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2 years ago

What is the energy equivalent of an object with a mass of 1.83 kg?

2 answers:

6 0

To determine the energy equivalent of an object, we use the famous equation of Einstein which is E=mc^2 where m is the mass of the object and c is the speed of light (3x10^8 m/s). We calculate as follows:

E = mc^2
E = 1.83 kg (3x10^8 m/s)^2
E = 1.647x10^17 J

Flauer [41]2 years ago

6 0

Answer:

1.65 x 10^17 J

Explanation:

took it on edge

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What happens to the pressure in all parts of a confined fluid if the pressure in one part is increased?

Sauron [17]

Answer:

The pressure will be transmitted equally to all other parts of the confined fluid causing a general increase in pressure throughout the container.

Explanation:

This is in line with pascal's law of pressure which states that the pressure exerted on a given mass of fluid is transmitted undiminished to other parts of the fluid.

4 0

3 years ago

If a wave has a frequency of 74 Hz and a speed of 444 m/s, what is the wave's wavelength?

DaniilM [7]

Answer:

6 m

Explanation:

velocity = wavelength x frequency

444 m/s = wavelength x 74 Hz

444 m/s / 74 Hz = wavelength

wavelength = 6m

4 0

2 years ago

A charged particle of mass 0.0050 kg is subjected to a 5.0 T magnetic field which acts at a right angle to its motion. If the pa

Irina18 [472]

Answer:

0.01 C

Explanation:

Applying,

F = qvBsinФ................ Equation 1

Where F = Force on the charged particle, q = charge on the particle, v = velocity, B = magnetic field, Ф = angle

Since the charged particle noves in a circle,

F = mv²/r................. Equation 2

Where m = mass of the particle, v = velocity of the particle, r = radius of the circle

Substitute equation 2 into equation 1

mv²/r = qvBsinФ

make q the subject of the equation

q = mv/(rBsinФ)............. Equation 3

Given: m = 0.005 kg, v = 2 m/s, r = 0.2 m, B = 5 T, Ф = 90° (Act at right angle)

Substitute these values into equation 3

q = (0.005×2)/(0.2×5×sin90°)

q = 0.01/(1)

q = 0.01 C

5 0

2 years ago

A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of

34kurt

Answer:

The centripetal acceleration of the runner is $1.73\ m/s^2$ .

Explanation:

Given that,

A runner completes the 200 m dash in 24.0 s and runs at constant speed throughout the race. We need to find the centripetal acceleration as he runs the curved portion of the track. We know that the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$