A meter stick hurtles through space at a speed of 0.95c relative to you, with its length perpendicular to the direction of motion. You measure its length to be equal to:_______
a. 0 m.
b. 0.05 m.
c. 0.95 m.
d. 1.00 m.
e. 1.05 m.

Answer:

Everytime you do an experiment you need something that is regular. For example if you try and measure how much germs spread in bread. you need 1 bread thats clean and 3 different breads for different molds. So thats called a CONTROL

AAAAAAAAAAAA is the answer

Answer:

They had the same speed.

Explanation:

It won't be velocity, because velocity is a vector quantity. Speed is scalar.

Velocity is the rate of change of displacement. During their periods of travel, the cars definitely had the same velocity.

Velocity is the directional speed of a moving object as an indicator of its rate of change in location as perceived from a certain frame of reference and measured by a specific time standard.

Given that the first car travels 40 meters due east in 5 seconds. Therefore, we can write,

Distance = 40 meters

Time = 5 seconds

Velocity = Distance / Time = 40 meter/ 5 sec = 40 m/sec

Also, given that the second car travels 64 meters due west in 8 seconds. Therefore, we can write,

Distance = 64 meters

Time = 8 seconds

Velocity = Distance / Time = 64 meter/ 8 sec = 8 m/sec

Hence, During their periods of travel, the cars definitely had the same velocity.

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Answer:

[tex]{0.51 \mathrm{m} / \mathrm{s}^{2}}[/tex]

Explanation:

Angular acceleration

[tex]\begin{aligned}

\alpha &=\frac{\left(\omega_{f}-\omega_{i}\right)}{t} \\

\omega_{i} &=0 \\

\omega_{f} &=4.30 \mathrm{rev} / \mathrm{s} \\

&=4.30 \times 2 \pi \mathrm{rad} / \mathrm{s} \\

&=27.02 \mathrm{rad} / \mathrm{s} \\

\alpha &=\frac{(27.02-0)}{3.15} \\

&=8.57 \mathrm{m} / \mathrm{s}^{2}

\end{aligned}[/tex]

a)Tangential acceleration

[tex]\begin{aligned}

a &=r \alpha \\

&=\frac{12}{2} \times 10^{-2} \times 8.57 \\

a &=0.51 \mathrm{m} / \mathrm{s}^{2}

\end{aligned}[/tex]

The tangential acceleration of the disc is [tex]{0.51 \mathrm{m} / \mathrm{s}^{2}}[/tex]

This question involves the concepts of the equations of motion for angular motion.

The tangential acceleration of a point on the outer rim of the disk at the moment when its angular speed reaches 2 rev/s will be "0.532 m/s²".

First, we will use the first equation of motion for the angular motion to find out the angular acceleration:

[tex]\alpha=\frac{\omega_f-\omega_i}{t}[/tex]

where,

[tex]\alpha[/tex] = angular acceleration = ?

[tex]\omega_f[/tex] = final angular speed = (4.3 rev/s)[tex](\frac{2\pi\ rad}{1\ rev})[/tex] = 27.02 rad/s

[tex]\omega_i[/tex] = initial angular speed = 0 rad/s

t = time taken = 3.05 s

Therefore,

[tex]\alpha =\frac{27.02\ rad/s-0\ rad/s}{3.05\ s}\\\\\alpha= 8.86\ rad/s^2[/tex]

Now, the tangential acceleration can be given as follows:

[tex]a=r\alpha\\a=(\frac{diameter}{2})(8.86\ rad/s^2)\\\\a=(\frac{0.12\ m}{2})(8.86\ rad/s^2)\\\\[/tex]

a = 0.532 m/s²

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The attached picture shows the angular equations of motion.

Answer:

hypothesis , hope it helps

Explanation:

Answer:

Inference

Explanation:

Inference is something you predict after testing that's a result after an hypothesis has been made. Hypothesis is an intelligent guess based on some observed phenomena which can be subjected to further testing.

Answer:

Cube temperature = 526.83 K

Explanation:

Volume of the cube and sphere will be the same.

