The diagram below shows a 5.00-kilogram block
at rest on a horizontal, frictionless table.
5.00-kg
block
Table
Which of the following is the correct name and strength of the force holding the block up?

(B) 1.00 m

Explanation:

Since the meter stick is traveling with Jill, it will have the same speed as she does so relative to Jill, the meter stick is stationary so its length remains 1.00 m as measured by her.

When astronaut Jill leaves Earth in a spaceship and is now traveling at a speed of 0.280c and measure the meter stick to be 1 meter. Hence, option B is correct.

Length contraction is defined as the phenomenon of the moving object being shorter than its appropriate length, measured in the object's rest frame.

When the object travels with the speed of light, the length of the object gets more contracted than its original length, relative to the observer. It is also known as the Lorentz-Fitgerald contraction.

Length contraction, L = L₀√(1-v²/c²), where L is the original length, L₀ is the contracted length. c² is known as the velocity of light. v² is the velocity of the speed of the object.

From the given,

speed of the spaceship = 0.280c      (c is the speed of the light)

Length contraction, L = L₀ √(1-v²/c²)

The stick also travels in the spaceship. Hence, the length of the meter stick does not change. It remains at its original length of one meter.  Thus, the ideal solution is option B.

To learn more about length contraction:

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Answer:

875 Watts

Explanation:

P = W/t = mgh/t = 700(10)/8 = 875 Watts

Answer:

[tex]\boxed {\boxed {\sf 2 \ m/s^2}}[/tex]

Explanation:

Acceleration is the rate of change in velocity with respect to time. It is calculated by dividing the change in velocity by the change in time. The formula is:

[tex]a= \frac{ \Delta v}{\Delta t}[/tex]         or          [tex]a= \frac{v_f-v_i}{\Delta t}[/tex]

The change in velocity is the difference between the initial velocity and the final velocity. The motorcycle starts at rest, or 0 meters per second and reaches 6 meters per second. The change in time is 3 seconds.

[tex]\bullet \ v_f= 6 \ m/s\\\bullet \ v_i= 0 \ m/s \\\bullet \ \Delta t = 3 \ s[/tex]

Substitute the values into the formula

[tex]a= \frac { 6 \ m/s - 0 m/s}{3 \ s}[/tex]

Solve the numerator.

[tex]a= \frac{6 \ m/s}{3 \ s}[/tex]

[tex]a= 2 \ m/s^2[/tex]

The motorcyclist's acceleration is 2 meters per second squared.

Answer:

When an object covers equal distance in an equal interval of time, it is uniform motion but when an object covers unequal distance in an equal interval of time, it is called non uniform motion.

Answer:

the electric field and the electric potential increase 5 times

Explanation:

The electric field created by a point charge is

         E = k q / r²

in this case the charge changes from q₁ = 1 10⁺⁰ C to q₂ = 5 10⁻⁹ C

with the electric field is proportional to the charge

         E₅ = 5 E₁

the electrical power for a point charge is

         V = k q / r

as the electric power is proportional to the charge

         V₅ = 5 V₁

consequently both the electric field and the electric potential increase 5 times

An electric field is produced by a charged object:

[tex]\to E = \frac{k q}{r^2}\\\\[/tex]

In this situation, the charge shifts:

[tex]\to q_1 = 1 10^{\circ}\ C \\\\ \to q_2 = 5 \times 10^{-9}\ C[/tex]

so with electric field remaining proportional to the charge

[tex]\to E_5 = 5 E_1[/tex]

The electrical power consumed for a point charge:

 [tex]\to V = \frac{k q}{ r}[/tex]

since the electric power is related to the charge

[tex]\to V_5 = 5 V_1[/tex]

As a result, both the electric field as well as the electric potential increase by 5 times.

Learn more:

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Answer:

The speed of bullet is 354.2 m/s

Explanation:

initial temperature, T = 30 degree C

specific heat, c = 128 J/kg K

Latent heat, L = 24.7 x 1000 J/kg

melting point = 600 K = 327 degree C

Let the mass is m and the speed is v.

Kinetic energy = heat used to increase the temperature + Heat used to melt

[tex]\frac{1}{2} mv^2 = m c (T' - T) + m L\\\\0.5 v^2 = 128 \times (327 - 30) + 24.7\times 1000\\\\0.5 v^2 = 38016 + 24700 \\\\0.5 v^2 = 62716\\\\v = 354.2 m/s[/tex]

This seems to be incomplete, as we do not have any information about the magnetic field surrounding the wire, but we can answer in a general way.

