. Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point? What is your displacement vector? What is the direction of your displacement? Assume the +x-axis is to the east.

Answer:

1800 m/s

Explanation:

The equation is v = fλ

λ= 0.75

f = 2400 Hz

V = 2400 × 0.75

V = 1800 m/s

[ you did not give units for wavelength, I assumed it would be m/s]

Answer:

meter

Explanation:

Answer: The International System of Units is a system of measurement based on 7 base units

Explanation: the metre, kilogram, second, ampere, Kelvin, mole, and candela. These base units can be used in combination with each other.

Answer:

In a nutshell, units of A and B are [tex]\frac{1}{[l]^{2}}[/tex] and [tex]\frac{[l]}{[t]}[/tex], respectively.

Explanation:

From Dimensional Analysis we understand that [tex]x[/tex] and [tex]y[/tex] have length units ([tex][l][/tex]) and [tex]t[/tex] have time units ([tex][t][/tex]). Then, we get that:

[tex][l] = A\cdot [l]^{3}[/tex] (Eq. 1)

[tex][l] = B\cdot [t][/tex] (Eq. 2)

Now we finally clear each constant:

[tex]A = \frac{[l]}{[l]^{3}}[/tex]

[tex]A = \frac{1}{[l]^{2}}[/tex]

[tex]B = \frac{[l]}{[t]}[/tex]

In a nutshell, units of A and B are [tex]\frac{1}{[l]^{2}}[/tex] and [tex]\frac{[l]}{[t]}[/tex], respectively.

Answer:

A) By applying a known force, and measuring it's acceleration.

Explanation:

This is actually something that astronauts do in space as a mathmatical exercise when calculating the mass of an object since F = m × a.

Once the force, and acceleration are applied, the only unknown is the mass which can be solved by dividing force over acceleration. This is because inertial mass is equal to gravitational mass.

Answer:

D. 10 m/s

Explanation:

Answer:

III) [tex]d_{1}+ d_{2}=d_{t}[/tex]

Explanation:

I) coordinate (0,5) is the head for [tex]d_{1}[/tex] I will put the tail coordinate as (0,0) but it could be any other number in the x just not in the 5  with the the y being any other value.

II) coordinate (5,5) is the head for [tex]d_{2}[/tex] the tail needs to be in the head of [tex]d_{1}[/tex] being (0,5)

III) coordinates for [tex]d_{t}[/tex] is connecting the tail from [tex]d_{1}[/tex] and the head of [tex]d_{2}[/tex] making it (0,0)[tex](tail)[/tex] and (0,5)[tex](head)[/tex] and is written as [tex]d_{1}+ d_{2}=d_{t}[/tex]

(i) using coordinate grid notation to represent d₁, d₁ = 5y

(ii) using coordinate grid notation  to represent d₂, d₂ = 5x + 5y

(ii) The sum of d₁ and d₂ is written as 5x + 10y

In order to show the horizontal direction of a vector, we will place x after the magnitude of the vector.

Also, to show the vertical direction of a vector, we will place a y after the magnitude of the vector.

(i) Using coordinate grid to represent d₁ = (0, 5)

[tex]d_1 = 0(x) + 5(y)\\\\d_1 = 5y[/tex]

(ii) Using coordinate grid to represent d₂ = (5, 5)

[tex]d_2 = 5x + 5y[/tex]

(iii) The total vector is written as;

[tex]d_1 + d_2 = 5y + (5x + 5y)\\\\d_1 + d_2 = 5y + 5x + 5y\\\\d_1 + d_2 = 5x + 10y[/tex]

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Answer:

Package 1 will land at 228.0 m, package 2 will land at 308.0 m, and the distance between them is 80.0 m.

