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riadik2000 [5.3K]
3 years ago
5

The amount of heat needed

Physics
1 answer:
Masteriza [31]3 years ago
4 0
J=joules, c=specific heat, q= energy, and the Tf and Ti are the final and initial temperatures cause I couldn't find a delta sign.

Q=mc(T_{f}-T_{I})  \\ 33000j=2kg*c*80 \\\frac{33000j}{2kg*80} =c \\ c=206.25jkg^{-1}
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Why isn’t a bird sitting on a high-voltage power line electrocuted? Contrast this with the situation in which a large bird hits
Grace [21]

Answer:

The reason the bird is not electrocuted is due to some facts about circuit:

1. Completeness of circuit- This circuit needs to be

complete in order for current to flow. The bird standing on only one wire has not completed the circuit.

2. A potential difference: Another factor deciding

the direction of flow of current is (electric)

potential. Current always flows from a higher

potential to a lower potential. In other words it

can be said that electrons flow from lower

potential to higher one. (the direction of electric

current is opposite to that of the electrons). So

we need the potential difference for current to

flow. The bird standing on only one wire has no potential difference.

3. Path of least Resistance- Factor that decides

the path a current will flow in case of parallel

paths is the (electric) resistance offered by the

path. Current will always flow in the path that

offers least resistance. The leg of a bird has high resistance.

Explanation:

It has no potential difference as both the legs of bird are touching the same wire at same constant potential. ... If the bird would touch the ground while sitting on the wire or flap its wings and touch another electric wire with a different voltage, then it would get shocked and likely die by electrocution.

while in the other hand, the Bird that touches two wires with it wings at the same time will get electrocuted because it has completed a circuit and the its feathers created a potential difference .

8 0
3 years ago
Read 2 more answers
A proton with a speed of 3.2 x 106 m/s is shot into a region between two plates that are separated by a distance of 0.23 m. As t
harina [27]

Answer:

<h2>The magnitude of the magnetic is 0.145 T</h2>

Explanation:

Given :

Speed of proton v = 3.2 \times 10^{6}\frac{m}{s}

Mass of proton m = 1.67 \times 10^{-27} Kg

The force on the proton in magnetic field is given by,

  F = q (v \times B)

  F = qvB \sin \theta

But \sin 90 = 1    (∵ Force is perpendicular to the velocity so \theta = 90)

  F = qvB

When particle enter in magnetic field at the angle of 90° so particle moves in circle

So force is given by,

  F = \frac{mv^{2} }{r}

Where r = radius but in our case 0.23 m, q = 1.6 \times 10^{-19} C

By comparing above two equation,

  B = \frac{mv}{qr}

  B = \frac{1.67 \times 10^{-27} \times 3.2 \times 10^{6}  }{1.6 \times 10^{-19} \times 0.23 }

  B = 0.145 T

5 0
3 years ago
One scientist suggests that out of the different possible locations, they should design the model and build it at the equator re
Anika [276]

Answer:

B.

I think.

Explanation:

Mars doesn't have that much of an atmosphere!

Have a great day!

7 0
3 years ago
Find the force necessary to pull a 6 kg object 3 m/s2
mars1129 [50]

Answer:

<h2>18 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 6 × 3

We have the final answer as

<h3>18 N</h3>

Hope this helps you

6 0
3 years ago
A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

Substituting (5) in (1):

x=(23 m/s) cos(0\°) 0.63 s (6)

Finally:

x=14.49 m

5 0
3 years ago
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