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Otrada [13]
3 years ago
11

A 3 column table with 5 rows labeled table A. The first column is labeled voltage in volts with entries 1.0, 5.0, 10, 20, 50. Th

e second column is labeled current: calculated in amps with no entries. The last column is labeled current: experimental in amps. According to Ohm’s law, determine the calculated current for these values in Table A. voltage = 5 V, resistance = 20 Ω: A voltage = 20 V, resistance = 20 Ω: A voltage = 50 V, resistance = 20 Ω: A
Physics
2 answers:
dimaraw [331]3 years ago
7 0

Answer:

<em>0.25A</em>

<em>1.0A</em>

<em>2.5A</em>

Explanation:

Volgvan3 years ago
6 0

Answer:

0.25A

1.0A

2.5A

Explanation:

this is only for the calculated current column on edge

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D). located out side the nucleus
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3 years ago
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A 19 g bullet is fired into the bob of a ballistic pendulum of mass 1.3 kg. When the bob is at its maximum height, the strings m
katovenus [111]

Answer:

217.43298 m/s

Explanation:

m_1 = Mass of bullet = 19 g

m_2 = Mass of bob = 1.3 kg

L = Length of pendulum = 2.3 m

\theta = Angle of deflection = 60°

u = Velocity of bullet

Combined velocity of bullet and bob is given by

v^2-u^2=2as\\\Rightarrow v=\sqrt{2aL(1-cos\theta)+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times (1-cos60)+0^2}\\\Rightarrow v=3.13209\ m/s

As the momentum is conserved

m_1u=(m_1+m_2)v\\\Rightarrow u=\frac{(m_1+m_2)v}{m_1}\\\Rightarrow v=\frac{(0.019+1.3)\times 3.13209}{0.019}\\\Rightarrow v=217.43298\ m/s

The speed of the bullet is 217.43298 m/s

5 0
3 years ago
Hank and Harry are two ice skaters whiling away time by playing 'tug of war' between practice sessions. They hold on to opposite
Alex73 [517]

Answer:

the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1

Explanation:

Given the data in the question;

Hank and Harry are two ice skaters, since both are on top of ice, we assume that friction is negligible.

We know that from Newton's Second Law;

Force = mass × Acceleration

F = ma

Since they hold on to opposite ends of the same rope. They have the same magnitude of force |F|, which is the same as the tension in the rope.

Now,

Mass_{Hank × Acceleration_{Hank = Mass_{Henry × Acceleration_{Henry

so

Mass_{Hank /  Mass_{Henry = Acceleration_{Henry / Acceleration_{Hank

given that; magnitude of Hank's acceleration is 1.26 times greater than the magnitude of Harry's acceleration,

Mass_{Hank /  Mass_{Henry = 1 / 1.26

Mass_{Hank /  Mass_{Henry = 0.7937 or [ 0.7937 : 1 ]

Therefore, the ratio of Hank's mass to Harry's mass is 0.7937 or [ 0.7937 : 1 ]

8 0
3 years ago
A sample of argon gas (molar mass 40 g) is at four times the absolute temperature of a sample of hydrogen gas (molar mass 2 g).
qaws [65]

To solve this problem, let us recall that the formula for gases assuming ideal behaviour is given as:

rms = sqrt (3 R T / M)

where

R = gas constant = 8.314 Pa m^3 / mol K

T = temperature

M = molar mass

Now we get the ratios of rms of Argon (1) to hydrogen (2):

rms1 / rms2 = sqrt (3 R T1 / M1) / sqrt (3 R T2 / M2)

or

rms1 / rms2 = sqrt ((T1 / M1) / (T2 / M2))

rms1 / rms2 = sqrt (T1 M2 / T2 M1)

Since T1 = 4 T2

rms1 / rms2 = sqrt (4 T2 M2 / T2 M1)

rms1 / rms2 = sqrt (4 M2 / M1)

and M2 = 2 while M1 = 40

rms1 / rms2 = sqrt (4 * 2 / 40)

rms1 / rms2 = 0.447

 

Therefore the ratio of rms is:

<span>rms_Argon / rms_Hydrogen = 0.45</span>

7 0
3 years ago
A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom w
nirvana33 [79]

Answer:

|D_{depth} |=19.697m

Explanation:

To find Depth D of lake we must need to find the time taken to hit the water.So we use equation of simple motion as:

Δx=vit+(1/2)at²

x_{f}-x_{i}=v_{i}t+(1/2)at^{2}\\  -5.0m=(o)t+(1/2)(-9.8m/s^{2} )t^{2}\\ -4.9t^{2}=-5.0\\ t^{2}=5/4.9\\t=\sqrt{1.02} \\t=1.01s

As we have find the time taken now we need to find the final velocity vf from below equation as

v_{f}=v_{i}+at\\v_{f}=0+(-9.8m/s^{2} )(1.01s) \\v_{f}=-9.898m/s

So the depth of lake is given by:

first we need to find total time as

t=3.0-1.01 =1.99 s

|D_{depth} |=|vt|\\|D_{depth} |=|(-9.898m/s)(1.99s)|\\|D_{depth} |=19.697m

6 0
3 years ago
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