At maximum height, the velocity is zero.

A woman exerts a horizontal force of 4 pounds on a box as she pushes it up a ramp that is 10 feet long and inclined at an angle

stealth61 [152]

Answer:

W = 34.64 ft-lbs

Explanation:

given,

Horizontal force = 4 lb

distance of push, d = 10 ft

angle of ramp, θ = 30°

Work done on the box = ?

We know,

W = F.d cos θ

W = 4 x 10 x cos 30°

W = 40 x 0.8660

W = 34.64 ft-lbs

Hence, work done on the box is equal to W = 34.64 ft-lbs

6 0

3 years ago

A wooden block is sitting on an inclined plane near the bottom. The student gave the block a flick and it moved up the inclined

sladkih [1.3K]

Answer:

The block didn't slide due to balancing of gravitational force with friction force

Explanation:

When the block was given a flick the force provided an acceleration to it and it moved up the inclined plane. when the block reached top it was expected that it would slide back but it didn't this happened because of the frictional force acting on the bottom the block which was balancing the gravitational force component along the plane and this prevented sliding back of the block.

static friction was balancing mg*sin(theta)

fs = mg*sin(theta)

6 0

3 years ago

A drag racer starts her car from rest and accelerates at 5.5 m/s2 for the entire distance of 523 m. How long did it take the car

ANEK [815]

Answer:

t = 13.7 s or t = 14 s with proper significant figures

Explanation:

The initial speed is 0 m/s since the car starts from rest, acceleration is 5.5 m/s2 and distance is 523 m.

Since we have initial speed, acceleration and distance we can use the following formula to find the time. We can now use algebra to work out our answer.

d = vt + $\frac{1}{2}$ at²

523 = (0)t + ( $\frac{1}{2}$ )(5.5)t²

523 = 2.8t²

186.8 = t²

13.7 s = t

(t = 14 s with proper significant figures)

3 0

3 years ago

A. Draw the electric field lines around a negative charge.

Alborosie

<h2>a. Answer:</h2>

We use Electric field lines for visualizing electric fields, so this helps us to see the problem more real. So an electric field line is an imaginary line or curve drawn through a region of space such that the tangent at any point comes from the direction of the electric-field vector at that point. The electric field lines around a negative charge is shown in the First figure below.

<h2>b. Answer:</h2>

Electric forces can be found by using the Coulomb Law's that states <em>that The magnitude of the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. </em>This can be expressed as follows:

$F=k\frac{\left | q_{1}q_{2} \right |}{r^2} \\ \\ Where: \\ \\ k=9\times 10^9Nm^2/c^2 \\ \\ q_{1}=0.00150 C \ and \ q_{2}=0.00240 C \\ \\ r=0.900 m$

Then:

$F=9\times 10^9\frac{\left | 0.00150 \times 0.00240 \right |}{(0.900)^2} \\ \\ \therefore \boxed{F=40000N}$

This force is repulsive because the two charges are positive and recall that two positive charges or two negative charges repel each other while a positive charge and a negative charge attract each other.

<h2>C. Answer:</h2>

From the statement, we have two charged objects. Let's say that this charges are:

$q_{1} \ and \ q_{2}$

If the amount of charge on one of the objects is tripled, let's say this is the charge $q_{2}$ , then the new charge is:

$q_{N}=3q_{2}$

In the formula of Coulomb:

$F=k\frac{\left | q_{1}q_{N} \right |}{r^2} \\ \\ \therefore F=k\frac{\left | q_{1}(3q_{2}) \right |}{r^2} \\ \\ \therefore \boxed{F=3k\frac{\left | q_{1}q_{2} \right |}{r^2}}$

<em>The conclusion is that if the amount of charge on one of the objects is tripled, the electric force between two charged objects is also tripled</em>

<h2>d. Answer:</h2>

Let's use the Coulomb's Law again to solve this problem. We want to know how the electric force between two charged objects changes if the charges are moved closer together:

$F=k\frac{\left | q_{1}q_{N} \right |}{r^2}$

<em>By saying that the charges are moved closer together, we want to express that r becomes smaller. Since r is in the denominator, this implies that the electric force between these two charged objects becomes greater.</em>

<h2>e. Answer:</h2>

From the figure, we can see a metal sphere on a stand. There we have both positive and negative charges. We can say that the positive charge of this sphere is +10q and the negative and the negative charge is -10q. Since the electric charge is conserved, then the algebraic sum of all the electric charges in any closed system is constant. In conclusion, <em>the sphere has no net charge.</em>

<h2>f. Answer:</h2>

Here we want to know how the negative charges in the same sphere are redistributed when a positively charged rod is brought near it. Therefore, positive charge on rod repels positive charges on the sphere, creating zones of negative and positive charge as indicated in the second Figure.