What is the relative velocity of two beta particles moving in opposite directions at a speed of 0.8c?

Answers

Answer 1

That depends on where YOU are when you measure it.

If you're motionless in the laboratory, then you measure the particles flying apart at 1.6c .

If you're riding on one of the particles, you measure the other one flying away from you at less than c


Related Questions

An m = 6.11 g bullet is fired into a 365 g block that is initially at rest at the edge of a table of h = 1.08 m height as shown in the figure.
1. The bullet remains in the block, and after the impact the block lands d = 1.84 m from the bottom of the table. How much time does it take the block to reach the ground once it flies off the edge of the table?
2. What is the initial horizontal velocity of the block as it flies off the table?
3. Determine the initial speed of the bullet.

Answers

(a) The time taken the block to reach the ground once it flies off the edge of the table is 0.47 s.

(b) The initial horizontal velocity of the block as it flies off the table is 3.91 m/s.

(c) The initial speed of the bullet is 237.5 m/s.

Time of motion of the block

The time of motion of the block is calculated as follows;

h = ¹/₂gt²

where;

h is height of the tablet is time of motion

t = √(2h/g)

t = √( (2 x 1.08) / 9.8)

t = 0.47 s

Initial horizontal velocity of the block

x = vt

v = x/t

v = (1.84)/(0.47)

v = 3.91 m/s

Initial speed of the bullet

m1u1 + m2u2 = v(m1 + m2)

where;

m1 is mass of bulletm2 is mass of blocku1 is initial velocity of the bulletu2 is the initial velocity of the blockv is initial horizontal velocity of the block as it flies off the table

0.00611(u1) + 0.365(0) = 3.91(0.00611 + 0.365)

0.00611u1 = 1.451

u1 = 1.451/0.00611

u1 = 237.5 m/s

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Question 2 of 10
Which two types of energy does a book have as it falls to the floor?

Answers

Gravitational energy and motion energy

In a small village in southern Italy, childbirth is considered?

Answers

Answer:

a community event

Explanation:

A community event is the right answer

A ball A of mass 0.5 kg moving with a Velacity of 10 m/s a head on Collision with a ball B of mass 2kg moving with a Velocity of 1ms in the oppoite direction. If A and B stick together after Collision, Calculate the Common Velocity in the direction of A

Answers

Answer:

The common velocity v after collision is 2.8m/s²

Explanation:

look at the attachment above ☝️

A skier (m=55.0 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally.
a. If h = 6.50 m and D = 11.4 m, find H.
b. Find her total kinetic energy as she reaches the ground.

Answers

a. The value of H is 11.51 m

b. The total kinetic energy as she reaches the ground is 6.21 kJ.

What is mechanical energy?

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.

M.E = KE +PE

Given is a skier (m=55.0 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. Value of h = 6.50 m and D = 11.4 m.

Time taken to reach the ground after take off is

t = √2h/g

t = √(2x6.50) /9.81

t =1.15 s

The speed at that time is

speed = distance /time

v = 11.4 /1.15

v = 9.913 m/s

a. From energy conservation principle,

v = √2g(H-h)

Plug the values, we get

9.913 = √(2 x 9.81 (H - 6.5)

Squaring both sides and solving, we have

H = 11.51 m

Thus, the value of H is 11.51 m.

b. The energy at the ground will be only potential energy

Total kinetic energy at the ground = mgH

T.K.E = 55 x 9.81 x 11.51

T.K.E =6.21 kJ

Thus, the total kinetic energy is 6.21 kJ.

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what is the output voltage of a transformer if the input is 240volts, and the number of turns on the primary coil is 1200 and on the secondary coil is 200.​

Answers

The output voltage of a transformer is 40V.

What is an output voltage?

An output voltage is the voltage released by a device, such as a voltage regulator or a generator.

The formula to calculate Output voltage of transformer is:

[tex]\frac{V_{s} }{V_{p} } = \frac{N_{s} }{N_{p} }[/tex]

here,  [tex]V_{s}[/tex] is the secondary voltage, [tex]V_{p}[/tex] is the primary voltage, [tex]N_{s}[/tex] is the number of secondary windings and [tex]N_{p}[/tex] is the number of primary windings.

According to the question.

