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olya-2409 [2.1K]
3 years ago
8

A cylinder has 500 cm3 of water. After a mass of 100 grams of sand is poured into the cylinder and all air bubbles are removed b

y a vacuum pump, the water level rises to 537.5 cm3. Determine the specific gravity of the sand.
Physics
1 answer:
Nina [5.8K]3 years ago
3 0

Answer: SG = 2.67

Specific gravity of the sand is 2.67

Explanation:

Specific gravity = density of material/density of water

Given;

Mass of sand m = 100g

Volume of sand = volume of water displaced

Vs = 537.5cm^3 - 500 cm^3

Vs = 37.5cm^3

Density of sand = m/Vs = 100g/37.5 cm^3

Ds = 2.67g/cm^3

Density of water Dw = 1.00 g/cm^3

Therefore, the specific gravity of sand is

SG = Ds/Dw

SG = (2.67g/cm^3)/(1.00g/cm^3)

SG = 2.67

Specific gravity of the sand is 2.67

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Answer:

Explanation:

Givens

m = 942

F = 6731

t = 21 seconds

vi = 0

vf = ?

Formula

F = m * (vf -  vi )  / t

Solution

6731 = 942*(vf - 0)/21          Multiply both sides by 21

6731 * 21 = 942*vf

141351 = 942*vf                   Divide by 942

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Which material is not a fluid? water cork gasoline air
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Explanation:

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1. A box contains 10 blue chips. 5 red chips, and 15 yellow chips. Find the odds of choosing the
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Answer:

Explanation:

Blue: 10/30

Red: 5/30

Yellow: 15/30

3 0
3 years ago
At a distance D from a very long (essentially infinite) uniform line of charge, the electric field strength is 1000 N/C. At what
Alla [95]

Answer:

The correct option is (a).

Explanation:

We know that, the E is inversely proportional to the distance as follows :

E=\dfrac{k}{d^2}

We can write it as follows :

\dfrac{E_1}{E_2}=(\dfrac{d_2}{d_1})^2

Put all the values,

\dfrac{1000}{2000}=(\dfrac{d_2}{d})^2\\\\\sqrt{\dfrac{1000}{2000}}=(\dfrac{d_2}{D})\\\\0.7071=\dfrac{d_2}{d}\\\\d_1=0.7071D\\\\d_1=\dfrac{D}{\sqrt2}

So, the correct option is (a).

5 0
3 years ago
Two long straight wires are parallel and 8.6 cm apart. They are to carry equal currents such that the magnetic field at a point
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Answer:

(a) The current should be in opposite direction

(b) The current needed is 39.8 A

Explanation:

Part (a)

Based, on right hand rule, the current should be in opposite direction

Part (b)

given;

strength of magnetic field, B = 370 µT

distance between the two parallel wires, d = 8.6 cm

B = \frac{\mu_oI}{2\pi R}

At the center, the magnetic field strength is twice

B_c = 2(\frac{\mu_oI}{2\pi R}) =\frac{ \mu_oI}{\pi R}

R = d/2 = 8.6/2 = 4.3 cm = 0.043 m

B_c = \frac{ \mu_oI}{\pi R}\\\\I = \frac{B_c\pi R}{\mu_o} = \frac{370 *10^{-6}* \pi *0.043}{4\pi *10^{-7}}\\\\I = 39.8 \ A

Therefore, current needed is 39.8 A

6 0
3 years ago
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