Ask question

Ask question

All categories

3 years ago

13

An orbiting satellite stays over a certain spot on the equator of(rotating) Earth. What is the altitude (in km) of the orbit (ca

lleda geosynchronous orbit)?

1 answer:

Leviafan [203]3 years ago

5 0

Answer: 35786km

Explanation: At 35786km it is an important point for monitoring weather, communication and surveillance.

You might be interested in

The Hi line of the Balmer series is emitted in the transition from n = 3 to n = 2. Compute the wavelength of this line for l H a

irina1246 [14]

Answer:

$\lambda=550\ nm$

Explanation:

The Hi line of the Balmer series is emitted in the transition from n = 3 to n = 2 i.e. $n_i=3$ and $n_f=2$

The wavelength of Hi line of the Balmer series is given by :

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_f}-\dfrac{1}{n_i})$

$\dfrac{1}{\lambda}=1.09\times 10^7\times (\dfrac{1}{2}-\dfrac{1}{3})$

$\lambda=5.50\times 10^{-7}\ m$

$\lambda=550\ nm$

So, the wavelength for this line is 550 nm. Hence, this is the required solution.

6 0

3 years ago

Calculate the pressure exerted on the floor when an

skad [1K]

P = F / A

P = 2400 / 4

P = 24 × 100 / 4

P = 6 × 100

P = 600 N/m^2

7 0

3 years ago

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.14 m. The mug

defon

Say the horizontal component of the velocity is vx and the vertical is vy.
Initially at t=0 (as the mug leaves the bar) the components are v0x and v0y.
Obviously (I hope!) v0y = 0.

The equations for horizontal and vertical projectile motion (with the positive direction up) are

x = x0 + v0x t
y = y0 + v0y t - 1/2 g t^2 = y0 - 1/2 g t^2

Now choose the origin to be the end of the counter. x0=0 and y0=0. The equations simplify to

x = v0x t
y = - 1/2 g t^2

You know that x = 1.20m when y = -0.88m
From the y equation (and g=10 m/s^2) you can calculate the time that the mug hits the floor.
t = 0.420s
From the x equation we get the initial horizontal velocity
v0x = x/t = 1.2/0.42 = 2.86 m/s

(b) x-component of velocity is constant since there are no horizontal forces so vx = 2.86 m/s
y-component is given by v = u+at with u=0 and a=-g
vy = -gt = -4.2m/s

Now tan(angle) = vy/vx so angle = arctan(vy/vx)

7 0

3 years ago

a 300kg motorboat is turned off as it approaches a dock and coasts towards it at .5 m/s. Isaac, whose mass is 62 kg jumps off th

Zolol [24]

-- Before he jumps, the mass of (Isaac + boat) = (300 + 62) = 362 kg,
their speed toward the dock is 0.5 m/s, and their linear momentum is

Momentum = (mass) x (speed) = (362kg x 0.5m/s) = <u>181 kg-m/s</u>

<u>relative to the dock</u>. So this is the frame in which we'll need to conserve
momentum after his dramatic leap.

After the jump:

-- Just as Isaac is coiling his muscles and psyching himself up for the jump,
he's still moving at 0.5 m/s toward the dock. A split second later, he has left
the boat, and is flying through the air at a speed of 3 m/s relative to the boat.
That's 3.5 m/s relative to the dock.

His momentum relative to the dock is (62 x 3.5) = 217 kg-m/s toward it.

But there was only 181 kg-m/s total momentum before the jump, and Isaac
took away 217 of it in the direction of the dock. The boat must now provide
(217 - 181) = 36 kg-m/s of momentum in the opposite direction, in order to
keep the total momentum constant.

Without Isaac, the boat's mass is 300 kg, so

(300 x speed) = 36 kg-m/s .

Divide each side by 300: speed = 36/300 = <em>0.12 m/s ,</em> <u>away</u> from the dock.
=======================================

Another way to do it . . . maybe easier . . . in the frame of the boat.

In the frame of the boat, before the jump, Isaac is not moving, so
nobody and nothing has any momentum. The total momentum of
the boat-centered frame is zero, which needs to be conserved.

Isaac jumps out at 3 m/s, giving himself (62 x 3) = 186 kg-m/s of
momentum in the direction <u>toward</u> the dock.

Since 186 kg-m/s in that direction suddenly appeared out of nowhere,
there must be 186 kg-m/s in the other direction too, in order to keep
the total momentum zero.

In the frame of measurements from the boat, the boat itself must start
moving in the direction opposite Isaac's jump, at just the right speed
so that its momentum in that direction is 186 kg-m/s.
The mass of the boat is 300 kg so
(300 x speed) = 186

Divide each side by 300: speed = 186/300 = <em>0.62 m/s</em> <u>away</u> from the jump.

Is this the same answer as I got when I was in the frame of the dock ?
I'm glad you asked. It sure doesn't look like it.

The boat is moving 0.62 m/s away from the jump-off point, and away from
the dock.
To somebody standing on the dock, the whole boat, with its intrepid passenger
and its frame of reference, were initially moving toward the dock at 0.5 m/s.
Start moving backwards away from <u>that</u> at 0.62 m/s, and the person standing
on the dock sees you start to move away <u>from him</u> at 0.12 m/s, and <em><u>that's</u></em> the
same answer that I got earlier, in the frame of reference tied to the dock.

yay !

By the way ... thanks for the 6 points. The warm cloudy water
and crusty green bread are delicious.

4 0

3 years ago

Across what potential difference does an electron have to be accelerated in order to reach the speed v=3e7 m/s?

zmey [24]

When an electron is accelerated through potential difference then the speed that it attain will be explained by energy conservation

here by energy conservation we can say that

change in kinetic energy of electron = electrostatic potential energy gained through given potential difference

kinetic energy is given as

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

electrostatic potential energy is given as

$PE = qV$

now by energy conservation

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

given that for electron

$m = 9.1 * 10^{-31} kg$

$v = 3 * 10^7 m/s$

$q = 1.6 * 10^{-19} C$

now by plug in values

$1.6 * 10^{-19} * V = \frac{1}{2}*9.1* 10^{-31} *(3*10^7)^2$

$V = \frac{4.095 * 10^{-16}}{1.6 * 10^{-19}}$

$V = 2559.4 Volts$

So here it is accelerated through potential difference of 2559.4 Volts

4 0

3 years ago