We know that for a wire of length L, with a current I, and in a magnetic field B, the force can be written as:

F = L*(IxB)

if we define the right as the positive x-axis, and knowing that the current flows to the right, we can write:

I = i*(1, 0, 0)

And the field will be some random vector that can't be parallel to the current because in that case, we do not have any force.

To find the direction of the force, which will tell us the direction in which the wire deflects or moves, first, we need to point with our thumb in the direction of the current, in this case, to the right.

Now, with the hand open, using the tip of our other fingers we point in the direction of the magnetic field.

For example, if the magnetic field is in the positive z-axis, we will point upwards.

Now the palm of our hand tells us in which direction the force is applied.

This is the right-hand rule.

For example, in the case that the current goes to the right and the magnetic field is upwards, we could see that the force is to the front.

Explanation:

The density of earth [tex]\rho_E[/tex] is given by

[tex]\rho_E = \dfrac{M_E}{\left(\frac{4\pi}{3}R_E^3\right)}[/tex]

and in terms of this density, we can write the acceleration due to gravity on earth as

[tex]g_E =G\dfrac{M_E}{R_E^2} = \dfrac{4\pi G}{3}\rho_ER_E[/tex]

Similarly, the acceleration due to gravity [tex]g_P[/tex] on this new planet is given by

[tex]g_P = G\dfrac{M_P}{R_P^2} = G\dfrac{\frac{4\pi}{3}R_p^3\rho_P}{R_P^2}[/tex]

[tex]\:\:\:\:\:= \dfrac{4\pi G}{3}\rho_PR_P[/tex]

We know that this planet has the same density as earth and has a radius 2 times as large. We can then rewrite [tex]g_P[/tex] as

[tex]g_P = \dfrac{4\pi G}{3}\rho_E(2R_E)[/tex]

[tex]\:\:\:\:\:= 2\left(\dfrac{4\pi G}{3}\rho_ER_E\right) = 2g_E[/tex]

[tex]\:\:\:\:\:= 2(9.8\:\text{m/s}^2) = 19.6\:\text{m/s}^2[/tex]

We know

[tex]\boxed{\sf E=hv}[/tex]

[tex]\\ \sf\longmapsto E=6.63\times 10^{-34}J\times 3.6\times 10^{15}s^{-1}[/tex]

[tex]\\ \sf\longmapsto E=23.86\times 10^{-19}J[/tex]

[tex]\\ \sf\longmapsto E=2.38\times 10^{-18}J[/tex]

[tex]\\ \sf\longmapsto E=2.4\times 10^{-18}J[/tex]

Answer:

D!!!!!

Explanation:

Complete Question

Complete Question is attached below

Answer:

a)  [tex]D=0.7[/tex]

b)  [tex]W=1787.5N[/tex]

c)  [tex]\rho'=998.19kg/m^3[/tex]  

Explanation:

From the question we are told that:

Hot air:

Temperature [tex]T_a=360K[/tex]

Pressure [tex]P_a=100kPa[/tex]

Distance [tex]d=12m[/tex]

Weight [tex]W=1400N[/tex]

Water:

Temperature [tex]T_w=300K[/tex]

Pressure [tex]P_w=100kPa[/tex]

Since we have The same Reynolds number

a)

Generally the equation for equal Reynolds number is mathematically given by

Re_{air}=Re_{water}

Therefore

[tex]\frac{\rho V D}{\mu_{air}}=\frac{p_{water}*V*D}{\mu_{water]}}[/tex]

[tex]\frac{100*12}{300*0.28*1.81*10^{-3}}}=\frac{998*D}{0.000890}[/tex]

[tex]D=0.7[/tex]

b)

Generally the equation for Weight of scale is mathematically given by

[tex]W=\rho*V*g[/tex]

[tex]W=998*\frac{4}{3}*\pi*0.35^3*9.81[/tex]

[tex]W=1787.5N[/tex]

c)

Generally the equation for Density of buoyant material is mathematically given by

[tex]\rho'=\frac{w}{g*V}[/tex]

[tex]\rho'=\frac{1781.5}{\frac{4}{3}*\pi*0.35^3*9.81}[/tex]

[tex]\rho'=998.19kg/m^3[/tex]

Answer:

true

Explanation:

The kinetic energy of a gas is directly proportional to the temperature of the gas.because temperature is the average kinetic energy of a substance

I hope this helps

Answer: They are equal in magnitude.

