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erma4kov [3.2K]
4 years ago
9

Which is an example of chemical energy from batteries to electromagnetic ( light ) energy?

Physics
1 answer:
TiliK225 [7]4 years ago
4 0

Answer:

B.

Explanation:

The batteries make it so the chemical energy is being passed into the flashlight allowing it to work as designed forming light.

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Help please i will mark brainlist!!
Dmitrij [34]
The answer is A.


p=m/v
p= 240/60
p= 4 g/cm^3
7 0
3 years ago
Read 2 more answers
You drag a crate across a floor with a rope. The force applied is 750 N and the angle of the rope is 25.0° above the horizontalH
frez [133]

a.

The work done by a constant force along a rectilinear motion when the force and the displacement vector are not colinear is given by:

W=F\cos\theta\cdot d

where F is the magnitude of the force, theta is the angle between them and d is the distance.

The problen gives the following data:

The magnitude of the force 750 N.

The angle between the force and the displacement which is 25°

The distance, 26 m.

Plugging this in the formula we have:

\begin{gathered} W=\left(750\right)\left(\cos25\right)\left(26\right) \\ W=17673 \end{gathered}

Therefore the work done is 17673 J.

b)

The power is given by:

P=\frac{W}{t}

the problem states that the time it takes is 6 s. Then:

\begin{gathered} P=\frac{17673}{6} \\ P=2945.5 \end{gathered}

Therefore the power is 2945.5 W

5 0
1 year ago
Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc
4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance D_{A} beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed v_{A}. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed v_{B}, which is greater than v_{A}.

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: t=\frac{D_{A}}{v_{B}-v_{A}}

              Part B: x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

x=x_{0}+vt

where

x_{0} is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is D_{A}.

The equation will be:

x_{A}=D_{A}+v_{A}t

Car B started at the starting line. So, its equation is

x_{B}=v_{B}t

Part A: When they meet, both car are at "the same position":

D_{A}+v_{A}t=v_{B}t

v_{B}t-v_{A}t=D_{A}

t(v_{B}-v_{A})=D_{A}

t=\frac{D_{A}}{v_{B}-v_{A}}

Car B meet with Car A after t=\frac{D_{A}}{v_{B}-v_{A}} units of time.

Part B: With the meeting time, we can determine the position they will be:

x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )

x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Since Car B started at the starting line, the distance Car B will be when it passes Car A is x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}} units of distance.

5 0
3 years ago
When driving at night, only use your high-beam headlights___
zzz [600]

Answer:

B

Explanation:

While answer C may sound correct, Answer B is makes more sense. We know you cant use High-beam lights when u cant see ongoing traffic because it could affect the other driver coming across from you. Its good to use it when legal and safe, but in that term I still don't believe there's no reason for HIGH-beamed. That's this leaves B, when you are on u lighted streets.

6 0
3 years ago
A shopping cart rolls from x=19.9m to x=5.40m with an average velocity of -0.418m/s. How much time did it take? (unit=s)
andrew11 [14]

Given

initial position = Xi= 19.9m

Final position Xf = 5.4m

Average velocity= Va = -0.418m/s

it shows displacement is reverse.

To find  t=?

As   Va = (Xf- Xi) / t

t = (Xf-Xi) / ( Va)

t = ( 5.4-19.9) / (-0.418)

t = (-14.5 ) / (-0.418)   (-ve sign cancel out at numerator and denominator)

t =34.69 s

8 0
4 years ago
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