A hydrogen atom that has lost its electron is moving east in aregion where the magnetic field is directed from south to north. Itwill be deflected:_____________
a. up
b. down
c. not at all

Answer:

Average force = 67 mn

Explanation:

Given:

Initial velocity u = 0 m/s

Final velocity v = 67 m/s

Time t = 1 ms = 0.001 sec.

Computation:

Using Momentum theory

Change in momentum  = F × Δt

 (v-u)/t =  F × Δt

F × 0.001 = (67 - 0)/0.001

F= 67,000,000

Average force = 67 mn

Answer:

Approximately [tex]1.69 \times 10^{30}\; \rm kg[/tex].

Explanation:

Let [tex]M[/tex] and [tex]m[/tex] denote the mass of the star and the planet, respectively.

Let [tex]G[/tex] denote the constant of universal gravitation ([tex]G \approx 6.67408 \times 10^{-11}\; \rm m^{3} \cdot kg^{-1}\cdot s^{-2}[/tex].)

Let [tex]r[/tex] denote the orbital radius of this planet (assuming that [tex]r\![/tex] is constant.) The question states that [tex]r = 1.65 \times 10^{10}\; \rm m[/tex].

The size of gravitational attraction of the star on this planet would be:[tex]\displaystyle \frac{G \cdot M \cdot m}{r^{2}}[/tex].

If attraction from the star is the only force on this planet, the net force on this planet would be [tex]\displaystyle \frac{G \cdot M \cdot m}{r^{2}}[/tex].

Let [tex]\omega[/tex] denote the angular velocity of this planet as it travels along its circular orbit around the star. The size of [tex]\omega\![/tex] could be found from the period [tex]T[/tex] of each orbit: [tex]\omega = (2\, \pi) / T[/tex].

In other words, this planet of mass [tex]m[/tex] is in a circular motion with radius [tex]r[/tex] and angular velocity [tex]\omega[/tex]. Therefore, the net force on this planet should be equal to [tex]m \cdot \omega^2 \cdot r[/tex].

Hence, there are two expressions for the net force on this planet:

Equate the right-hand side of these two equations:

[tex]\displaystyle \frac{G \cdot M \cdot m}{r^2} = {\left(\frac{2\pi}{T}\right)}^{2}\, m \cdot r[/tex].

Simplify this equation and solve for [tex]M[/tex], the mass of the star:

[tex]\displaystyle M = \frac{{(2\pi / T)}^2 \cdot r^3}{G}[/tex].

Notice that [tex]m[/tex], the mass of the planet, was eliminated from the equation. That explains why this question could be solved without knowing the exact mass of the observed planet.

Convert the orbital period of this star to standard units:

[tex]\begin{aligned}T &= 14.5\; \text{day} \times \frac{24\; \text{hour}}{1\; \text{day}} \times \frac{3600\; \text{second}}{1\; \text{hour}} \\ & = 1.2528 \times 10^{6}\; \rm \text{second}\end{aligned}[/tex].

Calculate the mass of the star:

[tex]\begin{aligned} M &= \frac{{(2\pi / T)}^2 \cdot r^3}{G} \\ &\approx \frac{\displaystyle {\left(\frac{2\pi}{1.2528 \times 10^{6}\; \rm s}\right)}^{2} \times \left(1.65 \times 10^{10}\; \rm m\right)^{3}}{6.67408 \times 10^{-11}\; \rm m^{3}\cdot kg^{-1} \cdot s^{-2}}\\ &\approx 1.69 \times 10^{30}\; \rm kg\end{aligned}[/tex].

Answer:

If a population exceeds carrying capacity, the ecosystem may become unsuitable for the species to survive. If the population exceeds the carrying capacity for a long period of time, resources may be completely depleted.

Answer:

a. Potential energy of the sled is increased when height of sled is increased.

b. Potential energy of the sled is increased when height of sled is increased.

c. P.E₂₀ = 2 P.E₁₀

d. P.E₂₀₀ = 2 P.E₁₀₀

Explanation:

The potential energy of the sled can be given by the following:

[tex]Potential\ Energy = P.E = mgh\\[/tex]

where,

m = mass of sled

g = acceleration due to gravity

h = height of sled

a.

It is clear from the formula that potential energy of sled is directly proportional  to the height of sled.

Therefore, potential energy of the sled is increased when height of sled is increased.

b.

It is clear from the formula that potential energy of sled is directly proportional to the mass of sled.

Therefore, potential energy of the sled is increased when mass of sled is increased.

c.

