A force of 20N acts upon of 5 kg. Calculate the acceleration of the object.

Two forces F1 and F2 are acting on the box shown in the drawing, causing the box to move across the floor. The two force vectors

SpyIntel [72]

Answer: The correct answer is (b) F1 does more work than F2.

The block is moving horizontally the direction is opposite to direction of motion. The two forces applied are F1 and F2.F1 is applied horizontally so its horizontal component is F1cos0 = F1

While F2 is at angle so horizontal component of F2 is F2cos(theta) and as we know cos is a decreasing in 0 to 90 degree. Therefore F1 does more work than F2.

Work is defined as the product of the magnitude of a force and the displacement of the object it is acting upon. In this case, the forces F1 and F2 are acting on the box, causing it to move across the floor. To calculate the work done by each force, we need to calculate the magnitude of the force and the displacement of the box.

The magnitude of F1 is given in the diagram, and it is 2 N. The magnitude of F2 is also given, and it is 4 N. For the displacement of the box, we will assume it is 1 m.

Now we can use the formula W = F x d to calculate the work done by each force.

For F1: W1 = 2 N x 1 m = 2 J

For F2: W2 = 4 N x 1 m = 4 J

Since F1 does 2 J of work, and F2 does 4 J of work, F1 does more work than F2 does. Therefore, the correct answer is (b) F1 does more work than F2 does.

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7 0

1 year ago

Why are metals sometimes added to gold?

Komok [63]

Because solo gold isn't strong material. But with admixture it's stronger.

7 0

3 years ago

Read 2 more answers

David moves a 350 N bucket with 175 N of force using a pulley. What mechanical advantage does he have when using the pulley?

NikAS [45]

Answer:

the mechanical advantage David had when using the pulley is 2.

Explanation:

Given;

load moved by David, L = 350 N

effort applied by David, E = 175 N

The mechanical advantage David had when using the pulley is calculated as;

$M.A = \frac{Load}{Effort} \\\\M.A = \frac{350}{175} \\\\M.A = 2$

Therefore, the mechanical advantage David had when using the pulley is 2.

7 0

3 years ago

Technician A says voltage will not exist between any two points in a circuit if the potential is zero. Technician B says corrosi

nordsb [41]

Answer:

<u>Option-(A):</u> If there is no potential or we can say the difference in total amount of charges present inside the system then the voltage's value wont be greater then "0" inside the circuit or system.

Explanation:

While, the circuit or system requires the difference in total amount of energy that each atom on that spot has, as this potential energy difference or the total number of charges difference contribute to having a more specified manner of voltage,v in that system. And thus the flow of charges,Q from a point of origination or source towards a region of lower energy is all due to this difference in potential.

5 0

3 years ago

Two loudspeakers are located 3.0 m apart on the stage of an auditorium. A listener at point P is seated 19.0 m from one speaker

Amanda [17]

Answer:

The frequencies that the listener at P will hear a maximum intensity are: 85.75 Hz, 171.5 Hz, 257.25 Hz and 343 Hz.

Explanation:

Path Difference Δx is given as: nλ

where λ = $\frac{v}{f}$

Δx can be re-written as: n× $\frac{v}{f}$

where;

n = integer

v = speed of sound = 343 m/s

f = frequency

Δx = 19.0 m - 15.0 m

Δx = 4.0 m

$4.0 m =$ $\frac{n*343}{f}$

f = $\frac{n*343}{4}$

f = n × 85.75 Hz

Now; n = 1, 2, 3, 4 ; the frequencies that the listener at P will hear the maximum intensity will be

when n= 1

f = 1 × 85.75 Hz

f = 85.75 Hz

when n= 2

f = 2 × 85.75 Hz

f = 171.5 Hz

when n= 3

f = 3 × 85.75 Hz

f = 257.25 Hz

when n= 4

f = 4 × 85.75 Hz

f = 343 Hz

∴ the frequencies that the listener at P will hear the maximum intensity will be : 85.75 Hz, 171.5 Hz, 257.25 Hz and 343 Hz for n =1,2,3, and 4 respectively.

7 0

3 years ago