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xz_007 [3.2K]
3 years ago
5

What will the resistance be for a lamp that draws 4.6 amps of current from a 120-volt outlet? A. 550 ohms B. 115 ohms C. 26 ohms

D. 0.038 ohms E. 5.7 ohms
Physics
2 answers:
Neporo4naja [7]3 years ago
5 0
To answer the question above, use the equation,
                                    V = I x R
where V is the potential difference or voltage, I is current, and R is the resistance. Substituting the known values,
                                  120 V = (4.6 amps) x R
The value of R is 26.09 ohms. Therefore, the answer is letter C. 
Paraphin [41]3 years ago
3 0
R = V/I = 120 / 4.6 ≈ 26 ohms.

Option C.
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Mkey [24]

Answer : Celestial or azimuth - altitude

Explanation :  

Celestial : The celestial coordinates that are analogous to longitude and latitude are called RA and Dec.

RA = Right Ascension

Dec = Declination

RA is the measured in unit of time and Dec is measured in degree.

The equatorial coordinate system is the projection of the latitude and longitude coordinate system on the celestial sphere.

Azimuth - altitude :  Azimuth - altitude define the location of an object in the sky.

The altitude is the distance of an object appears to be above the horizon.

The Azimuth of the object is the angular distance  along the horizon to the location of the object.

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3 years ago
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If the length of a ramp (an inclined plane) is 12 ft, and it rises 2 ft, what is its MA?
dangina [55]
The MA is 6! Hope This Helps!
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3 years ago
Say you want to make a sling by swinging a mass M of 1.7 kg in a horizontal circle of radius 0.048 m, using a string of length 0
DENIUS [597]

Answer:

Tension in the string is equal to 58.33 N ( this will be the strength of the string )

Explanation:

We have given mass m = 1.7 kg

radius of the circle r = 0.48 mF=\frac{mv^2}{r}=\frac{1.7\times 4.05^2}{0.48}=58.33N

Kinetic energy is given 14 J

Kinetic energy is equal to KE=\frac{1}{2}mv^2

So \frac{1}{2}\times 1.7\times v^2=14

v^2=16.47

v = 4.05 m/sec

Centripetal force is equal to F=\frac{mv^2}{r}=\frac{1.7\times 4.05^2}{0.48}=58.33N

So tension in the string will be equal to 58.33 N ( this will be the strength of the string )

3 0
3 years ago
A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p
kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2

V(t) = V_0 + a * t,

Where y_0 its our initial height, V_0  our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

y(t) =  38 \frac{m}{s} * t - \frac{1}{2} g t^2

V(t) = 38 \frac{m}{s} - g * t

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

0 m = 38 \frac{m}{s} - g * t

and obtain:

t = 38 \frac{m}{s} / g

t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}

t = 3.9 s

Plugin this time on our first equation we find:

y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2

y=73.67 m

6 0
3 years ago
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Kobotan [32]

Answer:

v_y = v_{oy} - g t

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Explanation:

In this exercise you are asked to observe the change in velocity in a projectile launch.

If we assume that the friction force is small, the velocity in the x-axis must be constant

         vₓ = v₀ₓ

Therefore, the arrow (red) that represents this movement must not change in magnitude.

In the direction of the y axis, the acceleration of gravity is acting, so the magnitude of the velocity in this axis changes

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where the upward direction is positive, so the arrow represents this speed (blue) must decrease, reach zero and grow in a negative direction as time progresses

7 0
3 years ago
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