Now, volume of cube = a³

And ,volume of sphere = (4/3)πr³

Thus;

a³ = (4/3)πr³

a³ = 4.1187r³

Taking cube root of both sides gives;

a = 1.6119r

Formula for surface area of sphere is;

As = 4πr²

Also,formula for surface area of cube is; Ac = 6a²

Thus, since a = 1.6119r,

Then, Ac = 6(1.6119r)²

Ac = 15.5893r²

The formula for radiant power is;

Q' = eσT⁴A

Where;

e is emissivity

σ is Stefan boltzman constant = 5.67 x 10^(-8) W/m²k

T is temperate in kelvin

A is Area

So, for the cube;

(Qc)' = eσ(Tc)⁴(Ac)

For the sphere;

(Qs)' = eσ(Ts)⁴(As)

We are told (Qc)' = (Qs)'

Thus;

eσ(Tc)⁴(Ac) = eσ(Ts)⁴(As)

eσ will cancel out to give;

(Tc)⁴(Ac) = (Ts)⁴(As)

Since we want to find the cube's temperature Tc,

(Tc)⁴ = [(Ts)⁴(As)]/Ac

Plugging in relevant figures, we have;

(Tc)⁴ = [556⁴ × 4πr²]/15.5893r²

r² will cancel out to give;

(Tc)⁴ = [556⁴ × 4π]/15.5893

Tc = ∜([556⁴ × 4π]/15.5893)

Tc = 526.83 K

Answer:

C = 44.75 x 10⁻⁸ F

Explanation:

Assuming no loss of energy between capacitor and inductor

energy in inductor initially = 1/2 Li₀² where L is inductance and i₀ is peak current .

= .5 x 6 x 10⁻³ x .57²

= .97 x 10⁻³ J .

This energy is transferred to capacitor .

energy of capacitor = 1/2 CV²

= .5 x C x 66²

= 2178 C

2178C = .97 x 10⁻³

C = 44.75 x 10⁻⁸ F .

The magnetic energy stored in the inductor is transformed into electrical energy stored in the capacitor. The value of capacitance for the given circuit is 44.75 x 10⁻⁸ F

According to the law of conservation of energy, the magnetic energy stored in the inductor will be gradually lost and this energy will be stored in the capacitor as electrical energy.

Initially, the energy in the inductor is:

E = 1/2 Li₀²

where L is inductance

and i₀ is peak current.

E = 0.5 × 6 × 10⁻³ × (0.57)²

E = 0.97 × 10⁻³J

This energy is transformed into electrical energy stored in the capacitor.

So the capacitor energy is:

E = 1/2 CV²

where C is the capacitance

E = 0.5 × C × 66²

E = 2178 C

0.97 x 10⁻³ = 2178 C

C = 44.75 x 10⁻⁸ F

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Answer:

The mass of the cargo is [tex]M = 188.43 \ kg[/tex]

Explanation:

From the question we are told that

    The radius of the spherical balloon is  [tex]r = 7.40 \ m[/tex]

     The mass of the balloon is  [tex]m = 990\ kg[/tex]  

The volume of the spherical balloon is mathematically represented as

     [tex]V = \frac{4}{3} * \pi r^3[/tex]

substituting values

      [tex]V = \frac{4}{3} * 3.142 *(7.40)^3[/tex]

      [tex]V = 1697.6 \ m^3[/tex]

The total mass  the balloon can lift is mathematically represented as

     [tex]m = V (\rho_h - \rho_a)[/tex]

where [tex]\rho_h[/tex] is the density of helium with a  value of

       [tex]\rho_h = 0.179 \ kg /m^3[/tex]

and  [tex]\rho_a[/tex] is the density of air with a value of

        [tex]\rho_ a = 1.29 \ kg / m^3[/tex]

substituting values

          [tex]m = 1697.6 ( 1.29 - 0.179)[/tex]

         [tex]m = 1886.0 \ kg[/tex]

Now the mass of the cargo is mathematically evaluated as

        [tex]M = 1886.0 - 1697.6[/tex]

        [tex]M = 188.43 \ kg[/tex]

Answer:

a. 42N

b. 11.8m/s

c. 1.69s

d. 160N

Explanation:

a)  The tension of the rope is 130.66 N.

b) The speed of the bucket while strike the water = 4.64 m/s.

c) The time of fall is  = 4.303 second.

d)  While the bucket is falling, what is the force exerted on the cylinder by the axle is  130.66 N.

