Erica (37 kg ) and Danny (45 kg ) are bouncing on a trampoline. Just as Erica reaches the high point of her bounce, Danny is moving upward past her at 4.7 m/s . At that instant he grabs hold of her. What is their speed just after he grabs her?

Answer:

D. digital

Explanation:

Digital signals are transmitted through electromagnetic waves.

I'm not sure

Answer:

A charge is a physical entity that has been quantized. The limits on its values are the value of a charged particle quantized in the state where 'n'...

Explanation:

One example of a physical entity that is quantized is:

The amount of money in your pocket.

The amount can't have any fraction of 1 cent.  

Its value must be an integer-multiple of cents, or 0.01 dollar.    

When it increases or decreases, it jumps from one integer number of cents to the next integer number.  It doesn't "slide" from one to the next.  It can never have a value between two integer numbers of cents.

Answer:

[tex]I_2=0.50 w/m^2[/tex]

Explanation:

From the question we are told that:

initial Intensity [tex]I_1=0.020 w/m^2[/tex]

Final Electric field [tex]E_2=5E[/tex]

Generally the equation for Relation ship between intensity and Electric field is mathematically given by

 [tex]\frac{I_1}{I_2}= \frac{E_1^2}{E_2^2}[/tex]

Therefore

 [tex]I_2=\frac{I_1}{ \frac{E_1^2}{E_2^2}}[/tex]

 [tex]I_2=\frac{0.020}{ \frac{E^2}{5E^2}}[/tex]

 [tex]I_2=0.50 w/m^2[/tex]

This question is incomplete, the complete question is;

In June 1985, a laser beam was sent out from the Air Force Optical Station on Maui and reflected back from the shuttle Discovery as it passed by 354 km overhead. The diameter of the central maximum of the beam at the shuttle position was said to be 9.1 m, and the beam wavelength was 500 nm.

What is the effective diameter of the circular laser aperture at the Maui ground station

Answer:

the effective diameter of the circular laser aperture at the Maui ground station is 4.747 cm

Explanation:

Given the data in the question;

Separation between observer and point L = 354 km = 354000 m

Linear separation D = 9.1 m

wavelength λ = 500 nm = 500 × 10⁻⁹ m

Now, for small angles;

θ = D / L

θ = 9.1 m /  354000 m

θ = 2.57 × 10⁻⁵ rad

For a circular aperture;

sinθ = ( 1.22 × λ ) / d

for small angles;

θ = ( 1.22 × λ ) / d

so

θ = 2 × θ

θ = 2 × [( 1.22 × λ ) / d]

we substitute

2.57 × 10⁻⁵ = 2 × [( 1.22 × 500 × 10⁻⁹ ) / d]

2.57 × 10⁻⁵ = 0.00000122 / d

d = 0.00000122 / 2.57 × 10⁻⁵

d = 0.04747 m

d = ( 0.04747 × 100 )m

d = 4.747 cm

Therefore, the effective diameter of the circular laser aperture at the Maui ground station is 4.747 cm

Answer:

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Explanation:

i hope this helps you

Answer:

The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. ... Velocity is a physical vector quantity; both magnitude and direction are needed to define it

Explanation:

hope it helps

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Answer:

kdfihiyfxnlL chapter kar kaam

Explanation:

off influx teenon Palin

Answer:

Explanation:

Reverse the second equation, and change the sign of the enthalpy

Answer:

the period of oscillation of the given object is 0.14 s

Explanation:

Given;

mass of the object, m = 3 kg

extension of the spring, x = 0.085 m

The spring constant is calculated as follows;

[tex]F = mg = \frac{1}{2} ke^2\\\\2mg = ke^2\\\\k = \frac{2mg}{e^2} \\\\k = \frac{2\times 3 \times 9.8}{(0.085)^2} \\\\k = 8,138.41 \ N/m[/tex]

The angular speed of a 4 kg object is calculated as follows;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\frac{2\pi }{T} = \sqrt{\frac{k}{m} } \\\\T= 2\pi \sqrt{\frac{m}{k} } \\\\T = 2\pi \sqrt{\frac{4}{8138.41} }\\\\T = 0.14 \ s[/tex]

Therefore, the period of oscillation of the given object is 0.14 s

Answer:

There are a couple of interesting things about kinetic energy that we can see from the equation.

Kinetic energy depends on the velocity of the object squared. This means that when the velocity of an object doubles, its kinetic energy quadruples. A car traveling at 60 mph has four times the kinetic energy of an identical car traveling at 30 mph, and hence the potential for four times more death and destruction in the event of a crash.

Kinetic energy must always be either zero or a positive value. While velocity can have a positive or negative value, velocity squared is always positive.

