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Alexandra [31]
3 years ago
10

________________________involves connecting a voltmeter set to read DC volts to the suspected high-resistance cable ends and cra

nking the engine.
Physics
1 answer:
AnnyKZ [126]3 years ago
4 0

Answer:

Voltage Drop Testing

Explanation:

The motives behind conducting Voltage Drop Testing is to recognize or identify the presence of unwanted resistance in the wire. or any other electrical appliance.  The appliance used to calculate the voltage drop is digital volt ohmmeters.  

voltage drop is the amount of loss of voltage due to resistance when electric current flow through the circuit. Thus measuring the voltage drop in respect to the amount of current flow, resistance can be measured.

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Earth is about 150 million kilometers from the Sun, and the apparent brightness of the Sun in our sky is about 1300 watts/m2. Us
nalin [4]

Answer:

13 W/m^2

Explanation:

The apparent brightness follows an inverse square law, therefore we can write:

I \propto \frac{1}{r^2}

where I is the apparent brightness and r is the distance from the Sun.

We can also rewrite the law as

\frac{I_2}{I_1}=\frac{r_1^2}{r_2^2} (1)

where in this problem, we have:

I_1 = 1300 W/m^2 apparent brightness at a distance r_1, where

r_1 = 150 million km

We want to estimate the apparent brightness at r_2, where r_2 is ten times r_1, so

r_2 = 10 r_1

Re-arranging eq.(1), we find I_2:

I_2 = \frac{r_1^2}{r_2^2}I_1 = \frac{r_1^2}{(10r_1)^2}(1300)=\frac{1}{100}(1300)=13 W/m^2

5 0
3 years ago
2 cannons fire projectiles upwards with the same velocity. The
dybincka [34]

Answer:

Explanation:

Given

Two projectile is fired vertically upward

One has 4 times the mass of other

When Projectile is fired their trajectory is independent of mass of object. Also if they launched with same speed then both achieved same maximum height in same time and will hit the ground at the same moment.

3 0
3 years ago
A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi
nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

W = F * d

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

d=h*N, for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

V_f^2-V_i^2 = 2a\Delta X

Where,

V_f = Final velocity

V_i = Initial Velocity

a = Acceleration

\Delta X = Displacement

PART A) For the particular case of work we know then that,

W = F*d

W = m*g*(h*N)

W = 50*9.8*(0.3*30)

W = 4.41kJ

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

V_f^2-V_i^2 = 2a\Delta X

Here,

V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow3 steps in one second

v_f = 0

Replacing,

V_f^2-V_i^2 = 2a\Delta X

0-0.9^2=2a(30*0.3)

Re-arrange for a,

a = -\frac{0.9^2}{2*30*0.3}

a = -45*10^{-3}m/s^2

At this point we can calculate the time, which is,

t = \frac{\Delta V}{a}

t = \frac{0-0.9}{-45*10^{-3}}

t = 20s

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

6 0
3 years ago
Using hooke's law find the elastic constant of a spring that stretches 2 cm when 4newton force is applied to it
Kay [80]

<u>Answer:</u>

2N/cm

<u>Step-by-step explanation:</u>

According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:

F=kx

where, F is the force which is stretching or compressing the spring,

k is the spring constant; and

x is the distance the spring is stretched.

Substituting the given values to find the elastic constant  k to get:

F=kx

4=k(2)

k=\frac{4}{2}

k=2

Therefore, the elastic constant is 2 Newton/cm.

3 0
3 years ago
What is the separation distance, in meters, between masses m1 = 15 x 107 kg and m2 = 62 x 107 kg when the gravitational force be
nalin [4]

Answer:

94.1 m

Explanation:

From Coulombs law,

F  = Gm1m2/r²................... Equation 1

where F = force, m1 = first mass, m2 = second mass, G = universal constant, r = distance of separation.

Make r the subject of the equation,

r = √(Gm1m2/F)................. Equation 2

Given: F = 7×10² N, m1 = 15×10⁷ kg, m2 = 62×10⁷ kg,

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute into equation 2

r = √( 6.67×10⁻¹¹×15×10⁷×62×10⁷/7×10²)

r √(886.16×10)

r √(88.616×10²)

r = 9.41×10

r = 94.1 m.

Hence the distance of separation = 94.1 m

3 0
4 years ago
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