electron and proton are projected with same velocity normal to the magnetic field which one will suffer greater deflection​? why

Answer:

Explanation:

I've been doing these types of problems for many years and I don't think I've ever seen an "efficiency" rating on a ramp.

I'm going to ASSUME that 20% efficient means that 80% of the Potential energy that gets converted becomes system internal heat energy.

Potential energy at the start of a 2.0 m slide

PE = mgh = mg2sin30 = mg2(½) = mg J

0.8mg J gets converted to heat and 0.2mg converts to kinetic energy

0.2mg = ½mv²

v² = 0.4g

v = √(0.4(9.8)) = 1.979898... ≈ 2.0 m/s

Answer:

Explanation:

F = GMm/d²

d = √(GMm/F)

d = √(6.674e-11(8.68e25)(6.59e19) / 2.28e19)

d = 1.29398e8 = 1.29 x 10^8 m  center to center

Answer:

1.29 x 10^8 m  apart

Explanation:

Works in Acellus!

Answer:

Instrument Use

Accelerometer Measures acceleration

Altimeter Measures altitude of an aircraft

Ammeter Measures electric current in ampere

Anemometer Measures wind speed

Answer:b

Explanation:

-29.4 m/s

The velocity of the ball dropped from 80 m if it reaches the ground within 3 seconds is 26.6 m/s. If it is in midway within this time, then the velocity will be 29.4 m/s.

Velocity of a moving object is the measure of its distance travelled per unit time. Velocity is a vector quantity having both magnitude and direction. Acceleration is the rate of change in velocity.

Given that, height of the building = 80 m

the ball is moving downwards by acceleration due to gravity g = 9.8 m/s².

Then after 3 seconds, the velocity of the ball is calculated as follows:

velocity = acceleration × time

v = 9.8 m/s²  × 3 s = -29.4 m/s

If the ball reaches the ground within the time of 3 s, then, the velocity is:

v = 80 m/3s = 26.6 m/s.

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Answer:

In this problem basically we will use directions to find displacement and distance .

(Keep in mind that displacement is vector while distance is scalar quantity)  

And then we will use displacement and distance to find average velocity and average speed.

Explanation:

Answer:

626m

Explanation:

Answer: A , D & F

Explanation:

Answer:

At the highest point when you toss a ball into the air.

Explanation:

At the higest point of a trajectory of a ball, the velocity is zero for a split second and there is no speed and direction. However, there still is acceleration of -10 m/s^2 because the force of gravity is still acting upon it at that point.

Hi there!

An example of this could be when a ball is thrown vertically into the air and reaches the TOP of its trajectory.

When an object is thrown with a vertical velocity, the acceleration due to gravity results in a decrease in its positive (upward) velocity until it reaches its highest point, where the instantaneous velocity = 0 m/s and the object begins to fall back down (negative velocity).

Additionally, throughout its entire trajectory, the ball experiences an acceleration due to gravity of g = 9.8 m/s², even at its highest point where there is a velocity = 0 m/s.

Answer:

Papillae is correct

Explanation:

hope it helps you

Answer:

Papillae is the correct answer of this question

Answer:

5m

Explanation:

x=?

[tex]sin30^{0} =\frac{x}{10}[/tex]

[tex]x=10sin30^{0} =10(0.5)=5[/tex]

Hope this helps

Find the distance the ball travels:

3.4 meters - 1.5 meters = 1.9 meters

Now divide the distance the ball travels by the speed:

1.9 meters / 4.4 m/s = 0.43 seconds

Answer:

Explanation:

s = s₀ + v₀t + ½at²

There are an infinite number of solutions to this question as posed because we are not told the direction of the initial velocity.

Assuming ground is level and origin and UP the positive direction

The shortest amount of time possible is when the initial velocity is straight down

1.5 = 3.4 - 4.4t + ½(-9.8)t²

0 = -4.9t² - 4.4t + 1.9

t = (4.4 ±√(4.4² - 4(-4.9)(1.9))) / (2(-4.9))

positive answer is

t = 0.32 s

The longest amount of time possible is when the initial velocity is straight up.

1.5 = 3.4 + 4.4t + ½(-9.8)t²

0 = -4.9t² + 4.4t + 1.9

t = (-4.4 ±√(4.4² - 4(-4.9)(1.9))) / (2(-4.9))

positive answer

t = 1.22 s

If the initial velocity is horizontal, meaning no vertical velocity

1.5 = 3.4 + 0t + ½(-9.8)t²

-4.9t² = -1.9

t² = 0.38775...

t = 0.62 s

Any angle between UP and Down will have a different initial vertical velocity and result in a different time to catch height.

