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2 years ago

12

an electron is moving with a speed of 0.85c in a direction opposite to that of photon. calculate the relative velocity of the el

ectron and photon.

1 answer:

ololo11 [35]2 years ago

5 0

Answer:

1.85c

Explanation:

a photon moves at c, the electron is moving at 0.85c, and since they are moving in opposing directions, the relative speed would be 1.85c

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The primary difference between a barometer and a manometer is

Nitella [24]

Answer:

a barometer is used to measure atmospheric pressure, and a manometer is used to measure gauge pressure.

Explanation:

A barometer measures air pressure at any locality with sea level as the reference.

However, a manometer is used to measure all pressures especially gauge pressures. Thus, if the aim is to measure the pressure at any point below a fluid surface, a barometer is used to determine the air pressure. The manometer may now be used to determine the gauge pressure

The algebraic sum of these two values gives the absolute pressure.

4 0

3 years ago

2013 Indianapolis 500 champion Tony Kanaan holds his hand out of his IndyCar while driving through still air with standard atmos

Citrus2011 [14]

Answer:

(a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

Explanation:

Given that,

Speed v₁= 60 mph

Speed v₂ = 225 mph

Speed v₃ = 235 mph

(a). We need to calculate the maximum pressure on his hand

Using equation of pressure

$P_{1}=P+\dfrac{1}{2}\rho v^2+\rho gh$

there, no vertical movement

So, on neglect of height term

$P_{1}=P+\dfrac{1}{2}\rho v_{1}^2$

Where, P= atmospheric pressure

$\rho$ = air density

v = speed

Put the value in the equation

$P_{1}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(60\times1.4667)^2)$

$P_{1}=2126.0\ lb/ft^2$

$P_{1}=\dfrac{2126.0}{144}$

$P_{1}= 14.76\ psi$

(b). Speed v₂ = 225 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

$P_{2}=P+\dfrac{1}{2}\rho v_{2}^2$

Put the value in the equation

$P_{2}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(225\times1.4667)^2)$

$P_{2}=2246.17\ lb/ft^2$

$P_{2}=\dfrac{2246.17}{144}$

$P_{2}= 15.59\ psi$

(c). Speed v₃ = 235 mph

We need to calculate the maximum pressure on his hand

Using equation of pressure

$P_{3}=P+\dfrac{1}{2}\rho v_{3}^2$

Put the value in the equation

$P_{3}=14.7\times144+\dfrac{1}{2}\times(0.002376\times(235\times1.4667)^2)$

$P_{3}=2257.93\ lb/ft^2$

$P_{3}=\dfrac{2257.93}{144}$

$P_{3}= 15.68\ psi$

According to bernoulli's equation,

If the car increases the velocity the the pressure on the surface of the driver's hand increases.

The pressure from P₁ to P₃ are all near the value of one atmosphere.

So, the pressure difference of one atmosphere is not enough to break the driver's hand.

Hence, (a). The pressure is 14.76 psi.

(b). The pressure is 15.59 psi.

(c). The pressure is 15.68 psi.

All answer are reasonable.

5 0

3 years ago

Plz help me asap !!!!!!!!!!

nlexa [21]

Answer:

B- 65km

C- 2.8km/h

D- average speed

6 0

3 years ago

A ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally aw

astraxan [27]

26.54 m/s is the magnitude of its velocity just before it strikes the ground

h=100m,v=20m/s,g=9.8m/s

time it takes to reach the ground,

$[t=\sqrt2h/g],[=\sqrt2*100/9.8=4.51s]$

x= 120m

t= 4.52

v= x/t

v= 120/4.52

v= 26.54 m/s

The "speed at which an object changes its location" can be expressed using a vector number called velocity. Consider a person who moves swiftly while taking two steps forward and two steps back while remaining in one location. Velocity is a vector quantity. Therefore, velocity is cognizant of direction. The direction must be taken into account when determining an object's velocity. A speed of 55 mph is not enough information. The direction must be used to appropriately depict the item's velocity. Simply said, the direction of the velocity vector indicates the direction of motion of an object.

To know more about velocity visit : brainly.com/question/16379705

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1 year ago

Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal

kirza4 [7]

At t =0, the velocity of A is greater than the velocity of B.

We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;

A: vA (t) = ť^2 – 5t + 20

B: vB (t) = t^2+ 3t + 10

Given that t = 0 in both cases;

vA (0) = 0^2 – 5(0) + 20

vA = 20 m/s

For vB

vB (0) = 0^2+ 3(0) + 10

vB = 10 m/s

We can see that at t =0, the velocity of A is greater than the velocity of B.

Learn more: brainly.com/question/24857760

Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?

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3 years ago