If the force on a spring is 2 N and it stretched 0.5 m, what is the spring constant?

When running a half marathon (13.1 miles), it took Kevin 8 minutes to run from mile marker 1 to mile marker 2, and 18 minutes to

Vsevolod [243]

Answer:

It took Kevin 26 minutes to run from markers 1 to 4

His average speed from mile markers 1 to 4 is 0.154 miles/minute

Kevin must run by average speed 0.1 miles/minute to finish the race

Explanation:

Lets explain how to solve the problem

A half marathon 13.1 miles

Kevin took 8 minutes to run from mile marker 1 to mile marker 2 and

18 minutes to run from mile marker 2 to mile marker 4

→ He took 8 minutes and 18 minutes to run from marker 1 to marker 4

→ The total time of the first 4 marker = 8 + 18 = 26 minutes

It took Kevin 26 minutes to run from markers 1 to 4

Average speed is total distance divided by total time

The average speed of Kevin as he ran from mile marker 1 to mile

marker 4 is the 4 miles divides by 26 minutes

→ Average speed = 4 ÷ 26 = $\frac{2}{13}$ = 0.154 miles/minute

His average speed from mile markers 1 to 4 is 0.154 miles/minute

It took Kevin 71 minutes to pass mile marker 9

Kevin need to complete the race in 112 minutes, then what must

Kevin's average speed be as he travels from mile marker 9 to the

finish line?

The total distance of the race is 13.1 miles, he ran 9 miles

→ The remaining distance = 13.1 - 9 = 4.1 miles

He must run 4.1 miles to complete the race

The total time is 112 minutes, he used 71 minutes to run the first 9 miles

→ The remaining time = 112 - 71 = 41 minutes

He must finish the 4.1 miles in 41 minutes

→ His average speed for the last part of the race = 4.1 ÷ 41 = 0.1 mi/min

Kevin must run by average speed 0.1 miles/minute to finish the race

7 0

3 years ago

Read 2 more answers

A 400g sample of water absorbs 500j of energy. how did the water temperature change if the specific heat of water is 4.18j/g©. S

Mashutka [201]

In general, the quantity of heat energy, Q, required to raise a mass m kg of a substance with a specific heat capacity of c J/(kg °C), from temperature t1 °C to t2 °C is given by:

Q = mc(t2 – t1) joules

So:

(t2-t1) =Q / mc

As we know:
Q = 500 J

m = 0.4 kg

c = 4180 J/Kg °c

We can take t1 to be 0°c

t2 - 0 = 500 / ( 0.4 * 4180 )

t2 - 0 = 0.30°c

6 0

3 years ago

A billiard ball moving at 0.5 m/s strikes another identical billiard ball, which is at rest, in an elastic head-on collision on

NikAS [45]

Answer: assuming that the billiard balls are of identical weight the impacted billiard ball will move forward at around 0.5m/s (not considering energy conservation). The ball impacting the 2nd one would stop because most of its Kinetic energy would have been transferred into the not moving ball.

Explanation: hope this helps!

4 0

3 years ago

Not only did Skid get shot out of a cannon when he was a clown in the circus, but he was also launched into the air by a vertica

tiny-mole [99]

Answer:

v = 16.11 m / s

Explanation:

For this exercise we must use the principle of conservation of energy. We set a reference system on the part of the platform without elongation

starting point. When the spring is compressed

Em₀ = K_e + U = ½ k x² + m g x ’

final point. The point where it leaves the platform

Em_f = K = ½ m v²

energy is conserved

Em₀ = Em_f

½ k x² + m g x ’= ½ m v²

v² = $\frac{k}{m}$ x² + g x

let's calculate

v² = $\frac{8700}{55}$ 1.25² + 9.8 1.25

v² = 247.159 + 12.25 = 259.409

v = 16.11 m / s

8 0

3 years ago

In a simple circuit a 6-volt dry cell pushes charge through a single lamp which has a resistance of 3 Ω. According to Ohms law t

RSB [31]

Answer:

The answers and workings is in the Explanation section

Explanation:

In a simple circuit a 6-volt dry cell pushes charge through a single lamp which has a resistance of 3 Ω. According to Ohms law the current through the circuit is _____________ Amps.

According to Ohm’s law V =I*R

Where V = Voltage, I = Current and R = Resistance

I = V/R =6/3 =2 Amps of current

Answer = 2 Amps

 If a second identical lamp is connected in series, the 6-volt battery must push a charge through a total resistance of ________ Ω. The current in the circuit is then __________ Amps.

Since the second lamp is connected in series with the first, the total resistance will R₂ = 3 + 3 = 6Ω

2 resistance in series R₂ = 6Ω

The current in the circuit with the two lamps connected in series is I₂ =V/R₂ =6/6 = 1 Amps

The current is 1 Amps

Answer = 6Ω and 1 Amps

If a third identical lamp is connected in series, the total resistance is now _________Ω. The current through all three lamps in series is now _________ Amps.

Since the third lamp is connected in series with the first and second, the total resistance will R₃ = 3 + 3 + 3 = 9Ω

total resistance of the 3 lamps R₃ = 9Ω

The current in the circuit with the three lamps connected in series is

I =V/R₃ =6/9 = 0.67 Amps

The current through the 3 lamps I₃ = 0.67 Amps

Answer = 9Ω and 0.67 Amps

The current through each individual lamp is __________ Amps.

Since all 3 lamps are connected in series, the same current will flow through each of the 3 lamps, and that current is I₃

The current through each individual lamp is 0.67 Amp

Answer = 0.67 Amp

What is the power when a voltage of 120 volts drives a current of 3 amps through a device?

The formula for power P = I*V =120*3 = 360 Watts

power P = 360 Watts

Answer = 360 Watts

What is the current when a 90-W light bulb is connected to 120 V?

From P =I*V, make I the subject of the formula, I = P/V =90/120 = 0.75

Current= 0.75 Amps

Answer = 0.75 Amps

How much current does a 75-W light bulb draw when connected to 120 V?

Current I =P/V = 75/120 = 0.625 Amps.

Answer = 0.625 Amps

If part of an electric circuit dissipates energy at 6 W when it draws a current of 3 A, what voltage is impressed across it?

Voltage V =P/I =6/3 =2 Volts

Answer = 2 Volts

If a 60 W light bulb at 120 V is left on in your house to prevent burglary, and the power company charges 10 cents per kilowatt-hour, how much will it cost to leave the bulb on for 30 days? Show your work.

24 hours make 1 day, so the number of hours the bulb was left on = 24 *30 = 720 hours

The power rating of the bulb is 60W = 60/1000 = 0.06 KiloWatt

Total power consumed in kilowatt-hour = 0.06 * 720 = 43.2 kilowatt-hour

Cost for 30 days = 0.1*43.2 = $4.32 ( note that 10 Cents = $0.1)

Answer = $4.32

4 0

3 years ago