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3 years ago

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The diagram shows a person holding a bow and arrow. Which will most likely increase the kinetic energy in this system? a not rel

easing the string b increasing the mass of the arrow c releasing the string d pulling back less on the string

1 answer:

Fantom [35]3 years ago

7 0

Answer:

b. Increasing the mass of the arrow.

Explanation:

The formula is K=1/2mv^2. Increasing the mass also increases the kinetic energy.

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The distance between earth and sun is 15000000km. Light takes 499 seconds to reach earth from sun. Calculate the speed of light

iVinArrow [24]

To solve the problem we must know about the relationship between Speed, distance, and Time.

<h3>What is the relationship between Speed, distance, and Time?</h3>

We know that sped, distance, and time all are in a relationship to each other. this relationship can be given as,

$\rm{Speed = \dfrac{Distance}{Time}$

The speed of the light is 30,060.12 km/sec.

Given to us

The distance between the earth and the sun is 15000000km
Light takes 499 seconds to reach earth from the sun.

We know that speed can be described as,

$\rm{Speed = \dfrac{Distance}{Time}$

Therefore,

<h3>What is the speed of the light?</h3>

$\text{Speed of light} = \dfrac{\text{Distance between the earth and the sun}}{\text{Time taken by the light to travel the distance}}$

Substitute the value,

$\text{Speed of light} = \dfrac{15,000,000\ km}{499\ seconds}$

$\text{Speed of light} = 30,060.12\ km/sec$

Hence, the speed of the light is 30,060.12 km/sec.

Learn more about Speed, distance, and Time:

brainly.com/question/15100898

7 0

2 years ago

Read 2 more answers

A woman in a sprint race accelerates from rest to 8.7 m/s in 2.7 s. What is her displacement?

Trava [24]

Answer:

23.49m

Explanation:

Distance = velocity x time

8.7 x 2.7 = 23.49m

8 0

2 years ago

Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit

Ede4ka [16]

Answer:

$386.2^{\circ}F$

Explanation:

We are given that

$P_1=200lbf/in^2$

$P_2=60lbf/in^2$

$v_1=200ft/s$

$v_2=1700ft/s$

$T_1=500^{\circ}F$

$Q=0$

$C_p=1BTU/lb^{\circ}F$

We have to find the exit temperature.

By steady energy flow equation

$h_1+v^2_1+Q=h_2+v^2_2$

$C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}$

$1BTU/lb=25037ft^2/s^2$

Substitute the values

$1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}$

$500+1.598=T_2+115.4$

$T_2=500+1.598-115.4$

$T_2=386.2^{\circ}F$

7 0

3 years ago

A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the

zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

$E= \frac{kqr}{R^3}$

The potential at a distance can be represented as:

V(r) - V(0) = $-\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2$

V(r) - V(0) = $[\frac{qr^2}{8 \pi E_0R^3 }]$ ₀

V(r) = $-[\frac{qr^2}{8 \pi E_0R^3 }]$ ₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

= 0.56 × 10⁻² m

R = 1.29 cm

= 1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

$V(r)$ $= -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}$

$V(r)$ = -2.15 × 10⁻⁴ V

The difference between the radial distance and center can be expressed as:

V(r) - V(0) = $-\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2$

V(r) - V(0) = $[\frac{qr^2}{8 \pi E_0R^3 }]^R$

V(r) = $-\frac{qR^2}{8 \pi E_0R^3 }$

V(r) = $-\frac{q}{8 \pi E_0R }$

V(r) $= -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}$

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0

2 years ago

What is the voltage of the electrochemical cell written as: Cu(s) | Cu2+(aq) || Mg2+(aq) | Mg(s)?

erma4kov [3.2K]

Thank you for posting your question here at brainly. Feel free to ask more questions.
<span><span>The best and most correct answer among the choices provided by the question is </span>B.-2.71 V.</span>
Mg2+(aq) + 2e- -> Mg(s) E=-2.37 V

Cu2+(aq) + 2e- -> Cu(s) E =+ 0.34 V

Since Cu is acting as the anode, the equation needs to be reversed.

Cu(s) -> Cu2+(aq) + 2e- E =- 0.34 V

Ecell= -2.37 V+ (- 0.34 V) = -2.71 V <span><span>

</span><span>Hope my answer would be a great help for you. </span> </span> <span> </span>

4 0

3 years ago