[tex]N_{p}[/tex] =1200 , [tex]N_{s}[/tex]=200 , [tex]V_{p}[/tex]=240V  ,[tex]V_{s}[/tex]= ?

By using the formula of Output voltage,

[tex]{V_{s} } = \frac{V_{p}N_{s} }{N_{p} }[/tex]

=[tex]\frac{240*200}{1200}[/tex]

[tex]=40V[/tex]

Therefore, 40V is the output voltage of a transformer.

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Which statement is a true catalyst in a reaction

Answers

Answer:

it alters the rate of chemical reaction

hope it is useful

what is the energy of a photon with a frequency of 8.34 x 10^14 hz, in joules?

Answers

Answer:

E = 5.52 x 10⁻¹⁹  J

Explanation:

The formula for energy of a photon, E, is:

[tex]E = hf[/tex]

where h is Plank's constant = 6.62 x 10 ⁻³⁴ .

Using the formula:

E = 6.62 x 10⁻³⁴   x   8.34 x 10¹⁴

  = 5.52 x 10⁻¹⁹  J

Which sentence uses the subjunctive mood correctly?
A. Americans take fewer vacations than people in many other
nations.
OB. It's best that Marci learns how to do her own laundry.
C. It is important that she apologize to her sister for what she said.
D. Be sure to put your dirty clothes in the hamper, not next to it.

Answers

The sentence that uses the subjunctive mood correctly is as follows:

Be sure to put your dirty clothes in the hamper, not next to it.

Thus, the correct option is D.

What is Subjunctive mood?

Subjunctive mood may be defined as statements or actions that are in the probable situation of doubt and suspicion.

Except for the sentence fourth, the rest all indicates the factual concept with the appropriate condition.

Therefore, the correct option for this question is D.

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Answer:

C. it is important that she apologize to her sister for what she said

Explanation:

test approved

a highway curve has a radius of 300 m. at what angle should it be banked for a traffic speed of 100 km/h

Answers

The angle at which it should be banked for a traffic speed of 100 km/h is

14.57°.

How do find the angle of Banking of road?Banking of roads is the practice of raising the outer borders of curving roads over the inner edge in order to give vehicles the necessary centripetal force to make a turn safely.Skidding can be prevented by banking.The banking of roadways aids in preventing toppling or overturning.

To find the angle of banking following formula can be used,

[tex]tan \alpha = v^{2} /rg[/tex]

Given v = 100km/h or

          = 100  x 1000 / 3600

          = 27.7 m/s

Hence,

[tex]tan \alpha = (27.7)^{2} / 300 * 9.81[/tex]

[tex]tan\alpha = 0.26[/tex]

[tex]\alpha = tan^{-1} 0.26[/tex]

[tex]\alpha = 14.57[/tex]°

Hence, the angle of banking of the road will be 14.57°

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state two factors that affect current carrying capacity of an accumulator

Answers

The two factors that affect current carrying capacity of an accumulator will be accumulator size and ambient temperature

What is the function of the accumulator?

Utilizing the compressible and decompressible properties of nitrogen gas, an accumulation vessel is used to store hydraulic pressure.

These key deciding elements are:

1. Accumulator Size: The current capacity increases as the circumference of the conductor increases.

2. Ambient Temperature: The greater the ambient temperature, the less heat is needed to raise the insulation's maximum operating temperature.

3. Accumulator identification:

4. Conditions for Installation:

Hence, the two factors that affect current carrying capacity of an accumulator will be accumulator size and ambient temperature

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A rocket is fired with an initial Velocity OF 100m/s at an angle OF 55 above the horizontal, It explodes on the mountain Side 12s after its Firing What is x-and y-coordinates of the rocket relative to its Firing point?​

Answers

Answer:

( 688.29 , 276.66 )   meters

Explanation:

Break into horizontal and vertical components

x = 100 cos 55  = 57.36 m/s      

   the x coordinate is then 12 seconds * 57.36 m/s = 688.29 meters

y coordinate is a bit more complex because of the curved path due to gravity

Original y velocity =  100 sin 55

y = vo t - 1/2 (9.81)(t^2)

y =  100 sin 55  (12)  - 1/2 (9.81) (12)^2

      yf = 276.66   meters

what are risk factors of an autoimmune disease

Answers

Explanation:

Your Sex. Overall, 78% of people affected by autoimmune disease are female (1). ...