Mass of the water bucket; M = 15.0 kg

Mass of the cylinder; m =  12.0 kg

Height of the bucket; h = 10.0 m.

They are connected by a rope and a pivots.

So, acceleration of them is same and let it be a.

So equation of motion of both of them be:

Mg - T = Ma

and, T - mg = ma

Hence, a = g(M-m)/(M+m)

= 9.8(15-12)/(15+12)

= 1.08 m/s²

And, T = m(g+a)

= 12.0(9.8+1.08)

= 130.66 N.

a) so tension of the rope is 130.66 N.

b) speed of the bucket while strike the water = √2ah =√(2×1.08×10.0) m/s = 4.64 m/s.

c) The time of fall is = √2h/a = √(2×10/1.08) second = 4.303 second.

d) While the bucket is falling, what is the force exerted on the cylinder by the axle is tension of the rope, that is, 130.66 N.

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Answer:

The velocity is  [tex]v_2= 0.45 \ m/s[/tex]

Explanation:

From the question we are told that

      The initial speed of the hot water is  [tex]v_1 = 0.85 \ m/s[/tex]

     The pressure from the heater  [tex]P_1 = 450 \ KPa = 450 *10^{3} \ Pa[/tex]

      The height of the hot water before flowing is  [tex]h_1 = 0 \ m[/tex]

      The height of bathtub above the heater is [tex]h_2 = 3.70 \ m[/tex]

       The pressure in the pipe is [tex]P_2 = 414 KPa = 414 *10^{3} \ Pa[/tex]

       The density of water is [tex]\rho = 1000 \ kg/m^3[/tex]

Apply Bernoulli equation

      [tex]P_1 + \rho gh_1 +\frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2}\rho v_2 ^2[/tex]

Substituting values

     [tex](450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) = (1000 * *9.8*3.70) + (0.5*1000*v_2^2 )[/tex]

=>   [tex]v_2^2 = \frac{ (450 *10^{3}) + (1000 * 9.8 *0 ) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}[/tex]

=>   [tex]v_2= \sqrt{ \frac{ (450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}}[/tex]

=>    [tex]v_2= 0.45 \ m/s[/tex]

Answer:

Explanation:

The equivalent resistance of resistor connected parallel in the circuit is [tex]66.66 ohm[/tex]

The equivalent resistance is the total resistance measured in a parallel or series circuit. If several resistors are connected together and connected to a battery, the current supplied by the battery depends on the equivalent resistance of the circuit.

Resistors are in series whenever the current flows through the resistors sequentially. It is given by

[tex]R_{eq} = R_{1} + R_{2} + ....[/tex]

Resistors are in parallel when one end of all the resistors are connected by a continuous wire and the other end of all the resistors are also connected to one another through a continuous wire.

The equivalent resistance is the total resistance measured in a parallel. It is given by

[tex]\frac{1}{R_{eq} } = \frac{1}{R_{1} } + \frac{1}{R_{2} }+ ....[/tex]

Given:

Resistor, [tex]R_{1} = 100 ohm[/tex]

Resistor, [tex]R_{2} = 200 ohm[/tex]

Voltage, [tex]V = 10 Volt[/tex]

Since, resistors are connected in parallel, the equivalent resistor is given by,

[tex]\frac{1}{R_{eq} } = \frac{1}{R_{1} } + \frac{1}{R_{2} }[/tex]

[tex]\frac{1}{R_{eq} } = \frac{1}{100 } + \frac{1}{200 }[/tex]

[tex]R_{eq} = \frac{100*200}{100+200} \\R_{eq} = 66.66 ohm[/tex]

Hence, the equivalent resistor is [tex]66.66 ohm[/tex].

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Answer:

It's true

Explanation:

Because the ship is mafe up of aluminium, which is a light metal.

