equal to 0.56 m/s

Explanation:

given,

velocity of swimmer w.r.t still water = 1 m/s

width of river = 73 m

he arrives to the point = 41 m

$times = (distance)/(speed)$

$times = (73)/(1)$

t = 73 s

$velocity = (distance)/(time)$

= $(41)/(73)$

= 0.56 m/s

velocity of the river is equal to 0.56 m/s

A vertical scale on a spring balance reads from 0 to 245 N . The scale has a length of 10.0 cm from the 0 to 245 N reading. A fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.55 Hz . Ignoring the mass of the spring, what is the mass m of the fish?

Answers

Answer:

mass m of the fish is 7.35 kg

Explanation:

Given data

spring balance reads = 0 to 245 N

length = 10.0 cm

scale reading = 0 to 245 N

frequency f = 2.55 Hz

to find out

mass m of the fish

solution

we know the relation that is

ω = √(k/m) ......................1

here k = spring reading / length = 245 / 0.135

k = 1814.81 N/m

and

ω = 2π × f

ω = 2π × 2.5 = 15.71 rad/s

so put all value in equation 1 we get m

ω = √(k/m)

15.71 = √(1814.81/m)

so m = 7.35

mass m of the fish is 7.35 kg

Answer:

9.55 kg

Explanation:

F = 245 N

Let K be te spring constant

F = K x

K = 245 / 0.1 = 2450 N/m

ω = 2 x π x f = 2 x 3.14 x 2.55 = 16.014 rad/s

$\omega =\sqrt{(K)/(m)}$

where m be the mass of fish

$16.014 =\sqrt{(2450)/(m)}$

m = 9.55 kg

The wind-chill index is modeled by the function W = 13.12 + 0.6215T − 11.37v0.16 + 0.3965Tv0.16 where T is the temperature (°C) and v is the wind speed (km/h). When T = 12°C and v = 18 km/h, by how much would you expect the apparent temperature W to drop if the actual temperature decreases by 1°C? (Round your answers to two decimal places.)

Answers

This question involves the concepts of derivative, apparent temperature, actual temperature,and wind speed.

The drop in apparent temperature will be "1.25°C".

The apparent temperature (W) is given in terms of actual temperature (T) and wind speed (v) is given by the following function:

$W = 13.12 + 0.6215\ T-11.37\ v^(0.16)+0.3965\ Tv^(0.16)$

Taking the derivative with respect to actual temperature, we get:

$(dW)/(dT)=0.6215+0.3965\ v^(0.16)\n\n$

where,

dW = drop in apparent temperatures = ?

dT = drop in actual temperature = - 1°C

v = wind speed = 18 km/h

Therefore,

$dW=(-1)(0.6215-0.3965(18)^(0.16))$

dW = - 1.25°C

Learn more about derivatives here:

brainly.com/question/9964510?referrer=searchResults

Answer:

Δw=1.25°C

Explanation:

Given that

$w=13.12 +0.6215 T-11.37 v^(0.16)+0.3965 T v^(0.16)$

Given that T= 12°C and v=19 km/h

Now to find the drop in the apparent temperature w

$(dw)/(dT)=0.6215 +0.3965v^(0.16)$

$(\Delta w)/(\Delta T)= 0.6215 +0.3965 v^(0.16)$

Now by putting the values v=19 km/hr and ΔT=1

$(\Delta w)/(1)=0.6215 +0.3965* 18^(0.16)$

Δw=1.25°C

So we can say that when temperature is decrease by 1°C then apparent temperature will decrease by 1.25°C at given velocity.

Ball 1 (1.5 kg) moves to the right at 2 m/s and ball 2 (2.5 kg) moves to the left at 1.5 m/s. The balls stick together after collision. What is the speed and direction of ball 2 after the collision?

Answers

Answer:

0.1875 m/s leftward

Explanation:

Taking rightwards as positive

We are given:

Ball 1:

Mass (m1) = 1.5 kg

velocity (u1) = 2 m/s

Ball 2:

Mass (m2) = 2.5 kg

velocity (u2) = -1.5 m/s [negative because it is in the opposite direction]

Speed and Direction of Ball 2:

We are told that the balls stick together after the collision

We can say that the balls have the same velocity since they are sticking together

So, Final velocity of Ball 1 (v1) = Final velocity of Ball 2 (v2) = V m/s

According to the law of conservation of momentum

m1u1 + m2u2 = m1v1 + m2v2

replacing the variables

1.5(2) + (2.5)(-1.5) = V (1.5 + 2.5) [v1 = v2 = V]

3 + (-3.75) = 4V

-0.75 = 4V

V = -0.75/4 [dividing both sides by 4]

V = -0.1875 m/s

Hence, the balls will move at a velocity of 0.1875 m/s in the Leftward direction

Why are chemical processes unable to produce the same amount of energy flowing out of the sun as nuclear fusion?

Answers

Answer:

Explanation:

One of the major differences between nuclear reactions and chemical reactions is that nuclear reactions involve larger amount of energy than chemical energy. This is because the force between the protons and neutrons in the nucleus of an atom is much higher than the force of attraction between electrons and the positively charged nucleus, hence nuclear reactions involves/requires a larger amount of energy (because it's reactions involve the nucleus) than chemical reactions (because it's reactions involve the electrons).

Thus, during nuclear fusion, two light nuclei are bombarded against one another to produce a larger/heavier nuclei with the release of large amount of energy (because the forces between the protons and neutrons are much higher) unlike when two atoms/molecules are chemically combined together to form a new molecule with the rearrangement of electrons in the valence shells of the participating molecules.

At an intersection of hospital hallways, a convex spherical mirror is mounted high on a wall to help people avoid collisions. The magnitude of the mirror's radius of curvature is 0.560 m.A) Locate the image of a patient10.6m from the mirror. B) Indicate whether the image is upright or inverted.C) Determine the magnification of the image.

Answers

Answer:

Explanation:

For a convex mirror, the value of its image distance and its focal length are negative.

using the mirror formula 1/f = 1/u+1/v

f is the focal length = Radius of curvature/2 = 0.560/2

f= 0.28m

u is the object distance = 10.6m

v is the position of the image = ?

On substitution;

1/0.28 = 1/10.6 + 1/-v

3.57 = 0.094 - 1/v

3.57 - 0.094 = -1/v

3.476 = -1/v

v = -1/3.476

v = -0.2877m

B) Since the image distance is negative, this means that the image is an upright and a virtual image. All Upright images has their image distance to be negative.

C) Magnification = Image distance/object distance

Magnification = 0.2877/10.6

Magnification = 0.0271

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A hollow sphere of radius 0.25 m is rotating at 13 rad/s about an axis that passes through its center. the mass of the sphere is 3.8 kg. assuming a constant net torque is applied to the sphere, how much work is required to bring the sphere to a stop?