Every object around you is attracted to you. In fact, every object in the galaxy is attracted to every other object in the galaxy.

a. True
b. False

Answer:

In 1 second the amount of charge flowing through the conductor is 2.5 Q.

Explanation:

Answer:

The phases of the cardiac action potential correspond to the surface ECG (ECG) (Figure). The P wave reflects atrial depolarization (phase 0), the PR interval reflects the conduction velocity through the AV node, the QRS complex the ventricular depolarization and QT interval the duration potential ventricular action.

Answer:

The other angle is 75⁰

Explanation:

Given;

velocity of the projectile, v = 10 m/s

range of the projectile, R = 5.1 m

angle of projection, 15⁰

The range of a projectile is given as;

[tex]R = \frac{u^2sin(2\theta)}{g}[/tex]

To find another angle of projection to give the same range;

[tex]5.1 = \frac{10^2 sin(2\theta)}{9.81} \\\\100sin(2\theta) = 50\\\\sin(2\theta) = 0.5\\\\2\theta = sin^{-1}(0.5)\\\\2\theta = 30^0\\\\\theta = 15^0\\\\since \ the \ angle \ occurs \ in \ \ the \ first \ quadrant,\ the \ equivalent \ angle \\ is \ calculated \ as;\\\\90- \theta = 15^0\\\\\theta = 90 - 15^0\\\\\theta = 75^0[/tex]

Check:

sin(2θ) = sin(2 x 75) = sin(150) = 0.5

sin(2θ) = sin(2 x 15) = sin(30) = 0.5

Answer:

Explanation:

From the information given:

The motional emf can be computed by using the formula:

[tex]E = L^{\to}*(V^\to*\beta^{\to})[/tex]

[tex]E = L^{\to}*((x+y+z)*\beta^{\to})[/tex]

[tex]E = 0.50*((18\hat i+24 \hat j +72 \hat k )*0.0800)[/tex]

[tex]E = 0.50*((18*0.800)\hat k +0j+(72*0.080) \hat -i ))[/tex]

[tex]E = 0.50*((18*0.800)[/tex]

E = 0.72 volts

According to the question, suppose the wire segment was parallel, there will no be any emf induced since the magnetic field is present along the y-axis.

As such, for any motional emf should be induced, the magnetic field, length, and velocity are required to be perpendicular to one another .

Then the motional emf will be:

[tex]E = 0.50 \hat j *((18*0.800)\hat k -(72*0.080) \hat i ))[/tex]

E = 0 (zero)

Answer:

[tex]$\frac{d}{\lambda} = 1.54$[/tex]

Explanation:

Given :

The first dark fringe is for m = 0

[tex]$\theta_1 = \pm 19^\circ$[/tex]

Now we know for a double slit experiments , the position of the dark fringes is give by :

[tex]$d \sin \theta=\left(m+\frac{1}{2}\right) \lambda$[/tex]

The ratio of distance between the two slits, d to the light's wavelength that illuminates the slits, λ :

[tex]$d \sin \theta=\left(\frac{1}{2}\right) \lambda$[/tex]     (since, m = 0)

[tex]$d \sin \theta=\frac{\lambda}{2}$[/tex]

[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin \theta}$[/tex]

[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin 19^\circ}$[/tex]

[tex]$\frac{d}{\lambda} = 1.54$[/tex]

Therefore, the ratio is [tex]$\frac{1}{1.54}$[/tex]  or 1 : 1.54

Answer:

22.1 years

Explanation:

Since the current in the wire is I = nevA where n = electron density = 8.50 × 10²⁸ electrons/cm³ × 10⁶ cm³/m³= 8.50 × 10³⁴ electrons/m³, e = electron charge = 1.602 × 10⁻¹⁹ C, v = drift velocity of electrons and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.00 cm = 2 × 10⁻² m

Making v subject of the formula, we have

v = I/neA

So, v = I/neπd²/4

v = 4I/neπd²

Since I = 1,015 A, substituting the values of the other variables into the equation, we have

v = 4I/neπd²

v = 4(1,015 A)/[8.50 × 10³⁴ electrons/m³ × 1.602 × 10⁻¹⁹ C × π ×(2 × 10⁻² m)²]

v = 4(1,015 A)/[8.50 × 10³⁴ electrons/m³ × 1.602 × 10⁻¹⁹ C × π × 4 × 10⁻⁴ m²]

v = (1,015 A)/[42.779 × 10¹¹ electronsC/m]

v = 23.73 × 10⁻¹¹ m/s

v = 2.373 × 10⁻¹⁰ m/s

Since distance d = speed, v × time, t

d = vt

So, the time it takes one electron to travel the full length of the cable is t = d/v