Answer:

False

Explanation:

Took The Quiz

Answer:

The correct option is;

- 4x

Explanation:

From the inverse square law, as the distance from the source of a physical quantity increases, the intensity of the source is spread over an area proportional to the square of the distance of the object from the source

The inverse square law can be presented as follows;

[tex]I = \dfrac{S}{4\times \pi \times r^2 }[/tex]

As the distance, r, increases, the surface it covers also increases by the power of 2

Therefore, where the distance increases from r to 2·r, we have;

When, I, remain constant

[tex]I = \dfrac{4\times S}{4\times \pi \times (2\cdot r)^2 } = I = \dfrac{4\times S}{4\times 4\times \pi \times r^2 } = \dfrac{S}{4\times \pi \times r^2 }[/tex]

The surface increases to 4·S by the inverse square law

Therefore, the correct option is 4 × x.

Answer:

a) t = 27.00 h

b) B = 6.84 MeV/nucleon

Explanation:

a) The time can be calculated using the following equation:

[tex] R = R_{0}e^{-\lambda*t} [/tex]

Where:

R: is the radiation measured at time t

R₀: is the initial radiation

λ: is the decay constant

t: is the time

The decay constant can be calculated as follows:

[tex] t_{1/2} = \frac{ln(2)}{\lambda} [/tex]

Where:

t(1/2): is the half life = 4.5 h

[tex] \lambda = \frac{ln(2)}{t_{1/2}} = \frac{ln(2)}{4.5 h} = 0.154 h^{-1} [/tex]

We have that the radiation measured is 64 times the maximum permissible level, thus R₀ = 64R:  

[tex] \frac{R}{64R} = e^{-\lambda*t} [/tex]                      

[tex] t = -\frac{ln(1/64)}{\lambda} = -\frac{ln(1/64)}{0.154 h^{-1}} = 27.00 h [/tex]            

b) The binding energy (B) can be calculated using the following equation:

[tex]B = \frac{(Z*m_{p} + N*m_{n} - M_{A})}{A}*931.49 MeV/u[/tex]

Where:

Z: is the number of protons = 2 (for [tex]^{4}_{2}He[/tex])

[tex]m_{p}[/tex]: is the proton mass = 1.00730 u

N: is the number of neutrons = 2 (for [tex]^{4}_{2}He[/tex])

[tex]m_{n}[/tex]: is the neutron mass = 1.00869 u  

[tex]M_{A}[/tex]: is the mass of the He atom = 4.002602 u

A =  N + Z = 2 + 2 = 4    

The binding energy of [tex]^{4}_{2}He[/tex] is:

[tex]B = \frac{(2*1.00730 + 2*1.00869 - 4.002602)}{4}*931.49 MeV/u = 7.35\cdot 10^{-3} u*931.49 MeV/u = 6.84 MeV/nucleon[/tex]

Hence, the binding energy per nucleon is 6.84 MeV.

I hope it helps you!

Answer:

107 N, option d

Explanation:

Given that

mass of the ball, m = 0.2 kg

initial velocity of the ball, u = 20 m/s

final velocity of the ball, v = -12 m/s

time taken, Δt = 60 ms

Solving this question makes us remember "Impulse Theorem"

It states that, "that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object"

Mathematically, it is represented as

FΔt = m(v - u), where

F = the average force

Δt = time taken

m = mass of the ball

v = final velocity of the ball

u = initial velocity of the ball

From the question we were given, if we substitute the values in it, we have

F = ?

Δt = 60 ms = 0.06s

m = 0.2 kg

v = -12 m/s

u = 20 m/s

F = 0.2(-12 - 20) / 0.06

F = (0.2 * -32) / 0.06

F = -6.4 / 0.06

F = -106.7 N

Thus, the magnitude is 107 N

Answer:

The velocity is  [tex]v_2 = 6.8 \ m/s[/tex]

The pressure is  [tex]P_2 = 204978 Pa[/tex]

Explanation:

From the question we are told that

 The speed at which water is travelling through is  [tex]v = 1.7 \ m/s[/tex]

  The pressure is  [tex]P_1 = 205 k Pa = 205 *10^{3} \ Pa[/tex]

   The diameter of the new pipe is [tex]d = \frac{D}{2}[/tex]

Where D is the diameter of first pipe

According to the principal of continuity we have that

       [tex]A_1 v_1 = A_2 v_2[/tex]    

Now  [tex]A_1[/tex] is the area of the first pipe which is mathematically represented as

       [tex]A_1 = \pi \frac{D^2}{4}[/tex]

and  [tex]A_2[/tex] is the area of the second pipe which is mathematically represented as  