Explanation:

To find the distance at which the first package will land we need to calculate the time:

[tex] Y_{f} = Y_{0} + V_{0y}t - \frac{1}{2}gt^{2} [/tex]

Where:

Y(f) is the final position = 0

Y(0) is the initial position = 160 m

V(0y) is initial speed in "y" direction = 0

g is the gravity = 9.81 m/s²

t is the time=?                                          

[tex] 0 = 160 m + 0t - \frac{1}{2}9.81 m/s^{2}t^{2} [/tex]

[tex] t = \sqrt{\frac{2*160 m}{9.81 m/s^{2}}} = 5.7 s [/tex]

Now we can find the distance of the first package:

[tex] X_{1} = V_{0x}*t = 40.0 m/s*5.7 s = 228.0 m [/tex]

Then, after 2 seconds the distance traveled by plane is (from the initial position):

[tex] X_{p} = V_{0x}*t = 40.0 m/s*2 s = 80.0 m [/tex]

Now, the distance of the second package is:

[tex] X _{2} = X_{1} + X_{p} = 228.0 m + 80.0 m = 308.0 m [/tex]

The distance between the packages is:

[tex] X = X_{2} - X_{1} = 308.0 - 228.0 m = 80.0 m [/tex]

Therefore, package 1 will land at 228.0 m, package 2 will land at 308.0 m and the distance between them is 80.0 m.

I hope it helps you!

Answer:

Explanation

According to Newton's second law of motion,

F = ma

m is the mass

a is the acceleration

If the acceleration is 3.2 times the acceleration due to gravity, then a = 3.2g

The formula becomes;

F = m(3.2g)

F = 3.2mg

m= 75kg

g = 9.81m/s²

F = 3.2(75)(9.81)

F = 2,354.4N

Hence the force exerted by the jumper is 2,354.4N

Answer:

12.5 m

Explanation:

The first thing we would do is to calculate the wavelength. To do this, we use the formula

v = fλ, where

v = wave speed

f = frequency

λ = wavelength

If we make wavelength the formula, we have

wavelength = speed / frequency

Now, we substitute the values we had been given and we have

wavelength = (3 * 10^8 m/s) / (85.7 * 10^6 Hz) wavelength = 3.50 m

half of this said wavelength will be

= 3.50 / 2

= 1.75 m

As a result of the engineering constraints with the towers being more than 10 m apart, the distance can't be 1.75 m and as such, it has to be a multiple of 1.75m. So we say,

(10 / 1.75) = 5.7

So the separation will have to be 7 half wavelengths

= (7 * 1.75) = 12.5 m

Answer:

The height of the cliff is 121.276 m

Explanation:

Given;

initial velocity of the projectile, v₁ = 75 m/s

final velocity of the projectile, v₂ = 90 m/s

spring compression = 5 m

Apply the law of conservation of energy;

mgh₀ + ¹/₂mv₁² = mgh₂ + ¹/₂mv₂²

gh₀ + ¹/₂v₁² = gh₂ + ¹/₂v²

gh₁  - gh₂ = ¹/₂v₂² - ¹/₂v₁²

g(h₀  - h₂) = ¹/₂ (v₂² - v₁²)

h₀  - h₂ = ¹/₂g (v₂² - v₁²)

h₀ = h(cliff) + 5m

when the projectile hits the ground, Final height, h₂ = 0

[tex]h_o - 0 = \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} + 5= \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} = \frac{1}{2g}(v_2^2-v_1^2) - 5\\\\h_{cliff} = \frac{1}{2*9.8}(90^2-75^2) - 5\\\\h_{cliff} = 121.276 \ m[/tex]

Therefore, the height of the cliff is 121.276 m

Answer:

a)  + Q charge is inducce that compensates for the internal charge

b) There is no excess charge on the external face q_net = 0

c) E=0

Explanation:

Let's analyze the situation when a negative charge is placed inside the cavity, it repels the other negative charges, leaving the necessary positive charges to compensate for the -Q charge. The electrons that migrated to the outer part of the sphere, as it is connected to the ground, can pass to the earth and remain on the planet; therefore on the outside of the sphere the net charge remains zero.

With this analysis we can answer the specific questions

a)  + Q charge is inducce that compensates for the internal charge

b) There is no excess charge on the external face q_net = 0

c) If we create a Gaussian surface on the outside of the sphere the net charge on the inside of this sphere is zero, therefore there is no electric field, on the outside

d) If it is very reasonable and this system configuration is called a Faraday Cage

e) We cannot apply this principle to gravity since there are no particles that repel, in all cases the attractive forces.

