Protons are launched with an initial speed 9550 m/s at 37° above the horizontal, into a region where a uniform vertical electric field is present. The protons hit a target at a horizontal distance 1.27 mm from the point where the protons are launched. Find the electric field, the time of flight, and the maximum height reached by the protons above the point where they were launched.

Answers

Answer 1

Answer:

E=720,5 N/C

H=0,24mm

Explanation:

Protons Are Launched With An Initial Speed 9550 M/s At 37 Above The Horizontal, Into A Region Where A

Related Questions

g uppose that 3 J of work is needed to stretch a spring from its natural length of 30 cm to a length of 45 cm. (a) How much work is needed to stretch the spring from 35 cm to 37 cm? (Round your answer to two decimal places.) .02 Incorrect: Your answer is incorrect. J (b) How far beyond its natural length will a force of 10 N keep the spring stretched? (Round your answer one decimal place.)

Answers

Answer:

W= 0.319992 J  and distance is 3.75 cm

Explanation:

The energy needed to stretch the spring from 30 cm to 45 cm = 3 J

Now we required to find the requirement of energy when the spring is stretched from 35 cm to 37 cm.

So first find the work done to stretch the spring from 35 cm to 45 cm.

Work done, w = (1/2) kx^2

3 = (1/2)k(0.45 – 0.30)^2

k = 266.66 N/m

now, x1 = 0.35 – 0.30 = 0.05m

x2 = 0.37 – 0.30 = 0.07m

Now the amount of work done to stretch from 35cm to 37.

w = (1/2) k (x2^2 – x1^2)

w = (1/2) (266.66) (0.07^2 – 0.05^2 )

w= 0.319992 J

(b). Given F = 10 N

F = kx

x = F / k

x = 10 / 266.66

x = 0.0375m

x = 3.75 cm

Thus, distance is 3.75cm

Suppose a radio signal (light) travels from Earth and through space at a speed of 3 × 10^8/ (this is the speed of light in vacuum). How far (in meters) into space did the signal travel during the first 10 minutes?

Answers

Answer:

Explanation:

we know that

s=vt here v is the speed and s is distance covered by the signals

given data

v=3*10^8

t=10 min we have to convert it into seconds

1 minute=60 seconds

so

10 minutes =10*60/1 =600 seconds

now putting the value of v and t we can find the value of s

s=vt

s=3*10^8*600

s=1.8*10^11m

i hope this will help you

Speed is the rate of distance over time.

The signal will travel [tex]1.8 \times 10^{11}[/tex] meters in the first 10 minutes

The given parameters are:

[tex]s = 3 \times 10^8\ ms^{-1}[/tex] ---the speed of light

[tex]t = 10\ min[/tex] -- the time of travel

The relationship between speed, distance (d) and time is:

[tex]s = \frac dt[/tex]

Make d the subject

[tex]d = s \times t[/tex]

Substitute values for t and s

[tex]d = 3 \times 10^8\ ms^{-1} \times 10\ min[/tex]

Convert minutes to seconds

[tex]d = 3 \times 10^8\ ms^{-1} \times 10 \times 60s[/tex]

So, we have:

[tex]d = 3 \times 10^8\ m \times 10 \times 60[/tex]

[tex]d = 18 \times 10^{10}m[/tex]

Rewrite as:

[tex]d = 1.8 \times 10^{11}m[/tex]

Hence, the signal will travel [tex]1.8 \times 10^{11}[/tex] meters in the first 10 minutes

Read more about speed at:

https://brainly.com/question/21791162

how to find range with height and velocity

Answers

Answer:

[tex]R= \frac{ v_ i\: ^2\:\times sin\:20}{g}[/tex]

Explanation:

Mathew has a filtration kit, which consists of a funnel, a flask, and filter papers. Which of these mixtures can he separate using filtration?

Answers

Answer:

C. Muddy Water

The bird that migrates the farthest is the Arctic tern. Each year, the Arctic tern travels
32,000 km between the Arctic Ocean and the continent of Antarctica. Most of the
migration takes place within two four-month periods each year. Assume an Arctic
tern completes the second half of its annual migration distance in 122 days. Also
assume that during this time the tern flies directly north. If the tern flies the same
distance each day, what is its velocity in kilometers per day?

