Someone pls help……….

Good Afternoon my son has a science project the category is Physical science can you'll give me science Ideas he can do I'm New

Naddik [55]

As a experiment you could use is The affect of light in plants. A study He can do is maybe about Gravity affects and how it works on weight. Also you can do a all time favorite the Mentos and Diet Coke Experiment.

Hopefully one of these ideas will inspire you!

7 0

4 years ago

Read 2 more answers

A 0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of co

Genrish500 [490]

Answer: When 1.0kg of aluminium block is used, the final temperature of the mixture will be T = 36.2∘C

If 1.0kg copper block is used, T of the mixture will be = 17.4∘C

If 100g (0.1kg) of ice at 0∘C is used, T will be = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be= 147.1∘C

Explanation:

H = mcΘ

heat lost by block = heat gained by water

m₁c₁Θ₁ = m₂c₂Θ₂ where m₁ is mass of aluminium, m₂ is mass of water, c₁ is cAluminium, c₂ is cWater, Θ₁ is temperature change for aluminium, Θ₂ is temperature change for water.

0.5*900*(200-20) = m₁*4186*(20-0)

m₁ = 450*180/83270

m₁ = 0.973kg

when 1.0kg of aluminium block is used, the final temperature of the mixture will be T

heat lost by block = heat gained by water

1.0*900*(200-T) = 0.973*4186*(T-0)

180000 - 900T = 4073T

4973T = 180000

T = 180000/4973 = 36.2∘C

If 1.0kg copper block is used, T of the mixture will be

heat lost by block = heat gained by water

1.0*387*(200-T) = 0.973*4186*(T-0)

77400 - 387T = 4073T

4460T = 77400

T = 77400/4460 = 17.4∘C

If 100g (0.1kg) of ice at 0∘C is used, T will be

heat lost by block = heat gained by water + heat used in melting ice to form water at 0∘C

heat used in melting 0.1kg of ice, H = ml, where l= 33600J/Kg

0.5*900*(200-T) = 0.1*4186*(T-0) + 0.1*33600J/Kg

90000 - 450T = 418.6T + 33600

418.6T + 450T = 90000 - 33600

868.6T = 56400

T = 56400/868.6 = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be

0.5*900*(200-T) = 0.025*4186*(T-0) + 0.025*33600J/Kg

90000 - 450T = 104.65T + 8400

104.65T + 450T = 90000 - 8400

554.65T = 81600

T = 81600/554.65 = 147.1∘C

7 0

3 years ago

How does applying oil to a bicycle help the bicycle to work better?

dmitriy555 [2]

It Reduces Friction. !

8 0

3 years ago

Read 2 more answers

The center of mass of a 0.30-kg (non-uniform) meter stick is located at its 45-cm mark. What is the magnitude of the torque (in

lbvjy [14]

Answer:

The magnitude of the torque due to gravity if it is supported at the 28-cm mark is 0.5 N-m.

Explanation:

Given that,

Mass of the meter stick, m = 0.3 kg

Center of mass is located at its 45 cm mark.

We need to find the magnitude of the torque due to gravity if it is supported at the 28-cm mark. Torque acting on the object is given by :

$\tau=r\times F\\\\\tau=(45-28)\times 10^{-2}\times 0.3\times 9.79\\\\\tau=0.499\ N-m$

or

$\tau=0.5\ N-m$

So, the magnitude of the torque due to gravity if it is supported at the 28-cm mark is 0.5 N-m.

6 0

4 years ago

If the clot breaks free and only experiences a force due to the blood pressure, what is the relationship between the pressure on

Bingel [31]

Answer:

The pressure on each side of the clot is same.

Explanation:

The clot is moving with a constant velocity which implies that the total force on the clot is zero.

The force F is given by the formula.

F=ΔPA

where ΔP is the difference in pressure and A is the area.

Since the force is same on either side, therefore,

ΔPA=0

Also, the area is same on both side of the clot.

Therefore the difference in the pressure of both side is zero which implies that the pressure on both sides are equal.

Check out similar questions.

brainly.com/question/14555505

#SPJ10

6 0

2 years ago