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mr_godi [17]
3 years ago
5

A heating element on an electric range operating at 200. v has a resistance of 32.0 ohms. the current drawn by the element is

Physics
1 answer:
LenaWriter [7]3 years ago
4 0

Answer:

6.25 A

Explanation:

We cant find the current drawn by the element by using Ohm's law:

V=RI

where

V is the potential difference across the resistance of the element

R is the resistance

I is the current

In this problem, we know

V = 200 V

R=32.0 \Omega

Therefore we can re-arrange the equation and solve for the current:

I=\frac{V}{R}=\frac{200}{32}=6.25 A

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A 28 kg student stand on the surface of the Earth. What is the magnitude of the
Contact [7]

Answer:

F = 274.68[N]

Explanation:

The gravitational force is equal to the weight of a body, or this case that of a person. Weight can be calculated by means of the product of mass by gravitational acceleration. In this way we have the following equation:

F=m*g

where:

F = force or weight [N]

m = mass = 28 [kg]

g = gravity acceleration = 9.81 [m/s²]

Now replacing:

F=28*9.81\\F=274.68[N]

4 0
3 years ago
a wave in the ocean has an amplitude of 3m. Winds pick up suddenly, increasing the wave's amplitude to 6m. How does this change
Paha777 [63]

Answer

Hi,

An increase in amplitude from 3m to 6 m increases the energy it transports. The frequency of the wave is not affected

Explanation

Amplitude is the height of a wave where as frequency is the number of waves that pass by each second. A wave with bigger amplitude has more energy than a wave with smaller amplitude. A point where more waves pass contains more energy that is transferred every second. The change in the amplitude of a wave does not change its frequency. However, frequency is inversely related to the wavelength of a wave.

Best Wishes!

7 0
3 years ago
PLEASE HELP! <br><br> how am i supposed to label a graph using the points from the protractor?
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6 0
2 years ago
A 5 kg wooden block sitson a flat straight-away12 meters fromthe bottom of an infinitely long ramp, which has an angle of 20 deg
saveliy_v [14]

Answer:

(a) 19.71801m/s Velocity just before going up the ramp.

(b) 74.56338m.

Explanation:

We will solve it in two parts, first we will calculate time that 5kg wooden block would take to just reach ramp and with this time we will calculate final velocity that the wooden block would have in this time.

Second, we will calculate the component of velocity vector along inclined plane and the time that it would take for velocity to be 0 meters/s then with this time we will calculate the distance that inclined plane would travel along inclined plane.

Following formulas will be used.

                                  x(t) = \frac{1}{2} t^2 = 12m =16.2m/s^2 t^2

                                 F =ma

                                 V(t) = V_{o} +at

                                 x(t) = x_{0} +v_{0}t+\frac{1}{2}a t^2

(a) Calculating velocity right before going up the ramp.

 Wooden block is going on a straightaway and has net for on it.

         F_{n} =F-F_{s} = F-uF_{n}  = 100N-0.4*9.8m/s^2*5kg =81N

     and this force produces acceleration of

      a = \frac{F}{m}=\frac{81}{5} =16.2m/s^2 .

With this acceleration, wooden block would reach at the foot of ramp in.

          x(t) = 12m = 16.2m/s^2*t^2

         t = 1.217s

and final velocity will be

v(t) = v_{0}+at = 0+16.2m/s^2*1.2171s = 19.7180m/s.

this velocity of wooden box just before going up the ramp.

(b) How far up the ramp will the wooden block go before stopping.

Ramp is at 20° relative to horizontal therefore velocity along the ramp that the wooden block would have will be.

                              V= V_{h}cos(20) = 18.5288m/s

and deceleration along the ramp is

                              a = \frac{F_{s} }{m}

 Where F_{s} force of friction along the inclined plane.

F_s =  uF_n = u*m*a

a = 9.8m/s^2*cos(20) = 9.2089m/s^2

is a component of g along normal of the inclined plane.

                               F_{s} = 0.25*5kg*9.2089m/s^2

                              = 11.5112N

                              a = \frac{11.5112N}{5kg} = 2.3022m/s^2

And with this deceleration time needed to get wooded block to stop is.

                     v(t) = v_o-at = 18.5288m/s-2.3022m/s^2*t = 0

                        t = \frac{18.5288m/s}{2.3022m/s^2} =8.04813s

 and in that time wooden block would travel

   x(8.04813s) = 18.52881m/s *8.04813s-\frac{1}{} 2.3022m/s^2*(8.0481)^2=74.56338m

This is how up wooden box will go before coming to stop.

3 0
3 years ago
How does the energy possessed by a ball bearing change as it travels along an incline ramp?
ruslelena [56]
Kinetic energy increases, potencial energy decreases,
kinetic energy + potential energy = energy, energy can not be destroyed, just transformed


7 0
3 years ago
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