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3 years ago

Which is heavier sand or clay? then specify why

Physics

2 answers:

JulsSmile [24]3 years ago

8 0

I would say clay, because its a solid

Ksju [112]3 years ago

6 0

Clay is denser than the sand, and as density is directly proportional to the mass, "Clay" would be Heavier.

Hope this helps!

You might be interested in

A coin is placed on a vinyl stereo record that is making 33 1/3 revolutions per minute on a turntable. (a) In what direction is

mamaluj [8]

Answer: a) The acceletarion is directed to the center on the turntable. b) 5 cm; ac= 0.59 m/s^2; 10 cm, ac=1.20 m/s^2; 14 cm, ac=1.66 m/s^2

Explanation: In order to explain this problem we have to consider teh expression of the centripetal accelartion for a circular movement, which is given by:

ac=ω^2*r where ω and r are the angular speed and teh radios of the circular movement.

w=2*π*f

We know that the turntable is set to 33 1/3 rev/m so

the frequency 33.33/60=0.55 Hz

then w=2*π*0.55=3.45 rad/s

Finally the centripetal acceleration at differents radii results equal:

r= 0.05 m ac=3.45^2*0.05=0.50 m/s^2

r=0.1 ac=3.45^2*0.1=1.20 m/s^2

r=0.14 ac=3.45^2*0.14=1.66 m/s^2

4 0

3 years ago

Compare the catching of two different water balloons.

Stels [109]

Answer:

a. The volume V₁ and V₂

b. The case that involves the greatest momentum change = Case B

c. The case that involves the greatest impulse = Case B

d. b. The case that involves the greatest force = Case B

Explanation:

Here we have

Case A: V₁ = 150-mL, v₁ = 8 m/s

Case B: V₂ = 600-mL, v₁ = 8 m/s

a. The variable that is different for the two cases is the volume V₁ and V₂

b. The momentum change is by the following relation;

ΔM₁ = Mass, m × Δv₁

The mass of the balloon are;

Δv₁ = Change in velocity = Final velocity - Initial velocity

Mass = Density × Volume

Density of water = 0.997 g/mL

Case A, mass = 150 × 0.997 = 149.55 g

Case B, mass = 600 × 0.997 = 598.2 g

The momentum change is;

Case A: Mass, m × Δv₁ = 149.55 g/1000 × 8 m/s = 1.1964 g·m/s

Case B: Mass, m × Δv₁ = 598.2/1000 × 8 = 4.7856 g·m/s

Therefore Case B has the greatest momentum change

The case that has the gretest momentum change = Case B

c. The momentum change = impulse therefore Case B involves the greatest impulse

d. Here we have;

Impulse = Momentum change = $F_{average}$ × Δt = mΔV

∴ $F_{average}$ = m·ΔV/Δt

∴ For Case A $F_{average}$ = 149.55×8/Δt = 1196.4/Δt N

For Case B $F_{average}$ = 598.2×8/Δt = 4785.6/Δt

Where Δt is the same for Case A and Case B, $F_{average}$ for Case B >> $F_{average}$ for Case B

Therefore, Case B involves the greatest force.

4 0

2 years ago

If an object force of 50 N is used to move an object a distance of 20 m, what distance must the object be moved if the input for

steposvetlana [31]

Answer:

$\ d_{out} = 100 \ m.$

Explanation:

Given data:

$F_{in} = 50 \ \rm N$

$F_{out} = 10 \ \rm N$

$d_{in} = 20 \ m$

Let the distance traveled by the object in the second case be $d_{out}.$

In the given problem, work done by the forces are same in both the cases.

Thus,

$W_{in} = W_{out}$

$F_{in}.d_{in} = F_{out}.d_{out}$

$\Rightarrow \ d_{out} = \frac{F_{in}.d_{in}}{F_{out}}$

$\ d_{out} = \frac{50 \times 20}{10}$

$\ d_{out} = 100 \ m.$

5 0

2 years ago

If the electrical field in some region or space is zero, does that imply that there is no electrical charge in that region?

mrs_skeptik [129]

No
It means the resultant electrical charges had canceled each other out so there is no field to be sensed .

7 0

2 years ago

An open-topped freight car with mass 24,000 kg is coasting without friction along a level track. It is raining very hard, and th

skelet666 [1.2K]

Answer:

(a) v = 3..6 m/s

(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.

Explanation:

from the question we have the following:

mass of the car (Mc) = 24,000 kg

initial velocity of the car (u) = 4 m/s

mass of water (Mw) = 3000 kg

final velocity of the car (v) = ?

(a) we can calculate the final momentum of the car by applying the conservation of momentum where

initial momentum = final momentum

Mc x U = (Mc + Mw) x V

24000 x 4 = (24000 + 3000) x v

96,000 = 27000v

v =3.6 m/s

(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.

7 0

2 years ago