A swimmer heading directly through a 200m wide river reaches the opposite shore in 6 min 40s. She is washed downstream 480 m. How fast can you swim in calm water?

Answer:

Letter A. [tex]y=4.55 m[/tex]

Explanation:

Let's use the wave equation:

[tex]y=Asin(kx-\omega t)[/tex]

Therefore the displacement y will be:

[tex]y=6.44*sin(2.34*1.21-2.88*0.71)[/tex]

[tex]y=4.55 m[/tex]

The answer is letter A.

I hope it helps you!

Answer:

Explanation:

Find the displacement of a simple harmonic wave of amplitude 6.44 m at t = 0.71 s. Assume that the wave number is 2.34 m-1, the angular frequency is 2.88 rad/s, and that the wave is propagating in the +x direction at x = 1.21 m.

Amplitude (A) of the simple harmonic wave = 6.44 m

wave number (k) of the given wave = 2.34 m-1

Angular frequency (ω) of the given wave = 2.88 rad/s

Displacement x = 1.21 m and time t = 0.71 s

Then the general equation for the displacement of the given simple harmonic wave at given x and time t is given by

y = Asin(kx - ωt)

= (6.44 m)sin[(2.34 m-1)(1.21 m) - (2.88 rad/s)(0.71 s)]

Y=6.44sin(0.7866 rad)

0.7866rad*(180 degrees/pi rad) =45.1

Y=6.44sin(45.1)

Y=4.55m

Answer:

Q = - 5264 W = - 5.26 KW

Here, negative sign indicates the outflow of heat

Explanation:

Fourier's Law of heat conduction, gives the following formula:

Q = - KAΔT/t

where,

Q = Rate of Heat Conduction out of window = ?

K = Thermal Conductivity of Glass = 0.84 W/m.°C

A =Surface Area of window = 2.82 m²

ΔT = Difference in Temperature of both sides of surface

ΔT = Inner Surface Temperature - Outer Surface Temperature= 5°C - (- 10°C)

ΔT = 15°C

t = thickness of window = 0.675 cm = 0.00675 m

Therefore,

Q = - (0.84 W/m.°C)(2.82 m²)(15°C)/0.00675 m

Q = - 5264 W = - 5.26 KW

Here, negative sign indicates the outflow of heat.

Answer:

Fx = - (1/h)( wL2 + wL3 - wLend )

Explanation:

Assuming The twins Gilles and Jean has a weight ( w ) each

The torque that would balance the equation would be = wL2 + wL3 -------- 1

THEREFORE the ccw torques are = wLend + Fh ----------- 2

hence equation 2 equals equation 1

= wLend + Fh = wL2 + wL3 --------- 3

equation 3 can as well be represented as

F = ( 1/h) ( wL2 + wL3 - wLend )---------- 4

From equation 4 it can be seen that F is on the left hand side therefore the value of Fx is negative

therefore equation 4 is represented as

 Fx = - (1/h)( wL2 + wL3 - wLend )

Answer:

Exothermic

Explanation:

Depending on the unit you are in, the answer may vary.

This is an exothermic reaction because it produces heat (the beaker gets warm).

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  [tex]Q =2.094 C[/tex]

Explanation:

From the question we are told that

    The diameter of the wire is  [tex]d = 0.205cm = 0.00205 \ m[/tex]

     The radius of  the wire is  [tex]r = \frac{0.00205}{2} = 0.001025 \ m[/tex]

     The resistivity of aluminum is [tex]2.75*10^{-8} \ ohm-meters.[/tex]

       The electric field change is mathematically defied as

         [tex]E (t) = 0.0004t^2 - 0.0001 +0.0004[/tex]

Generally the charge is  mathematically represented as

       [tex]Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt[/tex]

Where A is the area which is mathematically represented as

       [tex]A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2[/tex]

 So

       [tex]\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega[/tex]

Therefore

      [tex]Q = 120 \int\limits^{t}_{0} { E(t) } \, dt[/tex]

substituting values

      [tex]Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt[/tex]

     [tex]Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.[/tex]

From the question we are told that t =  5 sec

           [tex]Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.[/tex]

          [tex]Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }[/tex]

         [tex]Q =2.094 C[/tex]

The charge (Q) passing through a cross-section of the conductor between time 0 seconds and time 5 seconds is 2.094 Coulomb.