       [tex]A_2 = \pi \frac{d^2}{4}[/tex]

Recall   [tex]d = \frac{D}{2}[/tex]

        [tex]A_2 = \pi \frac{[ D^2]}{4 *4}[/tex]

        [tex]A_2 = \frac{A_1}{4}[/tex]

So    [tex]A_1 v_1 = \frac{A_1}{4} v_2[/tex]

substituting value

        [tex]1.7 = \frac{1}{4} * v_2[/tex]    

        [tex]v_2 = 4 * 1.7[/tex]    

       [tex]v_2 = 6.8 \ m/s[/tex]

According to Bernoulli's equation  we have that

     [tex]P_1 + \rho \frac{v_1 ^2}{2} = P_2 + \rho \frac{v_2 ^2}{2}[/tex]

substituting values

     [tex]205 *10^{3 }+ \frac{1.7 ^2}{2} = P_2 + \frac{6.8 ^2}{2}[/tex]

     [tex]P_2 = 204978 Pa[/tex]

Answer:

A. 4148 J/K/Kg

B. 4148 J/K/L

Explanation:

A. Heat capacity per unit mass is known as the specific heat capacity, c.

C = Heat capacity/mass(kg)

C = (523 J/K) / 0.125 Kg = 4148 J/K/Kg

B. Volume of water = mass/density

Density of water = 1 Kg/L

Volume of water = 0.125 Kg/ 1Kg/L

Volume of water = 0.125 L

Heat capacity per unit volume = (523 J/K) / 0.125 L

Heat capacity per unit volume = 4148 J/K/L

Answer:

Final Temperature = 71 °C

Explanation:

In this case, the ideal gas equation is written as;

PV = mRT

Where;

P is pressure

V is volume

m is mass

R is gas constant

T is temperature

We will take the volume to be constant.

So, in the initial state, we have;

P1•V = m1•R•T1 - - - eq(1)

In the final state, we have;

P2•V = m2•R•T2 - - - - eq(2)

Combining eq (1) and eq(2),we have;

P1•m2•R•T2 = P2•m1•R•T1

Dividing both sides by R gives;

P1•m2•T2 = P2•m1•T1

Making T2 the subject gives;

T2 = (P2•m1•T1)/(P1•m2)

Now, we are given;

m1 = 2kg

m2 = ½*2 = 1kg

P1 = 4 atm

P2 = 2.2 atm

T1 = 40°C = 273 + 40 K = 313K

Plugging in this values into the T2 equation, we have;

T2 = (2.2 × 2 × 313)/(4 × 1)

T2 = 344 K

Converting to °C, we have;

T2 = 344 - 273 = 71 °C

Answer:

0.981 rad/sec^2

Explanation:

mass that pulls on string = 5 kg

weight due to mass = 5 x 9.81 = 49.05 N

radius of rod = 0.1 m

torque produced by this force on the rod = force x radius

torque = 49.05 x 0.1 = 4.905 N-m

mass of disk = 125 kg

radius of disk = 0.2 m

moment of inertia of the disk I = m[tex]r^{2}[/tex]

I = 125 x [tex]0.2^{2}[/tex] = 5 kg-m^2

from the equation, T = Iα

where T is torque

I is moment of inertia

α is angular acceleration

imputing values,

4.905 = 5α

α = 4.905/5 = 0.981 rad/sec^2

Answer:

it takes the shape and size of the container that it is in

Explanation:

Answer:

it takes the shape and size of a container

Answer:

The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.

Explanation:

The Free Body Diagram associated with the experiment is presented as attachment included below.

Friction is a contact force that occurs as a reaction against any change in state of motion, which is fostered by gravity.

Normal force is another contact force that appears as a reaction to the component of weight perpendicular to the direction of motion. Let consider a framework of reference consisting in two orthogonal axes, one being parallel to the direction of motion (x-axis) and the other one normal to it (y-axis). Equations of motion are described herein:

[tex]\Sigma F_{x} = W \cdot \sin \theta - f = 0[/tex]

[tex]\Sigma F_{y} = N - W \cdot \cos \theta = 0[/tex]

Where:

[tex]W[/tex] - Weight of the eraser, measured in newtons.

[tex]f[/tex] - Friction force, measured in newtons.