It appears from the comments on the other answer, that I have shown you how to arrive at three of the four possible solutions.  The initial direction is very important.

Answer:

Explanation:

0.120 rev/s(2π rad/rev) = 0.24π rad/s

At the highest point of the arc, gravity must supply the required centripetal acceleration. As the normal force is 1/4 of his normal weight, then 3/4 of gravity acceleration must be used as centripetal acceleration

0.75g = ω²R

R = 0.75(9.81) / (0.24π)²

R = 12.942198...

R = 12.9 m

The radius of the circle is equal to 12.95m which is rotating with an angular velocity of 0.120 rev/s.

A body spins in a vertical circle so that its motion at different points is different from the motion of the body is said to be vertical circular motion.

The velocity and tension vary in maximum magnitude from the lowest to the highest position because of the effect of the gravitational force of the earth.

Given, the angular velocity of the Ferris wheel, ω = 0.120 rev/s

ω = 0.120 rev/s × 2π rad/rev

ω= 0.7536 rad/s

If r is the radius of the circle and 'm' is the mass of the jack.

From newton's second law of motion, the net force will be equal to

mg - N = mrω²

mg - (mg/4) = mrω²

r = 3g/4ω²

r = 3×9.81 / (4× 0.7536)

r = 12.95 m

Therefore, the radius of the circle in which the jack travels is equal to 12.95m.

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Your question is incomplete, most probably the complete question was,

Jack sits in the chair of a Ferris wheel that is rotating at a constant 0.120 rev/s. As Jack passes through the highest point of his circular part, the upward force that the chair exerts on him is equal to one-fourth of his weight. What is the radius of the circle in which Jack travels?

(a) The height of the roller coaster at 120,000 potential energy is 122.45 m.

(b) The velocity of the roller coaster at 0 J kinetic energy is 0.

(c) The height of the roller coaster at 60,000 potential energy is 61.23 m.

(d) The velocity of the roller coaster at 60,000 J kinetic energy is 34.64 m/s.

(e) The height of the roller coaster at 40,000 potential energy is 40.82 m.

(f) The velocity of the roller coaster at 80,000 J kinetic energy is 40 m/s.

The given parameters:

When the kinetic energy = 0 and potential energy = 120,000 J

The height of the roller coaster is calculated as follows;

P.E = mgh

[tex]h = \frac{P.E}{mg}\\\\h = \frac{120,000}{100 \times 9.8} \\\\h = 122.45 \ m[/tex]

Since the kinetic energy = 0, the velocity of the roller coaster = 0

When the potential energy, P.E = 60,000 J, the kinetic energy, K.E is calculated as;

P.E + K.E = M.A

P.E + K.E = 120,000

60,000 + K.E = 120,000

K.E = 120,000 - 60,000

K.E = 60,000 J

The height of the roller coaster at 60,000 potential energy is calculated as follows;

[tex]h = \frac{P.E}{mg} \\\\h = \frac{60,000}{100 \times 9.8} \\\\h =61.23 \ m[/tex]

The velocity of the roller coaster at 60,000 J kinetic energy is calculated as follows;

[tex]K.E = \frac{1}{2} mv^2\\\\v^2 = \frac{2K.E}{m} \\\\v = \sqrt{ \frac{2K.E}{m}} \\\\v = \sqrt{ \frac{2\times 60,000}{100}}\\\\v = 34.64 \ m/s[/tex]

When the potential energy, P.E = 40,000 J, the kinetic energy, K.E is calculated as;

P.E + K.E = M.A

40,000 + K.E = 120,000

K.E = 120,000 - 40,000

K.E = 80,000

The height of the roller coaster at 40,000 potential energy is calculated as follows;

[tex]h = \frac{P.E}{mg} \\\\h = \frac{40,000}{100 \times 9.8} \\\\h = 40.82 \ m[/tex]

The velocity of the roller coaster at 80,000 J kinetic energy is calculated as follows;

[tex]v = \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2\times 80,000}{100} } \\\\v = 40 \ m/s[/tex]

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Answer:

b Shiny

Explanation:

Trust me it's right

Answer:

A saturated vapour is one that is in balance with its own liquid. Space is considered to be unsaturated if it contains fewer vapours than the maximum amount it can retain at a given temperature. Saturated vapour pressure is unaffected by volume, although it rises as temperature rises.

Explanation:

Answer:

According Lisa, both the ball and the balloon have the same forward velocity of Vx.

(D) is correct

The self-inductance of the solenoid is 8.25 mH.