Genetics. Certain disorders, such as lupus and multiple sclerosis, tend to run in families (3, 4). ...

Having an autoimmune disease. ...

Obesity. ...

Smoking and Exposure to Toxic Agents. ...Certain Medications. ...

Infections.

A ball is projected with initial velocity 50m/s at an angle of elevation of 37 degrees from the top of a cliff 55 m high. Calculate the total time the ball is in the air and the maximum horizontal distance covered.

Answers

The total time the ball is in the air is 1.47 s and the maximum horizontal distance covered is 58.7 m.

Time of motion of the ball

The time of motion of the ball is calculated as follows;

h = vt + ¹/₂gt²

55 = (50sin37)t + ¹/₂(9.8)t²

55 = 30.1t + 4.9t²

4.9t² + 30.1t - 55 = 0

Solve using formula method;

t = 1.47 s

Horizontal distance of the ball

X = vxt

where;

vx is the horizontal velocityt is time of motion

X = (50 x cos37) x 1.47

X = 58.7 m

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A car is traveling along a straight road at a velocity of 30m/s when its engine cuts off. For the next ten seconds, the car slows down, and its average acceleration is a₁. For the next five seconds, the car slows down further, and its average acceleration is a₂. The ratio of the average acceleration values is a₁/a₂=1.5. Find the velocity of the car at the end of the initial ten-second interval.​

Answers

The velocity of the car at the end of the initial ten-second interval is 15a₂ + 30.

Initial velocity of the car

v = u + at

after 10 seconds;

v = 30 + 10a₁  --(1)

after 5 seconds

v₂ = v + 5a₂ ---- (2)

solve (1) and (2) together

v₂ = 30 + 10a₁ + 5a₂

but a₁/a₂=1.5

a₁ = 1.5a₂

v₂ = 30 + 10(1.5a₂) + 5a₂

v₂ = 30 + 15a₂ + 5a₂

v₂ = 30 + 20a₂

from equation (2), v₂ = v + 5a₂

v + 5a₂ = 30 + 20a₂

v = 15a₂ + 30

Thus, the velocity of the car at the end of the initial ten-second interval is 15a₂ + 30.

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The practice of science can answer only scientific questions. And scientific
questions guide the design of investigations. What must be true of the
possible answers to a scientific question?
OA. They could be shown to be false by evidence.
B. They are at odds with established scientific theories.
C. They support an established scientific theory.
OD. They do not require the use of physical measurements.
SUBMIT

Answers

Evidence might be used to demonstrate their falsity. must apply to all potential solutions to a scientific topic.Option A is correct.

What is a Scientific hypothesis?

A scientific hypothesis is the main element in the scientific method and experiment.

It is said that a hypothesis is a scientific hypothesis. The majority must agree on it, or it must be done after thorough experimentation.

Only scientific inquiries can have their answers provided by scientific endeavor. Scientific concerns have an impact on the research's design as well.

Evidence might be used to demonstrate their falsity. must apply to all potential solutions to a scientific question.

Hence option A is correct.

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Imagine that you're doing an experiment to determine how the running speed of a hamster depends on the weight of the hamster when plotting your data what variable should be on the y-axis

Answers

The variable on the y-axis should be the running speed of hamsters since it is the dependent variable.

What is the dependent variable?

The dependent variable is the variable whose value fluctuates depending on other variables in an experiment.

In this case, the running speed is the dependent variable because it depends on the weight of the hamster.

The dependent variable is generally graphed on the Y axes, whereas the independent variable is graphed on the X-axes.

In conclusion, the variable on the y-axis should be the running speed of hamsters since it is the dependent variable.

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A skateboarder starts from rest and maintains a constant acceleration of 0.50 m/s² for 8.4 s. What is the rider's displacement during this time
meters

Answers

Answer:

4.2m/s

Explanation:

0.50x8.4=4.2m/s

In infrared satellite images, land appears _________ and clouds appear _________. A. red, yellow B. white, black C. black, white D. yellow, red

Answers

In infrared satellite images, land appears black and clouds appear white. C

What are infrared satellite images?

Infrared satellite images are images produced as a result of electromagnetic radiations reflected or emitted from a target surface in the infrared position of the electromagnetic spectrum.

On infrared images, the clouds emit very bright white color because of the low temperature and the land appears to be black because of the hot temperature.

