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4 days ago

6

Two balls, each with a mass of 0.5 kg, collide on a pool table. Is the law of conservation of momentum satisfied in this collisi

on? Explain why or why not.

2 answers:

Maru [3.2K]4 days ago

6 0

The principle of momentum conservation<span> is a key law in the field of physics. It asserts that the </span>momentum<span> within a system remains unchanged unless there are </span>external forces influencing the system. In the case of two balls, each weighing 0.5 kg, colliding on a pool table<span>, this principle does not hold because external forces acted upon the balls during the collision. </span>

Softa [2.9K]4 days ago

4 0

Indeed, the conservation of momentum law is upheld. The cumulative momentum prior to the impact is 1.5 kg • m/s, and it remains 1.5 kg • m/s after the impact as well. Thus, the momentum is unchanged following the collision.

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In an inertial frame of reference, a series of experiments is conducted. in each experiment, two or three forces are applied to

Yuliya22 [3215]

Objects will stay in a stationary position if the total force acting on them amounts to zero; this occurs when equal forces are applied in opposite directions. According to Newton's second law, if the net force on an object is zero, it will not move.

8 0

1 month ago

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in va

serg [3462]

Answer:

10000 V

0.00225988700565 m²

$8\times 10^{-12}\ F$

Explanation:

E = Electric field = $4\times 10^6\ V/m$

d = Distance = 2.5 mm

Q = Charge = 80 nC

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

The potential difference is calculated as

$V=Ed\\\Rightarrow V=4\times 10^6\times 2.5\times 10^{-3}\\\Rightarrow V=10000\ V$

The potential difference across the plates amounts to 10000 V

Area is determined by

$A=\dfrac{Q}{\epsilon_0E}\\\Rightarrow A=\dfrac{80\times 10^{-9}}{8.85\times 10^{-12}\times 4\times 10^6}\\\Rightarrow A=0.00225988700565\ m^2$

The area of each plate measures 0.00225988700565 m²

Capacitance is determined by

$C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 0.00225988700565}{2.5\times 10^{-3}}\\\Rightarrow C=8\times 10^{-12}\ F$

The capacitance is $8\times 10^{-12}\ F$

4 0

15 days ago

Five metal samples, with equal masses, are heated to 200oC. Each solid is dropped into a beaker containing 200 ml 15oC water. Wh

ValentinkaMS [3332]

Part 1) Which metal will cool the fastest?
To determine this, we need to consider the heat flow rate formula, which indicates the speed at which a substance can gain or lose heat:
$\frac{\Delta Q}{\Delta t} = -k \frac{A \Delta T}{x}$
where:
$\Delta Q$ denotes the heat exchanged
$\Delta t$ indicates the duration
k represents the thermal conductivity of the material
A is the area over which heat transfer takes place
$\Delta T$ shows the change in temperature
x is the thickness of the substance
It is evident that the heat flow rate $\frac{\Delta Q}{\Delta t}$ is directly related to k, the thermal conductivity. Thus, a higher value of k means that the metal will cool more quickly.
Upon examining the thermal conductivity values for each metal, we observe:
- Aluminium: 237 W/(mK)
- Copper: 401 W/(mK)
- Gold: 314 W/(mK)
- Platinum: 69 W/(mK)
Consequently, copper has the highest heat flow rate, making it the metal that cools the fastest.

Part 2) Which sample of copper demonstrates the greatest increase in temperature
To address this part, we can examine how the heat exchanged Q correlates with temperature increase $\Delta T$ :
$Q=m C_S \Delta T$
where m indicates the mass and Cs represents the specific heat capacity of the material. By rearranging the equation, we derive
$\Delta T= \frac{Q}{m C_s}$
As a result, it becomes clear that the temperature increase is inversely related to the mass m. Thus, the block exhibiting the highest temperature rise will be the one with the least mass, hence the right choice is A) 0.5 kg.

4 0

19 days ago

A student solving a physics problem for the range of a projectile has obtained the expression r= v20sin(2θ)g where v0=37.2meter/

ValentinkaMS [3332]

The formula for range is:

$R = \frac{v_o^2 sin2\theta}{g}$

Given values are:

$v_0=37.2m/s$

where θ equals 14.1 degrees

$g=9.80m/s^2$

Using the equation above,

$R = \frac{37.2^2 sin2*14.1}{9.80}$

The calculated range is 66.7 meters.

Therefore, the range is approximately 66.1 meters.

5 0

1 month ago

Read 2 more answers

Calculate the buoyant force in air on a kilogram of titanium (whose density is about 4.5 grams per cubic centimeter). compare wi

ValentinkaMS [3332]

1) The buoyant force acting on an object submerged in a fluid can be described as:
$B=d_f V_d g$
where $d_f$ indicates the fluid's density, $V_d$ represents the volume of the fluid displaced, and $g=9.81~m/s^2$ signifies the gravitational acceleration.

2) To determine the volume of the displaced fluid, we note that the titanium object is entirely submerged in the fluid (air), thus this volume matches the volume of 1 Kg of titanium, which has a density of $d=4.5~g/cm^3 = 4.5\cdot10^3~Kg/m^3$ . Using the correlation between density, volume, and mass, we derive
$V_d= \frac{m}{d}= \frac{1~Kg}{4.5\cdot10^3Kg/m^3}=2.22\cdot10^{-4}~m^3$

3) We can now revisit the equation in step 1) to compute the buoyant force. Given that the air density is $d_f = 1~Kg/m^3$ , this provides us with
$B=d_f V_d g=1~Kg/m^3 \cdot 2.22\cdot10^{-4}~m^3 \cdot 9.81~m/s^2=2.22\cdot10^{-3}~N$

4) The weight of 1 Kg of titanium is:
$W=mg=1~Kg \cdot 9.81~m/s^2=9.81~N$
Therefore, the buoyant force is negligible when compared to the weight.

7 0

1 month ago