Answer:

Displacement is 3 km North

Explanation:

Answer:

The boat has more buoyancy

Explanation:

Answer:

The displacement of the canoe is 1.46 m

Explanation:

Given that,

Mass of canoe = 61 kg

Mass of man = 86.5 kg

Length = 4 m

Let the the displacement of the canoe is x'

We need to calculate the displacement of the man

Using formula of displacement

[tex]x=x_{2}-x_{1}[/tex]

Put the value into the formula

[tex]x=4-(0.75+0.75)[/tex]

[tex]x=2.5\ m[/tex]

We need to calculate the displacement of the canoe

Using conservation of momentum

[tex]M_{m}v_{m}=(M_{c}+M_{m})v_{c}[/tex]

[tex]M_{m}\dfrac{x}{t}=(M_{c}+M_{m})\dfrac{x'}{t}[/tex]

[tex]86.5\times2.5=(61+86.5)\times x'[/tex]

[tex]x'=\dfrac{86.5\times2.5}{61+86.5}[/tex]

[tex]x'=1.46\ m[/tex]

Hence, The displacement of the canoe is 1.46 m

Answer:

Thanks!!!!! adding this so it doesn’t get deleted.

Explanation:

1. Electrostatic forces are trillions of times stronger than gravitational forces. 2. normal force and friction 3. contact forces 4. The electrostatic forces from the contact of the hands with the paper causes the paper molecules to separate. 5. The electrostatic forces between the molecules of the board prevent the force of gravity from breaking the board apart.

The correct statement over here is that electrostatic forces are trillions of times stronger than gravitational forces. Hence, option C is correct.

One of the basic forces in the cosmos is electrostatic force. In the universe, there are four basic forces. These include gravitational force, electromagnetic force, weak nuclear force, and strong nuclear force. Under the umbrella of electromagnetic force is electrostatic force. Two charges placed apart are subject to the electrostatic force. The size of each charged and the separation between them determines how much electrostatic force there will be.

When two charges of the same type are brought together, whether positive or negative, they repel one another. It is known as the electrostatic force of repelling when it operates among two charges that are similar.

Therefore, the electrostatic forces are trillions of times stronger than the gravitational forces.

To know more about Electrostatic Force:

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#SPJ2

Answer: C.

Explanation:

For a parallel-plate capacitor where the distance between the plates is d.

The capacitance is:

C = e*A/d

You can see that the distance is in the denominator, then if we double the distance, the capacitance halves.

Now, the stored energy can be written as:

E = (1/2)*Q^2/C

Now you can see that in this case, the capacitance is in the denominator, then we can rewrite this as:

E = (1/2)*Q^2*d/(e*A)

e is a constant, A is the area of the plates, that is also constant, and Q is the charge, that can not change because the capacitor is disconnected.

Then we can define:

K = (1/2)*Q^2/(e*A)

And now we can write the energy as:

E = K*d

Then the energy is proportional to the distance between the plates, this means that if we double the distance, we also double the energy.

If the plates are pulled apart to twice their original separation, then this will double the stored energy. Hence, option (C) is correct.

The given problem is based on the concept of parallel plat capacitor. For a parallel-plate capacitor where the distance between the plates is d.

The capacitance is:

C = e*A/d

here.

e is the permittivity of free space.

Since, the distance is inversely proportional then if we double the distance, the capacitance halves.  Now, the stored energy can be given as,

E = (1/2)*Q^2/C

here,

Q is the charge stored in the capacitor.

Now you can see that in this case, the capacitance is in the denominator, then we can rewrite this as:

E = (1/2)*Q^2*d/(e*A)

e is a constant, A is the area of the plates, that is also constant, and Q is the charge, that can not change because the capacitor is disconnected.

Then we can define:

K = (1/2)*Q^2/(e*A)

And now we can write the energy as:

E = K*d

So, the energy is proportional to the distance between the plates.

Thus, we can conclude that if the plates are pulled apart to twice their original separation, then this will double the stored energy. Hence, option (C) is correct.