Therefore, in infrared satellite images, land appears black and clouds appear white. C

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A long thin solid rod lies along the positive x-axis. One end is at x = 1.80 m and the other at x = 3.40 m. The linear mass density is λ = a[tex]x^{3}[/tex] + bx, where λ is measured in kg/m, and the constants have the following values: a = 1.70 kg/[tex]m^{4}[/tex] and b = 2.20 kg/[tex]m^{2}[/tex].
1. Determine the total mass of the rod.
2. Calculate the x-coordinate of the center of the mass for this rod.

Answers

1. The total mass of the rod is 61.48 kg.

2. The x-coordinate of the center of the mass for this rod is  172.6021 m.

What is density?

The density is the ratio of the mass and the volume of the object. It is denoted by ρ.

ρ = mass/Volume = m/V

Given a long thin solid rod lies along the positive x-axis. One end is at x = 1.80 m and the other at x = 3.40 m. The linear mass density is

λ = ax³ + bx,

where λ is measured in kg/m, and the constants have the following values: a = 1.70 kg/m⁴ and b = 2.20 kg/m²

So, λ = 1.70x³ + 2.20x

1. The total mass of the rod is

m =₁.₈³⁴ ∫ [1.70x³ + 2.20x] dx

m = [ 1.70x⁴/4 + 2.20x² /2 ]₁.₈³⁴

Putting the limits, we have

m = 69.51 - 8.02548

m = 61.48 kg

Thus, the total mass of the rod is 61.48 kg

2. The x-coordinate of the center of the mass for this rod

Xcm =∫ xλdx/m

Xcm =∫[ x ( 1.70x³ + 2.20x) /m ]dx

Xcm =∫[( 1.70x⁴ + 2.20x²) /m ]dx

On integrating and putting the limits  x = 1.80 m to x = 3.40 m

Xcm = = [ {1.70(3.4)⁵/5 + 2.20(3.4)³ /3} - {1.70(1.8)⁵/5 + 2.20(1.8)³ /3} ]

Xcm = 183.3034 - 10.70133

Xcm = 172.6021 m

Thus,  the x-coordinate of the center of the mass for this rod is  172.6021 m.

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2. A man pushes on a piano with mass 180 kg; it slides at constant velocity down a ramp that is inclined at 19.0° above the horizontal floor. Neglect any friction acting on the piano. Calculate the magnitude of the force applied by the man if he pushes (a) parallel to the incline and (b) parallel to the floor.

Answers

Answer: See below

Explanation:

Part (a):

As the velocity of the piano is constant, the net force on the piano is zero. The friction is also zero.

Here, F is the force applied by the man towards the inclined plane, mg is the weight of the piano and N is the normal force.

Applying Newton's law we get,

[tex]F = mg\sin \theta[/tex]

Substituting we get,

[tex]F &= \left( {180\;{\rm{kg}}} \right) \times \left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \sin 19.0^\circ \\ &= 574.3\;{\rm{N}}[/tex]

Therefore, the force is 574.3 N

Part (b)

Here, the force F is applied parallel to the floor.

The friction is zero.

Applying Newton's law we get,

[tex]F\cos \theta &= mg\sin \theta \\ F &= mg\tan \theta[/tex]

Substituting we get,

[tex]F &= \left( {180\;{\rm{kg}}} \right) \times \left( {9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right) \times \tan 19.0^\circ \\ &= 607.4\;{\rm{N}}[/tex]

Therefore, the force is 607.4 N

A stone is dropped from the top of a tower which touches the ground after 2 seconds. Calculate the speed of the stone hitting the ground surface. (g = 9.8 m/s²)​

Answers

Answer:

  19.6 m/s

Explanation:

The velocity is the product of acceleration and time.

__

Here, the acceleration is due to gravity.

  v = at

  v = (9.8 m/s²)(2 s) = 19.6 m/s

The stone hits the ground with a speed of 19.6 m/s.

••••••••••••••••••••••••••••••••••••••••••••[tex] \mathbb{ QUESTION:}[/tex]

A stone is dropped from the top of a tower which touches the ground after 2 seconds.

Calculate the speed of the stone hitting the ground surface. (g = 9.8 m/s²).

