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Savatey [412]
3 years ago
9

If the mass of a material is 45 grams and the volume of the material is 10 cm^3, what would the density of the material be

Physics
1 answer:
umka2103 [35]3 years ago
4 0
81.4/m3 is my answer

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Copper has a specific heat of 0.386 J/g°C. How much heat is required to increase 5.00 g of copper from 0.0°C to 10.0°C?
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The answer is 19.3 j
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A young hockey player stands at rest on the ice holding a 1.3-kg helmet. The player tosses the helmet directly in front of him w
navik [9.2K]

Answer

given,

initial speed of hockey player= 0 m/s

mass of the helmet, m = 1.3 Kg

initial speed of the helmet, u = 0 m/s

final speed of the helmet, v = 6 m/s

recoil speed of the hockey player, v' = 0.25 m/s

we need to calculate the mass of the hockey player, M = ?

using conservation of momentum

m u + M u' = M v' + m v

initial speed of ice skater is zero

1.3 x 0 + M x 0 =  M x (-0.25) + 1.3 x 6

negative sign is taken because recoil velocity is in opposite direction

0 = -0.25 M + 7.8

0.25 M = 7.8

M = 31.2 Kg

Hence, the mass of the young hockey player is equal to 31.2 Kg

5 0
3 years ago
Read the description of the centrioles. What is their function?
KATRIN_1 [288]

Answer:

Typically found in eukaryotic cells, centrioles are cylindrical (tube-like) structures/organelles composed of microtubules. In the cell, centrioles aid in cell division by facilitating the separation of chromosomes. For this reason, they are located near the nucleus.

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Find electric field at point p which is a distance l away from the both +q and -q
denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

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How does tripling the distance between two objects affect the gravitational force between them?
Wewaii [24]
It weakened the gravitational force between the oblects.
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