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Elanso [62]
2 years ago
5

Find the weight of an object of mass 5 kg

Physics
2 answers:
san4es73 [151]2 years ago
4 0

Answer:

1

For instance, on Earth, a 5.0-kg object weighs 49 N; on the Moon, where g is 1.67 m/s2, the object weighs 8.4 N.

2

The weight of the object on the surface of earth would be 5kg. The weight of the object on the surface of moon would be 5/6= 0.83 as the weight of any object on earth is six times than that on moon.

Elis [28]2 years ago
3 0

Answer:

weight on earth is mg

which is 5*9.8

49 Newton

weight on moon is 1/6 th of weight on earth

1/6*49

8.166 Newton..

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The volume of a container is reported as 3.49 ft3. what is the volume in units of in3?
Kobotan [32]

The volume in units of 6.03x10^3 in3

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So to convert foot into inches we have to multiply by 12

The volume of container = 3.49ft^3

=3.49×12×12×12

= 6030.72 in^3

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7 0
2 years ago
A Cannonball is shot at an angle of 35.0 degrees and is in flight for 11.0 seconds before hitting the ground at the same height
nikdorinn [45]

Answer:

Explanation:

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Time\ of\ Flight = \frac{2vsin(x)}{g}

vsin(x) = 11 * 9.8 / 2 = 53.9 m/sec

(A) v (Initial velocity) = 11 * 9.8 / 2 * sin(35) = 94.56 m/sec

Maximum Height = \frac{(vsinx)^{2} }{2g}

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(C) Horizontal Range = 94.56 * 0.81 * 11 = 842.52 m

4 0
3 years ago
A closely wound, circular coil with radius 2.50 cmcm has 740 turns. Part A What must the current in the coil be if the magnetic
Vika [28.1K]

Answer:

The current in the coil is 4.086 A

Explanation:

Given;

radius of the circular coil, R = 2.5 cm = 0.025 m

number of turns of the circular coil, N = 740 turns

magnetic field at the center of the coil, B = 0.076 T

The magnetic field at the center of the coil is given by;

B = \frac{N\mu_o I}{2R}

where;

μ₀ is permeability of free space = 4 x 10⁻⁷ m/A

I is the current in the coil

R is radius of the coil

N is the number of turns of the coil

The current in the circular coil is given by

B = \frac{N\mu_o I}{2R} \\\\I = \frac{2BR}{N\mu_o} \\\\I =\frac{2*0.076*0.025}{740*4\pi*10^{-7}} \\\\I = 4.086 \ A

Therefore, the current in the coil is 4.086 A

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