[tex] \mathbb{SOLUTION:} [/tex]

[tex]v = u + at[/tex]

[tex]v = 9.8 m/s² x 2sec[/tex]

[tex]v = 19.6 \: m/ {s}^{2} [/tex]

••••••••••••••••••••••••••••••••••••••••••••

Hence, The stone hit the ground with the velocity speed of 19m/.

- When you're solving velocity, you're determining how fast an object moves from its original position, with respect to a frame of reference, and a function of time.

- That means an object's velocity will be equal to the object's speed and direction of motion.

"Problem has been solve"

(ノ^_^)ノ

How do we solve questions C and D? I already did A and B and I am confused on how to continue

Answers

(a) The work done in moving the unit charge from point C to A is 7.62 x 10⁻³ J.

(b) The work done in moving the unit charge from point D to B is 7.62 x 10⁻³ J.

Work done in moving the charge from C to A

W = Fd

W = Kq²/d

from 0 origin to C, d = √(5² + 5²) = 7.07 mfrom 0 origin to A, d = 5 m

W(C to A) = W(0 to C) + W(0 to A)

[tex]W(C \ to \ A) = - \frac{Kq^2}{7.07} + \frac{Kq^2}{5} \\\\ W(C \ to \ A) = 0.0586 \ Kq^2\\\\W(C \ to \ A) = 0.0586 \times 9 \times 10^9 \times (3.8\times 10^{-6})^2\\\\W(C \ to \ A) = 7.62 \times 10^{-3} \ J[/tex]

Work done in moving the charge from D to Bfrom 0 origin to D, d = √(5² + 5²) = 7.07 mfrom 0 origin to B, d = 5 m

W(D to B) = W(0 to D) + W(0 to B)

W(D to B) = 7.62 x 10⁻³ J

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Suppose a piano tuner stretches a steel piano wire 7.5 mm. The wire was originally 0.975 mm in diameter, 1.45 m long, and has a Young's modulus of 2.10 × 1011 N/m2. Calculate the force a piano tuner applies to stretch the steel piano wire in Newtons.

Answers

The forces a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.

What is force?

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.

The application of a force may be used to describe the steel as an elastic element within a certain range of applied force.

Given data;

Young modulus, E=2.10 × 10¹¹ N/m²

Cross-sectional area,A

Final length,[tex]\rm L_f = 1.45 m = 1450 \ mm[/tex]

Initial length,[tex]\rm L_i = 7.5 mm[/tex]

[tex]\rm F = \frac{(L_f-L_i)(E)(A)}{L_1} \\\\ \rm F = \frac{(1450 - 7.5)(2.0 \times 0^{11})(0.746)}{1450} \\\\ F = 1.4 \times 10^{11} \ N[/tex]

Hence, the force a piano tuner applies to stretch the steel piano wire willl be 1.4 × 10¹¹ N.

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3. Sarah is a world-class diver in the women's 3-m springboard competition. Her height
h (in metres), above the water t seconds after she leaves the board is given by
h = -4.9t2 + 8.8t + 3. How long is Sarah in the air before she reaches the water?
Make sure to round your answer to the nearest tenth.
Answer:
2011, 2013, 2016 Alberta Education

Answers

Answer:

1.163 seconds

Explanation:

see the attached. solve the quadratic equation.

pleaseeeeeee I need HELP in 20 mins,Asap!!!


A 6.00 -kg clay ball is thrown directly against a perpendicular brick wall at a velocity of 22m/s^2 and shatters into three pieces, which all fly backward. The wall exerts a force of 2640N on the ball of for 0.1s. One piece of mass 2kg travels backward at a velocity of 10m/s and an angle of 32° above the horizontal. A second piece of mass 1kg travels at a velocity of 8m/s and an angle of 28° below the horizontal. What is the velocity of the third piece?​

Answers

The velocity of the third piece is 124.02 m/s at 1.05⁰ below the horizontal.

