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Marina CMI [18]

1 year ago

9

Newton’s cradle is a contraption where metal balls hang from a frame. When one ball is pulled and released, the collision causes

one ball on the opposite end to be released.
Which type of force can decrease the total momentum of this system?

(1 point)
Responses

A friction force
B strong nuclear force
C electrostatic force
D magnetic force

1 answer:

Marysya12 [62]1 year ago

6 0

Answer: A

Explanation: the principle of friction with swinging and colliding balls.

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Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum

cestrela7 [59]

Answer:

$T_{2}=278.80 K$

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

$(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$ .

Now, let's use the ideal gas equation to the initial and the final state:

$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$

Let's recall that the term nR is a constant. That is why we can match these equations.

We can find a relation between the volumes of the initial and the final state.

$\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}$

Combining this equation with the first equation we have:

$(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$

$(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}$

Now, we just need to solve this equation for T₂.

$T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}$

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

$p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40$

Finally, T2 will be:

$T_{2}=278.80 K$

6 0

3 years ago

A 2kg block of which material would require 450 joules of thermal energy to increase its temperature by 1 degree Celsius?

12345 [234]

The block is made of A) Tin, as its specific heat capacity is $0.225 J/(g^{\circ}C)$

Explanation:

When an amount of energy Q is supplied to a sample of material of mass m, the temperature of the material increases by $\Delta T$ , according to the following equation :

$Q=mC_s \Delta T$

where $C_s$ is the specific heat capacity of the material.

In this problem, we have:

m = 2 kg = 2000 g is the mass of the unknown material

$Q = 450 J$ is the amount of energy supplied to the block

$\Delta T = 1^{\circ}C$ is the change in temperature of the material

Solving the equation for $C_s$ , we can find the specific heat capacity of the unknown sample:

$C_s = \frac{Q}{m \Delta T}=\frac{450}{(2000)(1)}=0.225 J/(g^{\circ}C)$

And by comparing with tabular values, we can find that this value is approximately the specific heat capacity of tin.

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

7 0

3 years ago

What is the equation for potential energy?

Mrrafil [7]

You can calculate potential energy by:
U = m.g.h

Where, U = potential energy
m = mass
g = acceleration due to gravity
h = height

Hope this helps!

7 0

3 years ago

Read 2 more answers

A student is creating a table with properties of electromagnetic and mechanical waves.

katen-ka-za [31]

Answer:

The property of the wave marked X is related to the source of the wave

Explanation:

The source of of origin of waves

Electromagnetic wave are waves that consists of varying electric and magnetic field that vibrate perpendicular to each other and to the direction of propagation of the wave and they are therefore transverse waves and transfer energy

Electromagnetic waves originate from the vibration of charged particles that gives off varying electric and magnetic fields

Mechanical waves are defined as waves that require a material medium such as air, water, metal, plastic, stretched leather, or wood to propagate

Mechanical waves originate from vibration of the particles of a medium

Sound waves which is a form of longitudinal mechanical waves that propagates by the vibration of the particles of a given medium about a point parallel to the direction of propagation of the wave.

8 0

3 years ago

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

$\mathbf{x}(t) = A\sin(\omega\cdot t)$ .

The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

Since $\displaystyle \omega = \frac{2\pi}{T}$ , it can be concluded that:

$\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}$ .

For the first mass $m_1$ , if the time period is $T_1$ , then the spring constant would be:

$\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2$ .

Similarly, for the second mass $m_2$ , if the time period is $T_2$ , then the spring constant would be:

$\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Since the two springs are the same, the two spring constants should be equal to each other. That is:

$\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

6 0

3 years ago