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3 years ago

9

Tony ran 600 meters in 60 seconds. What was Tony's speed during the race?

2 answers:

Ray Of Light [21]3 years ago

7 0

Tony's speed during the race was 10

NARA [144]3 years ago

4 0

10 meters per second.

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How do physicists distinguish one type of electromagnetic wave from another?

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A wave is characterized by the cyclic occurrences of crests and troughs. Wavelengthis defined as the distance between two consecutive troughs or two crests and the Frequency is defined as the number of cycles that pass through a point per second

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3 years ago

A 12-volt automotive circuit has a current of 3 amps. Technician A says the electric power in this circuit is 36 watts. Technici

skelet666 [1.2K]

Answer:

Technician A is right.

Explanation:

Given that,

Voltage of circuit, V = 12 volt

Current in the circuit, I = 3 A

Technician A says the electric power in this circuit is 36 watts. Technician B says the electric power in this circuit is 4 watts. We need to say that which technician is correct.

The power of any circuit is given by :

$P=V\times I$

$P=12\ V\times 3\ A$

P = 36 watts

So, technician A is right. Hence, this is the required solution.

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2 years ago

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$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

The net force acting on the block is ~

$\qquad \sf \dashrightarrow \: 10 + 5$

$\qquad \sf \dashrightarrow \: 15 \:N$

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$\boxed{ \sf15} \: \boxed{ \sf N}$

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2 years ago

Which of the following is likely to contribute to geological events that take place on Earth?

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3 years ago

Read 2 more answers

A single loop of nickel wire, lying flat in a plane, has an area of 7.40 cm^2 and a resistance of 2.40 Ω. A uniform magnetic fie

ale4655 [162]

Explanation:

It is given that,

Area of nickel wire, $A=7.4\ cm^2=7.4\times 10^{-4}\ m^2$

Resistance of the wire, R = 2.4 ohms

Initial value of magnetic field, $B_1=0.5\ T$

Final magnetic field, $B_2=3\ T$

Time, t = 1.12 s

Let I is the induced current in the loop of wire over this time. Te emf induced in the wire is given by Faraday's law as :

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{d(BA)}{dt}$

$\epsilon=-A\dfrac{d(B)}{dt}$

$\epsilon=-A\dfrac{B_2-B_1}{t}$

$\epsilon=-7.4\times 10^{-4}\times \dfrac{3-0.5}{1.12}$

$\epsilon=1.65\times 10^{-3}\ V$

Induced current in the loop of wire is given by :

$I=\dfrac{\epsilon}{R}$

$I=\dfrac{1.65\times 10^{-3}}{2.4}$

$I=6.87\times 10^{-4}\ A$

So, the induced current in the loop of wire over this time is $6.87\times 10^{-4}\ A$ . Hence, this is the required solution.

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3 years ago