Explain how you would have to change the experimental set-up (electric field strength, drift chamber length, and/or buf

Answer:

[tex]E=6.86\times 10^6\ N/C[/tex]

Explanation:

Given that,

Mass of the sphere, m = 2.1 g = 0.0021 kg

Charge, q = 3 nC

We need to find the magnitude of the electric field that balanced the weight of sphere. Let it is E. So,

qE = mg

[tex]E=\dfrac{mg}{q}[/tex]

Put all the values,

[tex]E=\dfrac{0.0021\times 9.8}{3\times 10^{-9}}\\\\E=6.86\times 10^6\ N/C[/tex]

So, the magnitude of the elecric field is [tex]6.86\times 10^6\ N/C[/tex].

Answer:

Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge

Explanation:

The electric field for a uniform line of charge is given by E = λ/2πε₀r where λ = charge density and r = distance from line of charge.

If λ is negative, E is negative so it points in the negative direction towards the line of charge.

Also, since for negative charges, electric field lines end up in them, the electric field for an infinitely long negative line of charge points towards the charge perpendicular to it.

Also as r increases, E decreases since E ∝ 1/r

So, the electric field decreases at larger distances from the line of charge.

So, the electric field of a negative infinite line of charge Points perpendicularly toward the line of charge and decreases in strength at larger distances from the line charge.

Answer:

C

Explanation:

Answer: 321 J

Explanation:

Given

Mass of the box [tex]m=3\ kg[/tex]

Force applied is [tex]F=25\ N[/tex]

Displacement of the box is [tex]s=15\ m[/tex]

Velocity acquired by the box is [tex]v=6\ m/s[/tex]

acceleration associated with it is [tex]a=\dfrac{F}{m}[/tex]

[tex]\Rightarrow a=\dfrac{25}{3}\ m/s^2[/tex]

Work done by force is [tex]W=F\cdot s[/tex]

[tex]W=25\times 15\\W=375\ J[/tex]

change in kinetic energy is [tex]\Delta K[/tex]

[tex]\Rightarrow \Delta K=\dfrac{1}{2}m(v^2-0)\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 6^2\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 36\\\\\Rightarrow \Delta K=54\ J[/tex]

According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy

[tex]\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J[/tex]

Therefore, the magnitude of work done by friction is [tex]321\ J[/tex]

Answer:

The answer is Newton's third law of motion states that every action has an equal and opposite reaction. This means that force always act in pairs

The pair of forces described by Newton third law must be in opposite direction.

Every action have equal and opposite reaction. for example when we fire bullet from a gun, the gun will recoil back and bullet moves forward. In case of rocket, rocket is fired, thrust is reaction of force applied by the gas on the floor.

The motion of lift from an airfoil in which the air is diverted downward by the airfoil's action and the wing is pushed upward in response.

When a spinning ball moves, the air is deflected to one side, and the ball responds by travelling in the other direction.

A jet engine's motion generates thrust, and hot exhaust gases rush out the back of the engine, producing thrust in the opposite direction.

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Answer:

The magnitude of the average force exerted by the club on the ball during contact = mv/t

Explanation:

Impulse exerted on the ball = Momentum of the ball = mass * velocity = m*v

As we know,  

m*v = Integration of F.dt with limits 0 to T

Ft = mv

F = mv/t

The magnitude of the average force exerted by the club on the ball during contact = mv/t

Answer:

[tex]a=4\ m/s^2[/tex]

Explanation:

Given that,

Initial speed of a body, u = 0

Final speed of the body, v = 20 m/s

Time, t = 5 s

We need to find the acceleration of the body. We know that the acceleration of an object is equal to the rate of change of velocity divided by time taken. So,

[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{20-0}{5}\\\\a=4\ m/s^2[/tex]

So, the body's acceleration is equal to [tex]4\ m/s^2[/tex].

Answer:

K.E = 2450 Joules

Explanation:

Given the following data;

Mass = 0.01 kg

Velocity = 700 m/s

To find the kinetic energy;

La energía cinética (K.E) se puede definir como una energía que posee un objeto o cuerpo debido a su movimiento.

