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kotykmax [81]
3 years ago
5

A 0.40 kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.0 m on a frictionless horizontal su

rface. if the cord will break when the tension in it exceeds 64 n, what is the maximum speed the ball can have?
Physics
1 answer:
DerKrebs [107]3 years ago
3 0
<span>The centripetal force for such an arrangement can be found through the equation Fc = mv^2/r where m is the mass of the rotating object, v is that object's velocity, and r is the radius of rotation. In this case, we know that the maximum Fc that can be tolerated by the cord is 64N. Thus we set the equation up and solve for the value of v for which Fc = 64.
   64 = 0.4*(v^2)/1
 64/0.4 = 160 =
v^2 v = sqrt(160) = 12.65 m/s

   At any speed faster than 12.65 m/s, the cord will break.</span>
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Which energy changes take place when a pedaling cyclist uses a generator (dynamo) to light his bicycle
Fynjy0 [20]

When the pedaling cyclist uses a generator (dynamo) to light his bicycle lamp, the energy change is from mechanical to electrical.

<h3>Law of conservation of energy</h3>

The law of conservation of energy states that energy can neither be created nor destroyed but can be converted from one form to another.

Thus, we can conclude the following as it relates to energy changes;

  • When the pedaling cyclist uses a generator (dynamo) to light his bicycle lamp, the energy change is from mechanical to electrical.

Learn more about conservation of energy here: brainly.com/question/166559

4 0
2 years ago
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What is the rate of heat transfer by radiation, with an unclothed person standing in a dark room whose ambient temperature is 22
SIZIF [17.4K]

Answer:

5.45\times 10^{-4} W

Explanation:

T_{r} = Temperature of the room = 22.0 °C = 22 + 273 = 295 K

T_{s} = Temperature of the skin = 33.0 °C = 33 + 273 = 306 K

A = Surface area = 1.50 m²

\epsilon = emissivity = 0.97

\sigma = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻² K⁻⁴

Rate of heat transfer is given as

R = \epsilon \sigma A (T_{s}^{2} - T_{r}^{2})

R = (0.97)(5.67\times 10^{-8}) (1.50) ((306)^{2} - (295)^{2})

R = 5.45\times 10^{-4} W

3 0
3 years ago
A ray of light travels from a glass-to-liquid interface at an angle of 35.0º. Indices of refraction for the glass and liquid are
UkoKoshka [18]

Answer:

The angle of refraction for the ray moving through the liquid is = 32.3°

Explanation:

Refractive index of liquid (n₁/n₂) = sini/sinr

∴                 n₁/n₂ = sini/sinr ................ equation 1

n₁ = index of refraction for glass, n₂ = index of refraction for liquid

Where i = incident angle of the first medium, r = angle of refraction or angle in the second medium.

Since the light ray is traveling from glass - to - liquid,  the first medium is glass and the second medium is liquid. and the refractive index will be that liquid with respect to glass.

using the equation,

n₁/n₂ = sini/sinr

i = 35° , n₁ = 1.52, n₂= 1.63

Making sinr the subject of the equation above,

sinr = sini/(n₁/n₂)

sinr = sin35(1.52)/1.63

sinr =0.574(1.52)/1.63

sinr = 0.535

Taking the sin inverse of both side of the equation

sin⁻¹(sinr) = sin⁻¹(0.535)

∴ r = 32.3°

The angle of refraction for the ray moving through the liquid is = 32.3°

The right option is (b). 32.3°

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A reaction in a solution would proceed slowest when -
Zinaida [17]

the answer is A *but I am not sure*

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Find the value of a x , the x-component of the object's acceleration.
jok3333 [9.3K]
To find the x component use the following formula, where Ф = theta = the angle 'a' makes with the x axis.

a_x = a*cos(theta)

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