[tex]N[/tex] - Normal force, measured in newtons.

[tex]\theta[/tex] - Angle of the incline, measured in degrees.

The maximum allowable static friction force is:

[tex]f = \mu_{s} \cdot N[/tex]

Where:

[tex]\mu_{s}[/tex] - Coefficient of static friction, dimensionless.

[tex]N[/tex] - Normal force, measured in newtons.

Likewise, the kinetic friction force is described by the following model:

[tex]f = \mu_{k} \cdot N[/tex]

Where:

[tex]\mu_{k}[/tex] - Coefficient of static friction, dimensionless.

[tex]N[/tex] - Normal force, measured in newtons.

And weight is equal to the product of the mass of eraser and gravitational constant ([tex]g = 9.807\,\frac{m}{s^{2}}[/tex])

In this exercise, coefficients of static and kinetic friction must be determined. First equation of equilibrium has to be expanded and coefficient of friction cleared:

[tex]m\cdot g \cdot \sin \theta - \mu\cdot N = 0[/tex]

[tex]\mu = \frac{m\cdot g \cdot \sin \theta}{N}[/tex]

But [tex]N = m\cdot g \cos \theta[/tex], so that:

[tex]\mu = \tan \theta[/tex]

Now, coefficients of static and kinetic friction are, respectively:

[tex]\mu_{s} = \tan 35.6^{\circ}[/tex]

[tex]\mu_{s} \approx 0.716[/tex]

[tex]\mu_{k} \approx \tan 30.8^{\circ}[/tex]

[tex]\mu_{k} \approx 0.596[/tex]

The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.

Answer:[tex]8.062\ m/s[/tex]

Explanation:

Given

masss of football player [tex]M=110\ kg[/tex]

Velocity of football player [tex]u_1=8\ m/s[/tex]

mass of football [tex]m=0.41\ kg[/tex]

velocity of football [tex]u_2=25\ m/s[/tex]

Final velocity will be given by applying conservation of linear momentum

After catching the ball Player and ball moves with same velocity

[tex]\Rightarrow Mu_1+mu_2=(M+m)v[/tex]

[tex]\Rightarrow 110\times 8+0.41\times 25=(110+0.41)v[/tex]

[tex]\Rightarrow 880+10.25=110.41\times v[/tex]

[tex]\Rightarrow v=\frac{890.25}{110.41}=8.063\ m/s[/tex]

So, final velocity will be [tex]8.062\ m/s[/tex]

Answer:

The  drag coefficient is  [tex]D_z = 1.30512[/tex]  

Explanation:

From the question we are told that

     The density of air is  [tex]\rho_a = 1.21 \ kg/m^3[/tex]

     The diameter of bottom part is  [tex]d = 0.15 \ m[/tex]

The  power trend-line  equation is mathematically represented as

      [tex]F_{\alpha } = 0.9226 * v^{0.5737}[/tex]

let assume that the velocity is  20 m/s

Then

      [tex]F_{\alpha } = 0.9226 * 20^{0.5737}[/tex]

       [tex]F_{\alpha } = 5.1453 \ N[/tex]

The drag coefficient is mathematically represented as

      [tex]D_z = \frac{2 F_{\alpha } }{A \rho v^2 }[/tex]

Where  

     [tex]F_{\alpha }[/tex] is the drag force

      [tex]\rho[/tex] is the density of the fluid

       [tex]v[/tex] is the flow velocity

       A is the area which mathematically evaluated as

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

substituting values

     [tex]A = 3.142 * \frac{(0.15)^2}{4}[/tex]

     [tex]A = 0.0176 \ m^2[/tex]

Then

   [tex]D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }[/tex]

   [tex]D_z = 1.30512[/tex]  

Answer:

F = ILB

Explanation:

To find the net force on the conducting bar you take into account the following expression:

[tex]\vec{F}=I( \vec{L}X \vec{B})[/tex]

I: current in the conducting bar

L: length of the bar

B: magnitude of the magnetic field

In this case the direction of the magnetic field and the motion of the bar are perpendicular between them. The direction of the bar is + i, and the magnetic field poits upward + k. The cross product of these vector give us the direction of the net force:

+i X +k = +j

The direction of the force is to the right and its magnitude is F = ILB

An object's resistance to any change in motion is the Inertia of the object.

