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
Natali5045456 [20]
3 years ago
12

Lena is playing with a remote-controlled car in her backyard. She knows that the car uses a wheel and axle to move. What is the

primary source of power for the wheel and axle in the remote-controlled car?
A.
friction between the ground and the tires

B.
radio waves from the remote control

C.
chemical energy from the batteries

D.
potential energy with the car sitting at the top of a small hill
Physics
1 answer:
melomori [17]3 years ago
7 0

Answer:

C. chemical energy from the batteries

Explanation:

Remote-controlled car or commonly called RC cars are toy vehicles that are controlled or propelled to move from a distance by a wire-connected remote that sends radiowaves to the receiver. According to this question involving the remote-controlled car Lena is playing with, a transmitter from the remote propels the wheel and axle to make the car move.

The car contains a battery cell that uses it's chemical energy as a source of power for the wheel and axle system. Hence, the chemical energy from the batteries of a remote-controlled car powers the wheel and axle to make the car move.

You might be interested in
A series of bright fringes appears on the viewing screen of a Young's double-slit experiment. Suppose you move from one bright f
goblinko [34]
Ahahaha fidjdsd skssjsjsbs SJSU’s
6 0
3 years ago
A 54 kg person stands on a uniform 20 kg, 4.1 m long ladder resting against a frictionless wall.
SVETLANKA909090 [29]

A) Force of the wall on the ladder: 186.3 N

B) Normal force of the ground on the ladder: 725.2 N

C) Minimum value of the coefficient of friction: 0.257

D) Minimum absolute value of the coefficient of friction: 0.332

Explanation:

a)

The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:

W=mg: weight of the ladder, with m = 20 kg (mass) and g=9.8 m/s^2 (acceleration of gravity)

W_M=Mg: weight of the person, with M = 54 kg (mass)

N_1: normal reaction exerted by the wall on the ladder

N_2: normal reaction exerted by the floor on the ladder

F_f = \mu N_2: force of friction between the floor and the ladder, with \mu (coefficient of friction)

Also we have:

L = 4.1 m (length of the ladder)

d = 3.0 m (distance of the man from point A)

Taking the equilibrium of moments about point A:

W\frac{L}{2}sin 21^{\circ}+W_M dsin 21^{\circ} = N_1 Lsin 69^{\circ}

where

Wsin 21^{\circ} is the component of the weight of the ladder perpendicular to the ladder

W_M sin 21^{\circ} is the component of the weight of the man perpendicular to the ladder

N_1 sin 69^{\circ} is the component of the normal  force perpendicular to the ladder

And solving for N_1, we find the force exerted by the wall on the ladder:

N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{mg}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+Mg\frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{(20)(9.8)}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+(54)(9.8)\frac{3.0}{4.1}\frac{sin 21^{\circ}}{sin 69^{\circ}}=186.3 N

B)

Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of N_2.

We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.

Therefore, we have:

\sum F_y = 0\\N_2 - W - W_M =0

And substituting and solving for N2, we find:

N_2 = W+W_M = mg+Mg=(20)(9.8)+(54)(9.8)=725.2 N

C)

Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.

The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.

Therefore, we can write:

\sum F_x = 0\\F_f - N_1 = 0

And re-writing the equation,

\mu N_2 -N_1 = 0\\\mu = \frac{N_1}{N_2}=\frac{186.3}{725.2}=0.257

So, the minimum value of the coefficient of friction is 0.257.

D)

Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.

From part C), we saw that the coefficient of friction can be written as

\mu = \frac{N_1}{N_2}

This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was

N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}

We see that this quantity is maximum when d is maximum, so when

d = L

Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:

N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{L}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{W}{2}+W_M)

And substituting, we get

N_1=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{(20)(9.8)}{2}+(54)(9.8))=240.8 N

And therefore, the minimum coefficient of friction in order for the ladder not to slip is

\mu=\frac{N_1}{N_2}=\frac{240.8}{725.2}=0.332

Learn more about torques and equilibrium:

brainly.com/question/5352966

#LearnwithBrainly

7 0
3 years ago
If a car travels 60 mph for a distance of 180 miles, how much time<br> did it take?
jolli1 [7]

Answer:

3 hours

Explanation:

180 divided by 60 (mph means miles per hours by the way)

6 0
3 years ago
The electric field direction is defined by the direction of the force felt by (select one of the following answers):A. A negativ
steposvetlana [31]

Answer:

B

Explanation:

The formula for the electric field is Force (N)/charge(Coulombs). The electric field direction is defined by the direction of the force felt by a positive charge.

4 0
3 years ago
Determine the binding energy of an F-19 nucleus. The F-19 nucleus has a mass of 18.99840325 amu. A proton has a mass of 1.00728
Anvisha [2.4K]

Answer:

Energy = 1.38*10^13 J/mol

Explanation:

Total number of proton in F-19 = 9

Total number of neutron in F-19 = 10

Expected Mass of F-19  

= 9*1.007 + 10*1.008 = 19.152 u

Actual  mass of F-19 = 18.998 u

Energy of one particle of F-19 = 931.5*Δm = 931.5*(19.152-18.998)

= 143.234 MeV

Energy of one mole of F-19 = 143.234*10^6*1.6*10^-19*6.022*10^23  

= 1.38*10^13 J/mol

8 0
3 years ago
Other questions:
  • A hollow spherical iron shell floats almost completely submerged in water. The outer diameter is 58.2 cm, and the density of iro
    9·1 answer
  • What is a main reason why the scientific model of atoms is considered a theory rather than a law?
    8·2 answers
  • Which statement best describes what energy transfer diagrams show? Energy can change form, but the total amount of energy stays
    12·2 answers
  • uring the investigation of a traffic accident, police find skid marks 89.9 m long. They determine the coefficient of friction be
    5·1 answer
  • An electrostatic paint sprayer has a 0.200-m-diameter metal sphere at a potential of 25.0 kV that repels paint droplets onto a g
    8·1 answer
  • 7. An automobile with a radio antenna 1.0 m long travels at 100.0 km/h in a location where the Earth’s horizontal magnetic field
    6·1 answer
  • The phosphorus cycle is important to ecosystems. Choose all of these statements that are true concerning the phosphorus cycle.
    7·2 answers
  • Question 8 (1 point)
    11·1 answer
  • Sharon is driving on a straight road. She is driving north, and her speed is
    6·1 answer
  • A light ray is incident on a smooth surface at an angle of 25 degrees with the surface. What is the angle of incidence?
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!