Kinetic energy is not a vector. So a tennis ball thrown to the right with a velocity of 5 m/s, has the exact same kinetic energy as a tennis ball thrown down with a velocity of 5 m

what? Is there a picture you can show us with this question

Answer:

a) the required time is 0.6283 μs

b) the inductor current is 0.5 mA

Explanation:

Given the data in the question;

The capacitor voltage has its maximum value of 25 V at t = 0

i.e V[tex]_m[/tex] = V₀ = 25 V

we determine the angular velocity;

ω = 1 / √( LC )

ω = 1 / √( ( 20 × 10⁻³ H ) × ( 8.0 × 10⁻¹² F) )

ω = 1 / √( 1.6 × 10⁻¹³  )

ω = 1 / 0.0000004

ω = 2.5 × 10⁶ s⁻¹

a) How much time does it take until the capacitor is fully discharged for the first time?

V[tex]_m[/tex] =  V₀sin( ωt )

we substitute

25V =  25V × sin( 2.5 × 10⁶ s⁻¹ × t )

25V =  25V × sin( 2.5 × 10⁶ s⁻¹ × t )

divide both sides by 25 V

sin( 2.5 × 10⁶ × t ) = 1

( 2.5 × 10⁶ × t ) = π/2

t = 1.570796 / (2.5 × 10⁶)

t = 0.6283 × 10⁻⁶ s

t = 0.6283 μs

Therefore, the required time is 0.6283 μs

b) What is the inductor current at that time?

[tex]I[/tex](t) = V₀√(C/L) sin(ωt)

{ sin(ωt) = 1 )

[tex]I[/tex](t) = V₀√(C/L)

we substitute

[tex]I[/tex](t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )

[tex]I[/tex](t) = 25 × 0.00002

[tex]I[/tex](t) = 0.0005 A

[tex]I[/tex](t) = 0.5 mA

Therefore, the inductor current is 0.5 mA

The time taken for the capacitor to fully discharge is 6.28 x 10⁻⁷ s.

The current in the inductor at the given time is 0.0005 A.

The angular velocity of the circuit is calculated as follows;

[tex]\omega = \frac{1}{\sqrt{LC} } \\\\\omega = \frac{1}{\sqrt{20 \times 10^{-3} \times 8 \times 10^{-12} } } \\\\\omega = 2.5 \times 10^6 \ rad/s[/tex]

V = V₀(sinωt)

sinωt = V/V₀

sinωt = = 25/25

sin(ωt) = 1

ωt = sin⁻¹ (1)

ωt = π/2

2.5 x 10⁶ t = π/2

t = 1.57 / (2.5 × 10⁶)

t = 6.28 x 10⁻⁷ s

The current in the inductor at the given time is calculated as follows

[tex]I(t) = V_0 \sqrt{\frac{C}{L} } \\\\I = 25 \times \sqrt{\frac{8\times 10^{-12}}{20 \times 10^{-3}} } \\\\I = 0.0005 \ A[/tex]

Learn more about inductor current here: https://brainly.com/question/4425414

Answer:

Because molecules are to small for us people to see with the bare eye, unless you use a telescope.

Explanation:

Explanation:

Because in solid form, the molecules stood in place, not moving. In liquid state, the molecules move slowly but not as fast as air molecules.

Answer:

F₂ = -7.3 N

Explanation:

Given that,

The mass of an object, m₁ = 3.7 kg

First force, F₁ = 11 N

The net acceleration of the object is 1 m/s².

We know that,

F₁+F₂ = ma

11+F₂ = (3.7)(1)

F₂ = 3.7-11

F₂ = -7.3 N

so, the other force is 7.3 N and it is acting in west direction.

Answer:

[tex]t=2413s[/tex]

Explanation:

From the question we are told that:

Velocity [tex]v=1.5m/s[/tex]

Time [tex]t=30min=>30*60=>1800[/tex]

Distance [tex]d=24.3km[/tex]

Generally the Newton's equation for Speed going down the stream is mathematically given by

 [tex]v + u = \frac{d}{t}[/tex]

 [tex]1.5+v=frac{24300}{1800}[/tex]

 [tex]v=12m/s[/tex]

Therefore

 [tex]v + u = \frac{d}{t}[/tex]

 [tex]t=\frac{24300}{12-1.5}[/tex]

 [tex]t=2413s[/tex]

Answer:

The second one says cant open file............

By the work-energy theorem, the total work done on the car is equal to the change in its kinetic energy:

W = ∆K

W = 1/2 (0.34 kg) (22.9 m/s)² - 1/2 (0.34 kg) (6.5 m/s)²

W ≈ 82 J

Answer:

electromagnetism....

Answer:

The resistance of A is 6 ohms and the resistance of B is 3 ohms

Explanation:

Step 1: For the first connection (parallel connection), the resistance of B will be calculated.

Note: in a parallel connection, the voltage through each resistor is the same.