- The signs of the two changes in potential energy are opposite

Explanation:

When the proton and electron both move through the same displacement in an electric field, the change in potential energy that is associated with the proton is equal in magnitude.

Also, it should be noted that the signs of the two changes in potential energy are opposite.

The intensity of radiation is the defined as amount of energy per surface angle which can be used to determine the amount of energy emitting from a source that will hit another surface.

The factors that affect the intensity of radiation are

Type of emission from the source :This  can be alpha, gamma, beta or electromagnetic rays etc

Distance of the detector from the source: The shorter the distance between the source and the detector, the more the effect and vice versa for the longer the distance.

Type of materials between the source and the detector: The type of material between the source and the detector will tell how absorbing and penetrating the radiation is.

Read more on Radiation Intensity here:  https://brainly.com/question/10148635

Answer:

10.78 s

Explanation:

The force on the charge is computed by using the equation:

[tex]F^{\to}= qE^{\to} +q (v^{\to} + B^{\to}) \\ \\ F^{\to} = (9.12 \times 10^{-6}) *278 (-\hat k) +9.12 *10^{-6} *2.1 *0.18 (\hat i * \hat k) \\ \\ F^{\to} = -2.535 *10^{-3} \hat k -3.447*10^{-6} \hat j[/tex]

F = ma

∴

[tex]a ^{\to}= \dfrac{F^{\to}}{m}[/tex]

[tex]a ^{\to}= \dfrac{-1}{3.57\times 10^{-3}}(2.535*10^{-3}\hat k + 3.447*10^{-6} \hat j)[/tex]

[tex]a ^{\to}=-0.710 \hat k -9.656*10^{-4} \hat j[/tex]

At time t(sec; the partiCle velocity becomes [tex]v(t) = 3.78 v_o[/tex]

The velocity of the charge after the time t(sec) is expressed by using the formula:

[tex]v^{\to}= v_{o \ \hat i} + a^{\to }t \\ \\ \implies (2.1)\hat i -0.710 t \hat k -9.656 \times 10^{-4} t \hat j = 3.78 v_o \\ \\ \implies (2.1)^2 +(0.710\ t)^2+ (9.656 *10^{-4}t )^2 = (3.78 *2.1^2 \\ \\ \implies 4.41 +0.5041 t^2 +9.324*10^{-7} t^2 = 63.012 \\ \\ \implies 4.41 +0.5041 t^2 = 63.012\\ \\ 0.5041t^2 = 63.012-4.41 \\ \\ t^2 = \dfrac{58.602}{0.5041} \\ \\ t^2 = 116.25 \\ \\ t = \sqrt{116.25} \\ \\ \mathbf{t = 10.78 \ s}[/tex]

Answer:

[tex]dF=2.5Hz[/tex]

Explanation:

From the question we are told that:

Time [tex]T=0.2sec[/tex]

Generally the Period is given as

 [tex]T= 2 * 0.2 = 0.4[/tex]

Therefore difference in frequency dF

 [tex]dF=\frac{1}{T}[/tex]

 [tex]dF=\frac{1}{0.4}[/tex]

 [tex]dF=2.5Hz[/tex]

Answer:

θ' =  14.44 × [tex]10^{6}[/tex]

Explanation:

given data

total distance is d = 4820

radius = 1.4 cm

solution

we get here total angle by which the wheel rotates traveling is express as

⇒  [tex]\theta=2.40\times10^6\ \rm{rev}=2.40\times 2\pi\times10^6\ \rm{rad}[/tex]     ................1

and

total angle (θ)  and the total distance (d) express as

⇒ d = r × θ       ...............2

here r is radius

and here rotated through some other angle θ' so put value in given equation and find revolutions    

⇒  d = (r+r)θ'      ........3

here  r = d/θ

so

⇒ [tex]d = ( \frac{d}{\theta}+r) \theta'[/tex]    

so put value and get θ'

⇒  θ' = 2.40 × 2π × [tex]10^{6}[/tex] × [tex]\frac{4820 \times 10^3}{4820 \times 10^3 +0.014 \times 2.40 \times 2 \times \pi \times 10^6}[/tex]  

Answer:

T = 1/f = 1/50(s)

ω = 2πf = 100π (rad/s)

(vote 5 sao nhó :3 )