Given the following data:

Conversion:

Diameter of wire = 0.205 cm to m = 0.00205 meter.

Radius = [tex]\frac{Diameter}{2} =\frac{0.00205}{2} =0.001025\;meter[/tex]

To determine the charge (Q) passing through a cross-section of the conductor between time 0 seconds and time 5 seconds, we would apply Gauss's law in an electric field for a surface charge:

First of all, we would find the area of the wire.

[tex]Area = \pi r^2\\\\Area = 3.142 \times 0.001025^2\\\\Area = 3.3 \times 10^{-6}\;m^2[/tex]

Mathematically, Gauss's law in an electric field for a surface charge is given by the formula:

[tex]Q = \int\limits^t_0 {\frac{A}{\rho } E(t)} \, dt[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]Q= \int\limits^t_0 {\frac{3.3 \times 10^{-6}}{2.75\times 10^{-8} } (0.0004t^2-0.0001t+0.0004)} \, dt\\\\Q=120\int\limits^t_0 1{ (0.0004t^2-0.0001t+0.0004)} \, dt[/tex]

[tex]Q=120(\frac{0.0004t^3}{3} -\frac{0.0001t^2}{2} +0.0004t |\left{5} \atop {0} \right[/tex]

When t = 5 seconds:

[tex]Q=120(\frac{0.0004[5]^3}{3} -\frac{0.0001[5]^2}{2} +0.0004[5])\\\\Q=120(\frac{0.03}{3} -\frac{0.0025}{2} +0.002)\\\\Q=120(0.0167-0.00125+0.002)\\\\Q=120(0.01745)[/tex]

Q = 2.094 Coulomb.

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Answer:

D. Presynaptic terminal button

explanation:

Terminal Buttons are small knobs at the end of an axon that release chemicals called neurotransmitters. The terminal buttons form the Presynaptic Neuron

hope this helped!

Answer:

Potential difference is measured in volts

Explanation:

The standard metric unit on electric potential difference is the volt, abbreviated V and named in honor of Alessandro Volta. One Volt is equivalent to one Joule per Coulomb.

Answer:

Your answer is A.) volts

Explanation:

Answer:

The correct answer is a

Explanation:

The cosine function is

      cos θ = ca / ​​H

done ca is the adjacent leg (x-axis) and H is the hypotenuse (vector module)

we clear

    H = ca / ​​cos θ

therefore, to find the magnitude of the vector, the cathete is divided into the cosine.

The correct answer is a

Answer:

5.1 rad/s

Explanation:

Mechanical energy of the system is conserved since no external work is done on the sphere.

[tex]mgh = mv^2/2 + I\omega^2/2[/tex]

Substituting v = ωr and I = 2 m r^2/5, we get,

=> [tex]mgh=m(\omega r)^2/2 + (2\omega r^2/5)\omega^2/2[/tex]

=> [tex]mgh = m\omega^2r^2/2 + m\omega^2r^2/5[/tex]

=> [tex]gh =\omega^2r^2/2+\omega^2r^2/5[/tex]

=>  [tex]gh = 7\omega^2 r^2/10[/tex]

=>  [tex]\omega r = (10gh/7)^{1/2}[/tex]

=> [tex]\omega = (1/r)(10gh/7)^{1/2} = (1 / 1.7)(10\times 9.8\times 5.3 / 7)^{1/2}[/tex] = 5.1 rad/s

Answer:

4.93 m

Explanation:

According to the question, the computation of the height is shown below:

But before that first we need to find out the speed which is shown below:

As we know that

[tex]Speed = \frac{Distance}{Time}[/tex]

[tex]Speed = \frac{5}{0.504}[/tex]

= 9.92 m/s

Now

[tex]v^2 - u^2 = 2\times g\times h[/tex]

[tex]9.92^2 = 2\times 9.98 \times h[/tex]

98.4064 = 19.96 × height

So, the height is 4.93 m

We simply applied the above formulas so that the height i.e H could arrive

Answer:

v = 25.45 m/s

Explanation:

In order to calculate the initial speed of the object, you take into account the formula for the maximum height reaches by the object. Such a formula is given by:

[tex]h_{max}=\frac{v_o^2}{g}[/tex]   (1)

vo: initial speed of the object = 18 m/s

g: gravitational acceleration = 9.8 m/s²

Furthermore you use the following formula for the final speed of the object:

[tex]v^2=v_o^2-2gh[/tex]       (2)

h: height

You know that the speed of the object is 18m/s when it reaches one fourth of the maximum height. You use this information, and you replace the equation (1) in to the equation (2), as follow:

[tex]v^2=v_o^2-2g(\frac{h_{max}}{4})=v_o^2-\frac{1}{2}g(\frac{v_o^2}{g})\\\\v^2=v_o^2-\frac{1}{2}v_o^2=\frac{1}{2}v_o^2[/tex]

Then, you solve the previous result for vo:

[tex]v_o=\sqrt{2}v=\sqrt{2}(18m/s)=25.45\frac{m}{s}[/tex]

The initial speed of the object was 25.45 m/s

Note: The complete question is attached as a file to this solution. The parallel plate mentioned can be seen in this picture attached.

Answer:

E = 225 N/C

Explanation:

Note: At any point on the parallel plates of a capacitor, the electric field is uniform and equal.

Therefore, Electric field at x = 14 cm equals the electric field at x = 7 cm

V(x) = 31.5 Volts

x = 14 cm = 0.14 m

The magnitude of the electric field at any point between the parallel plate of the capacitor is given by the equation:

E = V(x)/d

E(x = 0.14) = 31.5/0.14

E(x=0.14) = 225 N/C

E(x=0.14) = E(x=0.07) = 225 N/C

Answer:

Explanation:

Using the law of conservation of momentum which states that 'the sum of momentum of bodies before collision is equal to their sum after collision. The bodies will move together with a common velocity after collision.

Momentum = Mass * Velocity

Before collision;

Momentum of receiver m1u1= 0 kgm/s (since the receiver is standing still)

Momentum of the tackler

m2u2 = 2.60*122 = 317.2 kgm/s

where m2 and u2 are the mass and velocity of the tacker respectively.

Sum of momentum before collision = 0+317.2 = 317.2 kgm/s

After collision

Momentum of the bodies = (m1+m2)v

v = their common velocity

m1 = mass of the receiver

Momentum of the bodies = (122+m1)(1.30)

Momentum of the bodies = 158.6+1.30m1

According to the law above;

317.2 = 158.6+1.30m1

317.2-158.6 = 1.30m1

158.6 = 1.30m1

m1 = 158.6/1.30

m1 = 122kg

The mas of the receiver is 122kg

Answer:

1. c

2. e

3. d

Explanation:

1.

From Einstein's Photoelectric Equation, we know that:

Energy given up by photon = Work Function + K.E of Electron

hc/λ = φ + K.E

where,

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of light source = 200 nm = 2 x 10⁻⁷ m

φ = (5.1 eV)(1.6 x 10⁻¹⁹ J/eV) = 8.16 x 10⁻¹⁹ J

Therefore,

(6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2 x 10⁻⁷ m) - 8.16 x 10⁻¹⁹ = K.E

K.E = (9.939 - 8.16) x 10⁻¹⁹ J

K.E = 1.778 x 10⁻¹⁹ J

The positive answer shows that electrons will be emitted. Since it is clear from the equation the the K.E of electron decreases with the increase in work function. Therefore:

c. less energetic photo-electrons (on average)

2.

The increase in light sources means an increase in the intensity of light. The no. of photons are increased, due to increase of intensity. Thus, more photons hit the metal and they eject greater no. of electrons. Therefore,

e. more photo-electrons ejected

3.

X-rays have smaller wavelength and greater energy than ultraviolet rays. Thus, the photons with greater energy will strike the metal and as a result, electrons with higher energy will be ejected.

d. more energetic photo-electrons (on average)

Answer:

Hope this helps :)

Explanation:

1. A

2. G (because the basic definition of internal energy is, the sum of kinetic and potential energies of water molecules)

Answer:

Tilted forward to keep the driest.

Explanation:

The rain is falling vertically so there is no wind. In these circumstances the umbrella should be tilted vertically forward.

The situation is the same as if you would stand still and the rain would come under an angle from the front.

Answer:

89.11kg

Explanation:

Note an object weighs less when in a fluid and the weight of the volume of the fluid displaced is known as the upthrust.