Matemáticamente, la energía cinética viene dada por la fórmula;

[tex] K.E = \frac{1}{2}mv^{2}[/tex]

Sustituyendo en la fórmula, tenemos;

K.E = ½ * 0.01 * 700²

K.E = 0.005 * 490000

K.E = 2450 Joules

Explanation:

thankyou fo

r the poimts

Answer:

I think the particles transferred between brandi's body and the door knob causing the shock is the electrons.

Answer:

Explanation:

the force of the rocket engine pushing it up,  the force of gravity pulling it down,    maybe some force of air resistance as the rocket goes fast,   hmmm    Free Body Diagrams  (FBD)  should have any and all forces on the model,  unless they are negligible . or so slight they really make little difference in the total  outcome.  

Answer:

speed = 0.9 mm/s

Explanation:

time, t = 40 s

initial angular speed, wo = 0 rad/s

final frequency, f = 1/1.03 rps = 0.97 rps

final angular speed, w = 2 x 3.14 x 0.97 = 6.1 rad/s

time, t = 40 s

distance, r = 5.9 mm

The angular acceleration is given y the first equation of motion.

[tex]w =wo + \alpha t\\6.1 = 0 +\alpha \times 40\\\alpha = 0.1525 rad/s^{2}[/tex]

The linear velocity is

[tex]v =5.9\times 10^{-3}\times 0.1525 = 9\times 10^{-4} m/s[/tex]

speed, v = 0.9 mm/s

Answer:

the number of electrons should equal to the the number of protons in a neutral atom

if there is a inequality between the numbers it means the atom has a + or - charge

Answer:

both the same

Explanation:

When a ball is thrown vertically upwards, it experiences that same acceleration due to gravity as an object thrown directly downwards.

This means that if we ignore the effects of air resistance, and the two balls have the same initial speed, they are expected both expected to hit the ground at the same speed as a result of the principle of conservation of energy.

Answer:

Both balls will hit the ground at the same speed because air resistance is ignored.

Explanation:

Answer:

D

Explanation:

Because the y axis is meter. If it is straight line at time and meter graph then it velocity and speed is 0

Answer:

[tex]\frac{1}{2}mR^2\omega^2[/tex]

Explanation:

The rotational kinetic energy of an object is given by [tex]KE_r=\frac{1}{2}I\omega^2[/tex], where [tex]I[/tex] is the object's moment of inertia/rotational inertia and [tex]\omega[/tex] is the object's angular speed.

What we're given:

Since no numerical value is given for any of these, it is implied the desired answer will be an equation in terms of the variables given.

Substituting [tex]I=MR^2[/tex]:

[tex]KE_r=\boxed{\frac{1}{2}mR^2\omega^2}[/tex]

Answer:

1.07 × 10⁸ m/s

Explanation:

Using the relativistic Doppler shift formula which can be expressed as:

[tex]\lambda_o = \lambda_s \sqrt{\dfrac{c+v}{c-v}}[/tex]

here;

[tex]\lambda _o[/tex] = wavelength measured in relative motion with regard to the source at velocity v

[tex]\lambda_s =[/tex] observed wavelength from the source's frame.

Given that:

[tex]\lambda _o[/tex] = 656.3 nm

[tex]\lambda_s =[/tex] 953.3 nm

We will realize that [tex]\lambda _o[/tex] > [tex]\lambda_s[/tex]; thus, v < 0 for this to be true.

From the above equation, let's make (v/c) the subject of the formula: we have:

[tex]\dfrac{\lambda_o}{\lambda_s}=\sqrt{\dfrac{c+v}{c-v}}[/tex]

[tex]\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2=\dfrac{c+v}{c-v}[/tex]

[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2-1}{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2+1}[/tex]

[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{656.3}{953.3} \Big)^2-1}{\Big(\dfrac{656.3}{953.3} \Big)^2+1}[/tex]

[tex]\dfrac{v}{c} =0.357[/tex]

v = 0.357 c

To m/s:

1c = 299792458 m/s

∴

0.357c = (299 792 458 × 0.357) m/s

= 107025907.5 m/s

= 1.07 × 10⁸ m/s

Answer:

D

Explanation:

Potential energy involves the change of an object's position, which in this case a rocket is increasing its vertical displacement from the ground.