[tex]P.E\ at\ 10\ m:\\P.E_{10} = 10mg\\P.E\ at\ 20\ m:\\P.E_{20} = 20mg\\\frac{P.E_{20}}{P.E_{10}} = \frac{20mg}{10mg}[/tex]

P.E₂₀ = 2 P.E₁₀

d.

[tex]P.E\ at\ 100\ kg:\\P.E_{100} = 100gh\\P.E\ at\ 200\ m:\\P.E_{200} = 200gh\\\frac{P.E_{200}}{P.E_{100}} = \frac{200gh}{100gh}[/tex]

P.E₂₀₀ = 2 P.E₁₀₀

Answer:

Explanation:

we know that,

force = mass × acceleration

∴ since speed/velocity is constant, acceleration should be zero.

∴ f = m × 0

f = 0 N

∴ If we apply this to ,

work = force × displacement

we get ,

w = 0 × s

∴ we can say that the net work is zero.

and hence the answer is true!!!

Answer:

The answer to your question is given below

Explanation:

To solve this problem, we'll assume that the plane is initially at rest.

Hence, the kinetic energy of the plane at rest is zero i.e Initial kinetic energy (KE₁) = 0

Next, we shall determine the final kinetic energy of the plan when the force was applied. This can be obtained as follow:

Force (F) = 5000 N

Distance (s) = 500 m

Energy (E) =?

E = F × s

E = 5000 × 500

E = 2500000 J

Since energy an kinetic energy has the same unit of measurement, thus, the final kinetic energy (KE₂) of the plane is 2500000 J

Finally, we shall determine the change in the kinetic energy of the plane. This can be obtained as follow:

Initial kinetic energy (KE₁) = 0

Final kinetic energy (KE₂) = 2500000 J

Change in kinetic energy (ΔKE) =?

ΔKE = KE₂ – KE₁

ΔKE = 2500000 – 0

ΔKE = 2500000 J

Hence, the change in the kinetic energy of the plane is 2500000 J.

Answer:

54.5 kmph

Explanation:

From work-kinetic energy principles, work done by friction on both pavement and gravel shoulder = kinetic energy change of vehicle

ΔK = W = -(f₁d₁ + f₂d₂) where f₁ = frictional force due to pavement = μ₁mg where μ₁ = coefficient of friction of pavement = 0.35, m = mass of vehicle and g = acceleration due to gravity = 9.8 m/s² and d₁ = distance moved by vehicle across pavement = 30 m and

f₂ = frictional force due to gravel shoulder = μ₂mg where μ₂ = coefficient of friction of pavement = 0.50, m = mass of vehicle and g = acceleration due to gravity = 9.8 m/s² and d₂ = distance moved by vehicle across gravel shoulder = 60 m

ΔK = 1/2m(v₁² - v₀²) where v₀ = initial velocity of vehicle, v₁ = final velocity of vehicle = 20 kmph = 20 × 1000/3600 = 5.56 m/s and m = mass of vehicle

So,

ΔK = -(f₁d₁ + f₂d₂)

1/2m(v₁² - v₀²) = -(μ₁mgd₁ + μ₂mgd₂)

1/2(v₁² - v₀²) = -(μ₁gd₁ + μ₂gd₂)

v₁² - v₀² = -2g(μ₁d₁ + μ₂d₂)

v₀² = v₁² + 2g(μ₁d₁ + μ₂d₂)

v₀ = √[v₁² + 2g(μ₁d₁ + μ₂d₂)]

substituting the values of the variables into the equation, we have

v₀ = √[(5.56 m/s)² + 2 × 9.8 m/s²(0.35 × 30 m + 0.5 × 60 m]

v₀ = √[30.91 (m/s)² + 4.9 m/s²(10.5 m + 30 m]

v₀ = √[30.91 (m/s)² + 4.9 m/s²(40.5 m]

v₀ = √[30.91 (m/s)² + 198.45 (m/s)²]

v₀ = √[229.36 (m/s)²

v₀ = 15.14 m/s

v₀ = 15.14 × 3600/1000

v₀ = 54.5 kmph

So, the initial speed of the vehicle is 54.5 kmph

Answer:

20m

Explanation:

The two tens cancel each other out, as they are in opposite directions. Now we only care about the 20m, which if we have no 10's, will end up 20m away.

The distance covered by man is 40 m and the displacement is 20 m towards the east.

The distance can be defined as the measurement of the total ground covered by an object during its motion. It is a scalar quantity.