Velocity of the third piece

The velocity of the third piece is calculated from the principle of conservation of linear momentum.

mu = m₁u₁ + m₂u₂ + m₃u₃

where;

m is mass of the clayu is velocity of the clayu₁ is velocity of first pieceu₂ is velocity of second pieceu₃ is velocity of third piecem₃ is mass of the third piece = 6 kg - (2 kg + 1 kg) = 3 kgMomentum in y - direction

6(22)sin(0) = 2(10)sin32 - 1(8)sin(28) + 3u₃y

0 = 6.84 + 3u₃y

u₃y = -6.84/3

u₃y = -2.28 m/s

Change in momentum

ΔP = Pf - Pi = J

where;

Pf is final momentumPi is the initial momentumJ is impulse

2640(0.1) = 2(10)cos32 +  1(8)cos(28) + 3u₃x - 6(22)

264 = -108 + 3u₃x

3u₃x = 372

u₃x = 372/3

u₃x = 124 m/s

Resultant velocity

u₃ = √(124² + 2.28²)

u₃ = 124.02 m/s

Direction of the velocity

tanθ = u₃y/u₃x

tanθ = 2.28/124

tanθ = 0.018

θ = 1.05⁰ (below the horizontal)

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How to magnetic stripes provide evidence of seafloor spreading

Answers

Magnetic stripes provide evidence of seafloor spreading due to their occurrence in the region of sea floor.

How to magnetic stripes provide evidence of seafloor spreading?

When the Earth's magnetic field reverses, a new stripe starts to form. Such magnetic patterns enable the scientist to recognize the occurrence of sea-floor spreading and provides strongest evidence for the theory of plate tectonics.

So we can conclude that magnetic stripes provide evidence of seafloor spreading due to their occurrence in the region of sea floor.

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Two particles P and Q each moves towards ther along Straight Line (M)(N), 51m long. P starts from M with velocity 5m/s and constant acceleration Of 1m/s². Q Starts from N at the Same time with velocity 6m/s and at a Constant acceleration of 3m/s². find the time when the
a) Particles are 30m apart, 13
b) Particles meet
c) velocity of P is ¾ of the velocity of Q


Answers

The time required by the particles are as follows:

a. t = 1.5 seconds

b. t = 3 seconds

c. t = 0.4 seconds

What is the time required?

The time required for the particles to be at several distances apart is calculated using the equation of motion given below:

[tex]S = ut + \frac{1}{2}at^{2}[/tex]

a) Time required to be 30 m apart:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51 - 30

S1 + S2 = 21

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 21

2t^2 + 11t - 21 = 0

Solving for time, t by factorization, t = 1.5 seconds

b) Time required to meet:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 51

2t^2 + 11t - 51 = 0

Solving for time, t by factorization, t = 3 seconds

c) Time required for velocity of P is ¾ of the velocity of Q:

Using the equation of motion: V = u + at

Vp = 3/4 Vq

4Vp= 3Vq

Substituting the values:

4(5 + t) = 3(6 + 3t)

5t = 2

t = 0.4 seconds

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The volume of the water in the graduated cylinder rose as some of the water was displaced by the table tennis ball. Find the volume of the ball using the formula

Vb = Vf – Vi
where Vb is the volume of the ball, Vf is the final volume of the water, and Vi is the initial volume of the water. Record the volume of the ball in Table A of your Student Guide.

What is the volume of the table tennis ball?

cm3

Answers

The approximate volume of table tennis ball is  80 cm³

What is volume?

Volume is defined as the amount of space occupied by the three dimensional object. S I unit of volume is m³ or cm³.

To find the volume of tennis ball using graduated cylinder.

Step 1 - Fill the graduated cylinder half or full.

Step 2 - Mark the initial volume of the water i.e. 100 cm³ (Vi)

Step 3  - Put the tennis ball in the graduated cylinder. Some of the water was displaced by the table tennis ball.

Step 4 - Mark the Final volume of the water (Vf) i.e. 180 cm³

Step 5 = Calculate the volume by using Formula

Vb = Vf – Vi = 180 cm³ - 100 cm³ = 80 cm³

Hence the volume of tennis ball (Vb) is 80 cm³

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Answer:

40

Explanation:

the work function for magnesium is 3.70 ev. what is its cutoff frequency?

Answers

The cutoff frequency for magnesium is 8.93 x 10¹⁴ Hz.

What is cutoff frequency?

The work function is related to the frequency as

W0 = h x fo

where, fo = cutoff frequency and h is the Planck's constant

Given is the work function for magnesium is  3.70 eV.

fo = 3.7 x 1.6 x 10⁻¹⁹ / 6.626 x 10⁻³⁴

fo = 8.93 x 10¹⁴ Hz.

Thus, the cut off frequency is 8.93 x 10¹⁴ Hz.

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