A small object A, electrically charged, creates an electric field. At a point P located 0.250 m directly north of A, the field has a value of 40.0 N/C directed to the south. If a second object B with the same charge as A is placed at 0.250m south of A (so that objects A and B and point P follow a straight line), what is the magnitude of the total electric field produced by the two objects at P?

Answer:

59.26°

Explanation:

Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.

Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.

Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration =  acosθ.

So, y = ut + 1/2a't²

y = 0 × t + 1/2(acosθ)t²

y = 0 + 1/2(acosθ)t²

y = 1/2(acosθ)t²   (1)

Also, both particles must move the same horizontal distance to collide in time, t.

Let x be the horizontal distance,

x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision

Also,  using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration =  asinθ.

So, x = ut + 1/2a"t²

x = 0 × t + 1/2(ainsθ)t²

x = 0 + 1/2(asinθ)t²

x = 1/2(asinθ)t²  (3)

Equating (2) and (3), we have

vt = 1/2(asinθ)t²   (4)

From (1) t = √[2y/(acosθ)]

Substituting t into (4), we have

v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²  

v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)  

v√[2y/(acosθ)] = ytanθ

√[2y/(acosθ)] = ytanθ/v

squaring both sides, we have

(√[2y/(acosθ)])² = (ytanθ/v)²

2y/acosθ = (ytanθ/v)²

2y/acosθ = y²tan²θ/v²

2/acosθ = ytan²θ/v²

1/cosθ = aytan²θ/2v²

Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1

secθ = ay(sec²θ - 1)/2v²

2v²secθ = aysec²θ - ay

aysec²θ - 2v²secθ - ay = 0

Let secθ = p

ayp² - 2v²p - ay = 0

Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have

ayp² - 2v²p - ay = 0

0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0

10.85p² - 15.68p - 10.85 = 0

dividing through by 10.85, we have

p² - 1.445p - 1 = 0

Using the quadratic formula to find p,

[tex]p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125[/tex]

Since p = secθ

secθ = 1.95625 or secθ = -0.51125

cosθ = 1/1.95625 or cosθ = 1/-0.51125

cosθ = 0.5112 or cosθ = -1.9956

Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.

So, cosθ = 0.5112

θ = cos⁻¹(0.5112)

θ = 59.26°

So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.

Answer:

B = 0.118 T

Explanation:

From Biot-Savart Law:

[tex]B = \frac{\mu_o I}{2\pi r}[/tex]

where,

B = strength of magnetic field = ?

μ₀ = 4π x 10⁻⁷ Tm/A

I = current enclosed = 474.1 A

r = radius = 0.8 mm = 8 x 10⁻⁴ m

Therefore,

[tex]B = \frac{(4\pi\ x\ 10^{-7}\ Tm/A)(474.1\ A)}{2\pi(8\ x\ 10^{-4}\ m)}[/tex]

B = 0.118 T

Answer:

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Explanation:

The 2 capacitors in the middle are connected in parallel so simply add their capacitance together:

[tex]5.0\:\mu\text{F} + 8.0\:\mu\text{F} = 13.0\:\mu \text{F}[/tex]

Now we have 3 capacitors connected in series so their equivalent capacitance [tex]C_{eq}[/tex] is

[tex]\dfrac{1}{C_{eq}} = \dfrac{1}{10.0\:\mu \text{F}} + \dfrac{1}{13.0\:\mu \text{F}} + \dfrac{1}{9.0\:\ mu \text{F}} [/tex]

or

[tex]C_{eq} = 3.5\:\mu \text{F}[/tex]

Answer:

810.94 m/s

Explanation:

Applying,

v = √(2gR)............. Equation 1

Where v = escape velocity of the spherical asteroid, g = acceleration due to gravity, R = radius of the earth

From the question,

Given: g = 0.636 m/s², R = 517 km = 517000 m

Substitute these values into equation 1

v = √(2×0.636×517000)

v = √(657624)

v = 810.94 m/s

Hence, the escape velocity is 810.94 m/s

Answer:

-700

formula is heat gained - work done

The change in internal energy if A system gains 1500J of heat and 2200J of work is done by the system on its surroundings, is 700 joules.

Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc.

Additionally, there is heat and work, which is energy being transferred from one body to another. Energy is always assigned based on its nature once it has been transmitted. Thus, heat transmitted may manifest as thermal energy while work performed may result in mechanical energy.

Given:

A system gains 1500J of heat and 2200J of work is done by the system on its surroundings,

Calculate the change in internal energy as shown below,

The change in internal energy = heat gained - work done

The change in internal energy = 1500 - 2200

The change in internal energy = -700 J

Thus, the change in internal energy is 700 joules.

To know more about Energy:

https://brainly.com/question/8630757

#SPJ5

Answer:

sup qwertyasdfghjk

Explanation:

Answer:

Power = Energy/time

Energy = Power xtime.

Time= 20hrs

Power = 100Watt =0.1Kw

Energy = 0.1 x 20 = 2Kwhr.

This Answer is in Kilowatt-hour ...

If the one given to you is in Joules

You'd have to Change your time to seconds

Then Multiply it by the power of 100Watts.

Answer:

SUPER TYPHOON (STY), a tropical cyclone with maximum wind speed exceeding 220 kph or more than 120 knots.

Answer:

[tex]T=326.928K[/tex]

Explanation:

From the question we are told that:

Emissivity [tex]e=0.44[/tex]

Absorptivity [tex]\alpha =0.3[/tex]

Rate of solar Radiation [tex]R=0.3[/tex]

Generally the equation for Surface absorbed energy is mathematically given by

 [tex]E=\alpha R[/tex]

 [tex]E=0.3*950[/tex]

 [tex]E=285W/m^2[/tex]

Generally the equation for Emitted Radiation is mathematically given by

 [tex]\mu=e(\sigmaT^4)[/tex]

Where

T=Temperature

 [tex]\sigma=5.67*10^8Wm^{-2}K_{-4}[/tex]

Therefore

 [tex]\alpha*E=e \sigma T^4[/tex]

 [tex]0.3*(950)=0.44(5.67*10^-8)T^4[/tex]

 [tex]T=326.928K[/tex]

Answer:

7.84 m/s

Explanation:

Height, h = 2 m

Initial velocity, u = 10 m/s

Angle, A = 80°

(a) Let the time taken to go to the net is t.

Use second equation of motion

[tex]h = u t + 0.5 at^2\\\\- 2 = - 10 sin 80 t - 4.9 t^2\\\\4.9 t^2 + 9.8 t - 2 = 0 \\\\t= \frac{- 9.8\pm\sqrt{9.8^2 + 4\times 4.9\times 2}}{9.8}\\\\t = \frac{- 9.8 \pm 11.6}{9.8}\\\\t = - 2.2 s , 0.2 s[/tex]

Time cannot be negative.

So, t = 0.2 s

The vertical velocity at t = 0.2 s is

v = u + at

v = 10 sin 80 - 9.8 x0.2

v = 9.8 - 1.96 = 7.84 m/s

Answer:

The correct option is (E).

Explanation:

Given that,

Mass of object 1, m₁ = 1 kg

Mass of object 2, m₂ = 2 kg

They collides after the collision. We need to find the speed of the two boxes after the collision.

The initial speeds of both boxes is not given. So, we can't put the values of their speeds in the momentum conservation equation.

So, the information is not enough.

Answer:

It is a example of physical change

Answer:

9.25 m/s

Explanation:

Answer:

1.61

Explanation:

According to Oxford dictionary, refractive index is, ''the ratio of the velocity of light in a vacuum to its velocity in a specified medium.''

If the clear transparent solid disappears when dipped into the liquid, it means that the index of refraction of the solid and liquid are equal.