Answers

Answer: 131.14km per day

Explanation: since the second half of the terns migration takes 122 days we can assume that the full migration would take 244 days. using this we can divide the total distance by the total amount of days it takes (because speed = distance/time) which is 32,000/244, which would be 131.14

What happens if we increase the value of the resistor in forward bias connection?

Answers

A series resistor is necessary in forward biasing because it cannot overflow the current. If we give the maximum amount of current, then diode will be burnt. ... then if we increase the voltage against the current we will get the curve.

A car has an initial velocity of 50 m/s and a constant
acceleration of 5 m/s2. What is the car's velocity after 3
seconds?

Answers

Vf=vi+at
Vf= (50m/s)+ (5m/s2)(3s)
Vf=65m/s

A cheetah runs at a constant velocity of 7 m/s. What is it’s acceleration in m/s/s
PLEASE HELP

Answers

Answer:

0 m/s²

Explanation:

Acceleration is the change in velocity over change in time.  If the velocity is constant, then the acceleration is 0.

20. Convert 36 km/hr to m/s.

Answers

The answer would be 10

Where do you feel that you are traveling at the fastest speed when on the swing?

Answers

Answer:

C

Explanation:

I think it's C, because at that point, you are going fastest. Sorry if im wrong, hope this helps.

Answer:

In between and the middle one

Explanation:

Question
In order for work to be done, what three things are necessary

Answers

Perseverance, good mind set, and work ethic
you will need work ethic, self control, and a good mindset

What are the two main types of star clusters?

Answers

Fuzzy clustering
Model-based clustering

Answer:

Open and globular

Explanation:

The space shuttle fleet was designed with two booster stages. If the second stage provides a thrust of 73 ​kilo-newtons and the space shuttle has an acceleration of 16,000 miles per hour squared ​, what is the mass of the spacecraft in units of​ pounds-mass ​?

Answers

Answer:

m = 81281.5 pounds.

Explanation:

Given that,

Force, F = 73 kN

Acceleration of the space shuttle, a = 16000 mi/h²

1 miles/h² = 0.0001241 m/s2

16000 mi/h² = 1.98 m/s²

We need to find the mass of the spacecraft.

According to Newton's second law,

F = ma

m is mass of the spacecraft

[tex]m=\dfrac{F}{a}\\\\m=\dfrac{73\times 10^3\ N}{1.98\ m/s^2}\\\\m=36868.68\ kg[/tex]

Since, 1 kg = 2.20462 pounds

m = 81281.5 pounds

Hence, the mass of the spacecraft is 81281.5 pounds.

HELP!!!
An arrow is shot into the air at an angle of 30.0 above the horizontal with a speed of 20.0 m/s. What are the x and y components of the
velocity of the arrow 1.0 s after it leaves the bowstring?

Answers

Answer:

Y(1s) = [tex]10\sqrt{3}[/tex] - 10.1

X(1s) = 10m/s

Explanation:

In annex I've done the explanation for the equations that I will just present here.

Assuming that the arrow stars from the position (0 ; 0) in the Cartesian Graphic, and with Xo and Yo the initial speeds:

[tex]Yo =\frac{\sqrt{3} }{2} . Vr\\Yo = \frac{\sqrt{3} }{2} . 20\\Yo = 10\sqrt{3} m/s \\Xo = \frac{1}{2} Vr\\Xo = 10m/s[/tex]

Ignoring friction with air, Xo = Xf

So, Xo is the same during all the movement.

X(1s) = 10m/s

For Yo is different. That component is suferring reductions from gravity.

We can find Yo(1s) with one the basic functions of cinematics:

Vf = Vo + at

Vf = Final Velocity

Vo = Start Velocity

a = aceleration - gravity (g) is negative here

t = time

Yf = Yo + gt

Yf = [tex]10\sqrt{3}[/tex] - 10.1

If you prefere, can be: Yf = 10. ([tex]\sqrt{3} - 1[/tex])

You drop a ball from a window located on an upper floor of a building. It strikes the ground with speed v. You now repeat the drop, but you ask a friend down on the ground to throw another ball upward at speed v. Your friend throws the ball upward at the same moment that you drop yours from the window. At some location, the balls pass each other. Is this location.

