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kiruha [24]
3 years ago
6

How much would a person whose mass is 60kg weight on the moon​

Physics
1 answer:
Valentin [98]3 years ago
4 0
Approximately 10kg

Given that gravity on the Moon has approximately 1/6th of the strength of gravity on Earth, a man who weighs 60kg on Earth would weigh approximately 10kg on the Moon.
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An isotope has a half life of 2.5 minutes how many half -lives go by in 17.5 minutes
ale4655 [162]

Answer:

7 half-lives

Explanation:

7 0
4 years ago
What is the speed of a bird of mass 8kg which has kinetic energy of 8836J?
Kay [80]

Answer:

For the bird moving in a straight line, the kinetic energy is one-half the product of the mass and the square of the speed: Ek=12mu2.

4 0
3 years ago
1. You are designing a new solenoid and experimenting with material for each turn. The particular turn you are working with is a
svetoff [14.1K]

Answer:

Explanation:

1 ) Magnetic field due to a circular coil carrying current

= μ₀I / 2r

I is current , r is radius of the wire , μ₀ = 4π x 10⁻⁷

= 4π x 10⁻⁷ x 15 / (2 x 3.5 x 10⁻²)

= 26.9 x 10⁻⁵ T

2 )

Negative z direction .

The direction of magnetic field due to a circular coil having current is known  

with the help of screw rule or right hand thumb rule.

3 )

If we decrease the radius the magnetic field will:__increase _____.

A. Increase.

Magnetic field due to a circular coil carrying current

B = μ₀I / 2 r

Here r is radius of the coil . If radius decreases magnetic field increases.

So magnetic field will increase.

4 0
3 years ago
The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is:
ludmilkaskok [199]

Answer: 8.1 x 10^24

Explanation:

I(t) = (0.6 A) e^(-t/6 hr)

I'll leave out units for neatness: I(t) = 0.6e^(-t/6)

If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).

For neatness let k = 1/(6x3600) = 4.63x10^-5, then:

I(t) = 0.6e^(-kt)

Providing t is in seconds, total charge Q in coulombs is

Q= ∫ I(t).dt evaluated from t=0 to t=∞.

Q = ∫(0.6e^(-kt)

= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.

= -(0.6/k)[e^-∞ - e^-0]

= -0.6/k[0 - 1]

= 0.6/k

= 0.6/(4.63x10^-5)

= 12958 C

Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.

5 0
4 years ago
If half of the weight of a flatbed truck is supported by its two drive wheels, what is the maximum acceleration it can achieve o
Scilla [17]

Answer:

Maximum acceleration will be equal to 3.43m/sec^2

Explanation:

We have given coefficient of kinetic friction \mu _k=0.7

And coefficient of static friction \mu _s=1

Acceleration due to gravity g=9.8m/sec^2

When truck moves maximum force will be equal to F=\mu _kmg

It is given that half of the weight is supported by its drive wheels

So force required =\frac{\mu _kmg}{2}

From newtons law maximum acceleration will be equal to a=\frac{\frac{\mu _kmg}{2}}{m}=\frac{\mu _kg}{2}=\frac{0.7\times 9.8}{2}=3.43m/sec^2

8 0
4 years ago
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