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2 years ago

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A space vehicle is coasting at a constant velocity of 22.3 m/s in the y direction relative to a space station. The pilot of the

vehicle fires a RCS (reaction control system) thruster, which causes it to accelerate at 0.277 m/s^2 in the x direction. After 46.6 s, the pilot shuts off the RCS thruster. After the RCS thruster is turned off, find:
a. The magnitude.
b. The direction of the vehicle's velocity relative to the space station. Express the direction as an angle (in degrees) measured from the y direction.

1 answer:

yanalaym [24]2 years ago

8 0

Answer:

a) 25.76 m/s

b) 30°

Explanation:

See attachment

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In Millikan's oil-drop experiment, one looks ata small oil drop held motionless between two plates. Take the voltage betwee the

AleksandrR [38]

Answer:

The charge on the drop is

q = 1.741 x 10 ⁻²¹ C

Explanation:

Electric field due to plates

Ef = V/d

Ef = 2033 V / (2.0 * 10^-2 m )

Ef = 101650 V/m

So, we can write

Ef * q = m*g

q = m*g / E f

The mass can be equal using the density and the volume so:

m = ρ * v

The volume can be find as:

v = 2.298 x 10 ⁻ ¹⁶ m³

q = ρ * v * g / Ef

q = 81 x 10 ³ kg/ m³ * 2.2298 x 10 ⁻ ¹⁶ m³ * 9.8 m/s² / 101650 V/m

The charge on the drop is

q = 1.741 x 10 ⁻²¹ C

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2 years ago

Explain why people have different skin colors?

stiks02 [169]

I think it because of UV rays ultra violet ray which can make their colors different

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3 years ago

Explain the role of energy in changes <br> of state.

slava [35]

A liquid requires enthalpy of vaporization to transform into vapor or gas at its boiling point. Here the element absorbs heat from surroundings or heat source.

This energy is used in breaking the forces of attraction among the atoms and molecules of the element. The molecules get separated to higher distances. The energy is converted in to the kinetic energy of the molecules in gaseous form and into the internal energy in terms of the temperature of the gas.

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3 years ago

A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y=y'$

$z=z'$

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

First we calculate the expression in the denominator

$\frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}$

$\sqrt{1-\frac{v^{2}}{c^{2}}} =0,714$

then we calculate t

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}$

$t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}$

$t=\frac{190s+79,388s}{0,714}$

finally we get that

$t=377,29s$

then we calculate x

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+39872396914m}{0,714}}$

$x=\frac{73872396914m}{0,714}}$

$x=103462740775,91m$

finally we get that

$x \approx 103,46x10^{9} m$

5 0

3 years ago

Andy Petite pitches a 0.8 kg baseball with a velocity of 67 m/s. Josh Hamilton

kodGreya [7K]

The Impulse delivered to the baseball is 89 kgm/s.

To solve the problem above, we use the formula of impulse.

⇒ Formula:

I = m(v-u)................. Equation 1

Where:

I = Impulse delivered to the baseball
m = mass of the baseball
v = Final velocity of the baseball
u = initial speed of the baseball

From the question,

⇒ Given:

m = 0.8 kg
u = 67 m/s
v = -44 m/s

⇒ Substitute these values into equation 1

I = 0.8(-44-67)
I = 0.8(-111)
I = -88.8
I ≈ -89 kgm/s

Note: The negative tells that the impulse is in the same direction as the final velocity and therefore can be ignored.

Hence, The Impulse delivered to the baseball is 89 kgm/s.

Learn more about impulse here: brainly.com/question/7973509

7 0

1 year ago