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lisov135 [29]
4 years ago
15

The energy of a given wave in the electromagnetic spectrum is 2.64 × 10-21 joules, and the value of Planck’s constant is 6.6 × 1

0-34 joule·seconds
4.00 × 1012 hertz
2.34 × 10-12 hertz
1.30 × 1013 hertz
2.52 × 10-6 hertz
8.11 × 10-7 hertz
Physics
1 answer:
bixtya [17]4 years ago
6 0
We are given
E = <span>2.64 × 10-21 J
h = </span><span>6.6 × 10-34 J s

The options given below are frequencies, therefore, the question must be asking about the frequency fo the given wave

The equation is
E = h f
Simply substitute and solve for f which is the frequency
f = </span>2.64 × 10-21 J / 6.6 × 10-34 J s
f = <span>4.00 × 1012<span> hertz</span></span>
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A cat weighing 7 kg chases a mouse at a speed of 4 m/s. What is the kinetic energy of the cat?
jeka57 [31]
54J
You calculate kinetic energy by the equation.
KE= 1/2mv^2
m= mass
v= velocity
then we have
KE = 1/2 x 3 x 6^2
KE = 1/2 x 3 x 36 = 54J
8 0
3 years ago
One end of a horizontal spring with force constant 76.0 N/m is attached to a vertical post. A 5.00-kg can of beans is attached t
Jobisdone [24]

Answer:

a. The speed is 2.39 m/s

b. The acceleration of the block is 10.2\frac{m}{s^{2} }

Explanation:

First, we have to do the energy balance where we consider two states, the first where the spring remains still and the second when it is stretched 0.400m:

K_{1} +U_{1}+W_{ext}=K_{2}+U_{2}\\K_{1}=0\\U_{1}=\frac{1}{2} kx^{2} _{1} =0\\W_{ext}=FΔx=(0.400m)\\

W_{ext}=20.4 Nm

U_{2} =\frac{1}{2} kx^{2} =\frac{1}{2} (76.0N/m)0.400^{2}=6.08Nm\\k_{2} =\frac{1}{2}mv^{2} _{2}  \\\frac{1}{2} mv^{2} _{2}=W_{ext}-U_{2}\\v_{2}=\sqrt{\frac{W_{ext}-U_{2}}{m} } \\v_{2}=\sqrt{\frac{20.4Nm-6.08Nm}{2.5kg} } \\v_{2}=2.39 \frac{m}{s}

To determine, the acceleration we solve the following equation for a:

F=ma\\a=\frac{F}{m} =\frac{51.0N}{5.00kg}\\a=10.2\frac{m}{s^{2} }

8 0
4 years ago
A projectile is fired at an angle such that the vertical component of its velocity
mestny [16]

Answer:

<em>It takes 3 seconds to reach its high point</em>

Explanation:

<u>Projectile Motion</u>

In a projectile motion (or 2D motion), the object is launched with an initial angle θ and an initial velocity vo.

The components of the velocity are

v_{ox}=v_o\cos\theta

v_{oy}=v_o\sin\theta

The speed in the horizontal direction at any time t is:

v_y=v_o\sin\theta-g.t

The time taken to reach the maximum height is when vy=0, or:

\displaystyle t_m=\frac{v_o\sin\theta}{g}

We are given the y-component of the velocity, thus:

\displaystyle t_m=\frac{30~m/s}{10~m/s^2}

t_m=3~s

It takes 3 seconds to reach its high point

3 0
3 years ago
11. Newton's Third Law of Motion relates to action and reaction. Which of the following scenarios accurately names the
hoa [83]

Answer:

b

Explanation:

4 0
3 years ago
A cylinder of gas at room temperature has a pressure . To p_{1} what temperature in degrees Celsius would the temperature have t
grandymaker [24]

In order to calculate the temperature, we need to know that temperature and pressure are directly proportional, that is, if the pressure increases, the temperature (in Kelvin) also increases in the same proportion.

So, first let's convert the temperature from Celsius to Kelvin, by adding 273 units:

\begin{gathered} K=C+273 \\ K=20+273 \\ K=293 \end{gathered}

Then, let's calculate the proportion:

\begin{gathered} \frac{P_1}{T_1}=\frac{P_2}{T_2} \\ \frac{p_1}{293}=\frac{1.5p_1}{T_2} \\ \frac{1}{293}=\frac{1.5}{T_2} \\ T_2=1.5\cdot293 \\ T_2=439.5\text{ K} \end{gathered}

Now, converting back to Celsius, we have:

\begin{gathered} C=K-273 \\ C=439.5-273 \\ C=166.5\text{ \degree{}C} \end{gathered}

So the temperature would be 166.5 °C.

6 0
1 year ago
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