Now, the person is going to displace the volume 89/1025 =0.087m3 { from density D = mass(M)/volume(V)}

The weight of the fluid displaced is the density of the fluid × volume of fluid displaced.

The weight of the fluid=0.087m3× 1kg/me = 0.087kg

Now the weight of the fluid displaced is referred to as the upthrust.

Now the real weight - the apparent weight = the upthrust.

Hence the apparent weight = real weight - upthrust

Apparent weight = 89.2-0.087 = 89.11kg

Answer:

882 N

Explanation:

F = ma

45.0 N = m (0.500 m/s²)

m = 90.0 kg

mg = 882 N

Answer:

t = 402 years

Explanation:

To find the number of year that electrons take in crossing the complete transmission line, you first calculate the drift speed of the electrons. Then, you use the following formula for the current in a wire:

[tex]I=nqv_dA[/tex]  (1)

n: number of mobile charge carrier per volume = 8.50*10^28 e/m^3

q: charge of the electron = 1.6*10^-19 C

vd: drift velocity of electron in the metal = ?

A: cross sectional area of the wire = π r^2 = π (0.02m/2)^2 = 3.1415*10^-4 m^2

I: current in the wire = 1110 A

You solve the equation (1) for vd:

[tex]v_d=\frac{I}{nqA}=\frac{110A}{(8.50*10^{28}m^{-3})(1.6*10^{-19}C)(3.1415*10^{-4}m^2)}\\\\v_d=2.59*10^{-4}m/s[/tex]

Next, you calculate the time by using the information about the length of the line transmission:

[tex]x=v_dt\\\\x=330km=330000m\\\\t=\frac{x}{v_d}=\frac{330000m}{2.59*10^{-4}m/s}=1,270,184,865s\\\\1,270,184,865s*\frac{1\ year}{3,156,107}=402.45\ years[/tex]

hence, the electrons will take aproximately 402 years in crossing the line of transmission

Answer:

a population increase

Explanation:

During the 20th century, the world population increased from 1.65 billion to 6 billion. In 1970, the world's population was half that of today. In less than 15 years, 47% of the population will live in areas already under heavy water stress. In Africa, between 75 and 250 million people will face growing shortages in 2020 due to climate change. The scarcity of some arid and semi-arid regions will have a decisive impact on migration.

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

[tex]K=qV[/tex]          (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

[tex]K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J[/tex]

The kinetic energy of the particle is also:

[tex]K=\frac{1}{2}mv^2[/tex]       (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}[/tex]

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

[tex]|F_B|=qvBsin(\theta)[/tex]

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

[tex]|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N[/tex]

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

[tex]r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm[/tex]

the radius of the trajectory of the electron is 10.1 cm

The speed, magnetic force and radius are respectively; 10.75 * 10⁶ m/s; 7.57 * 10⁻¹² N; 0.101 m

A) We know that the formula for kinetic energy can be expressed as;

K = qV

where;

q is charge of the particle = 2e = 2(1.6 × 10⁻¹⁹ C) = 3.2 × 10⁻¹⁹ C

V is potential difference = 1.2 × 10⁶ V

K = 3.2 × 10⁻¹⁹ *  1.2 × 10⁶

K = 3.84 × 10⁻¹³ J

Also, formula for kinetic energy is;

K = ¹/₂mv²

where v is speed

Thus;

v = √(2K/m)

v = √(2 * 3.84 × 10⁻¹³)/(6.64 * 10⁻²⁷)

v = 10.75 * 10⁶ m/s

B) The magnetic force is given by the formula;

F_b = qvB

F_b = (3.2 × 10⁻¹⁹ * 10.75 * 10⁶ * 2.2)

F_b = 7.57 * 10⁻¹² N

C) The formula to find the radius is;

r = mv/qB

r = (6.64 * 10⁻²⁷ * 10.75 * 10⁶)/(1.6 × 10⁻¹⁹ * 2.2)

r = 0.101 m

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Answer:

(a) t = 3.74 s

(b) H = 136.86 m

(c) Vₓ = 41.83 m/s,  Vy = 0 m/s

(d) ax = 0 m/s²,  ay = 9.8 m/s²

Explanation:

(a)

