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2 years ago

If an astronaut is 66 kg, what is their weight in N on the moon?

2 answers:

6 0

Mass isn’t dependent on gravity, so the man would still be 66
66
k
g
. His weight, however, is calculated using the formula:

=
w
=
m
g

So, weight, equal mass times the acceleration due to gravity. On the moon, acceleration due to gravity is 1.62/²
1.62
m
/
s
²
. If we plug these “givens” in, we get:

=(66)(1.62/²)
w
=
(
66
k
g
)
(
1.62
m
/
s
²
)

Therefore, weight equals 106.92
106.92
N
e
w
t
o
n
s
(the unit of weight)

Blababa [14]2 years ago

5 0

The mass of the person would be 66 kg on both Earth and the Moon.

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What is an independent or manipulated variable?

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Is the variable you change, independent, I, something I change.

5 0

3 years ago

Which of the following would most likely cause a decrease in the quality supplied

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AWhich of the following would most likely cause a decrease in the quantity supplied? A decrease in price.

4 0

2 years ago

A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the

mylen [45]

Answer:

Angular acceleration, is $708.07\ rad/s^2$

Explanation:

Given that,

Initial speed of the drill, $\omega_i=0$

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, $\omega_f=28940\ rev/min=3030.58\ rad/s$

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

$\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2$

So, the drill's angular acceleration is $708.07\ rad/s^2$ .

4 0

2 years ago

The isomerization of cyclopropane to form propene is a first-order reaction. At 760 K, 85% of a sample of cyclopropane changes t

Nataliya [291]

Answer:

so rate constant is 4.00 x 10^-4 $s^{-1}$

Explanation:

Given data

first-order reactions

85% of a sample

changes to propene t = 79.0 min

to find out

rate constant

solution

we know that

first order reaction are

ln [A]/[A]0 = -kt

here [A]0 = 1 and (85%) = 0.85 has change to propene

so that [A] = 1 - 0.85 = 0.15.

that why

[A] / [A]0= 0.15 / 1

[A] / [A]0 = 0.15

here t = (79) × (60s/min) = 4740 s

k = - {ln[A]/[A]0} / t

k = -ln 0.15 / 4740

k = 4.00 x 10^-4 $s^{-1}$

so rate constant is 4.00 x 10^-4 $s^{-1}$

3 0

2 years ago

Using hooke's law find the elastic constant of a spring that stretches 2 cm when 4newton force is applied to it

erica [24]

<u>Answer:</u>

2N/cm

<u>Step-by-step explanation:</u>

According to the Hooke's Law, the force required to extend or compress a spring is directly proportional distance you can stretch it, which is represented as:

$F=kx$