Since d = distance moved by free charge = length of transmission line = 165 km = 165 × 10³ m and v = drift velocity of charge = 2.373 × 10⁻¹⁰ m/s

t = 165 × 10³ m/2.373 × 10⁻¹⁰ m/s

t = 69.54 × 10⁷ s

t = 6.954 × 10⁸ s

Since we have 365 × 24 hr/day × 60 min/hr × 60 s/min = 31536000 s in a year = 3.1536 × 10⁷ s

So,  6.954 × 10⁸ s =  6.954 × 10⁸ s × 1yr/3.1536 × 10⁷ s = 2.21 × 10 yrs = 22.1 years

It will take one electron 22.1 years to travel the full length of the cable

Answer:

Explanation:

Answer:

a) A = 1.50 m / s²,  B = 1.33 m/s³,  b) a = 12.1667 m / s²,

c)  I = M (1.5 t + 1.333 t²) ,  d)  ΔI = M 2.833   N

Explanation:

In this exercise give the expression for the speed of the rocket

         v (t) = A t + B t²

and the initial conditions

         a = 1.50 m / s² for t = 0 s

         v = 2.00 m / s for t = 1.00 s

a) it is asked to determine the constants.

Let's look for acceleration with its definition

         a = [tex]\frac{dv}{dt}[/tex]

         a = A + 2B t

we apply the first condition t = 0 s

         a = A

         A = 1.50 m / s²

we apply the second condition t = 1.00 s

          v = 1.5 1 + B 1²

          2 = 1.5 + B

          B = 2 / 1.5

          B = 1.33 m/s³

the equation remains

           v = 1.50 t + 1.333 t²

b) the acceleration for t = 4.00 s

           a = 1.50 + 1.333 2t

           a = 1.50 + 2.666 4

           a = 12.1667 m / s²

c) The thrust

           I = ∫ F dt = p_f - p₀

Newton's second law

          F = M a

          F = M (1.5 + 2 1.333 t) dt

we replace and integrate

         I = M ∫ (1.5 + 2.666 t) dt

         I = 1.5 t + 2.666 t²/2

         I = M (1.5 t + 1.333 t²) + cte

in general the initial rockets with velocity v = 0 for t = 0, where we can calculate the constant

         cte = 0

         I = M (1.5 t + 1.333 t²)

d) the initial push

For this we must assume some small time interval, for example between

t = 0 s and t = 1 s

        ΔI = I_f - I₀

        ΔI = M (1.5 1 + 1.333 1²)

        ΔI = M 2.833   N

Answer:

Explanation:

From the given information:

From the rotational axis, the distance of the force of gravity is:

d_g = 25+5.0 cm

d_g = 30.0 cm

d_g = 30.0 × 10⁻² m

However, the relative distance of FB  cos 75.9° from the axis is computed as:

d_B = 5.0 cm

d_B = 5.0 × 10⁻² m

The net torque rotational equilibrium = zero (0)

i.e.

[tex]\tau_g -\tau_B = 0 \\ \\ F_gd_g -F_gcos 75.9^0 d_B = 0 \\ \\ F_B = \dfrac{F_g d_g}{F_g cos 65.6} \\ \\ F_B = \dfrac{(3.2)(9.8)(30*10^{-2})}{(5.0*10^{-2} * cos 75.9)} \\ \\ \mathbf{F_B = 772.4 N}[/tex]

= 772.4 N

Thus, the force exerted = 1772.4 N

I see six (6) loops.

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

Answer:

1.67 N

Explanation:

Applying,

F = u(dm/dt)+m(du/dt)................ Equation 1

Where F = force, m = mass of the vehicle, u = speed.

Since u is constant,

Therefore, du/dt = 0

F = u(dm/dt)............... Equation 2

From the question,

Given: u = 10 m/s, dm/dt = 10 kg/min = (10/60) kg/s

Substitute these values into equation 2

F = 10(10/60)

F = 100/60

F = 1.67 N

Explanation:

Here,we've been given that,

We've to find the acceleration of the object. By using the third equation of motion,

→ v² - u² = 2as

→ (420)² - (0)² = 2 × a × 145

→ 176400 - 0 = 290a

→ 176400 = 290a

→ 176400 ÷ 290 = a

→ 608.275862 m/s² = a

If you know initial speed and final speed, you can find the average speed.  Then, knowing distance, you can find the time.

