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gulaghasi [49]
3 years ago
9

PRELIMINARY QUESTIONS

Physics
1 answer:
zimovet [89]3 years ago
3 0
French fries were on your car
You might be interested in
A through shaped tank is located 3 meters underground (The top of the tank is 3 meters away from ground level.) The dimensions o
vovangra [49]

Answer:

900kJ

Explanation:

Work done hereis lifting that water mass on to the surface level

v = volume of the tank

d= density of water (1000kg/m3)

g = gravotational acceleration

workdone=Weight * distance

                = (v*d*g)*distance

                ==((\frac{8+4}{2} )*5*10)*1000*10*3\\=900000J\\=900kJ

4 0
3 years ago
Compare the weight of a mountain climber when she is at the bottom of a mountain with her weight when she is at the top of the m
kakasveta [241]

Answer:

The correct answer is  a. Both are the same

Explanation:

For this calculation we must use the gravitational attraction equation

    F = G m M / r²

Where M will use the mass of the Earth, m the mass of the girl and r is the distance of the girl to the center of the earth that we consider spherical

To better visualize things, let's repair the equation a little

     F = m (G M / r²)

The amount in parentheses called acceleration of gravity, entered the force called peos

     g = G M / r²

     F = W

    W = m g

When analyzing this equation we see that the variation in the weight of the girl depends on the distance, which is the radius of the earth plus the height where the girl is

    r = Re + h

    Re = 6.37 10⁶ m

    r² = (Re + h)²

    r² = Re² (1 + h / Re)²

Let's replace

    W = m (GM / Re²)   (1+ h / Re)⁻²

    W = m g   (1+ h / Re)⁻²

This is the exact expression for weight change with height, but let's look at its values ​​for some reasonable heights h = 6300 m (very high mountain)

     h / Re = 10 ⁻³

     (1+ h / Re)⁻² = 0.999⁻²

Therefore, the negligible weight reduction, therefore, for practical purposes the weight does not change with the height of the mountain on Earth

The correct answer is a

4 0
3 years ago
A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii. If the three of them are released simultaneou
Roman55 [17]

Answer:

The solid sphere will reach the bottom first.

Explanation:

In order to develop this problem and give it a correct solution, it is necessary to collect the concepts related to energy conservation. To apply this concept, we first highlight the importance of conserving energy so we will match the final and initial energies. Once this value has been obtained, we will concentrate on finding the speed, and solving what is related to the Inertia.

In this way we know that,

\Delta KE = - \Delta PE

KE_t + KE_r = mgh

We know as well that the lineal and angular energy are given by,

KE_r = \frac{1}{2}I\omega^2

And the tangential kinetic energy as

KE_t = \frac{1}{2} mv^2

Where\omega = \frac{v}{R}

Replacing

\frac{1}{2}mv^2 + \frac{1}{2}I\frac{v}{R} = mgh

Re-arrange for v,

v=\sqrt{\frac{2mgh}{m+I/R^2}}

We have here three different objects: solid cylinder, hollow pipe and solid sphere. We need the moment inertia of this objects and replace in the previous equation found, then,

For hollow pipe:

I_{hp}=mR^2

v_{hp}=\sqrt{\frac{2mgh}{m+(mR^2)/R^2}}

v_{hp}=\sqrt{\frac{2mgh}{m+m)}

v_{hp}=\sqrt{gh}

For solid cylinder:

I_{sc}=\frac{1}{2}mR^2

v_{sc}=\sqrt{\frac{2mgh}{m+(1/2mR^2)/R^2}}

v_{sc}=\sqrt{\frac{2mgh}{m+1/2m}}

v_{sc}=\sqrt{\frac{3}{4}gh}

For solid sphere,

I_{ss}=\frac{2}{5}mR^2

v_{ss}=\sqrt{\frac{2mgh}{m+(2/5mR^2)/R^2}}

v_{ss}=\sqrt{\frac{2mgh}{m+2/5m}}

v_{ss}=\sqrt{\frac{10}{7}gh}

Then comparing the speed of the three objects we have:

v_{hp}

\sqrt{gh}

3 0
4 years ago
Imagine you’re an engineer making a string of battery powered holiday lights. If a bulb burns out current cannot flow through th
Anit [1.1K]

Answer:

The 2 light bulbs can be connected in parallel to each other to avoid disconnection when one bulb burns out.

Explanation:

The parallel connection is required not series. A parallel connection is the connection of electronic components (e.g bulbs, LED, resistors, capacitors etc) in such a way that the same voltage is supplied across the ends of the components.  While in a series connection, the components are connected to each other end-to-end.

As regard the question, parallel connection ensures that the brightness any of the bulbs is not affected with respect to the other bulbs. And other bulbs continue to function when any burns out. The 2 light bulbs should be connected in parallel to the baterry to avoid disconnection of all the bulbs.

4 0
4 years ago
The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .
ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is W=1.20 \cdot 10^{-3}J

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: W_e=q\Delta V

where

q is the charge

\Delta V is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

K_f - K_i = W + W_e = W+q\Delta V

and since the charge starts from rest, K_i = 0, so the formula becomes

K_f = W+q\Delta V

In this problem, we have

W=1.20 \cdot 10^{-3}J is the work done by the external force

q=-6.70 \mu C=-6.7\cdot 10^{-6}C is the charge

K_f = 4.72\cdot 10^{-4}J is the final kinetic energy

Solving the formula for \Delta V, we find

\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V

4 0
3 years ago
Read 2 more answers
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