All categories

3 years ago

An ice block measuring 5cm by 5cm by 5cm has a desity of 5g/cm3. What is the mass of the ice block?

2 answers:

4 0

Density = (mass) / (volume)

Volume of your block = (5cm x 5cm x 5cm) = 125 cm³

5 gm/cm³ = (mass) / (125 cm³)

Multiply each side by (125 cm³): Mass = 625 gm

Note:
I don't know what kind of frozen substance you're working with,
but maybe you ought to be careful handling it ... I know that's
not water ice you have there. The density of water ice is not
5 gm/cm³. In fact, it's a little less than 1 gm/cm³. That's why
the ice floats in your soda.

maksim [4K]3 years ago

4 0

5cm×5cm×5cm=125cm^3
125cm^3×5g/cm^3=625g
The cm^3 cancel out

You might be interested in

Hawks and gannets soar above the ground and, when they spot prey, they fold their wings and essentially drop like a stone. They

denis-greek [22]

Answer:

v = 54.2 m / s

Explanation:

Let's use energy conservation for this problem.

Starting point Higher

Em₀ = U = m g h

Final point. Lower

$Em_{f}$ = K = ½ m v²

Em₀ = Em_{f}

m g h = ½ m v²

v² = 2gh

v = √ 2gh

Let's calculate

v = √ (2 9.8 150)

v = 54.2 m / s

3 0

3 years ago

When Simon grows (let’s say he doubles his mass) what happens to his GPE when he is at the balcony?

DerKrebs [107]

Answer:

His gravitational potential energy will increase as well.

Explanation:

Let gpe represent gravitational potential energy.

gpe = mass × gravitational field strength × height

From the formula above, we can conclude that as the mass of a body increases, it's gpe increases too.

5 0

2 years ago

A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b

jonny [76]

Answer:

Part a)

$I = 3.16 \times 10^{-5} W/m^2$

Part b)

$P_o = 0.162 Pa$

Explanation:

Part a)

Level of sound = 75 dB

now we know that

$L = 10 Log\frac{I}{I_0}$

here we know that

$I_0 = 10^{-12} W/m^2$

now we have

$75 = 10 Log(\frac{I}{10^{-12}})$

$I = 3.16 \times 10^{-5} W/m^2$

Part b)

Intensity of sound wave is given as

$I = \frac{1}{2}\rho A^2\omega^2 c$

here we know that

$A = \frac{P_o}{Bk}$

so we have

$I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c$

$I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}$

now we know

$\rho = 1.2 kg/m^3$

$c = 340 m/s$

$B = 1.4 \times 10^5 Pa$

now we have

$3.16 \times 10^{-5} = \frac{1}{2}(1.2)P_o^2\frac{340^3}{(1.4\times 10^5)^2}$

$P_o = 0.162 Pa$

5 0

3 years ago

A Bullet Off mass 100 gm is fired From A Gun Off mass 5 Kg. If the backward velocity of the gun's 5 m / s, what is forward veloc

Elena L [17]

Answer:

250 m/s

Explanation:

The mass of the bullet, m₁ = 100 g = 0.1 kg

The mass of the gun, m₂ = 5 kg

The backward velocity of the gun, v₂ = -5 m/s

Given that the momentum is conserved, we have;

The total initial momentum = The total final momentum

The gun and the bullet are at rest, therefore, we have;

The initial momentum = 0

The total final momentum = m₁·v₁ + m₂·v₂

Where;

v₁ = The forward velocity of the bullet

Therefore, we get;

m₁·v₁ + m₂·v₂ = 0

0.1 kg × v₁ + 5 kg × (-5 m/s) = 0

0.1 kg × v₁ = 5 kg × 5 m/s

v₁ = (5 kg × 5 m/s)/(0.1 kg) = 250 m/s

The forward velocity of the bullet, v₁ = 250 m/s

6 0

3 years ago

1.) if you are sitting still are you accelerating? How do you know?

castortr0y [4]

yes. gravity is working on you and everything on you.

also earth is rotating

no not in respect ofimmediate surroundings

object could move at constant velocity = no change in either magnitude or direction

change of position "proves" this

5 0

3 years ago