Time to reach maximum height by the projectile is given as:

t = V₀ Sinθ/g

where,

V₀ = Launching Speed = 55.6 m/s

Angle with Horizontal = θ = 41.2°

g = 9.8 m/s²

Therefore,

t = (55.6 m/s)(Sin 41.2°)/(9.8 m/s²)

t = 3.74 s

(b)

Maximum height reached by projectile is:

H = V₀² Sin²θ/g

H = (55.6 m/s)² (Sin²41.2°)/(9.8 m/s²)

H = 136.86 m

(c)

Neglecting the air resistance, the horizontal component of velocity remains constant. This component can be evaluated by the formula:

Vₓ = V₀ₓ = V₀ Cos θ

Vₓ = (55.6 m/s)(Cos 41.2°)

Vₓ = 41.83 m/s

Since, the projectile stops momentarily in vertical direction at the highest point. Therefore, the vertical component of velocity will be zero at the highest point.

Vy = 0 m/s

(d)

Since, the horizontal component of velocity is uniform. Thus there is no acceleration in horizontal direction.

ax = 0 m/s²

The vertical component of acceleration is always equal to the acceleration due to gravity during projectile motion:

ay = 9.8 m/s²

Answer:

19.872 degrees

Explanation:

Mathematically;

Using Snell’s law

n1 sin A = n2 sinB

Where ;

n1 = refractive index in air = 1

n2 is refractive index in water = 1.33

A = ?

B = 45

Substituting the values in the equation;

1 sin A = 1.33 sin45

Sin A = 1.33 sin 45

A = arc sin (1.33 sin 45)

A = 70.12

Thus, the actual direction of the Sun with respect to the horizon = 90-A = 19.872 degrees

Explanation:

The relation between resistance and resistivity is given by :

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

[tex]\rho[/tex] is resistivity of material

l is length of wire

A is area of cross section of wire

Resistivity of a material is the hidden property. If one wire has 3 times the length of the other, then it doesn't affect its resistivity. Hence, the resistivity of two wires is

Answer:

The average velocity of Jackson is 18.056 m/s South

The average velocity of Hunter is 10.65 m/s East

Explanation:

initial velocity of Jackson, u = 25 km/h east = 6.944 m/s east

time for this motion, [tex]t_i[/tex] = 20 minutes = 1200 seconds

⇒initial displacement of Jackson, [tex]x_i[/tex] = (6.944 m/s) x (1200 s) = 8332.8 m

Final velocity of Jackson, v =  45 km/h South = 12.5 m/s South

time at Jackson's final position, [tex]t_f[/tex] = 20 minutes + [tex]t_i[/tex] = 20 minutes + 20 minutes

time at Jackson's final position, [tex]t_f[/tex] = 40 minutes = 2400 s

⇒Final displacement of Jackson,[tex]x_f[/tex] = (12.5 m/s) x (2400 s) = 30,000m

Average velocity of Jackson;

[tex]= \frac{x_f-x_i}{t_f-t_i} \\\\= \frac{30,000-8332.8}{2400-1200} \\\\= 18.056 \ m/s \ South[/tex]

initial velocity of Hunter, u = 45 km/h South = 12.5 m/s South

time for this motion, [tex]t_i[/tex] = 10 minutes = 600 seconds

⇒initial displacement of Hunter, [tex]x_i[/tex] = (12.5 m/s) x (600 s) = 7500 m

Final velocity of Hunter, v =  40 km/h east = 11.11 m/s east

time at Hunter's final position, [tex]t_f[/tex] = 30 minutes + [tex]t_i[/tex] = 30 minutes + 10 minutes

time at Hunter's final position, [tex]t_f[/tex] = 40 minutes = 2400 s

⇒Final displacement of Hunter,[tex]x_f[/tex] = (11.11 m/s) x (2400 s) = 26,664m

Average velocity of Hunter;

[tex]= \frac{x_f-x_i}{t_f-t_o} \\\\= \frac{26,664-7500}{2400-600} \\\\= 10.65 \ m/s \ east[/tex]

Answer:

A) E = 278925.62 N/C with direction; radially out.

B) E = 43048.47 N/C with direction radially out.

C) E = -3214.29 N/C with direction radially in.

Explanation:

From Gauss' Law, the Electric field for any spherically symmetric charge or charge distribution is the same as the point charge formula. Thus;

E = kQ/r²

where;

Q is the net charge within the distance r.