Answer:

P = -1 D

Explanation:

For this exercise we must use the equation of the constructor

       / f = 1 / p + 1 / q

where f is the focal length, p and q is the distance to the object and the image, respectively

The far view point is at p =∞  and its image must be at q = -100 cm = 1 m, the negative sign is because the image is on the same side as the image  

        [tex]\frac{1}{f} = \frac{1}{infinity} + \frac{1}{-1}[/tex]

         f = 1 m

         P = 1/f

          P = -1 D

Answer:

the work done on the ball is 0

Explanation:

Given the data in the question;

Katie swings a ball around her head at constant speed in a horizontal circle with circumference 2.1 m.

circle circumference = 2πr = 2.1 m

radius r will be; r = 2.1 m / 2π = 0.33 m

Tension force = 34.4 N

one half revolution means, displacement of the ball is;

d = 2r = 2 × 0.33 = 0.66 m

Now, Work done = force × displacement × cosθ

we know that, the angle between the tension force on string and displacement of object is always 90.

so we substitute

Work done = 34.4 N × 0.66 m × cos(90)

Work done = 34.4 N × 0.66 m × 0

Work done = 0 J

Therefore,  the work done on the ball is 0

Answer:

30ms

Explanation:

you need to multiple the 10ms by 3s which gives you 30ms

Answer:

The rms voltage of new generator is 3.4 V.

Explanation:

f = 3200 Hz

rms voltage, V = 2 V

rms current, i = 28 mA

Now

f' = 4700 Hz

rms current, i' = 70 mA

let the new rms voltage is V'.

[tex]i = \frac{V}{Xc} = V \times 2\pi fC....(1)\\\\i' = V' \times 2 \pi f' C..... (2)\\\\\frac{i}{i'} =\frac{V f}{V' f'}\\\\\frac{28}{70}=\frac{2\times 3200}{V'\times 4700}\\\\V' = 3.4 V[/tex]

Answer:

the force happens on the wall and couch

Explanation:

she is using her arm strength to lift and hold

Answer:

[tex]u=34cm[/tex]

Explanation:

From the question we are told that:

Far point is [tex]V=34 cm[/tex]

Near point is [tex]u=17 cm[/tex]

Therefore

Focal Length

[tex]f=-34cm[/tex]

Generally the equation for the Lens is mathematically given by

[tex]\frac{1}{u}=\frac{1}{f}-\frac{1}{v}[/tex]

[tex]\frac{1}{u}=\frac{1}{-34}-\frac{1}{-17}[/tex]

[tex]u=34cm[/tex]

Speed:

Acceleration:

The work done by [tex]\vec F[/tex] along the given path C from A to B is given by the line integral,

[tex]\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r[/tex]

I assume the path itself is a line segment, which can be parameterized by

[tex]\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath[/tex]

with 0 ≤ t ≤ 1. Then the work performed by F along C is

[tex]\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}[/tex]

Answer:

1231

Explanation:

Answer:

F' = 128 N

Explanation:

The electrostatic force of attraction between two charges is given by Colomb's Law, as follows:

[tex]F = \frac{kq_1q_2}{r^2}\\\\[/tex]

where,

F = Force of attraction = 1 N

G = universal gravitational constant

q₁ = magnitude of the first charge

q₂ = magnitude of the second charge

r = distance between charges

Therefore,

[tex]1\ N = \frac{kq_1q_2}{r^2}[/tex] --------------------- eq(1)

Now, we apply the changes given in the question:

[tex]F' = \frac{k(2q_1)(4q_2)}{(\frac{1}{4}r)^2}\\\\F' = 128(\frac{kq_1q_2}{r^2})[/tex]

using eq (1):

F' = 128(1 N)

F' = 128 N

Answer:

5.71×10¹⁴ Hz

Explanation:

Applying,

v = λf................. Equation 1

Where v = speed of the electromagnetic radiation, λ = wavelength of the electromagnetic radiation, f = frequency

make f the subject of the equation

f = v/λ............. Equation 2

From the question,

Given: λ = 525 nm = 5.25×10⁻⁷ m,

Constant: Speed of electromagnetic wave (v) = 3.0×10⁸ m/s

Substitute these values into equation 2

f = (3.0×10⁸)/(5.25×10⁻⁷)

f = 5.71×10¹⁴ Hz

Hence the frequency of light is 5.71×10¹⁴ Hz