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LuckyWell [14K]
3 years ago
12

Enter an expression in the box to Write the equation of the line perpendicular to y=-3x-1 that passes through the point (3,4).

Physics
1 answer:
Dafna11 [192]3 years ago
4 0

Answer: -3x+13

Explanation:

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When a 10 V battery is connected to a resistor, 5 A of current flows through the resistor. What is the resistor's value?
SashulF [63]

Given data

*The value of battery voltage is V = 10 V

*The current flows through the resistor is I = 5 A

The formula for the resistor is given by the Ohm's law as

R=\frac{V}{I}

Substitute the values in the above expression as

\begin{gathered} R=\frac{10}{5} \\ =2\text{ ohm} \end{gathered}

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1 year ago
If a canadian plane is traveling at 500 miles/hr, how fast is this in m/s?
patriot [66]
223.52 meters per second
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3 years ago
How strong is the repulsive force exerted on two point charges that each carry 1.0 E-6 C of negative charge and are 0.30 meters
Brrunno [24]

Answer:

Match the scenario to the type of movement that caused it.

Ted is in his home when a strong vibration occurs that shakes the dishes out of his cupboards.

Whitney feels a slight tremor three weeks subsequent to an earthquake in a nearby town.

Sam pulls off the side of a road after feeling the road move unexpectedly while driving to work.

Foreshock

Mainshock

Aftershock

Explanation: PLZZ HELP ME I NEED TO PASS STEM!

6 0
3 years ago
Read 2 more answers
A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertica
Radda [10]

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed \omega =40rad/s

mass of m_2=0.8 kg dropped at r=0.3 m from center

let \omega _2 be the final angular velocity of cylinder

Conserving Angular momentum

L_1=L_2

\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2

\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2

26.01\times 40=26.082\times \omega _2

\omega _2=39.88 rad/s

3 0
3 years ago
How long does it take for a 3.5 kW electric water heater to heat 40 kg of water? from 20 ° C to 75 ° C? The specific heat capaci
iren [92.7K]

Answer:

2633.7 s

Explanation:

From the question,

Heat lost by the water heater = Heat gained by the water

Applying,

P = cm(t₂-t₁)/t.................. Equation 1

Where P = power of the heat, c = specific heat capacity of water, m = mass of water, t₁ = initial temperature, t₂ = final temperature, t = time

make t the subject of the equation

t = cm(t₂-t₁)/P.............. Equation 2

From the question,

Given: c = 4190 J/kgK, P = 3.5 kW = 3500 W, m = 40 kg, t₁ = 20°C, t₂ = 75°C

Substitute these values into equation 2

t = 4190×40(75-20)/3500

t = 9218000/3500

t = 2633.7 s

4 0
3 years ago
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