Hence, when a transparent solid is immersed in a liquid having the same refractive index, there is no refraction at the boundary between the two media. As long as there is no refraction between the two media, the solid can not be seen because the solid and liquid will appear to the eye as one material.

Answer:

The maximum height risen by the bullet-baseball system after the collision is 81.76 m.

Explanation:

Given;

mass of the bullet, m₁ = 0.033 kg

mass of the baseball, m₂ = 0.15 kg

initial velocity of the bullet, u₁ = 222 m/s

initial velocity of the baseball, u₂ = 0

let the common final velocity of the system after collision = v

Apply the principle of conservation of linear momentum to determine the common final velocity.

m₁u₁  +  m₂u₂  = v(m₁  + m₂)

0.033 x 222   +  0.15 x 0     = v(0.033 + 0.15)

7.326  =  v(0.183)

v = 7.326 / 0.183

v = 40.03 m/s

Let the height risen by the system after collision = h

Initial velocity of the system after collision = Vi = 40.03 m/s

At maximum height, the final velocity, Vf = 0

acceleration due to gravity for upward motion, g = -9.8 m/s²

[tex]v_f^2 = v_i^2 +2gh\\\\0 = 40.03^2 - (2\times 9.8)h\\\\19.6h = 1602.4\\\\h = \frac{1602.4}{19.6} \\\\h = 81.76 \ m[/tex]

Therefore, the maximum height risen by the bullet-baseball system after the collision is 81.76 m.

Answer:

a) [tex] \alpha = 1.28 rad/s^{2} [/tex]  

b) Option ii. The angular acceleration would increase

Explanation:

a) The angular acceleration is given by:

[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]

Where:

[tex] \omega_{f} [/tex]: is the final angular speed = v/r

v: is the tangential speed = 6.35 m/s

r: is the radius = 45.0 cm = 0.45 m

[tex]\omega_{0}[/tex]: is the initial angular speed = 0 (the hoop starts from rest)

t: is the time = 11.0 s

α: is the angular acceleration

Hence, the angular acceleration is:

[tex] \alpha = \frac{\omega}{t} = \frac{v}{r*t} = \frac{6.35 m/s}{0.45 m*11.0 s} = 1.28 rad/s^{2} [/tex]  

b) If the radius were smaller, the angular acceleration would increase since we can see in the equation that the radius is in the denominator ([tex] \alpha = \frac{v}{r*t} [/tex]).

Therefore, the correct option is ii. The angular acceleration would increase.

I hope it helps you!  

Answer:

Explanation:

Moment of inertia of solid sphere = 2/5 m R²

m is mass and R is radius of sphere.

Putting the values

Moment of inertia of solid sphere I₁

Moment of inertia of hollow  sphere I₂

Kinetic energy of solid sphere ( both linear and rotational )

= 1/2 ( m v² + I₁ ω²)                [ ω is angular velocity of rotation ]

= 1/2 ( m v² + 2/5 m R² ω²)

= 1/2 ( m v² + 2/5 m v²)

=1/2 x 7 / 5 m v²

= 0.7 x 5 x 4² = 56 J .

This will be equal to work to be done to stop it.

Kinetic energy of hollow sphere ( both linear and rotational )

= 1/2 ( m v² + I₂ ω²)  [ ω is angular velocity of rotation ]

= 1/2 ( m v² + 2/3 m R² ω²)

= 1/2 ( m v² + 2/3 m v²)

=1/2 x 5 / 3 m v²

= 0.833 x 5 x 4² = 66.64 J .

This will be equal to work to be done to stop it.

Answer:

0.000064 cubic meters.

Explanation:

Given the following data;

Length of side = 4 centimeters

Conversion:

100 centimeters = 1 meters

4 cm = 4/100 = 0.04 meters

To find the volume of cuboid;

Mathematically, the volume of a cuboid is given by the formula;

Volume of cuboid = length * width * height

However, when all the sides are equal the formula is;

Volume of cuboid = L³

Volume of cuboid = 0.04³

Volume of cuboid = 0.000064 cubic meters.

