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3 years ago

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When the position of the mass is farthest from the equilibrium position, what is the velocity of the mass?

1 answer:

Llana [10]3 years ago

5 0

At the most distant point, the size of the speed is zero (0 m/s). This is a direct result of preservation of vitality. PE = KE. The most distant far from the harmony position is the maximum PE. Hence it can have no KE. No KE implies no speed since KE = .5mv2

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Question 3. AP PHYSICS. Currently learning about torque so I assume you apply torque. I don't get it at all

8090 [49]

wow that is confusing

8 0

3 years ago

g Estimate the number of photons emitted by the Sun in a second. The power output from the Sun is 4 Ã— 1026 W and assume that th

vagabundo [1.1K]

Answer:

The value is $N = 1.107 *10^{45 } \ photons$

Explanation:

From the question we are told that

The power output from the sun is $P_o = 4 * 10^{26} \ W$

The average wavelength of each photon is $\lambda = 550 \ nm = 550 *10^{-9} \ m$

Generally the energy of each photon emitted is mathematically represented as

$E_c = \frac{h * c }{ \lambda }$

Here h is the Plank's constant with value $h = 6.62607015 * 10^{-34} J \cdot s$

c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

So

$E_c = \frac{6.62607015 * 10^{-34} * 3.0 *10^{8} }{ 550 *10^{-9} }$

=> $E_c = 3.614 *10^{-19} \ J$

Generally the number of photons emitted by the Sun in a second is mathematically represented as

$N = \frac{P }{E_c}$

=> $N = \frac{4 * 10^{26} }{3.614 *10^{-19}}$

=> $N = 1.107 *10^{45 } \ photons$

5 0

2 years ago

lesantik [10]

Answer:

It is B

it is called a jumper cable because it jumps the power from one car to the other

3 0

3 years ago

A technician wearing a brass bracelet enclosing area 0.00500 m2 places her hand in a solenoid whose magnetic field is 3.10 T dir

aliya0001 [1]

Explanation:

Given that,

Area enclosed by a brass bracelet, $A=0.005\ m^2$

Initial magnetic field, $B_i=3.1\ T$

The electrical resistance around the circumference of the bracelet is, R = 0.02 ohms

Final magnetic field, $B_f=0.93\ T$

Time, $t=16\ ms=16\times 10^{-3}\ s$

The expression for the induced emf is given by :

$\epsilon=-\dfrac{d\phi}{dt}$

$\phi$ = magnetic flux

$\epsilon=-\dfrac{d(BA)}{dt}$

$\epsilon=-A\dfrac{d(B)}{dt}$

$\epsilon=-A\dfrac{B_f-B_i}{t}$

$\epsilon=-0.005 \times \dfrac{0.93-3.1}{16\times 10^{-3}}$

$\epsilon=0.678\ volts$

So, the induced emf in the bracelet is 0.678 volts.

Using ohm's law to find the induced current as :

V = IR

$I=\dfrac{V}{R}$

$I=\dfrac{0.678}{0.02}$

I = 33.9 A

or

I = 34 A

So, the induced current in the bracelet is 34 A. Hence, this is the required solution.

5 0

3 years ago

PLEASE HEEEEEEELP

Zepler [3.9K]

<h2> $\large{\underbrace{\underline{\fcolorbox{White}{pink}{\bf{ANSWER♥︎}}}}}$ </h2><h3>kinetic energy is given as</h3>

KE = (0.5) m v²

given that : v = speed of the bottle in each case = 4 m/s when m = 0.125 kg

KE = (0.5) m v² = (0.5) (0.125) (4)² = 1 J

when m = 0.250 kg KE = (0.5) m v² = (0.5) (0.250) (4)² = 2 J

when m = 0.375 kg KE = (0.5) m v² = (0.5) (0.375) (4)² = 3 J

when m = 0.0.500 kg KE = (0.5) m v² = (0.5) (0.500) (4)² = 4 J

8 0

3 years ago