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MrRa [10]
1 year ago
6

How much mercury most placed inside the glass vessel of capacity 500 cc so that the volume of space unoccupied by mercury always

remains constant?
Physics
1 answer:
Tema [17]1 year ago
4 0

Answer: 500cc of mercury should be placed

Explanation:

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Suppose two astronauts on a spacewalk are floating motionless in space, 3.0 m apart. Astronaut B tosses a 15.0 kg IMAX camera to
marta [7]

Answer:

\frac{ 112.5}{15+m_{A}}=v_{f}

(we need the mass of the astronaut A)

Explanation:

We can solve this by using the conservation law of the linear momentum P. First we need to represent every mass as a particle. Also we can simplify this system of particles by considering only the astronaut A with an initial speed v_{iA} of 0 m/s and a mass m_{A} and the IMAX camera with an initial speed v_{ic} of 7.5 m/s and a mass m_{c} of 15.0 kg.

The law of conservation says that the linear momentum P (the sum of the products between all masses and its speeds) is constant in time. The equation for this is:

P_{i}=p_{ic}+p_{iA}\\P_{i}=m_{c}v_{ic}+m_{A} v_{iA}\\P_{i}=15*7.5 + m_{A}*0\\P_{i}=112.5 \frac{kg.m}{s}

By the law of conservation we know that P_{i} =P_{f}

For P_{f} (final linear momentum) we need to treat the collision as a plastic one (the two particles stick together after the encounter).

So:

P_{i} =P_{f}=112.5\\

112.5=(m_{c}+m_{A})v_{f}\\\frac{ 112.5}{m_{c}+m_{A}}=v_{f}\\\frac{ 112.5}{15+m_{A}}=v_{f}

3 0
1 year ago
On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he
Liula [17]

Answer:

correct is d) a ’= g / 2

Explanation:

For this exercise let's use the kinematics equations

On earth

      v = v₀ - a t

     a = (v₀- v) / T

On planet X

    v = v₀ - a' t’

    a ’= (v₀-v) / 2T

Let's substitute the land values ​​in plot X

     a’= a / 2

Now let's use Newton's second law

       W = ma

      m g = m a

      a = g

We substitute

      a ’= g / 2

So we see that on planet X the acceleration is half the acceleration of Earth's gravity

4 0
1 year ago
An 1876 N crate is being pushed across a level force at a constant speed by a force of 747 N. What is the coefficient of kinetic
nekit [7.7K]

The crate only moves horizontally, so its net vertical force is 0. The only forces acting in the vertical direction are the crate's weight (pointing downward) and the normal force of the surface on the crate (pointing upward). By Newton's second law, we have

∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0   →   <em>n</em> = <em>mg</em> = 1876 N

where <em>n</em> is the magnitude of the normal force.

In the horizontal direction, the crate is moving at a constant speed and thus with no acceleration, so it's completely in equilibrium and the net horizontal force is also 0. The only forces acting on it in this direction are the 747 N push (pointing in the direction of the crate's motion) and the kinetic friction opposing it (pointing in the opposite direction). By Newton's second law,

∑ <em>F</em> (horizontal) = 747 N - <em>f</em> = 0   →   <em>f</em> = 747 N

The frictional force is proportional to the normal force by a factor of the coefficient of kinetic friction, <em>µ</em>, such that

<em>f</em> = <em>µn</em>   →   <em>µ</em> = <em>f</em> / <em>n</em> = (747 N) / (1876 N) ≈ 0.398188 ≈ 0.40

8 0
2 years ago
A vertical cylinder is divided into two parts by a movable piston of mass m. The piston and cylinder system is well insulated (t
Mekhanik [1.2K]

Answer:

Final temperature will be 438.076 K

Explanation:

We have given temperature T_1=323K

Volume V_1=V\ and\ V_2=\frac{V}{2}

As there is no heat transfer so this is an adiabatic process

For and adiabatic process TV^{\gamma -1}=constant

Here \gamma =1.4

So T_1V_1^{\gamma -1}=T_2V_2^{\gamma -1}

T_2=\left ( \frac{V_1}{V_2} \right )^{\gamma -1}\times T_1

T_2=\left ( \frac{V}{\frac{V}{2}} \right )^{1.4 -1}\times 332=2^{0.4}\times 332=438.076K

4 0
2 years ago
A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in t
stiv31 [10]

Answer:

D) 117 rad/s

Explanation:

We can treat this system as a circular motion where the origin is the elbow joint, the ball rotation velocity v is 35 m/s, the rotation radius is r = 0.3m.

As the ball is leaving the pitcher hand at such speed and rotation radius. Its angular velocity is:

\omega = \frac{v}{r} = \frac{35}{0.3} = 117 rad/s

3 0
1 year ago
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