Ask question

Ask question

All categories

2 years ago

10

susan and joe push a table 18 meters down a hall by exerting a force of 200 newtons (n). what is the total amount of work done o

n the table? 1600 joules 1800 joules 2000 joules 3600 joules

1 answer:

prisoha [69]2 years ago

3 0

Answer:

3600joules

Explanation:

formula :W=FS

W=work done (J)

F=force (N)

S=displacement moved in the direction of force (m)

200N×18m

=3600J

You might be interested in

The wavelength of yellow sodium light in air is 589 nm.

julsineya [31]

Explanation:

answer is attached, kindly see answer

3 0

3 years ago

Consider Newton's Law of Universal Gravitation: FG= G (m1 m2)/d2 .

bixtya [17]

Answer:

C

Mass is directly proportional to the Force of Gravity. If Mass increases, then the Force of Gravity increases; however, Distance is indirectly (or inversely) proportional to the Force of Gravity. If Distance increases, then the Force of Gravity decreases.

Explanation:

The formula for the force of gravity between two objects is

$F=G\frac{m_1 m_2}{d^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two objects

d is the separation between the two objects

We notice the following:

- F is directly proportional to the masses, $F\propto m_1, m_2$ . This means that if one of the masses increases, then the force between them, F, increases in a proportional way

- F is inversely proportional to the square of the distance, $F\propto \frac{1}{d^2}$ . This means that if the distance between the two objects is increased, the force between them will decrease, and vice-versa.

So, the correct answer is

C

Mass is directly proportional to the Force of Gravity. If Mass increases, then the Force of Gravity increases; however, Distance is indirectly (or inversely) proportional to the Force of Gravity. If Distance increases, then the Force of Gravity decreases.

7 0

3 years ago

Read 2 more answers

An inattentive driver is traveling 18.0 m/s when he notices a red light ahead. his car is capable of decelerating at a rate of 3

ryzh [129]

Speed of the car given initially

v = 18 m/s

deceleration of the car after applying brakes will be

a = 3.35 m/s^2

Reaction time of the driver = 0.200 s

Now when he see the red light distance covered by the till he start pressing the brakes

$d_1 = v* t$

$d_1 = 18* 0.200 = 3.6 m$

Now after applying brakes the distance covered by the car before it stops is given by kinematics equation

$v_f^2 - v_i^2 = 2 a s$

here

vi = 18 m/s

vf = 0

a = - 3.35

so now we will have

$0^2 - 18^2 = 2*(-3.35)(s)$

$s = 48.35 m$

So total distance after which car will stop is

$d = d_1 + d_2$

$d = 48.35 + 3.6 = 51.95 m$

So car will not stop before the intersection as it is at distance 20 m

3 0

3 years ago

A 6.00-kg box sits on a ramp that is inclined at 37.0° above the horizontal. the coefficient of kinetic friction between the box

forsale [732]

We need to see what forces act on the box:

In the x direction:

Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.

In the y direction:

N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force.

From N-Gcosα=0 we get:

N=Gcosα, we will need this for the force of friction.

Now to solve for Fh:

Fh=ma + Ff + Gsinα,

Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²

Fh=ma + μmgcosα+mgsinα

Now we plug in the numbers and get:

Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N

The horizontal force for pulling the body up the ramp needs to be Fh=71 N.

8 0

3 years ago

How are points made in soccer ?

Anit [1.1K]

A- by kicking the ball into the net/goal

5 0

3 years ago

Read 2 more answers