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Paraphin [41]
3 years ago
11

The specific gravity of mantle rock is about 3.3. True False

Physics
1 answer:
Alla [95]3 years ago
5 0
True is the correct answer
You might be interested in
What force is necessary to accelerate a 2500 kg care from rest to 20 m/s over 10 seconds?
EleoNora [17]
Force = mass x acceleration
force = 2500kg x (20m/s / 10m/s)
force = 2500kg x 2m/s^2
force = 5000kg m/s^2 = 5kN

i hope this is right (^^)
4 0
3 years ago
A 12-kg piece of metal displaces 1.6 L of water when submerged. Part A Find its density. Express your answer to two significant
Tatiana [17]

Answer:

ρ = 7500 kg/m³

Explanation:

Given that

mass ,m = 12 kg

Displace volume ,V= 1.6 L

We know that

1000 m ³ = 1 L

Therefore V= 0.0016 m ³

When metal piece is fully submerged

We know that

mass = Density x volume

m=\rho \times V

Now by putting the values in the above equation

\rho=\dfrac{12}{0.0016}\ kg/m^3

ρ = 7500 kg/m³

Therefore the density of the metal piece will be  7500 kg/m³.

6 0
3 years ago
Question 1 of 10
ivanzaharov [21]
The answer would be 46.482 because you multiply 18.3 by 2.54 because for every inch you get 2.54 centimeters
6 0
2 years ago
Read 2 more answers
4:list one food chain that is part of the food web.
Sophie [7]
4. Grass - Caterpillar - Hedgehog - Fox
5. Caterpillar, Rabbit, Mouse.
6. Cougar and Fox.
7. Bacteria
8. The bird, hedgehog, Fox and cougar would be effected since the Hedgehogs and birds would soon die out due to the loss of their food. Once they die out, the cougar and Fox would have no predators left to eat.
6 0
3 years ago
An electron and a 0.033 0-kg bullet each have a velocity of magnitude 495 m/s, accurate to within 0.010 0%. Within what lower li
lara31 [8.8K]

Answer:

1.170*10^-3 m

3.23*10^-32 m

Explanation:

To solve this, we apply Heisenberg's uncertainty principle.

the principle states that, "if we know everything about where a particle is located, then we know nothing about its momentum, and vice versa." it also can be interpreted as "if the uncertainty of the position is small, then the uncertainty of the momentum is large, and vice versa"

Δp * Δx = h/4π

m(e).Δv * Δx = h/4π

If we make Δx the subject of formula, by rearranging, we have

Δx = h / 4π * m(e).Δv

on substituting the values, we have

for the electron

Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 4.95*10^-2

Δx = 6.63*10^-34 / 5.67*10^-31

Δx = 1.170*10^-3 m

for the bullet

Δx = (6.63*10^-34) / 4 * 3.142 * 0.033*10^-31 * 4.95*10^-2

Δx = 6.63*10^-34 / 0.021

Δx = 3.23*10^-32 m

therefore, we can say that the lower limits are 1.170*10^-3 m for the electron and 3.23*10^-32 for the bullet

7 0
3 years ago
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