Answers

Answer:

 y = y₀ (1 - ½ g y₀ / v²)

Explanation:

This is a free fall problem. Let's start with the ball that is released from the window, with initial velocity vo = 0 and a height of the window i

          y = y₀ + v₀ t - ½ g t²

          y = y₀ - ½ g t²

for the ball thrown from the ground with initial velocity v₀₂ = v

         y₂ = y₀₂ + v₀₂ t - ½ g t²

     

in this case y₀ = 0

         y₂2 = v t - ½ g t²

at the point where the two balls meet, they have the same height

         y = y₂

         y₀ - ½ g t² = vt - ½ g t²

         y₀i = v t

         t = y₀ / v

since we have the time it takes to reach the point, we can substitute in either of the two equations to find the height

         y = y₀ - ½ g t²

         y = y₀ - ½ g (y₀ / v)²

         y = y₀ - ½ g y₀² / v²

        y = y₀ (1 - ½ g y₀ / v²)

with this expression we can find the meeting point of the two balls

How much power is needed to lift the 200-N object to a height of 10 m in 4 s?

Answers

Answer:

500 watts

Explanation:

Recall that the definition of power is the amount of energy delivered per unit of time.

In our case, the energy delivered is potential energy which we can estimate as the product of the weight of the object times the distance it is lifted above ground:

200 N x 10 m = 2000  Nm

then the power is the quotient of this potential energy divided the time it took to lift the object to that position:

Power = 2000 / 4   Nm/s = 500 Nm/s = 500 watts

If a 500kg elephant is sliding across a frictionless patch of ice, how much force is needed to keep the elephant from slowing down?

Answers

Answer:

4905N

Explanation:

The force needed to keep the elephant from slowing down is expressed as shown according to Newtons second law of motion.

Force = mass * acceleration due to gravity

Given

Mass of elephant = 500kg

acceleration due to gravity = 9.81m/s²

Force = 500*9.81

Force = 4905N

Hence the force needed to keep the elephant from slowing down is 4905N

NEED HELP FAST!!
As the air on the surface of the Earth warms, what happens to the density of the air?

A.It decreases
B.It increases
C.It remains constant
D.It decreases, then increase

Answers

Answer:

B

Explanation:

because when the air rises the density increases

We want to predict what will happen to the density of the air on the surface of the Earth when it warms up.

We will see that the correct option is D: "It decreases, then increases"

We know that the temperature of a given object (in this case a mass of air) is related to the kinetic energy of the particles that conform it.

As the temperature increases, the kinetic energy also increases, thus, the amount of motion of each particle increases, thus, the volume of the object increases.

Now remember that:

density = Mass/Volume.

So if the volume of something increases, we will see that the density decreases (as the volume is in the denominator).

Then if the temperature of the air increases, we will see that the density of the air in the surface decreases.

But it does not end there, as only the air near the surface suffers this change of density, we will have a denser mass of air (colder air) above it. And because it is denser (has more mass in the same volume) we can say that it is heavier.

Then eventually the hot air will rise, and the cold air will fall down, thus the density of the air in the surface increases again, as the colder and denser air comes near the surface.

This is one way of how wind currents are born.

Concluding we can see that the correct option is D: "It decreases, then increases"

If you want to learn more, you can read:

https://brainly.com/question/14482731

At a time of 30 seconds a runner passes a distance marker labeled "125 meters." If the velocity of the runner is +5.0 m/s, when did the runner pass the distance marker for 75 meters?

Answers

Answer:

Explanation

He runs at 5m/s, so in 30 s he should be at 150m.  So you have to do 125m - 150m and you'll get -25m, this is his initial position. They want to know the time when he hits 75m, so you would do 75 + 25, and get 100. Then do 100m / 5m/s, and you will get 25 seconds.

Thermal effects refers to the:

Answers

Answer:

removal of heat by cooling towers

Yup I agree with that answer

please help on magnitude !

Answers

Answer:

Yes, the answer is 100 N

Explanation:

Car is pushed right with 150 N and obviously frictional force will act in the opposite direction of the fore provided and that is 50 N

Hence, 150 - 50 is 100 N

Magnitude is 100 N

Direction is Towards the right

Since they asked for Magnitude the answer should be 100 N

Which inference about Arthur is supported by the text?

