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4 years ago

When work is done on an object, where does the energy used to do the work go?

1 answer:

3 0

-- If the work is done to make the object move faster, then
the work done becomes kinetic energy of the object.

-- If work is done on the object but it doesn't move any faster,
then there must be friction holding it back. In that case, the work
that's done just to keep the object moving becomes heat, in the
places where the friction acts on it.

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John throws a baseball from a standing position. Which of the three possible paths will the ball follow?

Marysya12 [62]

Answer: C

Explanation:

Bro this is common sense. Throw something across the room and watch it. It isnt going to be repulsed and fly up. It also isnt going to continue on forever. It will clearly fall and hit the ground

3 0

3 years ago

Read 2 more answers

A laser beam with a frequency of 180 Hz forms an 8 m standing wave with 10 nodes.

DIA [1.3K]

Answer:33

Explanation:

F = frequency

N = Node count

w = wave lenght

v = wave velocity

L = distance wave traveled

First find wave length of laser

w = (2/(N))*(L)

w = (2/(10))*(8)

w = 1.6

then using (w), find velocity

V = (w)(F)

V = (1.6)*(108)

V = 288

Plug in V and the new frequency to solve for new node count

F = NV/2L

(600) = (N)*(288) / 2 * (8)

(N) = 33.33

there are 33 nodes

8 0

3 years ago

Consider the sections of two circuits illustrated above. Select True or False for all statements. After connecting c and d to a

Basile [38]

Answer:

Follows are the solution to the given question:

Explanation:

In this question the missing file of the circuit is not be which is defined in attached file please find it.

In Option 1, this statement is true because the current is on $R_3 \ and \ R_4$ , that is the same.

In option 2, this statement is false because $Rcd=R_3+R_4$ therefore it implies that Rcd is always larger then $R_3$ .

In option 3, this statement is true because the voltage of $R_1 \ and \ R_2$ is always equal.

In option 4, this statement is true because $Rab= \frac{1}{(\frac{1}{R1}+\frac{1}{R2})}=\frac{R1R2}{(R1+R2)}. \frac{R2}{(R1+R2)}$ is always smaller then 1 therefore, $R_1 \times (\frac{R2}{(R1+R2)})$ is always equal to R1.

5 0

3 years ago

I run around a circular track, with radius 25m, for 4 and times before stopping. It takes me 16

Sergio039 [100]

Answer:

13m/s. is the answer probably but do u have ms gallup too?

4 0

3 years ago

An unknown mass of each substance, initially at 25.0 ∘C, absorbs 1920 J of heat. The final temperature is recorded. Find the mas

zhannawk [14.2K]

Answer: mass for Pyrex glass 84.21g

mass for sand 61.6g

mass for ethanol 41.32g

mass for water 62.07g

Explanation

By definition specific heat is the amount of heat required to change the temperature of 1 kg mas by 1°C

Q=mcΔT is formula for specific heat

Q is heat transfer

m is mass

ΔT is change in temperature

c is specific heat

c of Pyrex glass= 0.75 j/g°C

c of sand = 0.84 j/g°C

c of ethanol= 2.42 j/g°C

c of water = 4.18 j/g°C

now we will make M(mass) the subject, so equation becomes

m=Q/cΔT

for

pyrex glass Tf=55.4°C

m=1920/(55.4-25)*0.75

m=84.21g {after cutting J(joules) and °C we are left with g(grams)}

for

sand Tf=62.1°C

m=1920/(62.1-25)*0.84

m=61.6g {after cutting J(joules) and °C we are left with g(grams)}

for

ethanol Tf=44.2°C

m=1920/(44.2-25)*2.42

m=41.32g {after cutting J(joules) and °C we are left with g(grams)}

for

water Tf=32.4°

m=1920/(32.4-25)*4.18

m=62.07g {after cutting J(joules) and °C we are left with g(grams)}

i hope you understand the solution, thank you.

7 0

3 years ago