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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>

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Two long, parallel wires are separated by 2.2 mm. Each wire has a 32-AA current, but the currents are in opposite directions. Pa

Alex

Answer:

$B=1.1636*10^{-3}T$

Explanation:

Given data

$d_{wires}=2.2mm=0.022m\\ I_{current}=32A\\$

To find

Magnitude of the net magnetic field B

Solution

The magnitude of the net magnetic field can be find as:

$B=2*u\frac{I}{2\pi r}\\ B=2*(4\pi*10^{-7} )\frac{32}{2\pi (0.022/2)} \\ B=1.1636*10^{-3}T$

3 0

3 years ago

The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio

Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

$\gamma = 5/3$

T1 = 525 K

T2 = 275 K

We know that

$P_1 = \frac{nRT_1}{v}$

$P_2 = \frac{nrT_2}{v}$

n and v remain same at both side. so we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}$

$P_1 = \frac{21}{11} P_2$ ..............1

let final pressure is P and temp $T_1 {f} and T_2 {f}$

$P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}$

$P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}$ ..................2

similarly

$P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}$ .............3

divide 2 equation by 3rd equation

$\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}$

$T_1 {f} = 1.48 T_2 {f}$

thus, temperature on left side is 1.48 times the temperature on right

6 0

2 years ago

A person carries a plank of wood 2 m long with one hand pushing down on it at one end with a force F1 and the other hand holding

sveta [45]

Answer:

F1= 588 N

F2= 784 N

Explanation:

Please see the attached file.

5 0

3 years ago

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

5 0

2 years ago

Out of aluminum,copper,steel,and glass.Which material do you think will be the best thermal conductor?

Nastasia [14]

I believe it is copper

6 0

3 years ago