[tex]V = I_AR_A = I_BR_B\\\\R_B = \frac{V}{I_B} = \frac{6}{2} = 3 \ ohms[/tex]

Step 2: The resistance of A will be calculated from the second connection (series connection)

Note: in series connection, the current flowing in each resistor is the same

[tex]V = V_A + V_B\\\\V = IR_A + IR_B\\\\The \ voltage \ drop \ in \ B; \ V_B = V- V_A\\\\V_B = 6 - 4 = 2 \ V\\\\IR_B = 2\ V\\\\I = \frac{2 \ V}{R_B}= \frac{2}{3} \ A\\\\The \ resistance \ of \ A \ is \ calculated \ as ;\\\\IR_A = 4 \ V\\\\R_A = \frac{4}{I} = \frac{4 \times 3}{2} = 6 \ ohms[/tex]

Answer:

The maximum magnetic force is 2.637 x 10⁻¹² N

Explanation:

Given;

Power, P = 8.25 m W = 8.25 x 10⁻³ W

charge of the radiation, Q = 1.12 nC = 1.12 x 10⁻⁹ C

speed of the charge, v = 314 m/s

area of the conecntration, A = 1.23 mm² = 1.23 x 10⁻⁶ m²

The intensity of the radiation is calculated as;

[tex]I = \frac{P}{A} \\\\I = \frac{8.25 \times 10^{-3} \ W}{1.23 \ \times 10^{-6} \ m^2} \\\\I = 6,707.32 \ W/m^2[/tex]

The maximum magnetic field is calculated using the following intensity formula;

[tex]I = \frac{cB_0^2}{2\mu_0} \\\\B_0 = \sqrt{\frac{2\mu_0 I}{c} } \\\\where;\\\\c \ is \ speed \ of \ light\\\\\mu_0 \ is \ permeability \ of \ free \ space\\\\B_0 \ is \ the \ maximum \ magnetic \ field\\\\B_0 = \sqrt{\frac{2 \times 4\pi \times 10^{-7} \times 6,707.32 }{3\times 10^8} } \\\\B_0 = 7.497 \times 10^{-6} \ T[/tex]

The maximum magnetic force is calculated as;

F₀ = qvB₀

F₀ = (1.12 x 10⁻⁹) x (314) x (7.497 x 10⁻⁶)

F₀ = 2.637 x 10⁻¹² N

Answer:

460.52 s

Explanation:

Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that

dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = [tex]e^{kt}[/tex]

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

k = -0.01

So, V/V₀ = [tex]e^{-0.01t}[/tex]

V = V₀[tex]e^{-0.01t}[/tex]

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

Thus

V = V₀[tex]e^{-0.01t}[/tex]

0.1V₀ = V₀[tex]e^{-0.01t}[/tex]

0.1V₀/V₀ = [tex]e^{-0.01t}[/tex]

0.1 = [tex]e^{-0.01t}[/tex]

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

㏑(0.01) = -0.01t

So, t = ㏑(0.01)/-0.01

t = -4.6052/-0.01

t = 460.52 s

Answer:

The advantages of Hovercraft:

They can travel over almost any non-porous surface.

They can operate to and from any unprepared beach or slipway.

They take fast, direct routes compared to a conventional marine vessel.

Answer:

[tex]d =3.7*10^{-3} m[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=1.2kg[/tex]

Spring constant [tex]\mu=22Nm^{-1}[/tex]

Amplitude [tex]A=5cm=0.05m[/tex]

Generally the equation for displacement d is mathematically given by

 [tex]d = Asin(\omega t)[/tex]

Where

 [tex]\omega=angular\ velocity[/tex]

 [tex]\omega=\sqrt{k/m}[/tex]

 [tex]\omega=\sqrt{22/1.2}[/tex]

 [tex]\omega=4.2817rads^{-1}[/tex]

Therefore  

 [tex]d = 0.05*sin(4.2817*1)[/tex]  

 [tex]d =3.7*10^{-3} m[/tex]

Answer:

the is metallurgy .....

Answer:

The maximum speed is 18.86 m/s.

Explanation:

initial  radius, r = 150 m

maximum speed, v = 26.5 m/s

new radius, r' = 76 m

Let the new maximum speed is v'.

The formula of the maximum speed is

[tex]tan\theta = \frac{v^2}{rg}[/tex]

So,

[tex]\frac{v'^2}{v^2}=\frac{r'}{r}\\\\\frac{v'^2}{26.5\times 26.5}=\frac{76}{150}\\\\v=18.86 m/s[/tex]

Answer:

We want to solve the sum:

6.74*10⁴ + 8.95*10⁴

first, we take the common factor 10⁴ out, so we get:

(6.74 + 8.95)*10⁴

Now we solve the sum:

(15.66)*10⁴

Now we want to rewrite it in exponential form, wo we can rewrite it as:

(15.66)*10⁴ = (1.566*10)*10⁴ = (1.566)*10*10⁴ = (1.566)*10⁴⁺¹ = 1.566*10⁵

k = 5.

Answer:

Tx not but mybe

Explanation:

for that reason its just trying to help

According to Coulomb's law, rank the interactions between charged particles from highest potential energy to lowest potential energy.

Highest potential energy to lowest potential energy.

b. 1+ charge and 1+ charge seperated by 100 pm

a. 1+ charge and 1- charge seperated by 200 pm

c. 1+ charge and 1- charge seperated by 100 pm

d. 2+ charge and 1- charge seperated by 100 pm

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