Please hurry! 30 points

A block of mass 10.0 kg is pulled to the right along a rough horizontal surface with a constant horizontal force of 20.0 N. The

Yakvenalex [24]

Answer:

the magnitude of acceleration will be 1.50m/s^2

Explanation:

To calculate your acceleration, you can use your formula that states that the net force on an object is equal to the mass of the object multiplied by the acceleration of the object. Fnet=ma

if you draw out this situation and label the forces you will have your vector towards the right with a magnitude of 20.0N and then your friction vector will be pointing to the left (in other words, in the negative direction) (opposing the direction of movement) with a magnitude of 5.00N, with the 10.0 kg box in the middle.

The net force will be calculated using F1+F2=Fnet where your F1=20.0N and F2= -5.00N (since it is towards the negative direction).

you will find that Fnet=15.0N

With that, plug in the values you know to calculate the acceleration of the block:

Fnet=ma

(15.0N)=(10.0kg)a from her you can divide both sides by 10 to isolate a:

1.50=a (and now make sure to label the units of your answer)

a=1.50m/s^2 (which is the typical unit for acceleration)

7 0

3 years ago

A clinical psychologist is one type of pure basic research psychologist . True are false

adoni [48]

false. clinical deals with patients and treats the.

research looks at root causes which clinical applies

8 0

3 years ago

Help please it's a test

UNO [17]

Unbalanced. _______ and not equal. in not sure about the 2nd choice from the info im looking at. the balanced version is 2NaOH + H2SO4 = Na2SO4 + 2H2O
(i hope this helps a little)

7 0

4 years ago

Read 2 more answers

A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th

Nina [5.8K]

Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(4)}{2}(2(0)+4)$

$F = 25000 lb$

b) Force on deep end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(10)}{2} (2(0)+10)$

$F = 187500 lb$

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

$F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)$