KimYurii posted the first answer to this question.  

That answer is well organized, well presented, elegant and correct, and it deserves to be awarded "Brainliest" and several merit badges.

My problem is that I can never remember all the different formulas.  I guess I had to work with so many uvum in all the Physics, Geometry, and Calculus classes that I took, I filled up all the memory slots with formulas, and over the years they all eventually merged into a big glob of goo.  Now, the only formulas I can remember are the ones I had to use as an Electrical Engineer.

When I see this kind of question, I can only remember one or two simple formulas, and I reason it out like this:

Starting speed . . . zero

Ending speed . . . 420 m/s

Formula:  Average speed . . . (1/2)·(0 + 420) = 210 m/s

Distance covered . . . 145 m

Formula: Time taken = (distance) / (average speed) = (145/210) second

(Now you have the time.)

Formula: Distance = (1/2)·(acceleration)·(time²)

145 m = (1/2)·(acceleration)·(145/210 sec)²

Acceleration = 290 m / (145/210 s)²

Acceleration = 608.28 m/s²

Explanation:

[tex]\underline{\boxed{\large{\bf{Option \; A!! }}}} [/tex]

Here,

We have to find the equivalent resistance of the circuit.

Here, [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] are connected in series, so their combined resistance will be given by,

[tex]\longrightarrow \rm { R_{(1,2)} = R_1 + R_2} \\ [/tex]

[tex]\longrightarrow \rm { R_{(1,2)} = (2 + 2) \; Omega} \\ [/tex]

[tex]\longrightarrow \rm { R_{(1,2)} = 4 \; Omega} \\ [/tex]

Now, the combined resistance of [tex]\rm { R_1} [/tex] and [tex]\rm { R_2} [/tex] is connected in parallel combination with [tex]\rm { R_3} [/tex], so their combined resistance will be given by,

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \dfrac{1}{R_{(1,2)}} + \dfrac{1}{R_3} } \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1}{4} + \dfrac{1}{2} \Bigg ) \;\Omega} \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{1 + 2}{4} \Bigg ) \;\Omega} \\ [/tex]

[tex]\longrightarrow \rm {\dfrac{1}{ R_{(1,2,3)}} = \Bigg ( \dfrac{3}{4} \Bigg ) \;\Omega} \\ [/tex]

Reciprocating both sides,

[tex]\longrightarrow \rm {R_{(1,2,3)}= \dfrac{4}{3} \;\Omega} \\ [/tex]

Now, the combined resistance of [tex]\rm { R_1} [/tex], [tex]\rm { R_2} [/tex] and [tex]\rm { R_3} [/tex] is connected in series combination with [tex]\rm { R_4} [/tex]. So, equivalent resistance will be given by,

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= R_{(1,2,3)} + R_4} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4}{3} + 2 \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{4 + 6}{3} \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \rm {R_{(1,2,3,4)}= \Bigg ( \dfrac{10}{3} \Bigg ) \; \Omega} \\ [/tex]

[tex]\longrightarrow \bf {R_{(1,2,3,4)}= 3.33 \; \Omega} \\ [/tex]

Henceforth, Option A is correct.

[tex]\underline{\boxed{\large{\bf{Option \; B!! }}}} [/tex]

Here, we have to find the amount of flow of current in the circuit. By using ohm's law,

[tex] \longrightarrow [/tex] V = IR

[tex] \longrightarrow [/tex] 3 = I × 3.33

[tex] \longrightarrow [/tex] 3 ÷ 3.33 = I

[tex] \longrightarrow [/tex] 0.90 Ampere = I

Henceforth, Option B is correct.

[tex] \tt \purple{Hope \; it \; helps \; you, Army! \heartsuit } \\ [/tex]

Answer:

Answer to the following question is as follows;

Explanation:

The puck's real altitude is lower than ones projection. That's because the mechanism may not be completely frictionless. Electricity is nevertheless wasted owing to particle interactions such as friction, which might explain why the present the results is lower than predicted.

Answer: 2.4 millimeters = 0.0024 meters

Explanation: A millimeter is 1/1000 of a meter. By diving 2.4 by 1000, you get 0.0024.

Answer:

frequency

Explanation:

The phenomenon of apparent change in frequency due to the relation motion between the source and the observer is called Doppler's effect.

