_______ are nonmetals that react with metals to form salts.

El. A horizontally directed force of 40 N is used to pull a box a distance of 2.5 m across a tabletop. How much work is done by

PIT_PIT [208]

The same formula of work can be applied to all the questions. The answers are:

E1. 100J

E2. 40N

E3. 5m

E4. a.) 360 J b.) 240 J c.) 120 J

El. If a horizontally directed force of 40 N is used to pull a box a distance of 2.5 m across a tabletop. The formula to get the much work that will be done by the 40 - N force will be

Work done = force x distance

Work done = 40 x 2.5

Work done = 100 J

E2. If a woman does 160 J of work to move a table 4 m across the floor. We will use the same formula to calculate the magnitude of the force that the woman applied to the table assuming the force is applied in the horizontal direction.

Work done = force x distance

160 = 4F

F = 160/4

Force F = 40 N

E3. Given that a force of 60 N used to push a chair across a room does 300 J of work. Same formula to get how far the chair move in this process.

Work done = force x distance

300 = 60 x distance

distance = 300/60

Distance = 5 m

Therefore, the chair moved 5m away.

E4. Given that a rope applies a horizontal force of 180 N to pull a crate a distance of 2 m across the floor. And a frictional force of 120 N opposes this motion.

a. The work done by the force applied by the rope can be found by

W = F x S

W = 180 x 2

W = 360 J

b. What is the work done by the frictional force?

W = $F_{r}$ x s

W = 120 x 2

W = 240 J

c. What is the total work done on the crate?

W = (F - $F_{r}$ ) x distance

Where $F_{r}$ = frictional force

Substitute all the parameters

W = (180 - 120) x 2

W = 60 x 2

W = 120 J

Learn more about work here: brainly.com/question/8119756

4 0

2 years ago

An electron, moving south, enters a magnetic field of certain strength. Because of this field the electron curves upward. What i

e-lub [12.9K]

Answer: The magnetic field points to the west.

Explanation:

If we take the plane as:

North = positive y-axis

East = positive x-axis.

We have that the electron is moving south, so the velocity of the electron can be written in vector form as:

V = (0, -v, 0)

Now, when the electron interacts with the magnetic field, the electron moves upwards.

We know that the interaction between an electron and a magnetic field is:

F = q*VxB

So we have that this force acts on the z-direction.

Now, to solve this we can use the righ hand rule.

First, we point wit our hand to the direction of the velocity (negative y-axis) now, we want that our thumb points up (the direction of the force) so the side where our palm faces is the direction of the field B.

But, remember that an electron has a negative charge, so the actual equation is:

F = -q*VxB

So the magnetic field actually points in the opposite direction of our palm, to the west.

Now, we can also solve it mathematically as:

F = (0, 0, f) = -q*(0, -v,0)x(a, b, c)

where B = (a, b, c) is the vector of the magnetic field.

(0, 0, f) = -q*(-v*b -0, -0 + 0, 0 -(- a*v)) = -q*(-v*b, 0, a*v)

then we have that b must be equal to zero, and that:

f = -q*a*v

and f is positive, then we have:

a = -(f/q*v)

then the vector of the magnetic field is:

B = (-(f/q*v), 0, 0)

so it points in the negative x-axis, that is the West, as we found earlier.

4 0

3 years ago

A 15 m ladder with a mass of 51 kg is leaning against a frictionless wall, which makes an angle of 60 degrees with the horizonta

timofeeve [1]

Answer:

Explanation:

Given that, .

Mass of ladder is 51kg

Then, it weight is

WL = mg = 51 × 9.81 = 500.31N

This weight will act at the midpoint of the ladder

Length of ladder is 15m

The ladder makes an angle 55°C with the horizontal

An object whose mass is 81kg is at 4m from the bottom of the ladder

Then, weight of object

Wo = mg = 81 × 9.81 = 794.61 N

Using newton second law

Check attachment

Ng is normal force on the ground

Ff is the horizontal frictional force

Nw Is the normal force on the wall

ΣFy = 0

Ng = Wo + WL

Ng = 794.61 + 500.31

Ng = 1294.92 N

Also

ΣFx = 0

Ff — Nw = 0

Then,

Ff = Nw

Now taking moment about point A.

Check attachment

using the principle of equilibrium

Sum of clockwise moment equals to sum of anti-clockwise moment

Also note that the Normal force on the wall is not perpendicular to the ladder, so we will resolve that and also the weights of ladder and weight of object

Clockwise = Anticlockwise

Wo•Cos60 × 4 + WL•Cos60 × 7.5 = Nw•Sin60 × 15

794.61Cos60 × 4 + 500.31Cos60 × 7.5 = Nw × Sin60 × 15

1589.22 + 1876.163 = 12.99•Nw

3465.383 = 12.99•Nw

Nw = 3465.383 / 12.99

Nw = 266.77 N

Since, Nw = Ff

Then, Ff = 266.77N

the horizontal force exerted by the ground on the ladder is 266.77 N