1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
White raven [17]
3 years ago
9

Question 13 (1 point)

Physics
1 answer:
KonstantinChe [14]3 years ago
3 0

Answer:

Reflective clothing. So vehicles can see them and stuff.

You might be interested in
Would u rather/ be able to fly. or be able to turn invisable
mojhsa [17]

Answer:

fly......................

4 0
3 years ago
Read 2 more answers
The amount of kinetic energy an object has depends on its.
dalvyx [7]
<span>The amount of kinetic energy an object has
depends on its mass and speed.</span>
5 0
3 years ago
Read 2 more answers
A 330.-ohm resistor is connected to a 5.00-volt battery. What is the current through the resistor?
Gelneren [198K]

I = 0.0152 A = 15.2 mA

Explanation:

Using Ohm's law,

V = IR or

I = V/R

= (5.00 V)/(330. ohms)

= 0.0152 A

= 15.2 mA

8 0
3 years ago
Read 2 more answers
A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind
Marat540 [252]

Answer:

Part a)

F_v = 4.28 N

Part B)

L = 1.02 m

Part C)

v = 1.25 m/s

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

Tcos\theta = mg

T sin\theta = F_v

\frac{F_v}{mg} = tan\theta

F_v = mg tan\theta

F_v = 1.2\times 9.81 (tan20)

F_v = 4.28 N

Part B)

Here we can use energy theorem to find the distance that it will move

-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2

(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2

(-3.5 + 2.54)L = - 0.98

L = 1.02 m

Part C)

At terminal speed condition we know that

F_v = mg

bv^2 = mg

2.5 v^2 = 3.9

v = 1.25 m/s

7 0
3 years ago
A weightlifter exerts an upward force on a 1000-N barbell and holds it at a height of 1 meter for 2 seconds. Approximately how m
Alex73 [517]

Amount of work done is zero and so power = 0 watts.

<u>Explanation:</u>

Power is the rate at which work is done, or W divided by delta t. Since the barbell is not moving, the weightlifter is not doing work on the barbell.Therefore, if the work done is zero, then the power is also zero.It may seem unusual that the data given in question is versatile i.e. A weightlifter exerts an upward force on a 1000-N barbell and holds it at a height of 1 meter for 2 seconds. But, still the answer is zero watts , this was a tricky question although conceptual basis of question was good! Power is dependent on amount of work done which is further related to displacement and here the net displacement is zero ! Hence, amount of work done is zero and so power = 0 watts.

6 0
3 years ago
Other questions:
  • A proton moves at constant velocity in the direction, through a region in which there is an electric field and a magnetic field.
    10·1 answer
  • PLEASE ANSWER QUICKLY!!! 4. Identify the part or item associated with the internal-combustion engine that the statement is
    9·2 answers
  • according to adam smith and other classical economists, why is the economic theory supporting market economies(or capitalism) mu
    14·1 answer
  • Item 13You have four $10 bills and eighteen $5 bills in your piggy bank. How much money do you have?
    11·1 answer
  • Does a resistor resist current or voltage
    6·1 answer
  • BRAINLEST HURRY PLZZ HELP<br> the options for number 1 2 and 3 are A B C D
    9·2 answers
  • Whats the mass of the sun times pi?
    13·1 answer
  • How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and
    5·1 answer
  • Im bored help me join this if u want with friends
    5·1 answer
  • Which location focuses its use on a nonrenewable energy source? a wind mill farm a natural gas power plant a fruit orchard a com
    7·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!