Learn more about the energy stored in a capacitor here:

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Answer:

T₂ = 49.3°C

Explanation:

Applying law of conservation of energy to the system we get the following equation:

Energy Supplied by Resistor = Energy Absorbed by Tank + Energy Absorbed by Water

E = mC(T₂ - T₁) + m'C'(T'₂ - T'₁)

where,

E = Energy Supplied by Resistor = 100 KJ = 100000 J

m = mass of copper tank = 13 kg

C = Specific Heat of Copper = 385 J/kg.°C

T₂ = Final Temperature of Copper Tank

T₁ = Initial Temperature of Copper Tank = 27°C

T'₂ = Final Temperature of Water

T'₁ = Initial Temperature of Water = 50°C

m' = Mass of Water = 4 kg

C' = Specific Heat of Water = 4179.6 K/kg.°C

Since, the system will come to equilibrium finally. Therefor:  T'₂ = T₂

Therefore,

(100000 J) = (13 kg)(385 J/kg.°C)(T₂ - 27°C) + (4 kg)(4179.6 J/kg.°C)(T₂ - 50°C)

100000 J = (5005 J/°C)T₂ - 135135 J + (16718.4 J/°C)T₂ - 835920 J

100000 J + 135135 J + 835920 J = (21723.4 J/°C)T₂

(1071055 J)/(21723.4 J/°C) = T₂

T₂ = 49.3°C

Complete Question

The complete  question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

     The original voltage is  [tex]V_o[/tex]

     The new voltage is [tex]V  =\frac{V_o}{2}[/tex]

     The capacitance is  [tex]C = 150\ nF = 150 *10^{-9} \  F[/tex]

     The first resistance is  [tex]R_i =  26 \Omega[/tex]

      The second resistance is [tex]R_E =  200 \Omega[/tex]

Generally the equivalent resistance is  

        [tex]R_e =  R_1 + R_E[/tex]

=>     [tex]R_e =  26 +200 [/tex]

=>     [tex]R_e = 226 \ \Omega [/tex]

Generally the time constant is mathematically represented as

     [tex]\tau  =  RC[/tex]

=>  [tex]\tau  =  226 * 150 *10^{-9}[/tex]

=>  [tex]\tau  =  3.39 *10^{-5} \  s [/tex]

Generally the voltage is mathematically represented as

    [tex]V =  V_o  e^{-\frac{t}{\tau} }[/tex]

=>   [tex]\frac{V_o}{2} =  V_o  e^{-\frac{t}{\tau} }[/tex]

=>   [tex]0.5 =    e^{-\frac{t}{\tau} }[/tex]

=>   [tex]ln(0.5) =    {-\frac{t}{ 3.39 *10^{-5} } }[/tex]

=>  [tex]ln(0.5)   * 3.39 *10^{-5}  =   -t [/tex]

=>  [tex]t = 2.35*10^{-5} \  s  [/tex]

Answer:

when the electron transferred permanently to another atom

Answer:

Density is calculated by dividing the mass of an object by its volume. The Sun has a mass of 1.99×1030 kg and a radius of 6.96×108 m. What is the average density of the Sun?

Answer:

102 m

Explanation:

Given that It takes a minimum distance of 48.96 m to stop a car moving at 12.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.

Let the stopping distance be equal to S.

According to the definition of speed,

Speed = distance / time.

make time the subject of the formula

Time = distance / speed

then, the equivalent time is:

48.96 / 12 = S / 25

Cross multiply

12S = 48.96 x 25

12S = 1224

S = 1224 / 12

S = 102 m

Therefore, the stopping distance is 102 m

Answer:

A group of protons and neutrons that are surrounded by electrons  I think that's the answer...

Explanation:

Answer:

false

Explanation:

Answer:

V = 331.59m/s

Explanation:

First we need to calculate the time taken for the shell fire to hit the ground using the equation of motion.