The given parameters;

The area of the solenoid is calculated as follows;

[tex]A = \pi r^2\\\\A = \pi \times (0.02)^2\\\\A = 0.00125 \ m^2[/tex]

The self-inductance of the solenoid is calculated as follows;

[tex]L = \frac{\mu_o N^2 A }{l} \\\\L = \frac{(4\pi \times 10^{-7}) \times 1700^2 \times 0.00125}{0.55} \\\\L = 0.00825 \ H\\\\L = 8.25\ mH[/tex]

Thus, the self-inductance of the solenoid is 8.25 mH.

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Answer:

not sure i need points

Explanation:

Answer:

how many feet?

Explanation:

Answer:

Explanation:

F = ma

a = F/m

a = mBg / (mB + mA)

a = 3.5(9.8)/(3.5 + 3.4)

a = 4.971014...

a = 5.0 m/s²

If you want to use individual Free Body Diagrams

mass A will have downward weight and upward normal forces equal at mAg

and a horizontal force of string tension T

F = ma

T = mAa

mass B will have a downward force of mBg and an upward force of T

mBg - T = mBa

substitute for T

mBg - mAa = mBa

mBg = a(mB + mA)

a = mBg / (mB + mA)   which is identical to the above answer.

Answer:

The Aluminum

Explanation:

With a larger specific heat, water requires more heat to raise its temperature by a temperature degree.

In this system, with equal masses of water and aluminum, the heat moving from the aluminum lowering its temperature by one degree is not sufficient to raise the water temperature by one degree.

The energy in the system is given by the initial potential energy at the point 1.

The linear velocity at point 3, is approximately 33.59 m/s.

Reasons:

The parameters are;

Height at point 1, h₁ = 83 m

Radius of the ring = 8 cm

Mass of the ring, M = 8 kg

Height at point 2, h₂ = 32 m

At point 2, we have;

Change in potential energy = Kinetic energy

Which gives;

(83 - 32) × 9.81 × 8 = 0.5 × 8 × v² + 0.5 × 8 × 0.08² × (v/0.08)²

Which gives;

v ≈ 22.37 m/s

At point 3, the rotational kinetic energy remains constant while the

translational kinetic energy increases as follows;

K.E. at point 3 = Initial kinetic energy + Change in potential energy

Which gives;

K.E. at point 3 = 0.5 × 8 × v₃³ ≈ 0.5×8×22.37² + 32×9.81×8

[tex]v_3^2 = \dfrac{0.5 \times 8 \times 22.37^2 + 32 \times 9.81 \times 8}{0.5 \times 8} = 1128.15[/tex]

v₃ ≈ √(1128.15) ≈ 33.59

The linear velocity at point 3, v₃ ≈ 33.59 m/s

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The probable question parameters as obtained from a similar question online are;

Height at point 1, h₁ = 83 m

Radius of the ring = 8 cm

Mass of the ring, M = 8 kg

Height at point 2, h₂ = 32 m

This problem is describing a pump from which water is discharged at 1 MPa and 165 °C and is asking for the specific volume and internal energy at those conditions, thus, we can use the steam tables for resolving this requirement.

First of all, we need to remember that water can be a saturated liquid, vapour or liquid-vapour mixture, and this is determined for the temperature and pressure it is at.

In this case, we find that at 165 °C the saturation pressure is about 0.6178 MPa; this means we are referring to a saturated liquid so that both the specific volume and internal energy can be simply read from the steam tables as vf and uf as follows:

[tex]v=0.001127\frac{m^3}{kg}\\\\u=761.67 \frac{kJ}{kg}[/tex]

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. Power

Patintero

, also known as harangang-taga or tubigan, (Intl. Translate: Escape from the hell or Block the runner) is a traditional Filipino children's game. Along with tumbang preso, it is one of the most popular outdoor games played by children in the Philippines.[1]

is an ultramarathon trail race held in Frozen Head State Park near Wartburg, Tennessee. If runners complete 60 miles (97 km) this is known as a "fun run." The full course is about 100 miles (160 km). The race is limited to a 60-hour period and takes place in late March or early April of each year.

The angular speed of the merry-go-round reduced more as the sandbag is

placed further from the axis than increasing the mass of the sandbag.