The displacement can be defined as the measurement of the shortest path over the ground is covered by an object during its motion. It is a vector quantity.

Given that a man starts walking towards the north and covered 10 m. he turns east and walks for 20 m then turns right and walks for 10 m. The attachment shows the total ground area covered by the man.

The total distance covered by the man is given below.

[tex]D = 10 + 20 + 10[/tex]

[tex]D = 40 \;\rm m[/tex]

The displacement of the man is the difference between its final position and initial position.

[tex]d = 20 - 0[/tex]

[tex]d = 20 \;\rm m[/tex]

Hence we can conclude that the distance covered by man is 40 m and the displacement is 20 m towards the east.

To know more about distance and displacement, follow the link given below.

The mass m of the object = 5.25 kg

Given

k = spring constant = 3.5 N/cm

Δx= 30 cm - 15 cm = 15 cm

Required

the mass m

Solution

F=m.g

Hooke's Law

F = k.Δx

[tex]\tt m.g=k.\Delta x\\\\m.10=3.5\times 15\\\\m=5.25~kg[/tex]

Answer:

the correct one is the a,       U₂ = 2 U₁

Explanation:

The gravitational potential energy is

            U = m g h

if the stones reach the same height, the energy of the first stone is

            U₁ = m₁ g h

The second stone is twice the mass and reaches the same height

             m₂ = 2 m₁

potential energy is

           U₂ = m₂ g h

           U₂ = 2 m₁ g h

           U₂ = 2 U₁

therefore the energy is double.

When reviewing the statements, the correct one is the a

A battery of 9 volts is connected to a resistor of 1000 ohms, then the value of current flowing through it will be 0.009 A.

Any flow of electrical charge carriers, such as ions, holes, or subatomic charged particles, is referred to as electrical current.

When electrons serve as the charge transfer in a wire, the amount of charge moving through any point of wire in a given amount of time is measured as electric current. Electric charges' motion is intermittently reversed in the alternating current, but not in direct current.

In many situations, the direction of movement in electric circuits is assumed to be the direction of positive charge flow, which is the way opposite of the real particle drift. When properly specified, the current is known as conventional current.

From the question,

V = ir

i = v/r

i = 9/1000

i = 0.009 A

Therefore, the value of the current flowing through it is  0.009 A.

To know more about Current :

https://brainly.com/question/13076734

#SPJ2

Answer:

d. 1.4 m/s

Explanation:

Given;

initial velocity of the `basketball, u = 4.0 m/s

acceleration of the basketball, a = -0.5 m/s² (leftward)

distance the basketball traveled, d = 14 m

the final velocity of the basketball, v = ?

(this final velocity should be less than the initial velocity since the ball is slowing down at a constant rate)

Apply the following the kinematic equation;

v² = u² + 2ad

v² = 4² + 2(-0.5)14

v² = 16 - 14

v² = 2

v = √2

v = 1.41 m/s

Therefore, the final velocity of the basketball is 1.4 m/s.

Answer:

k. e= 1/2 mv^2

Ke = 1/2 * 1.5 * 10^5 * 70^2

3.675 *10^8 joules

Answer:

if the angular momentum is constant, the skater is closer, going faster.  

  w₂ = r₁² / r₂² w₁

Explanation:

For the skaters to perform in the turn in l, the angular number of the same must be equal, so we stammer the concept of conservation of angular momentum

             L₁ = L₂

the angular momentum is

           L = I w

we substitute

             I₁ w₁ = I₂ w₂

             w₂ = [tex]\frac{I_{1} }{I_{2} }[/tex] w₁

if we consider the skater as a particle, his moment of inertia is

             I = m r²

we substitute

              w₂ = r₁² / r₂² w₁

therefore, if the angular momentum is constant, the skater is closer, going faster.

Answer:

A volcano fed by highly viscous magma is likely to be a greater threat to life and property than a volcano supplied with very fluid magma because with high viscous magma gas is trapped more in the magma so the gas will build up and then eventually explode, whereas with fluid magma the gas can escape allowing the magma.

HOPE THIS HELPS!!!

Explanation:

Less fluid magma done great damaged to the property and life as compared to highly viscus magma.

No, volcanoes that are fed by highly viscous magma are not a greater threat to life and property than volcanoes supplied with very fluid magma because the highly viscous magma can't move to a large distance due to its large viscosity.

While on the other hand, those volcanoes that supplied with very fluid magma do great damaged to the property due to its easily flowing on the surface of earth so we can conclude that less fluid magma done great damaged to the property and life as compared to highly viscus magma.