S = ut + 1/2at²

Given height of the cliff S = 80m

initial velocity u = 0m/s²

a = g = 9.81m/s²

Substitute

80 = 0+1/2(9.81)t²

80 = 4.905t²

t² = 80/4.905

t² = 16.31

t = √16.31

t = 4.04s

Next is to get the vertical velocity

Vy = u + gt

Vy = 0+(9.81)(4.04)

Vy = 39.6324

Also calculate the horizontal velocity

Vx = 1330/4.04

Vx = 329.21m/s

Find the magnitude of the velocity to calculate speed of the shell as it hits the ground.

V² = Vx²+Vy²

V² = 329.21²+39.63²

V² = 329.21²+39.63²

V² = 108,379.2241+1,570.5369

V² = 109,949.761

V = √ 109,949.761

V = 331.59m/s

Hence the speed of the shell as it hits the ground is 331.59m/s

Answer:

2.83kg

Explanation:

Answer:

2.83kg

Explanation:

Given

initial density = 1.18 kg/m3

Final density in the tank = 4.95 kg/m3.

Let us write the mass balance first.

Change in the mass of the system=mass of the air entering the system - Mass of air out the system

Mass that entered= M2 - M1

But DENSITY= MASS/ VOLUME

Mass= volume × Density

We can expressed the mass in terms of density since density is given in the question.

Mass that entered= (volume × density)2 - ( volume × density)1

= (V ρ)2 - (V ρ)1

But V1= V2 the volume remains the same

= ( ρ2 - ρ1)v

= (4.95 kg/m3 - 1.18 kg/m3) 0.75 m3

= 3.77× 0.75

= 2.8275kg

Mass that entered= 2.83kg

therefore, mass of air that has entered the tank= 2.83kg

Explanation:

The question is incomplete. Here is the complete question.

A toy rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 98 m and acquired a velocity of  30m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground. The upward acceleration of the rocket during the burn phase is closest to...

Given

initial velocity of rocket u = 0m/s

final velocity of rocket = 30m/s

Height reached by the rocket = 98m

Required

upward acceleration of the rocket

Using the equation of motion below to get the acceleration a:

[tex]v^2 = u^2+2as\\30^2 = 0^2 + 2(a)(98)\\900 = 196a\\a = \frac{900}{196}\\a = 4.59m/s^2[/tex]

Hence  upward acceleration of the rocket during the burn phase is closest to 5m/s²

Note that the velocity used in calculation was assumed.

Answer:

i. The magnetic force of repulsion.

ii. The magnetic force of attraction.

Explanation:

A magnet is a material that has the attraction and repulsion capability. Magnets has two poles, north and south, thus would attract or repel another magnet in its neighborhood. It can either be a permanent or temporal magnet, and attracts ferrous metals.

i. In the case of two combinations where two ends are the same, it could be observed that the two ends (poles) repels each other. Thus since like poles repels, magnetic force of repulsion is felt.

ii. In the case of one combination in which the two ends are different, the two ends (poles) attract. Since unlike poles attracts, magnetic force of attraction is observed.

Answer:

T = 1.4696 10⁴ years

Explanation:

For this exercise we must use Kepler's laws, specifically the third law which is the application of the universal law of gravitation to Newton's second law

               F = ma

               G m M / r² = m a_c = m v² / r

                G M / r = v²

the speed of the circular orbit is

               v = 2π r / T

we substitute

               G M / r = 4π² r² / T²

               T² = (4π² / G M)  r³

Kepler proved that this expression is the same if the radius is changed by the semi-major axis of an ellipse

               T² = (4π² /GM)  a³

the constant is worth

                (4π² / GM) = 2.97 10⁻¹⁹    s² / m³

let's reduce the distance to SI units

AU is the distance from the Earth to the Sun

               a = 600 AU = 600 AU (1.496 10¹¹ m / 1 AU)

               a = 8.976 10¹³ m

               T² = 2.97 10⁻¹⁹ (8.976 10¹³)³

               T² = 21.4786 10²²

               T = 4.63 10¹¹ s

Let's reduce to years

               T = 4.63 10¹¹s (1 h / 3600s) (1 day / 24 h) (1 year / 365 days)

               T = 1.4696 10⁴ years