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uysha [10]
3 years ago
11

A puck of mass m 0.085 kg is going in a circle on a horizontal frictionless surface. It is held in its orbit by massless string

of length L = 0.84 m. The puck makes one revolution every t= 0.55 s. 50%
What the magnitude of the tension in the string, F, in Newtons, while it spins?
Physics
1 answer:
attashe74 [19]3 years ago
8 0

Answer:

Explanation:

Magnitude of tension

= centripetal force

= m ω²R

= m (2π/T)² R , T is time period of revolution.

= .085 x (2 x 3.14 / .55)² x .84

= 9.3  N .  

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the average power of the Sun is 3.79x1026 Watts, answer the following questions: 13. What is the average intensity of light at E
Lemur [1.5K]

Explanation:

Given that,

Average power of sun P=3.79\times10^{26}\ Watt

We need to calculate the intensity of light at Earth's position

Using formula of intensity

I=\dfrac{P_{avg}}{4\pi r^2}

Where, I = intensity

P = power

Put the value into the formula

I=\dfrac{3.79\times10^{26}}{4\pi\times(1.496\times10^{11})^2}

I=1347.616\ W/m^2

So, The intensity is 1347.616 W/m².

(A). We need to calculate the pressure on a solar sail due to the light of the sun if it's fully reflective

Using formula for fully reflective

P = \dfrac{2I}{c}

Put the value into the formula

P=\dfrac{2\times1347.616}{3\times10^{8}}

P=8.984\times10^{-6}\ N/m

(B).  We need to calculate the pressure on a solar sail due to the light of the sun if it's fully reflective

Using formula for fully absorptive

P=\dfrac{I}{c}

P=\dfrac{1347.616}{3\times10^{8}}

P=4.492\times10^{-6}\ N/m

Hence, This is the required solution.

6 0
3 years ago
A 50N girl pushes a 10,000 N car with force of 200N. What is the force the car pushes back at the girl? *
miss Akunina [59]

Answer:

200N

Explanation:

EVERY FORCE IS OPPOSED BY AN EQUAL FORCE, REGARDLESS OF THE WEIGHT OF THE OBJECTS APPLYING THE FORCE.

8 0
3 years ago
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lesantik [10]

Answer:

240 meter

Explanation:

d= s*d=60m/s * 4sec=240 meter

7 0
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3 years ago
From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum up
NemiM [27]

Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

K.E = (0.5)(mv₀²)

where,

K.E = initial kinetic energy of projectile = 1430 J

m = mass of projectile = 0.41 kg

v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ  = (83.5 m/s) Cos(40.5°)

<u>v₀ₓ = 63.5 m/s</u>

y- component = v₀y = v₀Sinθ  = (83.5 m/s) Sin(40.5°)

<u>v₀y = 54.2 m/s</u>

7 0
3 years ago
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