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2 years ago

How much force is required to drag a 90 lb. box up this "frictionless" inclined plane? 109 lb. 10 lb. 81 lb. 9 lb.

Physics

2 answers:

MrRissso [65]2 years ago

5 0

I believe is 10 lb if not it's 9 lb.

Alex Ar [27]2 years ago

4 0

The correct answer is b) 10 lbs.

I hope that I was of help.

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There are lots of examples of ideal gases in the universe, and they exist in many different conditions. In this problem we will

elena-14-01-66 [18.8K]

Answer:

P = ρRT/M

Explanation:

Ideal gas equation is given as follows generally:

PV = nRT (1)

P = pressure in the containing vessel

V = volume of the containing vessel

n = number of moles

R = gas constant

T = temperature in K

n = m/M

m = mass of the gas contained in the vessel in g

M = molar mass in g/mol

ρ = m/V

Density of the gas = ρ

Substituting for n in (1)

PV = mRT/M. (2)

Dividing equation (2) through by V

P = m/V ×RT/M

P = ρRT/M

5 0

1 year ago

What are the magnitude and direction of the force the pitcher exerts on the ball? (enter your magnitude to at least one decimal

murzikaleks [220]

Details are missing in the question. Complete text of the problem:

"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.

(a) Through what distance does it move before its release? (m)
(b) What are the magnitude and direction of the force the pitcher exerts on the ball? (Enter your magnitude to at least one decimal place.)"

Solution

(a) The pitcher accelerates the baseball from rest to a final velocity of $v_f = 16.5 m/s$ , so $\Delta v=16.5 m/s$ , in a time interval of $\Delta t = 181 ms=0.181 s$ . The acceleration of the ball in the horizontal direction (x-axis) is therefore

$a_x = \frac{\Delta v}{\Delta t}= \frac{16.5 m/s}{0.181 s}=91.2 m/s^2$

And the distance covered by the ball during this time interval, before it is released, is:

$S= \frac{1}{2} a_x (\Delta t)^2 = \frac{1}{2} (91.2 m/s^2)(0.181 s)^2=1.49 m$

(b) For this part we need to consider also the weight of the ball, which is $W=mg=2.28 N$

From this, we find its mass: $m= \frac{W}{g}= \frac{2.28 N}{9.81 m/s^2}=0.23 Kg$

Now we can calculate the magnitude of the force the pitcher exerts on the ball. On the x-axis, we have

$F_x = m a_x = (0.23 kg)(91.2 m/s^2)=20.98 N$

We also know that the ball is moving straight horizontally. This means that the vertical component of the force exerted by the pitcher must counterbalance the weight of the ball (acting downward), in order to have a net force of zero along the y-axis, and so:

$F_y=W=mg=2.28 N$ (upward)