Answers

There is no picture

A runner completes the 300-meter dash in 38 seconds. What is the speed of the runner? Round your answer the answer to the nearest tenth.

Answers

Answer:

speed = 7.9 m/s

Explanation:

speed = total distance / time taken

speed = 300 / 38

speed = 7.89473684 m/s

to the nearest tenth

speed = 7.9 m/s

three short square wood blocks measuring 3.5 per side support a machine weighing 29500 lbs. What is the compressive stress in the blocks
A. 798lbs
B. 421lbs
C. 1404lbs
D. 803lb

Answers

Answer:

D. 803 lbs

Explanation:

In order to find the compressive stress on all three blocks we first need to find the normal surface area of each:

Surface Area of 1 Block = 3.5 x 3.5

Surface Area of 1 Block = 12.25

Surface Area of all 3 Blocks = A = 3 x 12.25

Area = 36.75

Now, the stress is given by the following formula:

Stress = Force/Area

Stress = 29500 lbs/36.75

Stress = 802.72 lbs

Hence, the correct option will be:

D. 803 lbs

A large negatively charged object is placed on a wooden table. A neutral metallic ball rolls straight towards the object but stops before it touches it. A second neutral metallic ball rolls up along the path followed by the first ball, strikes the first ball driving it a bit closer to the negatively charged object and stops. After all balls have stopped rolling, the first ball is closer to the negatively charged object than is the second ball. At no time did either ball touch the charged object. Which statement is correct concerning the final charge on each ball

Answers

Answer:

the charge of each small sphere, which is + Q / 2

Explanation:

For this exercise we must use the fact that a charged object induces charges on nearby bodies

Induced charge comes from the fact that charges of the same sign repel and charges of different signs attract.

In this case the large, fixed ball with a -Q charge induces a positive charge in the nearest part and the negative charges are repelled to the furthest point, but the net charge on the metallic sphere remains zero. It should be emphasized that since the charges are of different signs, there is an attractive force between the two spheres.

This first metallic sphere now has a negative charge on the back, this charge induces a positive charge on the second sphere, as the charges are of a different sign, they attract each other, which is why the force is attractive.

When the first sphere stops the second sphere hits it, at this moment the charge of the two spheres is equal, therefore the induced charge in the two spheres is + Q. When the two spheres are separated, the charge on each of them is half, that is, the sphere has a charge + Q / 2 and the second sphere has a charge + Q / 2.

Therefore the first sphere is subjected to two forces: an attractive force with the large sphere of charge -Q and a repulsive force with the second sphere of charge + Q / 2.

 

So the first sphere must approach the big ball and the second sphere must move away from the big sphere.

This is the process of the movement of this exercise, unfortunately the statements with which to compare this process do not appear, but one of the most common questions of what is the charge of each small sphere, which is + Q / 2

Using the equation for for Newton's Second law, F=ma, solve the following problem. You have been given an object with a mass of 6g and an acceleration of 2 m/s2, what is the force?
Group of answer choices
A. 12N

B. 3N

C. 8N

D. 120N

Answers

Answer:

F = 0.012 N

Explanation:

Given that,

Mass of the object, m = 6 g

Acceleration, a = 2 m/s²

1 kg = 1000 grams

6 g = 0.006 kg

Force, F = ma

So,

[tex]F=0.006\ kg\times 2\ m/s^2\\F=0.012\ N[/tex]

So, the force is 0.012 N.

Two identical pellet guns are fired simultaneously from the edge of a cliff. These guns impart an initial speed of 23.7 m/s to each pellet. Gun A is fired straight upward, with the pellet going up and then falling back down, eventually hitting the ground beneath the cliff. Gun B is fired straight downward. In the absence of air resistance, how long after pellet B hits the ground does pellet A hit the ground

Answers

Answer:

Ta - Tb = 4.84 sec (approx)

Explanation:

Time taken by A = Time taken By B + Extra time

Ta - Tb = T

We know that

Time (T) = [v-u] /a ...................[a=9.8 m/s²]

So,

Ta - Tb = [v-u] /a

Ta - Tb = [-23.7-23.7] /9.8

Ta - Tb = 4.8367 sec

Ta - Tb = 4.84 sec (approx)

A Van de Graaff generator is charged so that a proton at its surface accelerates radially outward at 1.35 ✕ 1012 m/s2. Find the following. (a) the magnitude of the electric force on the proton at that instant magnitude N direction ---Select--- (b) the magnitude and direction of the electric field at the surface of the generator magnitude N/C direction ---Select---

Answers

Answer:

(a).

