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Mashcka [7]
4 years ago
13

Can someone please help!

Physics
1 answer:
Ludmilka [50]4 years ago
8 0
It would be 24 candelas because since you have 6 lux and 4 meters all you do is multiply 6x4=24
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A discus thrower (Fig. P4.27, page 97) accelerates a discus from rest to a speed of 25.0 m/s by whirling it through 1.25 rev. As
Aleksandr-060686 [28]

The time interval needed for the disk to rev from leave to 25.0 m/s

= 0.57 s

<h3>What is the time interval?</h3>
  • A larger period of time can be split up into multiple shorter, equal-length segments.
  • These are referred to as time periods. Consider the scenario where you wished to gauge a car's speed over an hour-long trip. You could break up an hour into ten-minute segments.
  • In Hz, frequency is defined as the number of cycles per second. Simply divide 1 by the frequency to determine the time interval for a known frequency (e.g., a frequency of 100 Hz has a time interval of 1/(100 Hz) = 0.01 seconds; 500 Hz has a time interval of 1/(500Hz) = 0.002 seconds, etc.).

Given:

vo = 0

v = 25.4 m/s

r = 0.95 m

n = 1.21 rev

A.

r*ω = v

ω = v/r

= 25.4/.95

= 26.74 rad/s

B.

θ = 2πn

= 2π × (1.21)

= 7.6 m

Using equation of motion,

ωf^2 = ωi^2 + 2αθ

(26.74)^2 = 2 × α × (7.6)

α = 715.03/15.2

= 47.04 rad/s^2

C.

Using equation of motion,

θ = ωi*t + 1/2*α*t^2

θ = 0 + 1/2*α*t^2

7.6 = 47.04/2 × t^2

= sqrt(0.323)

= 0.57 s

Therefore the correct answer is 0.57 s

To learn more about time interval, refer to:

brainly.com/question/479532

#SPJ4

6 0
1 year ago
a body weights 28N at a height of 3200km from the earth surface.What will be the gravitational force on that body if its lies on
alekssr [168]

Answer:

The object would weight 63 N on the Earth surface

Explanation:

We can use the general expression for the gravitational force between two objects to solve this problem, considering that in both cases, the mass of the Earth is the same. Notice as well that we know the gravitational force (weight) of the object at 3200 km from the Earth surface, which is (3200 + 6400 = 9600 km) from the center of the Earth:

F_G=G\,\frac{M_E\,m}{d^2} \\28\,\,N=G\,\frac{M_E\,m}{9600000^2}

Now, if the body is on the surface of the Earth, its weight (w) would be:

F_G=G\,\frac{M_E\,m}{d^2} \\w=G\,\frac{M_E\,m}{6400000^2}

Now we can divide term by term the two equations above, to cancel out common factors and end up with a simple proportion:

\frac{w}{28} =\frac{9600000^2}{6400000^2} \\\frac{w}{28} =\frac{9}{4} \\\\ \\w=\frac{9\,*\,28}{4}\,\,\,N\\w=63\,\,N \\

4 0
3 years ago
A series RL circuit includes a 6.05 V 6.05 V battery, a resistance of R = 0.655 Ω , R=0.655 Ω, and an inductance of L = 2.55 H.
Ivenika [448]

Answer:

The induced emf 1.43 s after the circuit is closed is 4.19 V

Explanation:

The current equation in LR circuit is :

I=\frac{V}{R} (1-e^{\frac{-Rt}{L} })    .....(1)

Here I is current, V is source voltage, R is resistance, L is inductance and t is time.

The induced emf is determine by the equation :

V_{e}=L\frac{dI}{dt}

Differentiating equation (1) with respect to time and put in above equation.

V_{e}= L\times\frac{V}{R}\times\frac{R}{L}e^{\frac{-Rt}{L} }

V_{e}=Ve^{\frac{-Rt}{L} }

Substitute 6.05 volts for V, 0.655 Ω for R, 2.55 H for L and 1.43 s for t in the above equation.

V_{e}=6.05e^{\frac{-0.655\times1.43}{2.55} }

V_{e}=4.19\ V

5 0
3 years ago
What is most often given a value of zero to describe an object position on a srtaight line
Vsevolod [243]
Reference point is most often given a value of zero to describe an object's position on a straight line
6 0
3 years ago
Which solid-state component can be used as a switch to turn current on or off?
Papessa [141]
The answer is Transistor. Its a semiconductor device used to amplify or switch electronic signals and electrical power. It is composed of semiconductor material usually with at least three terminals for connection to an external circuit.
4 0
3 years ago
Read 2 more answers
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