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3 years ago

How much work is done when a 1-kg mass is raised a vertical distance of 1 m

1 answer:

5 0

Weight = mass times gravity. . . . Work = (weight) x (height) = (mass) x (gravity) x (height) = (1) x (9.8) x (1) = 9.8 joules.

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A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/

s344n2d4d5 [400]

Answer:

At $t = (70 / 3) \; {\rm s}$ (approximately $23.3 \; {\rm s}$ .)

Explanation:

Note that the acceleration of the car between $t = 0\; {\rm s}$ and $t = 20\; {\rm s}$ ( $\Delta t = 20\; {\rm s}$ ) is constant. Initial velocity of the car was $v_{0} = 0\; {\rm m\cdot s^{-1}}$ , whereas $v_{1} = 35\; {\rm m\cdot s^{-1}}$ at $t = 20\; {\rm s}\!$ . Hence, at $t = 20\; {\rm s}\!\!$ , this car would have travelled a distance of:

$\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}$ .

At $t = 20\; {\rm s}$ , the truck would have travelled a distance of $x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}$ .

In other words, at $t = 20\; {\rm s}$ , the truck was $400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m}$ ahead of the car. The velocity of the car is greater than that of the truck by $35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}$ . It would take another $(50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s}$ before the car catches up with the truck.

Hence, the car would catch up with the truck at $t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}$ .

3 0

2 years ago

This instrument can used to visualize atoms

Lisa [10]

Answer:

can someone follow me

Explanation:

Plqae

3 0

2 years ago

Read 2 more answers

A 1,000 kg car has 50,000 joules of kinetic energy what is its speed??

Vaselesa [24]

Using KE=1/2mv^2

v= $\sqrt{50000/500$

so v=10

3 0

3 years ago

Read 2 more answers

En la figura los émbolos son de masa despreciable y están en reposo, como se ve en la figura, siendo M=300 kg. Determine: a) la

Leni [432]

The image mentioned is in the attachment

Answer: a) P = 2450 Pa;

b) P = 2940 Pa;

c) F = 4.9 N

Explanation:

a) Pressure is a force applied to a surface of an object or fluid per unit area.

The image shows a block applying pressure on the large side of the piston. The force applied is due to gravitation, so:

P = $\frac{F}{A}$

P = $\frac{m.g}{A}$

P = $\frac{300.9.8}{1.2}$

P = 2450 Pa

The pressure generated by the block is P = 2450 Pa.

b) A static liquid can also exert pressure and can be calculated as:

$P_{staticfluid} =$ ρ.g.h

where

ρ is the density of the fluid

h is the depth of the fluid

g is acceleration of gravity

$P_{staticfluid} =$ 600.9.8.0.5

$P_{staticfluid} =$ 2940 Pa

The pressure in the fluid at 50 cm deep is $P_{staticfluid} =$ 2940 Pa.

c) For the system to be in equilibrium both pressures, pressure on the left side and pressure on the right side, have to be the same:

$P_{s} = P_{b}$

$\frac{F}{A_s}$ = $\frac{F_b}{A_b}$

F = $\frac{F_b}{A_b}.A_s$

Adjusting the units, $A_{s}$ = 0.002 m².

F = $\frac{300.9.8.0.002}{1.2}$

F = 4.9 N

The force necessary to be equilibrium is F = 4.9 N.

4 0

3 years ago

A person's center of mass is very near the hips, at the top of the legs. Model a person as a particle of mass mm at the top of a

muminat

Answer:

$\mathbf{v_{max} = \sqrt{gL}}$

Explanation:

Considering an object that moving about in a circular path, the equation for such centripetal force can be computed as:

$\mathbf {F = \dfrac{mv^2}{2}}$

The model for the person can be seen in the diagram attached below.

So, along the horizontal axis, the net force that is exerted on the person is:

$mg cos \theta = \dfrac{mv^2}{L}$

Dividing both sides by "m"; we have :

$g cos \theta = \dfrac{v^2}{L}$

Making "v" the subject of the formula: we have:

$v^2 = g Lcos \theta$

$v=\sqrt{ gL cos \theta$

So, when $\theta$ = 0; the velocity is maximum

∴

$v_{max} = \sqrt{gL \ cos \theta}$

$v_{max} = \sqrt{gL \ cos (0)}$

$v_{max} = \sqrt{gL \times 1}$

$\mathbf{v_{max} = \sqrt{gL}}$

Hence; the maximum walking speed for the person is $\mathbf{v_{max} = \sqrt{gL}}$

3 0

3 years ago