Answer:

Q = 3877 KJ

Explanation:

Since, the system is closed and isolated. Therefore, the law of conservation of energy can be written as:

Heat Absorbed By Water (Q) = Heat required to raise the temperature of water (Q₁) + Heat required to convert water to steam (Q₂)

Q = Q₁ + Q₂   ----- equation (1)

Now, for Q₁:

Q₁ = m C ΔT

where,

m = Mass of Water = 1.5 kg

C = Specific Heat of Water = 4187 J/kg.°C

ΔT = Change in Temperature of Water = T₂ - T₁ = 100°C - 22°C = 78°C

Therefore,

Q₁ = (1.5 kg)(4187 J/kg.°C)(78°C)

Q₁ = 490 x 10³ J =490 KJ

Now, for Q₂:

Q₂ = m H

where,

m = Mass of Water = 1.5 kg

H = Heat of Vaporization of Water = 2258 KJ/kg

Therefore,

Q₂ = (1.5 kg)(2258 KJ/kg)

Q₂ = 3387 KJ

Substituting the values in equation (1), we get:

Q = Q₁ + Q₂

Q = 490 KJ + 3387 KJ

Q = 3877 KJ

Answer:

by using it's buoyant or floating effect by Archimedes.

the buoyant force act on the astronauts body and make he/ she feels like in low gravity.

the buoyant force equation is

F = Density of liquid x earth gravitational field x volume of astronauts body and suit.

the Weight of astronauts in the pools will be less than in the land or air.

Weight in water = weight in air/land - buoyant force

so the astronauts will feel like in the outer space with low gravity.

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:

Explanation:

Volume of asteroid = 4/3 x π x 160³

= 17.15 x 10⁶

mass = volume x density

= 17.15 x 10⁶ x 2600

= 445.9 x 10⁸ kg

kinetic energy

= 1/2 x 445.9 x 10⁸  x( 12.6 )² x 10⁶

= 35.4 x 10¹⁷ J .

2 )

energy of 15 megaton

= 4.184 x 10¹⁵ x 15 J

= 62.76 x 10¹⁵ J

No of bombs required

= 35.4 x 10¹⁷ / 62.76 x 10¹⁵

= 56.4 Bombs .

Answer: B

Explanation:

I would say the shrinking of the polar ice caps because in order for ice caps to shrink, they would have to obviously melt. This will cause the sea level and total volume of sea water to rise and cover up the land bridges

Answer:B :)

Explanation:

Answer:

a. 200 m

b. 100 m

Explanation:

Solution:-

- We will first draw three points marked A,B and Q from left most to right most.

- We are told that the antennas at A and B radiate in phase. This means the radio-waves emitted by each antenna are synchronous in terms of ( frequency and wavelength ).

- We will denote the common wavelength of coherent sources of radio-waves ( A and B ) with λ.

- The relation between the wavelength ( λ ) and the path difference between the source and observation point ( Q ) for the case of destructive interference is:

                             AQ - BQ = n*λ/2

Where,

             n: The order of wavelength

             AQ: The distance between antenna A and point Q

             BQ: The distance between antenna B and point Q

- The point Q is positioned ( 100 + 50 ) m away from antenna A and 50 m from antenna B. Hence,

                            150 - 50 = n*λ/2

- To determine the longest wavelength ( λ ) to meet destructively at point Q with the given path difference. The order of wavelength ( n ) must be minimum ( 1 ). Therefore,

                           100 = λ/2

                           λ = 200 m  .... Answer

- The relation between the wavelength ( λ ) and the path difference between the source and observation point ( Q ) for the case of constructive interference is:

                             AQ - BQ = n*λ

Where,

             n: The order of wavelength

             AQ: The distance between antenna A and point Q

             BQ: The distance between antenna B and point Q

- The point Q is positioned ( 100 + 50 ) m away from antenna A and 50 m from antenna B. Hence,

                            150 - 50 = n*λ

- To determine the longest wavelength ( λ ) to meet constructively at point Q with the given path difference. The order of wavelength ( n ) must be minimum ( 1 ). Therefore,

                           100 = λ

                           λ = 100 m  .... Answer