The rank from largest to smallest angular speed is presented as follows;

[m = 10 kg, r = 0.25·R]

              [tex]{}[/tex] ⇩

[m = 20 kg, r = 0.25·R]

              [tex]{}[/tex] ⇩

[m = 10 kg, r = 0.5·R]

              [tex]{}[/tex] ⇩

[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]

              [tex]{}[/tex] ⇩

[m = 10 kg, r = 1.0·R]

Reasons:

The given combination in the question as obtained from a similar question online are;

1: m = 20 kg, r = 0.25·R

2: m = 10 kg, r = 1.0·R

3: m = 10 kg, r = 0.25·R

4: m = 15 kg, r = 0.75·R

5: m = 10 kg, r = 0.5·R

6: m = 40 kg, r = 0.25·R

According to the principle of conservation of angular momentum, we have;

[tex]I_i \cdot \omega _i = I_f \cdot \omega _f[/tex]

The moment of inertia of the merry-go-round, [tex]I_m[/tex] = 0.5·M·R²

Moment of inertia of the sandbag = m·r²

Therefore;

0.5·M·R²·[tex]\omega _i[/tex] = (0.5·M·R² + m·r²)·[tex]\omega _f[/tex]

Given that 0.5·M·R²·[tex]\omega _i[/tex] is constant, as the value of  m·r² increases, the value of [tex]\omega _f[/tex] decreases.

The values of m·r² for each combination are;

Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²

Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²

Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²

Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²

Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²

Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²

Therefore, the rank from largest to smallest angular speed is as follows;

Combination 3 > Combination 1 > Combination 5 = Combination 6 >

Combination 2

Which gives;

[m = 10 kg, r = 0.25·R] > [m = 20 kg, r = 0.25·R] > [m = 10 kg, r = 0.5·R] > [m =

10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R] > [m = 10 kg, r = 1.0·R].

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Answer:

everything will float up and go up to space and die

Explanation:

gravity keeps us down and once it stops everything will float up. And if it were to stop everywhere everything and everyone will die and everything will be destroyed.

The answer is C) kilogram

The SI unit in the given options is kilogram, the correct option is C.

The most common unit of measurement is the standard unit. These units are standardized, which means that everyone understands an object's size, weight, and other properties.

Standard Units of Measurement have fixed values that cannot be changed.

The International System of Units, abbreviated SI in all languages and sometimes pleonastically as the SI system, is the modern version of the metric system and the world's most widely used measurement system.

The SI is a coherent system of units of measurement that begins with seven base units: the second (symbol s, the unit of time), the metre (m, length), the kilogram (kg, mass), the ampere (A, electric current), and so on.

Thus, the correct option is C.

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Answer:

Approximately [tex]2\; \rm s[/tex], assuming that the floor of this parking lot is level, [tex]\mu_{\rm k} = 0.60[/tex], and [tex]g = 9.81\; \rm m\cdot s^{-2}[/tex].

Explanation:

Let [tex]m[/tex] denote the mass of this vehicle. Weight of this vehicle: [tex]m\, g[/tex].

If the floor of this parking lot is level, the normal force on this vehicle would be equal to its weight: [tex]N = m \, g[/tex].

Given that [tex]\mu_{\rm k}[/tex], the kinetic friction between this vehicle and the ground would be consistently [tex]\mu_{\rm k} \, N = \mu_{\rm k} \, m \, g[/tex] until the vehicle comes to a stop.

Assuming that all forces on this vehicle other than friction are balanced. The net force of this vehicle during braking would be [tex](-\mu_{\rm k} \, m \, g)[/tex] (negative because this force is opposite to the direction of the motion.)

By Newton's second law of motion, the acceleration of this vehicle would be:

[tex]\begin{aligned}a &= \frac{F_\text{net}}{m} \\ &= \frac{-\mu_{\rm k} \, m \, g}{m} \\ &= -\mu_{\rm k}\, g \\ &= -0.60 \times 9.81\; \rm m\cdot s^{-2} \\ &= -5.886\; \rm m\cdot s^{-2}\end{aligned}[/tex].

In other words, braking would reduce the velocity of this vehicle by a constant [tex]5.886\; \rm m\cdot s^{-1}[/tex] every second until the vehicle comes to a stop. Calculate the time it would take to reduce the velocity of this vehicle from [tex]v_{0} = 12\; \rm m\cdot s^{-1}[/tex] to [tex]v_{1} = 0\; \rm m\cdot s^{-1}[/tex]:

[tex]\begin{aligned}t &= \frac{v_{1} - v_{0}}{a} \\ &= \frac{0\; \rm m\cdot s^{-1} - 12\; \rm m\cdot s^{-1}}{-5.886\; \rm m \cdot s^{-2}} \\ &\approx 2.0\; \rm s \end{aligned}[/tex].

Acceleration is the slope of the velocity-time graph. Since the acceleration here is constant, the velocity-time graph of this vehicle would be a line with a negative slope.