I see six (6) loops.

I attached a drawing to show where I get six loops from.

Answer:

F = m g sin theta      force accelerating block

m a = m g sin theta

a = 9.8 sin 24 = 3.99 m/sec^2

The question is incomplete. The complete question is :

A high-speed bullet train accelerates and decelerates at the rate of 4 ft/s^2. Its maximum cruising speed is 90 mi/h. What is the maximum distance the train can travel if it accelerates from rest until it reaches its cruising speed and then runs at that speed for 15 minutes?

Solution :

Given :

Speed of the bullet train, v = 90 mi/h

                                            = [tex]$90 \times \frac{5280}{3600}$[/tex]

                                            = 132 ft/s

Time = 15 minutes

        = 15 x 60

        = 900 s

Acceleration from rest,

[tex]$a(t) = 4 \ ft/s^2$[/tex]

[tex]$v(t) = 4t + C$[/tex]

Since, v(0) = 0, then C = 0, so velocity is

v(t) = 4t ft/s

Then find the position function,

[tex]$s(t) = \frac{4}{2}t^2 + C$[/tex]

      [tex]$=2t^2+C$[/tex]

It is at position 0 when t = 0, so C = 0, and the final position function for only the time it is accelerating is :

[tex]$s(t) = 2t^2$[/tex]

Time to get maximum cruising speed is :

4t = 132

t = 33 s

Distance travelled (at cruising speed) by speed to get the remaining distance travelled.

[tex]$900 \ s \times 132 \ \frac{ft}{s} = 118800 \ ft$[/tex]

Total distance travelled, converting back to miles,

[tex]$2178 + 118800 = 120978\ ft . \ \frac{mi}{5280 \ ft}$[/tex]

                      = 22.9125 mi

Therefore, the distance travelled is 22.9125 miles

Answer:

Look at work

Explanation:

Series:

I is the same for all resistors so just find the value of Req. In series Req= R1+R2+...+Rn. So here it will be 4.5+2.3=6.8ohms. Ieq=Veq/Req=4.41A. And since current is the same across all resistors the current to the lightbulb is 4.41A.

Parallel:

V is the same for all resistors so start of by finding Req. In parallel, Ieq=I1+I2+...+In. So I1= 30/4.5= 6.67A and I2= 13.04A. Ieq= 6.67+13.04= 19.71A.

The question is incomplete. The complete question is :

A mass is attached to the end of a spring and set into oscillation on a horizontal frictionless surface by releasing it from a compressed position. The record of time is started when the oscillating mass first passes through the equilibrium position, and the position of the mass at any time is described by x = (4.7 cm)sin[(7.9 rad/s)πt].

Determine the following:

(a) frequency of the motion

(b) period of the motion

(c) amplitude of the motion

(d) first time after t = 0  that the object reaches the position x = 2.6 cm

Solution :

Given equation : x = (4.7 cm)sin[(7.9 rad/s)πt].

Comparing it with the general equation of simple harmonic motion,

 x = A sin (ωt + Φ)

  A = 4.7 cm

  ω = 7.9 π

a). Therefore, frequency, [tex]$f=\frac{\omega}{2 \pi}$[/tex]

                                             [tex]$=\frac{7.9 \pi}{2 \pi}$[/tex]

                                             = 3.95 Hz

b). The period, [tex]$T=\frac{1}{f}$[/tex]

                        [tex]$T=\frac{1}{3.95}[/tex]

                            = 0.253 seconds

c). Amplitude is A = 4.7 cm

d). We have,

    x = A sin (ωt + Φ)

    [tex]$x_t=4.7 \sin (7.9 \pi t)$[/tex]

    [tex]$2.6 = 4.7 \sin (7.9 \pi t)$[/tex]

     [tex]$\sin (7.9 \pi t) = \frac{26}{47}$[/tex]

     [tex]$7.9 \pi t = \sin^{-1}\left(\frac{26}{47}\right)$[/tex]

          Hence, t = 0.0236 seconds.

