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3 years ago

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el of a Shs when places in water. The same piece of day ficas ftis made into the stage of a boat. Why does the day ficat when ma

de into the shape of a boat? Once ses dese when sterec re me seace of a beet. O mesec staze disclasg a greater volume of water than the ball shape. The bevant force is greater on the day when it has a spherical shape.

1 answer:

SCORPION-xisa [38]3 years ago

8 0

If my truck loses a ski, and the sky is yellow, then how many ounces of milk will I need to make a house?

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A soccer player takes a corner kick, lofting a stationary ball 33.0° above the horizon at 15.0 m/s. If the soccer ball has a mas

Alexxandr [17]

Explanation:

It is given that,

Mass of the soccer ball, m = 0.425 kg

Speed of the ball, u = 15 m/s

Angle with horizontal, $\theta=33^{\circ}$

Time for which the player's foot is in contact with it, $\Delta t = 5.1\times 10^{-2}\ s$

Part A,

The x component of the soccer ball's change in momentum is given by :

$\Delta p_x=mv\ cos\theta$

$\Delta p_x=0.425\times 15\ cos(33)$

$p_x=5.34\ kg-m/s$

The y component of the soccer ball's change in momentum is given by :

$\Delta p_y=mv\ sin\theta$

$\Delta p_y=0.425\times 15\ sin(33)$

$p_y=3.47\ kg-m/s$

Hence, this is the required solution.

3 0

3 years ago

The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu

Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

$E=\phi + K$

where

$E=\frac{hc}{\lambda}$ is the energy of the incident light, with h being the Planck constant, c being the speed of light, and $\lambda$ being the wavelength

$\phi$ is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

$\lambda=190 nm=1.9\cdot 10^{-7}m$ , so the energy of the incident light is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J$

Converting in electronvolts,

$E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV$

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

$\phi = E-K=6.5 eV-4.0 eV=2.5 eV$

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

$\lambda=510 nm=5.1\cdot 10^{-7} m$

So its energy is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J$

Converting in electronvolts,

$E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV$

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: $\phi = 4.5 eV$

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

$E=\phi+K=4.5 eV+2.7 eV=7.2 eV$

Then the copper is replaced with sodium, which has work function of

$\phi = 2.3 eV$

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

$K=E-\phi = 7.2 eV-2.3 eV=4.9 eV$

7 0

3 years ago

An important consideration in materials physics is the number of defects that are unavoidable due to statistical considerations.

goblinko [34]

Answer:

Explanation:

find attached the solution

8 0

3 years ago

You plug a string of 100 lights in series into a 120 V power outlet, and each light has a resistance of 3.00 _. If each light ha

UNO [17]

V = IR

By completing the equation, i found that the total power equation is : 4.8,
Which means that it's not exceed the power rating.

So i believe the answer would be : The string will remain lit

hope this helps

6 0

4 years ago

How can goal setting help with academic performance?

san4es73 [151]

Answer:

Setting goals keeps students focused on desired outcomes and provides a clear direction for success

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3 years ago