Learn more about magma here: https://brainly.com/question/23661578

Answer: The red at the bottom represents the base of the tube, not the red liquid.

Explanation:

The densest materials have more weight per unit of volume.

This means that those elements will always flow to the bottom of the containers, like the one in the image.

The red liquid being the least dense one, can not go to the bottom by its own means.

There could be some cases, like:

The red liquid when solid, is way denser than in its liquid phase, and then the red at the bottom could be solid phase of the red liquid, but there is no mention of this in the question, then we can discard this idea.

Another trivial idea is that the red liquid at the bottom could be trapped by some kind of wall, but again, there is no mention of this, so again we can discard this idea.

The thing that makes sense is that the red at the bottom represents the base of the tube and not the red liquid.

Answer:

The value is [tex]R = 0.321 \ \Omega[/tex]

Explanation:

From the question we are told that

   The diameter is  [tex]d = 8.252 \ mm = 0.008252 \ m[/tex]

    The length of the wire is  [tex]l = 1.0 \ km = 1000 \ m[/tex]

   Generally the cross sectional area of the copper wire is mathematically represented as

           [tex]A = \pi * \frac{d^2}{4}[/tex]

=>        [tex]A = 3.142 * \frac{ 0.008252^2}{4}[/tex]

=>         [tex]A = 5.349 *10^{ - 5} \ m^2[/tex]

Generally the resistance is mathematically represented as

      [tex]R = \frac{\rho * l }{A }[/tex]

Here [tex]\rho[/tex] is the resistivity of copper with the value  [tex]\rho = 1.72*10^{-8} \ \Omega \cdot m[/tex]

=>    [tex]R = \frac{1.72 *10^{-8} * 1000 }{5.349 *10^{ - 5} }[/tex]

=>    [tex]R = 0.321 \ \Omega[/tex]

Answer:

v₀ = 677.94 m / s ,   θ = 286º

Explanation:

We can solve this exercise using the kinematic expressions, let's work on each axis separately.

X axis

has a relation of aₓ = 5.10 m / s², the motor is on for a time of t = 675 s, reaching the speed vₓ = 3630 m / s, let's use the relation

         vₓ = v₀ₓ + aₓ t

         v₀ₓ = vₓ - aₓ t

let's calculate

         v₀ₓ = 3630 - 5.10 675

         v₀ₓ = 187.5 m / s

Y Axis

        [tex]v_{y}[/tex] = v_{oy} - a_{y} t

         v_{oy} = v_{y} - a_{y} t

let's calculate

        v_{oy}  = 4276 - 7.30 675

         v_{oy} = -651.5 m / s

we can give the speed starts in two ways

a)   v₀ = (187.5 i ^ - 651.5 j ^) m / s

b) in the form of module and angle

Let's use the Pythagorean theorem

            v₀ = [tex]\sqrt{v_{ox}^{2} + v_{oy}^{2} }[/tex]

            v₀ = [tex]\sqrt{187.5^{2} +651.5^{2} }[/tex]

            v₀ = 677.94 m / s

we use trigonometry

            tan θ = [tex]\frac{v_{oy} }{v_{ox} }[/tex]

            θ = tan⁻¹ \frac{v_{oy} }{v_{ox} }

            θ = tan⁻¹ ([tex]\frac{-651.5}{187.5}[/tex])

            θ = -73.94º

This angle measured from the positive side of the x-axis is

            θ‘ = 360 - 73.94

            θ = 286º

Answer:

12345678901234567890

Answer:

√2

Explanation:

From the question, we're given that the

Acceleration of the leaf is 1 m/s²

Change in displacement of the leaf is 1 m/s.

Again, from the question, we can tell that the initial velocity u = 0, since the object starts at rest

Now, to solve this, we don't the equation of motion to ur

S = ut + 1/2at², substituting the whole parameters, we then have

1 = 0 * t + 1/2 * 1 * t²

1 = 1/2 * t²

t²/2 = 1

t² = 2

t = √2 seconds

Therefore the time it takes the leaf to dislodge is 2 seconds

Answer:

See explanation below

Explanation:

For two vectors A and B to have a positive resultant, they must both move in the positive direction i.e along the positive x axis

For two vectors A and B to have a zero resultant, they must both have the same magnitude but move in the opposite direction i.e one in the positive direction while the other in the negative direction.

Fie example if A = +5N, B = -5B

So that A+B = +5-5 = 0N

Answer:

Surface tension is the downward force acting on the surface of liquid due to presence of inter molecular forces or cohesive forces between the particles of liquid.

