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inna [77]
3 years ago
10

Calculate the density 1.33*10⁻⁷ gcm⁻³ into kgm⁻³.

Physics
1 answer:
Alex Ar [27]3 years ago
5 0

1 Kg = 1,000g

Therefore 1.33 × 10^-7 g = 1.33 × 10^-10 kg

1m³ = 1,000,000cm³

Therefore 1 cm³ = 1 × 10^-6 m³

Dividing the mass per unit volume you get:

(1.33 × 10^-10 kg) ÷ (1 × 10^-6 m³)

= 1.33 × 10^(-10--6) = 1.33 × 10^(-10 + 6) = 1.33 × 10^-4 kg/m³

Density = 1.33 × 10^-4 kg/m³

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Why is weathering slow in cold, dry places?
Alexxandr [17]

Rate of weathering depends on temperature and moisture. Cold dry places have less water to weather things


Hope this helped!

Don't forget: Mark Brainliest!

Have a good day :)

6 0
3 years ago
We all have a tendency to make illusory correlations from time to time. Try to think of an illusory correlation that is held by
Artist 52 [7]

Answer:

Thats a personal question

Explanation:

It asks about you

(illusion is something you think is there but is not)

5 0
2 years ago
A star has an absolute magnitude of 4 and a surface temperature of 5,000 degrees C. According to the HR diagram, list the type o
Ierofanga [76]
On sources it says it would  just be the super giant star 
3 0
3 years ago
A ball of moist clay falls 17.3 m to the ground. It is in contact with the ground for 24.0 ms before stopping. (a) What is the a
gizmo_the_mogwai [7]

Answer:

Acceleration,  767.08\ m/s^2

Explanation:

Given that,

Height from a ball falls the ground, h = 17.3 m

It is in contact with the ground for 24.0 ms before stopping.

We need to find the average acceleration the ball during the time it is in contact with the ground.

Firstly, find the velocity when it reached the ground. So,

v^2=u^2+2ah

u = initial velocity=0 m/s

a = acceleration=g

v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 17.3} \\\\v=18.41\ m/s

It is in negative direction, u = -18.41 m/s

Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.

a=\dfrac{v-u}{t}\\\\a=\dfrac{0-18.41}{24\times 10^{-3}}\\\\a=767.08\ m/s^2

So, the average acceleration of the ball during the time it is in contact is 767.08\ m/s^2.

4 0
3 years ago
Calculate the maximal friction force for a parked car between the rubber tires and a wet street. Assume the car’s mass is 1600 k
NikAS [45]

Answer:

Fr=12544N

Explanation:

1. Find the equation of eht maximal friction force:

The maximal friction force is given by the equation Fr=usmg, where μs is the static friction coefficient, m is the car´s mass and g is the gravitational force.

2. Replace values in the equation to find the answer:

Fr=0.8*1600kg*9.8\frac{m}{s^{2}}

Fr=12544N

5 0
3 years ago
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