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3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Physics

1 answer:

Yuki888 [10]3 years ago

4 0

7. 1.47 x 10^5mm to kilometers

Answer: 0.147

Shown answer:

8.4.7 kg to pounds

Answer:

Shown answer:

9.138.4oz to grams

Answer: 3923.574

Shown Answer:

10. 65.5km to miles

Answer: 40.69981

Shown answer:

11. 23.6ft to cm

Answer: 719.328

Shown answer:

12. 2.36 x 10^4 s to days

Answer: 0.273148148 days

Shown answer:

13. 13.6 L to U.S. Liquid quarts

Answer:

Shown answer:

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A coil has an inductance of 5 H and a resistance of 20 Ω. If a DC voltage of 100 V is applied to the coil, find the energy store

nadya68 [22]

Answer:62.5 J

Explanation:

Given

Inductance(L)=5 H

resistance(R) $=20 \Omega$

Voltage(V)=100 V

$Current=\frac{100}{20}=5 A$

Current in L-R circuit is given by

$I=I_0\left [ 1-e^{-\frac{Rt}{L}}\right ]$

and Power=i^2R+Li\frac{di}{dt}[/tex]

For Steady state i.e. at $t=\infty$

$I=I_0$

Energy Stored is $E=\frac{Li^2}{2}$

$E=\frac{5\times 5^2}{2}=62.5 J$

6 0

3 years ago

How to determine my slope in physics graph

Hatshy [7]

Answer:

Determine the coordinates of two points on the line. Calculate the difference between these two locations' y-coordinates (rise). Calculate the x-coordinate difference between these two places (run). Divide the y-coordinate difference by the x-coordinate difference (rise/run or slope).

5 0

2 years ago

What was the maximum speed of the car in your experiment?

MA_775_DIABLO [31]

The measurements used in the experiment is the amount of speed over time.

The measurement of speed is indicated along the “y” axis.

Upon viewing the graph, the highest point along the “y” axis shown is 25 m/s. This would be the maximum.

The maximum speed of the car would be 25 m/s.

7 0

3 years ago

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 38 ft/s2. what is the dist

e-lub [12.9K]

Convert 38 ft/s^2 to mi/h^2. Then we se the conversion factor > 1 mile = 5280 feet and 1 hour = 3600 seconds.

So now we show it > $38 \frac{ft}{s^2} x \frac{1mi}{5280ft} x \frac{(3600s)^2}{(1h)^2} = 93272.27 \frac{mi}{h^2}$

Then we have to use the formula of constant acceleration to determine the distance traveled by the car before it ended up stopping.

Which the formula for constant acceleration would be > $v_2^2=v_1^2 + 2as$

The initial velocity is 50mi/h $(v_1=50)$

When it stops the final velocity is $(v_2=0)$

Since the given is deceleration it means the number we had gotten earlier would be a negative so a = -93272.27

Then we substitute the values in....

$0^2 = 50^2 + 2(-93272.27)s

0 = 2500 - 186544.54s

Isolate S next.

185644.54s= 2500

s = 2500/(185644.54)

s=0.0134
$

So we can say the car stopped at 0.0134 miles before it came to a stop but to express the distance traveled in feet we need to use the conversion factor of 1 mile = 5280 feet in otherwards > $0.0134 mi * \frac{5280ft}{1mi} = 70.8 ft$
So this means that the car traveled in feet 70.8 ft before it came to a stop.

4 0

3 years ago

Describe the motionof an object that gas acceleration of 0 m/s

Maru [420]

If an object has acceleration of zero meters per second², then the object is traveling at a constant speed.

Note: The constant speed doesn't have to be zero, so the object is not necessarily at rest. It can be moving at any speed, as long as the speed doesn't change. (If the speed changed, then the acceleration wouldn't be zero.)

3 0

4 years ago