All categories

3 years ago

Upward pull of 850 N on a 81.6 kg bale of hay. What is the magnitude of the bales acceleration?

1 answer:

5 0

Weight = (mass) x (acceleration of gravity).

When I calculate the weight of the 81.6 kg, the number I use for gravity
is 9.807 m/s². That gives a weight of 800.25 N, so I think that's where the
question got the crazy number of 81.6 kg ... whoever wrote the problem
wants the hay to weigh 800 N, and that's what I'll use for the weight.

The forces on the bale of hay are gravity: 800N downward, and the
guy on the truck with the pitchfork pulling upward on it with 850 N.
The net force on the bale is (850 - 800) = 50 N upward.

Use Newton's second law of motion: (Net force) = (mass) x (acceleration)

Divide each side by 'mass' :

Acceleration = (net force)/(mass)

On the hay wagon,

Acceleration = (50 N upward) / (81.6 kg) = <em>0.613 m/s² upward</em>

You might be interested in

There's more to motion than simply changing position. True Or False

prohojiy [21]

Answer: I think that it is False, if its wrong I am sorry.

Explanation:

8 0

2 years ago

Read 2 more answers

Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the

Doss [256]

Answer:

$v_{su} = 19.44 m/s$

Explanation:

$m_{su}=5.68x10^{29}kg\\m_{sa}=5.68x10^{26}kg$

$T=9.29x10^8\\r_{o}=1.43x10^{12}$

If the sun considered as x=0 on the axis to put the center of the mass as a:

$m_{su}*r_{o}=(m_{sa}+m_{su})*r_{1}$

solve to r1

$r_1=\frac{m_{sa}*r_{o}}{m_{sa}+m_{su}}=\frac{5.68x10^{26}*1.43x10^{12}}{5.68x10^{26}+5.68x10^{26}}$

$r_1=1.428x10^9m$

Now convert to coordinates centered on the center of mass. call the new coordinates x' and y' (we won't need y'). Now since in the sun centered coordinates the angular momentum was

$L = \frac{m_{sa}*2*pi*r_1^2}{T}$

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum. So that means

$L_{sun}=\frac{m_{sa}*2*\pi *( 2r_{o}*r_1 -r_1^2)}{T}$

Since

$L_{su}= m_{su}*v_{su}*r_1$

then

$v_{su}=\frac{m_{sa}*2*pi*(2r_{o}*r_{1}-r_{1}^2)}{T*m_{sa}*r_1}$

$v_{su} = 19.44 m/s$

7 0

3 years ago

A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The

IgorC [24]

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

$v=u+at\\\Rightarrow 0=4.5-32.1\times t\\\Rightarrow \frac{-4.5}{-32.1}=t\\\Rightarrow t=0.14 \s$

Time taken by the ball to reach the highest point is 0.14 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.5\times 0.14+\frac{1}{2}\times -32.1\times 0.14^2\\\Rightarrow s=0.315\ ft$

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

$s=ut+\frac{1}{2}at^2\\\Rightarrow 4.315=0t+\frac{1}{2}\times 32.1\times t^2\\\Rightarrow t=\sqrt{\frac{4.315\times 2}{32.1}}\\\Rightarrow t=0.518\ s$