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____ [38]
3 years ago
5

A 12-kg piece of metal displaces 1.6 L of water when submerged. Part A Find its density. Express your answer to two significant

figures and include the appropriate units.
Physics
1 answer:
Tatiana [17]3 years ago
6 0

Answer:

ρ = 7500 kg/m³

Explanation:

Given that

mass ,m = 12 kg

Displace volume ,V= 1.6 L

We know that

1000 m ³ = 1 L

Therefore V= 0.0016 m ³

When metal piece is fully submerged

We know that

mass = Density x volume

m=\rho \times V

Now by putting the values in the above equation

\rho=\dfrac{12}{0.0016}\ kg/m^3

ρ = 7500 kg/m³

Therefore the density of the metal piece will be  7500 kg/m³.

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A boxer can hit a heavy bag with great force. Why can't he hit a piece of tissue paper in midair with the same amount of force?
12345 [234]

Answer:

This is due to impulse

Explanation:

Impulse equal to mΔv and FΔt

You can set these equal as mΔv = FΔt

When a boxer punches a tissue, it is like punching a cushion or a pillow. The time that the hit takes is much grater than if they were to hit something solid. In addition, the change in velocity of the boxer's arm would be much greater when they hit a punching bag. In this equation, the greater the time, the less force that is needed.

6 0
3 years ago
An electron has an initial velocity of (19.0 j + 18.0 k) km/s and a constant acceleration of (3.00 ✕ 1012 m/s2)i in a region in
asambeis [7]

Answer:

E=(-17.08 i +7.2 j -7.6 k )N/C

Explanation:

v= (19.0j+18.k)km/s

a=3.0x10^{13}m/s^2

\beta= 400x10{-6}T

Electron information needed to solve the question:

m_e=9.11x10^{-31}kg

q=-1.6x10{-19}C

F=F_E+F_B=q*(E+Vx \beta)

F=m*a

m*a=q*(E+Vx\beta)

E=\frac{m*a}{q}-(Vx\beta )

E=\frac{9.11x10{-31}kg*3.0x10^{12}m/s^2}{-1.6x10{-19}C}-[(19.0x10^3mj+18.0x10^3m)xi(400x10^{-6}T)]

E=-i17.08N/C-[7.6(-k)+7.2(j)]N/C

E=(-17.08 i +7.2 j -7.6 k )N/C

6 0
2 years ago
Object A of mass 0.70 kg travels horizontally on a frictionless surface at 20 m/s. It collides with object B, which is initially
AnnZ [28]

Answer:

25.71 kgm/s

Explanation:

Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.

Given that K₂ = 0.7K₁

1/2mv₂² = 0.7(1/2mv₁²)

v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s

Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.

Th magnitude of object A's momentum change is thus 25.71 kgm/s

6 0
2 years ago
How strongly the planet you're on pulls on you is your
ladessa [460]
Is the answer you are looking for Gravity? Gravity is what pulls us down to earth.
4 0
2 years ago
Read 2 more answers
It is estimated that uranium is relatively common in the earth's crust, occurring in amounts of 4 g / metric ton. A metric ton i
timama [110]

Answer:

The mass of Uranium present in a 1.2mg sample is 4.8 \cdot 10^{-6}\,mg

Explanation:

The ration between Uranium mass and total sample mass is: \frac{4g}{1000kg} =\frac{4g}{1000000g}=\frac{1}{250000}

For a sample of mass 1.2 mg, the amount of uranium is:

1.2\, mg \cdot \frac{1}{250000}=4.8 \cdot 10^{-6}\,mg

7 0
3 years ago
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