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An archer pulls the bowstring back for a distance of 0.470 m before releasing the arrow. The bow and string act like a spring wh

never [62]

Answer:

(a) 46.94 J.

(b) 55.95 m/s

Explanation:

(a)

Potential Energy: This is the energy of a body, due to its position. The S.I unit of potential energy is Joules (J).

The formula of potential energy in a stretched spring is

Ep = 1/2ke² .......................... Equation 1

Where Ep = potential energy of the spring, k = Force constant of the spring, e = extension or compression.

Given: k = 425 N/m, e = 0.47 m.

Substitute into equation 1

Ep = 1/2(425×0.47²)

Ep = 46.94 J.

(b)

at the instant When the arrow leaves the bow, the potential energy of the arrow is converted kinetic energy of the bow.

I.e,

Ep = 1/2mv² ............. Equation 2

Where m = mass of the arrow, v = velocity of the arrow.

make v the subject of the equation

v = √(2Ep/m)............. Equation 3

Given: Ep = 46.94 J, m = 0.03 Kg.

Substitute into equation 3

v = √(2×46.96/0.03)

v = √(93.92/0.03)

v = √(3130.67)

v = 55.95 m/s

5 0

2 years ago

A black widow spider hangs motionless from a web that extends vertically from the ceiling above. If the spider has a mass of 1.2

leva [86]

Answer:

Tension = 0.012 N

Explanation:

If the black widow spider is hanging vertically motionless from the ceiling above. Then, the weight of the spider must be balancing the tension in the spider web. Therefore,

Tension = Weight

Tension = mg

where,

m = mass of spider = 1.27 g = 0.00127 kg

g = acceleration due to gravity = 9.8 m/s²

Therefore,

Tension = (0.00127 kg)(9.8 m/s²)

Tension = 0.012 N

8 0

3 years ago

A scientist notices that an oil slick floating on water when viewed from above has many different colors reflecting off the surf

Sati [7]

Answer:

a) The minimum thickness of the oil slick at the spot is 313 nm

b) the minimum thickness be now will be 125 nm

Explanation:

Given the data in the question;

a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?

t $_{min$ = λ/2n

given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20

we substitute

t $_{min$ = 750 / 2(1.20)

t $_{min$ = 750 / 2.4

t $_{min$ = 312.5 ≈ 313 nm

Therefore, The minimum thickness of the oil slick at the spot is 313 nm

b)

Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?

minimum thickness of the oil slick at the spot will be;

t $_{min$ = λ/4n

given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50

we substitute

t $_{min$ = 750 / 4(1.50)

t $_{min$ = 750 / 6

t $_{min$ = 125 nm

Therefore, the minimum thickness be now will be 125 nm

4 0

2 years ago

Why does the entropy of a gas increase when it expands into a vacuum? Why does the entropy of a gas increase when it expands int

Mrrafil [7]

Answer:

The entropy of a gas increases when it expands into a vacuum because the number of possible states increases .

Explanation:

When a gas expand in a vacuum, the molecules of the gases vibrates very fast and starting moving with higher velocity in random directions which means the level of disorder in the gases increases.

Now the possible state of the gas molecule increases such as the particle can be located at different position due to increased randomness.

Entropy is the measure of this randomness and thus with this increased randomness entropy also increases.

6 0

3 years ago

A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the

aleksandrvk [35]

Answer:

θ= 5 radian

Explanation:

Given data:

Radius r = 0.70 m

Initial angular speed ω_i = 2rev/s

Time t = 5 s

Final angular speed ω_f =0

so we have angular displacement

$\theta= \frac{\omega_f-\omega_i}{2}\times t$

putting values

$\theta= \frac{0-2}{2}\times5$ = 5 rad

8 0

3 years ago