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3 years ago

A man weighing 490 N on earth weighs only 81.7 N on the moon. His mass on the moon is ____ kg. (Use g=9.8 m/s²)

Physics

1 answer:

PSYCHO15rus [73]3 years ago

4 0

Answer:

mass= 50kg

since the mass is constant everywhere

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Acceleration problem <br> Show work plz

Dennis_Churaev [7]

Answer:

The answer to your question is: vo = 25 m/s

Explanation:

data

a = -7.5 m/s²

d = 42 m

vf = 0 m/s

vo = ?

Formula

vf² = vo² - 2ad

Substitution

0² = vo² - 2(7.5)(42)

We clear vo from the equation

vo² = 2(7.5)(42)

vo² = 630 simplifying

vo = 25 m/s result

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3 years ago

Evaluate ( 64800 ms ) 2 to three significant figures and express the answer in Si units. Express your answer using three signifi

Sergeu [11.5K]

Answer:

42.0×10² second²

Explanation:

Here, time is given in milisecond

(64800 ms)²

= 4199040000 ms²

The SI unit is seconds

1 second = 1000 milisecond

$1\ milisecond=\frac{1}{1000}\ second$

$\\\Rightarrow 1\ milisecond^2=\left(\frac{1}{1000}\right)^2\ second^2$

$4199040000\ ms^2=4199040000\times \left(\frac{1}{1000}\right)^2\ second^2=4199.04\ second^2$

42.0×10² second²

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3 years ago

What do you understand by family health?<br>

AleksandrR [38]

Answer:

Family health provides, besides the cure of disease, various activities and programs in prevention, rehabilitation, and health promotion for all members of families/households. pls mark me branilest

3 0

3 years ago

Read 2 more answers

A new planet has been discovered and given the name Planet X . The mass of Planet X is estimated to be one-half that of Earth, a

harina [27]

Answer:

vₐ = v_c $( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )$

Explanation:

To calculate the escape velocity let's use the conservation of energy

starting point. On the surface of the planet

Em₀ = K + U = ½ m v_c² - G Mm / R

final point. At a very distant point

Em_f = U = - G Mm / R₂

energy is conserved

Em₀ = Em_f

½ m v_c² - G Mm / R = - G Mm / R₂

v_c² = 2 G M (1 /R - 1 /R₂)

if we consider the speed so that it reaches an infinite position R₂ = ∞

v_c = $\sqrt{\frac{2GM}{R} }$

now indicates that the mass and radius of the planet changes slightly

M ’= M + ΔM = M ( $1+ \frac{\Delta M}{M}$ )

R ’= R + ΔR = R ( $1 + \frac{\Delta R}{R}$ )

we substitute

vₐ = $\sqrt{\frac{2GM}{R} } \ \frac{\sqrt{1+ \frac{\Delta M}{M} } }{ \sqrt{1+ \frac{ \Delta R}{R} } }$

let's use a serial expansion

√(1 ±x) = 1 ± ½ x +…

we substitute

vₐ = v_ c ( $(1 + \frac{1}{2} \frac{\Delta M}{M} ) \ ( 1 - \frac{1}{2} \frac{\Delta R}{R} )$ )

we make the product and keep the terms linear

vₐ = v_c $( \ 1 + \frac{1}{2} ( \frac{\Delta M}{M} - \frac{\Delta R}{R}) \ )$

5 0

3 years ago

Identify the type of particle that would be given off by each of the following nuclear reactions:

dsp73

Answer:

c.Beta (1 e-) is the answer.

3 0

3 years ago