We know that force is

F = m a

So

F = (1.67 x 10^(-27) x (1.38 x10^12)

F = 2.3 x 10^-15 N facing the radially outward direction

(b).

Similarly Force for charge is

F = q E = m a

So relating

E = F/q = 2.3x 10^-15 /(1.6 10^ -19

E = 144.75 N/C facing the radially outward direction

The answer to the question is

(a) The force on the proton is [tex]2.254\times10^{-15}\,N[/tex] directed radially outwards.

(b) The electric field at the surface of the generator is [tex]14031.25\, N/C[/tex] directed radially outwards.

The answer can be explained as shown below.

Given that a proton accelerates radially outward at [tex]1.35\times 10^{12}\,m/s^2[/tex].

ie; [tex]a=1.35\times 10^{12}\,m/s^2[/tex]We know the mass of a proton, [tex]m_p = 1.67\times10^{-27}\,kg[/tex].

From Newton's second law we have,

[tex]F=mg[/tex]

But here, [tex]m=m_p[/tex]

So, the electric force on the proton is;

[tex]F = m_p\, a= (1.67\times10^{-27}\,kg)\, \times(1.35\times 10^{12}\, m/s^2)=2.254\times10^{-15}\,N[/tex]The force on the proton is [tex]2.254\times10^{-15}\,N[/tex] directed radially outwards.

Also, we know that, in electrostatics,

[tex]F=Eq\\ \\\implies E=\frac{F}{q}[/tex]The charge of a proton is, [tex]q=1.6\times 10^{-19}\,C[/tex]

Therefore, the electric field is given by,

[tex]E=\frac{2.254\times 10^{-15}N}{1.6\times 10^{-19}\,C} = 14031.25\, N/C[/tex]The electric field at the surface of the generator is [tex]14031.25\, N/C[/tex] directed radially outwards.

Learn more about the electric force here:

https://brainly.com/question/8960054

I'm confused, and I can't seem to get this question right. someone please help. Only two forces act on an object (mass = 5.00 kg), as in the drawing. (F = 55.0 N.) Find the magnitude and direction (relative to the x axis) of the acceleration of the object. (Drawing: 45 degrees, Fx is 40 N)

Answers

Answer:

17.6 m/s², 26.2° above x axis

Explanation:

Apply Newton's second law.

Sum of forces in the x direction:

∑Fₓ = ma

40 N + 55.0 N cos 45° = (5.00 kg) aₓ

aₓ = 15.8 m/s²

Sum of forces in the y direction:

∑Fᵧ = ma

55.0 N sin 45° = (5.00 kg) aᵧ

aᵧ = 7.78 m/s²

Use Pythagorean theorem to find the magnitude of the resultant.

a² = aₓ² + aᵧ²

a² = (15.8 m/s²)² + (7.78 m/s²)²

a = 17.6 m/s²

Use trigonometry to find the angle.

tan θ = aᵧ / aₓ

tan θ = (7.78 m/s²) / (15.8 m/s²)

θ = 26.2°

You must exert a force of 5N on a book to slide it across a table. If you do 2.5 J of work in the process, how far has the book moved?

Answers

Answer:

0.5 m

Explanation:

The following data were obtained from the question:

Force (F) applied = 5 N

Work done (W) = 2.5 J

Distance (s) =?

Work done (W) can be defined as the product of force (F) and distance (s) moved in the direction of the force. From the above definition, work done (W) can be represented mathematically as:

Work done (W) = Force (F) × Distance (s)

W = F × s

With the above formula, we can obtain the distance moved by the book as shown below:

Force (F) applied = 5 N

Work done (W) = 2.5 J

Distance (s) =?

W = F × s

2.5 = 5 × s

Divide both side by 5

s = 2.5/5

s = 0.5 m

Therefore, the book moved a distance of 0.5 m.

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