When a rocket is increasing its vertical displacement from the ground, it exhibits both potential and kinetic energy. Therefore option D is correct.

At the initial stage, when the rocket is on the ground and not moving, it possesses potential energy. This potential energy is in the form of stored energy due to its elevated position above the ground.

As the rocket launches and gains altitude, it continues to accumulate potential energy because it is moving higher against the force of gravity.

Simultaneously, as the rocket moves upward, it also gains kinetic energy. Kinetic energy is the energy associated with the rocket's motion.

The faster the rocket moves, the greater its kinetic energy becomes. As the rocket ascends, its speed increases, resulting in an increase in kinetic energy.

Therefore, in the context of a rocket increasing its vertical displacement from the ground, both potential energy (due to its height) and kinetic energy (due to its motion) are present.

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Answer:

you can simply answer p1v1=p2v2 p2 = p1v1÷ p2 then inter what given in the formula you get it pl=1.20x 105pa vl= 0.40m3 v2=0.025m3

Answer:

39.40 MeV

Explanation:

Determine the minimum possible Kinetic energy

width of region = 5 fm

From Heisenberg's uncertainty relation below

ΔxΔp ≥ h/2 , where : 2Δx = 5fm ,  Δpc = hc/2Δx = 39.4 MeV

when we apply this values using the relativistic energy-momentum relation

E^2 = ( mc^2)^2 + ( pc )^2 = 39.4 MeV ( right answer ) because the energy grows quadratically in nonrelativistic approximation,

Also in a nuclear confinement ( E, P >> mc )

while The large value will portray a Non-relativistic limit  as calculated below

K = h^2 / 2ma^2 = 1.52 GeV

Answer:

A. the change in smell

Explanation:

sry if its wrong it's what I know

Answer:

The answer is " because it is in the opposite direction of the bullet".

Explanation:

Given:

[tex]m_1= 70 \ kg\\\\m_2 = 0.01\ kg\\\\v_2 = 500\ \frac{m}{s}\\\\v_1 = ?[/tex]

Using formula:

It recoils the velocity of the rifle [tex]= -0.07142 \ \ \frac{m}{s}[/tex]  

Its negative sign displays the opposite direction of the rifle.

speed = 40 m/s

Explanation:

Since the object is dropped, V0y = 0.

Vy = V0y - gt

= -(10 m/s^2)(4 s)

= -40 m/s

This means that its velocity is 40 m/s downwards. Its speed is simply 40 m/s.

The speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.

There are three equations of motion given by  Newton

The first equation is given as follows

v = u + at

the second equation is given as follows

S = ut + 1/2×a×t²

the third equation is given as follows

v² - u² = 2×a×s

Keep in mind that these calculations only apply to uniform acceleration.

As given in the problem, we have to find the speed acquired by a freely falling object 4 seconds after being dropped from a rest position,

By using the first equation of motion,

v = u + at

initial velocity(u) = 0 m/s

acceleration(a) = 10 m/s²

v = 0 + 10×4

v = 40 meters/seconds

Thus, the speed acquired by a freely falling object 4 seconds after being dropped from a rest position would be 40 meters/seconds.

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Answer:

hydrogen

Explanation:

hope it helps bye.

Answer:

Newtons third law of motion: Balanced forces

Every action has a corresponding and opposing response, according to Newton's third law of motion. As a result, forces always work in pairs.   Once more, tug-of-war is a prime illustration.

The third law of motion by Newton states that equal, but diametrically opposed forces always act in pairs. There is an equal but opposite reaction to every action, to put it another way.

The forces are balanced if the pullers are exerting equal force but going in the opposite direction on either side of the rope. There is hence no motion.

Although equal and opposite in nature, action and reaction forces cannot be balanced since they act on separate things and do not cancel one another out.

Therefore, This means that when you push against a wall, the wall pushes back against you with an equal amount of force.

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Answer:

The cost of energy is $ 0.34.

Explanation:

The energy is the capacity to do work.

The energy is a scalar quantity and its SI unit is Joule.

The commercial unit of energy is kWh.

Cost of 1 kWh energy = $ 0.17

energy loss by standard window is 2 kWh .

So, the cost of lost of energy is

Cost = $ 0.17 x 2 = $ 0.34