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
son4ous [18]
2 years ago
12

If a boy weighs 490 N and stands on a flat rough surface with a

Physics
1 answer:
OleMash [197]2 years ago
8 0

The force of static friction acting on the boy is 333 Newtons.

Hence, option (d) 333N is the correct answer.

Given the data in the question;

  • Weight of boy; F_n = 490N
  • Coefficient of static friction; u_s = 0.68

Force of static friction acting on the boy; F_s = \ ?

<h3>Friction</h3>

Friction is simply referred as the resistance an object experience when moving or trying to move over another object. It can either by dynamic or static.

Static friction is the resistance experienced by a body at rest.

Coefficient of friction is the measure of how easily a body moves in relation to the other.

Static friction is expressed as;

F_s = u_s * F_n

Where F_s is the normal force and u_s is the coefficient of static friction.

We substitute our given values into the equation above

F_s = u_s * F_n\\\\F_s = 0.68 * 490N\\\\F_s = 333.2N\\\\F_s = 333N

The force of static friction acting on the boy is 333 Newtons.

Hence, option (d) 333N is the correct answer.

Learn more friction: brainly.com/question/17237604

You might be interested in
What is a moment of a force
kirza4 [7]

Answer:

In physics and mechanics, torque is the rotational equivalent of linear force. It is also referred to as the moment, moment of force, rotational force or turning effect, depending on the field of study. The concept originated with the studies by Archimedes of the usage of levers

7 0
2 years ago
One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a
Maurinko [17]

This question is not complete.

The complete question is as follows:

One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?

Explanation:

a. Using the expression;

T = 2π√R/g

where R = radius of the space = diameter/2

R = 800/2 = 400m

g= acceleration due to gravity = 9.8m/s^2

1/T = number of revolutions per second

T = 2π√R/g

T = 2 x 3.14 x √400/9.8

T = 6.28 x 6.39 = 40.13

1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute

6 0
3 years ago
What kind of energy change occurs when a battery is operating a remote control toy?
solniwko [45]

Answer:

Explanation:

The correct answer is option C.

When the battery is operating a remote control toy the energy is converted from potential energy to the kinetic energy.

A battery stores electrical potential from the chemical reaction.

When the circuit is connected to the potential energy of the battery helps in the movement of the toy.

The energy produced by the movement of the control toy is kinetic energy.

Hence, we can say that Potential energy is changed to kinetic energy

7 0
3 years ago
A bumper cart has a mass of 200 kg and has a protective bumper around it that behaves like a spring. The spring constant is 5000
34kurt
Part A:
For this part we’re assuming all the kinetic energy of the moving bumper car is converted into elastic potential energy in the spring since the car is brought to rest. Therefore you can find the total kinetic energy to get your answer:

KE = ½ mv^2
KE = ½ (200)(8)^2
KE = 6400 J

Part B:
Now you can use Hooke’s law to find the force:

F = kx
F = (5000)(0.2)
F = 1000 N
4 0
3 years ago
A camera is equipped with a lens with a focal length of 34 cm. When an object 2.4 m (240 cm) away is being photographed, what is
puteri [66]

Answer:

The magnification is -6.05.

Explanation:

Given that,

Focal length = 34 cm

Distance of the image =2.4 m = 240 cm

We need to calculate the distance of the object

\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}

Where, u = distance of the object

v = distance of the image

f = focal length

Put the value into the formula

\dfrac{1}{u}=\dfrac{1}{34}-\dfrac{1}{240}

\dfrac{1}{u}=\dfrac{103}{4080}

u =\dfrac{4080}{103}

The magnification is

m = \dfrac{-v}{u}

m=\dfrac{-240\times103}{4080}

m = -6.05

Hence, The magnification is -6.05.

6 0
2 years ago
Other questions:
  • A construction worker runs a jack hammer for 1.5 hours. The jack hammer has 2.4 kilowatts of power. How much electrical energy d
    14·1 answer
  • Two long straight wires carry currents perpendicular to the xy plane. One carries a current of 50 A and passes through the point
    11·1 answer
  • A series AC circuit contains a resistor, an inductor of 150 mH, a capacitor of 5.00 mF, and a generator with DVmax 5 240 V opera
    12·1 answer
  • DEFINE HEAT CAPACITY
    5·2 answers
  • 1. Compare and Contrast microwaves with visible light using wavelength, frequency and energy
    9·1 answer
  • It takes a photon 8 minutes and 25 seconds ro reach earth. What is the distance (labeled 1 au) in meter between the sun and eart
    13·1 answer
  • A red cart has a mass of 4 kg and a velocity of 5 m/s. There is a 2-kg blue cart that is parked and not moving, thus its velocit
    10·1 answer
  • Is burning trash a physical change or chemical change?
    15·1 answer
  • What happens when a glass rod and silk cloth are rubbed with each other?
    13·2 answers
  • The main source of information used by astronomers to learn about objects in space is.
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!