Answer:

(b)

Explanation:

The voltage always lags the current by 90°, regardless of the frequency.

Answer:

because the mass of the apple is very less compared to the mass of earth. Due to less mass the apple cannot produce noticable acceleration in the earth but the earth which has more mass produces noticable acceleration in the apple. thus we can see apple falling on towards the earth but we cannot see earth moving towards the apple.

Answer:

   a = w² r

Explanation:

In this exercise, indicate that the wheel has angular velocity w, the worm experiences the same angular velocity if it does not move, and has an acceleration towards the center of the circle, according to Newton's second law, called the centripetal acceleration.

           a = v² / r

angular and linear variables are related

           v = w r

we substitute

          a = w² r

where r is the radius of the wheel

Answer:

W = Q * V     work done on charge Q

A. W = .5 C * 1.5 V = .75 Joules

B. P = W / t = .75 J / 1 sec = .75 Watts

Answer:

The correct answer is "1.2 J".

Explanation:

Seems that the given question is incomplete. Find the attachment of the complete query.

According to the question,

Now,

⇒ [tex]W=\int_{x_1}^{x_2}F \ dx[/tex]

⇒      [tex]=\int_{x_1}^{x_2}-kx \ dx[/tex]

⇒      [tex]=-k \int_{-0.20}^{0}x \ dx[/tex]

By putting the values, we get

⇒      [tex]=-(60)[\frac{x^2}{2} ]^0_{-0.20}[/tex]

⇒      [tex]=-60[\frac{0}{2}-\frac{0.04}{2} ][/tex]

⇒      [tex]=1.2 \ J[/tex]

Answer:

The hot temperature is 157.5 K

The cold temperature is 48.8 K

Explanation:

Step 1: Data given

The working substance of a certain Carnot engine is 1.50 mol of an ideal monatomic gas.

The volume increases by a factor of 5.7

The work output of the engine is 940 J in each cycle.

During the isothermal expansion portion of this engine's cycle, the volume of the gas doubles. This means V2 = 2*V1 (and V4 = 2*V3)

Step 2:For a carnot engine:

V2/V1 = V4/V3

Work = nR((T1)ln(V2/V1) - (T2)ln(V4/V3))

⇒with Work = the work done in the cycle = 940J

⇒with n = the number of moles = 1.50 moles

⇒with R = the gas constant = 8.314 J/mol*K

⇒with T1 = the hot temperature

⇒With T2⇒ the cold temperature

where R = 8.31 J/mol K Gas Constant

940J = 1.5moles * 8.314 J/mol*K * (T1*ln(2) - T2*ln(2)))

940 = 1.5 * 8.314 ln(2) * (T1-T2)

(T1-T2) = 940 / (1.5*8.314*ln(2))

(T1-T2) = 108.7K

For the reversible adiabatic expansion: T2 = T1*(V1/V2)^(R/Cv). Where V2/V1 = 5.7 (Because during the adiabatic expansion the volume increases by a factor of 5.7)

For a monatomic ideal gas, Cv = 3/2R

When we combine both, we'll have:

T2 = T1*(1/5.7)^(R/3/2R)

T2 = T1*(1/5.7)^(2/3)

T2= T1 * 0.31

Since we know that (T1-T2) = 108.7K

we have:

T1 - 0.31T1= 108.7K

0.69T1 = 108.7K

T1 = 157.5K

T2 = 157.5*0.31 = 48.8K

Answer:

E = m c^2 = 2.5 * (3 * 10E8)^2 = 2.25 * 10E17 Joules

Answer:

The answer is D. 2.25 × 1017 J

Explanation:

got it right on edge 2021