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myrzilka [38]
3 years ago
11

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

Physics
1 answer:
makkiz [27]3 years ago
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
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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
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