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neonofarm [45]
3 years ago
15

How long will it take, in minutes, for a transformer to transfer 2.3 X 10^6 J of energy from a 120-V circuit to a 345-V circuit

with a current of 1.5 A?
Physics
1 answer:
masya89 [10]3 years ago
3 0

Answer:

e

Explanation:

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Please help! I especially need help with the second question but help with the first one would be most appreciated!
lara31 [8.8K]

Answer:

a) Team A will win.

b) The losing team will accelerate towards the middle line with 0.01 m/s^{2}

Explanation:

Given that Team-A pulls with a force , F_{1} = 50N

and Team-B pulls with a force , F_{1} = 45N

∵ F_{1} > F_{2}

The rope will move in the direction of force F_{1}.

∴ Team-A will win.

b) Considering both the teams as one system of total mass , m = 246+253 = 499 kg

Net force on the system , F = F_{1} - F_{2} = 50-45 = 5N

Applying Newtons first law to the system ,

F = ma , where 'a' is the acceleration of the system.

Since , both the teams are connected by the same rope , their acceleration would be the same.

∴ 5 = 499×a

∴ a = 0.01 m/s^{2}

4 0
3 years ago
What is the lowest level of waves on the electromagnetic energy spectrum?
Juli2301 [7.4K]
Radio waves are the lowest level of waves
3 0
3 years ago
The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away fr
WARRIOR [948]
In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
y_n= \frac{n \lambda D}{a} (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
\lambda is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit

In our problem, 
D=37.0 cm=0.37 m
\lambda=530 nm=5.3 \cdot 10^{-7} m
while the distance between the first and the fifth minima is
y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m (2)

If we use the formula to rewrite y_5, y_1, eq.(2) becomes
\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m
Which we can solve to find a, the width of the slit:
a= \frac{4 \lambda D}{0.5 \cdot 10^{-3} m}= \frac{4 (5.3 \cdot 10^{-7} m)(0.37 m)}{0.5 \cdot 10^{-3} m}=  1.57 \cdot 10^{-3} m=1.57 mm

7 0
3 years ago
Is baking soda less dense or more dense than water?
boyakko [2]
It’s more dense than air and less dense than liquid!
6 0
3 years ago
A car has a mass of 1000 kg and accelerates at 2 meters per second per second. what is the magnitude of the net force exerted on
9966 [12]
F=ma
Therefore the net force = 1000kg × 2 metres per second per second
So F=2000 N
4 0
3 years ago
Read 2 more answers
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