Surface tension decreases with increase in temperature as the forces among particles decrease due to increase in kinetic energy and thus the cohesive nature decreases and thus surface tension also decreases.

Surface tension may decrease or increase with increase in soluble impurities .Insoluble impurities decrease the surface tension.

Answer:

C) The spring constant of each half will be twice the spring constant of the original long spring since it will stretch only half as much under the same tension.

Explanation:

Hooke's law states that the force needed to extend or compress a spring by a distance is proportional to that distance. If is given as:

F = ke, where F is the force applied, k is spring constant and e is the extension.

If a force f is applied to a spring with a spring constant k and by a distance stretched (x) then:

k = F / x

For half the spring, if the same force F is applied, the stretch would be half (x/2), hence the spring constant C is:

C = F / (x/2)

C = 2 (F / x) = 2 * spring constant of original spring

Answer: the child's centripetal acceleration=24.50 m/s²

Explanation:

Given that mass of child= 50 kg

radius of merry go round= 2.25m

angular speed = 3.30 rad/s

Centripetal Acceleration  = v²/ r

  But  V= ωr

So Centripetal Acceleration  = v²/ r =  (ωr)²/ r

=(3.30)² x  (2.25)²/ 2.25 = (3.30)² x  2.25

=24.5025m/s²

=24.50 m/s²

Answer:

Explanation:

Given;

output power of the incandescent lightbulb , P = 120 W

input voltage, V = 200 V

The resistance of the filament is calculated as;

[tex]P = \frac{V^2}{R}[/tex]

Where;

R is the resistance of the filament

[tex]R = \frac{V^2}{P} \\\\R = \frac{200^2}{120} \\\\R = 333.33 \ ohms[/tex]

The rms current is given as;

[tex]P = I_{rms} V_{rms}\\\\I_{rms} = \frac{P}{V_{rms}} \\\\I_{rms} =\frac{120}{200}\\\\ I_{rms} = 0.6 \ A[/tex]

Complete Question

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute).Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light

Answer:

The diameter is  [tex]D = 0.59 \ m[/tex]    

Explanation:

From the question we are told that

      The best resolution is  [tex]\theta = 0.3 \ arcsecond[/tex]

       The  wavelength is  [tex]\lambda = 700 \ nm = 700 *10^{-9 } \ m[/tex]

Generally the

         1 arcminute  = >  60 arcseconds

=>      x arcminute =>   0.3 arcsecond

So

       [tex]x = \frac{0.3}{60 }[/tex]

=>    [tex]x = 0.005 \ arcminutes[/tex]

Now

         60 arcminutes  =>  1 degree

          0.005 arcminutes = >  z degrees  

=>       [tex]z = \frac{0.005}{60 }[/tex]

=>      [tex]z = 8.333 *10^{-5} \ degree[/tex]

Converting to radian  

           [tex]\theta = z = 8.333 *10^{-5} * 0.01745 = 1.454 *10^{-6} \ radian[/tex]

Generally the resolution is mathematically represented as

            [tex]\theta = \frac{1.22 * \lambda }{ D}[/tex]

=>    [tex]D = \frac{1.22 * \lambda }{\theta }[/tex]

=>     [tex]D = \frac{1.22 * 700 *10^{-9} }{ 1.454 *10^{-6} }[/tex]    

=>     [tex]D = 0.59 \ m[/tex]    

Answer:

The value is  [tex]f_1 = 828 \ Hz[/tex]

Explanation:

From the question we are told that

    The speed is [tex]v = 25 \ m/s[/tex]

     The speed of the ambulance is [tex]u = 35 \ m/ s[/tex]

     The frequency of the siren is  [tex]f = 850 \ Hz[/tex]

Generally from Doppler effect equation we have that

        [tex]f_1 = \frac{v_s - v }{ v_s + u } * f[/tex]

Here [tex]v_s[/tex] is the velocity of sound with the value  [tex]v_s = 343 \ m/s[/tex]

  =>   [tex]f_1 = \frac{343 - 25 }{ 343 + 35 } * 850[/tex]

  =>   [tex]f_1 = 828 \ Hz[/tex]

Answer:

Where is the picture

Explanation:

WHERE IS THE PICTURE

alpha is the excess return on an investment after adjusting for market related volatility and random fluctuations.

beta is a measure of volatility relative to a benchmark ,such as the S&P 500.

Explanation:

alpha and beta are two different parts of an equation used to explain the performance of stocks and investments funds. But in maths alpha and beta is the Greek alphabet