So, when we move farther, the frequency of sound decreases. The formula of the Doppler's effect is  

[tex]f' = \frac{v + v_o}{v+ v_s} f[/tex]

where, v is the velocity of sound, vs is the velocity of source and vo is the velocity of observer, f is the true frequency. f' is the apparent frequency.

Answer:

978.6084 Newton

Explanation:

Given the following data;

To find the weight in Newtown;

Conversion:

1 lb = 4.448220 N

220 lb = 220 * 4.448220 = 978.6084 Newton

220 lb = 978.6084 Newton

Therefore, the weight of the astronaut in Newton is 978.6084.

Weight can be defined as the force acting on a body or an object as a result of gravity.

Mathematically, the weight of an object is given by the formula;

Weight = mg

Where;

Note:

Answer:

4.86 m

Explanation:

Given that,

The frequency produced by a humming bird, f = 70 Hz

The speed of sound, v = 340 m/s

We need to find how far does the sound  travel between wing flaps. Let the distance is equal to its wavelength. So,

[tex]v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{340}{70}\\\\\lambda=4.86\ m[/tex]

So, the sound travel 4.86 m between wings flaps.

Answer:

a = 6.67 m/s²

Explanation:

F = 10.0 N

m = 1.50 kg

a = ?

F = ma

10.0 = (1.50)a

6.67 = a

Answer:

b)

Explanation:

GCMs (general circulation models) are useful instruments for gaining a quantitative knowledge of climate processes. Physical processes in the atmosphere, cryosphere, and land surface are represented by them. They are used for modeling the global climate system's reaction to rising greenhouse gas concentrations available at the moment by utilizing spectral models based on the energy emitted by the biosphere and clouds.

10N/m

Explanation:

f=kx

k=f/x

k=20N/0.2m

k=10N/m

Answer:

b)

Explanation:

Answer:

B. Is its acceleration constant

Explanation:

Uniform circular motion can be described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. ... An object undergoing uniform circular motion is moving with a constant speed. Nonetheless, it is accelerating due to its change in direction.

Answer:

A (3 seconds)

Explanation:

Well here we have a type of motion called projectile motion and it is pretty similar to an upside down parabola. The top of the trajectory is the vertex of the parabola and is also when v=0.

Lets identify our givens.

Givens:

Horizontal speed= 30m/s

Vertical Speed= 30 m/s

Since the ball is in freefall after being launched ay=-g(take up to be positive) and ax=0

The ball is launched from the ground so y0=0

Final vertical velocity= 0

This problem is now relatively easy because we only need to find the vertical distance so we can ignore horizontal speed and use

vy=vy0+ayt

Plug in our givens

0=30-10t

solve for t

t=3 seconds

Answer:The answer is C

Explanation:

Answer:

13.94 rpm

Explanation:

Given that,

The diameter of the pulley, d = 14 foot

Radius, r = 7 foot

The linear velocity of the pulley, v = 14 mph = 20.53 ft/s

We need to find the angular velocity in rpm.

We know that, the relation between the linear velocity and the angular velocity is as follows :

[tex]v=r\omega\\\\\omega=\dfrac{v}{r}\\\\\omega=\dfrac{20.53}{14}\\\\\omega=1.46\ rad/s[/tex]

or

[tex]\omega=13.94\ rpm[/tex]

So, the angular velocity of the pulley is 13.94 rpm.

Answer: hello your question is incomplete below is the missing part

A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.

answer:

- q

Explanation:

Since the spherical cavity was carved out of a neutral conducting sphere hence the electric field inside this conductor = zero

given that there is a point charge +q at the center of the spherical cavity hence for the electric field inside the conductor to be = zero the total surface charge qint on the wall of the cavity will be -q

Answer:

P = 0.14 hp

Explanation:

The power required by the ship is given as:

[tex]P = \frac{Work}{Time} = \frac{Potential\ Eenrgy}{t}\\\\P = \frac{mgh}{t}[/tex]

where,

P = Power = ?

m = mass to pump = (12 lb)(1 kg/2.20 lb) = 5.44 kg

g = acceleration due to gravity = 9.81 m/s²

h = height = 2 m

t = time = 1 s

Therefore,

[tex]P = \frac{(5.44\ kg)(9.81\ m/s^2)(2\ m)}{1\ s}\\\\P = 106.8\ W[/tex]

Converting to horsepower (hp):

[tex]P = (106.8\ W)(\frac{1\ hp}{746\ W})[/tex]

P = 0.14 hp