We are given the charge Q = 15-nC and

spherical shell of radius 10cm

A) The distance r = 2.2 cm = 0.022 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C

While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²

E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.022)²

E = 278925.62 N/C

This will be radially out ,since the net charge is positive.

B) The distance r = 5.6 cm = 0.056 m is between the surface and the point charge, so only the point charge lies within this distance and Q = 15 nC = 15 x 10^(-9) C

While k is coulombs constant with a value of 9 × 10^(9) N.m²/C²

E = ((9 x 10^(9) × (15 x 10^(-9)))/(0.056)²

E = 43048.47 N/C

This will be radially out ,since the net charge is positive.

C) The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;

Q = 15 nC - 22 nC

Q = -7 nC = -7 x 10^(-9) C

and;

E = (9 x 10^(9)*(-7 x 10^(-9))/(0.14)²

E = -3214.29 N/C

This will be radially in, since the net charge is negative. You can indicate this with a negative answer.

A) When The distance r is = 2.2 cm = 0.022 m is between the surface and also the point charge, also that so only the point charge lies within this distance and also Q = 15 NC = 15 x 10^(-9) C

B) When The distance r = 5.6 cm = 0.056 m is between the surface and also the point charge, so only the point charge lies within this distance and also Q = 15 nC = 15 x 10^(-9) C

C) Then when The distance r = 14 cm = 0.14 m is outside the sphere so the "net" charge within this distance is due to both given charges. Thus;

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Answer:

B. 9.4 s

Explanation:

In order to calculate the total time taken by the ball to hit the ground, we first analyze the upward motion. We will use subscript 1 for upward motion. Now, using 1st equation of motion:

Vf₁ = Vi₁ + gt₁

where,

Vf, = Final Velocity in upward motion = 0 m/s (ball stops at highest point)

Vi = Initial Velocity in upward motion = 36.2 m/s

g = - 9.8 m/s² (negative due to upward motion)

t₁ = Time taken in upward motion = ?

Therefore,

0 m/s = 36.2 m/s + (-9.8 m/s²)(t₁)

t₁ = (36.2 m/s)/(9.8 m/s²)

t₁ = 3.7 s

Now, using 2nd equation of motion:

h₁ = (Vi₁)(t₁) + (0.5)(g)(t₁)²

where,

h₁ = distance from top of building to highest point ball reaches = ?

Therefore,

h₁ = (36.2 m/s)(3.7 s) + (0.5)(-9.8 m/s²)(3.7 s)²

h₁ = 133.58 - 66.86 m

h₁ = 66.72 m

No, considering downward motion and using subscript 2, for it.

Using 2nd equation of motion:

h₂ = (Vi₂)(t₂) + (0.5)(g)(t₂)²

where,

h₂ = height of the highest point from ground = h₁ + height of building

h₂ = 66.72 m + 90 m = 156.72 m

Vi₂ = Initial Speed during downward motion = 0 m/s (ball stops for a moment at highest point)

t₂ = Time Taken in downward motion = ?

g = 9.8 m/s²

Therefore,

156.72 m = (0 m/s)(t₂) + (0.5)(9.8 m/s²)(t₂)²

t₂² = (156.72 m)/(4.9 m/s²)

t₂ = √31.98 s²

t₂ = 5.7 s

Now, the total time taken by ball to reach the ground is"

Total Time = T = t₁ + t₂

T = 3.7 s + 5.7 s

T = 9.4 s

Therefore, the correct answer is:

B. 9.4 s

Answer:

30 minutes

Explanation:

Energy per time is constant, so:

E₁ / t₁ = E₂ / t₂

m₁C₁ΔT₁ / t₁ = m₂C₂ΔT₂ / t₂

(1 kg) C (70°C − 25°C) / 15 min = (1.5 kg) C (80°C − 20°C) / t

(1 kg) (45°C) / 15 min = (1.5 kg) (60°C) / t

3/min = 90 / t

t = 30 min

Answer:

The mass of the block is 1250g.

Explanation:

Given that the formula for density is ρ = mass/volume. Then you have to substitute the values into the formula :

[tex]ρ = \frac{mass}{volume} [/tex]

Let density = 250,

Let volume = 5,

[tex]250 = \frac{m}{5} [/tex]

[tex]m = 250 \times 5[/tex]

[tex]m = 1250g[/tex]