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3 years ago

Juan lives 100m away from bill whats juan`s average speed if he reaches bill home in 50 s

1 answer:

6 0

<span>Distance = Speed x Time, we are going to derived our equation.
You can compute for the average speed using the equation: speed = distance ÷ time.
So plugging in our data, it will look like:
Speed = distance / time
= 100 m / 50 s
= 2 m/sec
Juan average speed is 2 m / sec. </span>

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The international space station makes 15.65 revolutions per day in its orbit around the earth. assuming a circular orbit, how hi

sweet-ann [11.9K]

<span>373.2 km The formula for velocity at any point within an orbit is v = sqrt(mu(2/r - 1/a)) where v = velocity mu = standard gravitational parameter (GM) r = radius satellite currently at a = semi-major axis Since the orbit is assumed to be circular, the equation is simplified to v = sqrt(mu/r) The value of mu for earth is 3.986004419 Ă— 10^14 m^3/s^2 Now we need to figure out how many seconds one orbit of the space station takes. So 86400 / 15.65 = 5520.767 seconds And the distance the space station travels is 2 pi r, and since velocity is distance divided by time, we get the following as the station's velocity 2 pi r / 5520.767 Finally, combining all that gets us the following equality v = 2 pi r / 5520.767 v = sqrt(mu/r) mu = 3.986004419 Ă— 10^14 m^3/s^2 2 pi r / 5520.767 s = sqrt(3.986004419 * 10^14 m^3/s^2 / r) Square both sides 1.29527 * 10^-6 r^2 s^2 = 3.986004419 * 10^14 m^3/s^2 / r Multiply both sides by r 1.29527 * 10^-6 r^3 s^2 = 3.986004419 * 10^14 m^3/s^2 Divide both sides by 1.29527 * 10^-6 s^2 r^3 = 3.0773498781296 * 10^20 m^3 Take the cube root of both sides r = 6751375.945 m Since we actually want how far from the surface of the earth the space station is, we now subtract the radius of the earth from the radius of the orbit. For this problem, I'll be using the equatorial radius. So 6751375.945 m - 6378137.0 m = 373238.945 m Converting to kilometers and rounding to 4 significant figures gives 373.2 km</span>

7 0

3 years ago

Read 2 more answers

The skater is has a mass of 75 kg. Find the total Potential energy of the skater at the top of the ramp at 6 m.

olga_2 [115]

Epot= mgh= 75*9,81*6= 4415 j

5 0

3 years ago

At the moment t = 0, a 20.0 V battery is connected to a 5.00 mH coil and a 6.00 Ω resistor. (a) Immediately thereafter, how does

insens350 [35]

(a) On the coil: 20 V, on the resistor: 0 V

The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:

$V = V_R + V_L$

The potential difference across the inductance is given by

$V_L(t) = V e^{-\frac{t}{\tau}}$ (1)

where

$\tau = \frac{L}{R}=\frac{0.005 H}{6.00 \Omega}=8.33\cdot 10^{-4} s$ is the time constant of the circuit

At time t=0,

$V_L(0) = V e^0 = V = 20 V$

So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:

$V_R = V-V_L = 20 V-20 V=0$

(b) On the coil: 0 V, on the resistor: 20 V

Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

$t >> \tau$

Since $\tau$ is at the order of less than milliseconds.

Using eq.(1), we see that for $t >> \tau$ , the exponential becomes zero, and therefore the potential difference across the coil is zero:

$V_L = 0$

Therefore, the potential difference across the resistor will be

$V_R = V-V_L = 20 V- 0 = 20 V$

(c) Yes

The two voltages will be equal when:

$V_L = V_R$ (2)

Reminding also that the sum of the two voltages must be equal to the voltage of the battery:

$V=V_L +V_R$

And rewriting this equation,

$V_R = V-V_L$

Substituting into (2) we find

$V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V$

So, the two voltages will be equal when they are both equal to 10 V.

(d) at $t=5.77\cdot 10^{-4}s$

We said that the two voltages will be equal when

$V_L=\frac{V}{2}$

Using eq.(1), and this last equation, this means

$V e^{-\frac{t}{\tau}} = \frac{V}{2}$

And solving the equation for t, we find the time t at which the two voltages are equal:

$e^{-\frac{t}{\tau}}=\frac{1}{2}\\-\frac{t}{\tau}=ln(1/2)\\t=-\tau ln(0.5)=-(8.33\cdot 10^{-4} s)ln(0.5)=5.77\cdot 10^{-4}s$

(e-a) -19.2 V on the coil, 19.2 V on the resistor

Here we have that the current in the circuit is

$I_0 = 3.20 A$

The problem says this current is stable: this means that we are in a situation in which $t>>\tau$ , so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law

$V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V$

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

$V_L(0)=.V$

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

(e-b) 0 V on both

After several seconds,

$t>>\tau$

If we use this approximation into the formula

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

We find that

$V_L = 0$

And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.

7 0

3 years ago

You serve a volley ball with a mass of 1.5 kg. The ball leaves your hand at 15m/s. The ball has how much kinetic energy?

pav-90 [236]

Answer:

the answer is 168.75 J

5 0

3 years ago

A proton traveling at 17.6° with respect to the direction of a magnetic field of strength 3.28 mT experiences a magnetic force o

umka2103 [35]

Answer:

a) The proton's speed is 5.75x10⁵ m/s.

b) The kinetic energy of the proton is 1723 eV.

Explanation:

a) The proton's speed can be calculated with the Lorentz force equation:

$F = qv \times B = qvBsin(\theta)$ (1)

Where:

F: is the force = 9.14x10⁻¹⁷ N

q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C

v: is the proton's speed =?

B: is the magnetic field = 3.28 mT

θ: is the angle between the proton's speed and the magnetic field = 17.6°

By solving equation (1) for v we have:

$v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s$

Hence, the proton's speed is 5.75x10⁵ m/s.

b) Its kinetic energy (K) is given by:

$K = \frac{1}{2}mv^{2}$

Where:

m: is the mass of the proton = 1.67x10⁻²⁷ kg

$K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV$

Therefore, the kinetic energy of the proton is 1723 eV.

I hope it helps you!

3 0

3 years ago