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dimulka [17.4K]
2 years ago
8

Matter is anything that has mass and takes up space. Which units of measurement can you use to describe these two properties?

Physics
1 answer:
Papessa [141]2 years ago
4 0

Answer: Matter is anything that has mass and takes up space . Mass is amount of mass that an object contains. Mass is measured by scale . Volume is amount of space that matter takes up .

Explanation:

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94 points if you hurry <br><br><br><br> The phase of matter with the least kinetic energy:
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solid

Explanation:

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A frictionless pendulum is made with a bob of mass 19.7 kg. The bob is held at height = 0.934 meter above the bottom of its traj
Alika [10]

Answer:

265 J

Explanation:

Energy=PE+KE=mgh+ 0.5mv^{2} where KE is kinetic energy, PE is potential energy, m is the mass of an object, v is the speed, h is the height and g is acceleration due to gravity.

Substituting 19.7 Kg for mass, 0.934 for h, 2.93 for v and 9.81 for g then

Energy=19.7(9.81*0.934+0.5*2.93^{2})=265.063303\approx 265 J

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2 years ago
A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.77
Vikki [24]
The formula for the rotational kinetic energy is

KE_{rot} = \frac{1}{2}(number \ of\ propellers)( I)( omega)^{2}

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

I=m L^{2}=(2.67 \ kg) (0.777 \ m)^{2}  =2.07459 \ kgm^{2}

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

KE_{rot} =( \frac{1}{2} )(5)(2.07459 \ kgm^{2}) (60\ rad/s)^{2}

KE_{rot} =18,671.31 \ J

6 0
3 years ago
To determine the muzzle velocity of a bullet fired from a rifle, you shoot the 2.47-g bullet into a 2.43-kg wooden block. The bl
Elza [17]

Answer:

The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 2.47 g = 0.00247 kg

mass of the wooden block, m₂ = 2.43 kg

initial velocity of the wooden block, u₂ = 0

height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m

let the initial velocity of the bullet on leaving the gun's barrel = v₁

let final velocity of the bullet-wooden block system after collision = v₂

Apply the principle of conservation of linear momentum;

Total initial momentum = Total final momentum

m₁v₁ + m₂u₂ = v₂(m₁ + m₂)

0.00247v₁  + 2.43 x 0  =  v₂(2.43 + 0.00247)

0.00247v₁ = 2.4325v₂ -------(1)

The kinetic energy of the bullet-block system after collision;

K.E = ¹/₂(m₁ + m₂)v₂²

K.E = ¹/₂ (2.4325)v₂²

The potential energy of the bullet-block system after collision;

P.E = mgh

P.E = (2.4325)(9.8)(0.00295)

P.E = 0.07032

Apply the principle of conservation of mechanical energy;

K.E = P.E

¹/₂ (2.4325)v₂² = 0.07032

1.21625 v₂²  = 0.07032

v₂²  = 0.07032  / 1.21625

v₂² = 0.0578

v₂ = √0.0578

v₂ = 0.24 m/s

Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;

0.00247v₁ = 2.4325v₂

0.00247v₁ = 2.4325 (0.24)

0.00247v₁ = 0.5838

v₁ = 0.5838 / 0.00247

v₁ = 236.36 m/s

Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

5 0
2 years ago
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