What is the relationship between resistance and current in a circuit with no change in voltage?
A. Current and resistance must be equal in a circuit.
B. A circuit that has more resistance will have smaller current.
C. Current does not depend on resistance in a circuit.
D. A circuit that has more resistance will have a greater current.

Answer:

They hit the ground at the same time

Explanation:

Mass doesn't matter in free fall. As long as they were released with the same velocity and the same height, they'll hit the ground at the same time

Answer:

13.8 N

[tex]41.44^{\circ}[/tex]

Explanation:

[tex]F_1=6\ \text{N}[/tex]

[tex]F_2=8\ \text{N}[/tex]

[tex]F_1\cos\theta_1\hat{i}+F_1\sin\theta_1\hat{j}\\ =6\cos30^{\circ}+6\sin30^{\circ}\hat{j}\\ =5.2\hat{i}+3\hat{j}[/tex]

[tex]F_2\cos\theta_2\hat{i}+F_2\sin\theta_2\hat{j}\\ =8\cos50^{\circ}+8\sin50^{\circ}\hat{j}\\ =5.14\hat{i}+6.13\hat{j}[/tex]

[tex]F_R=F_1+F_2=10.34\hat{i}+9.13\hat{j}[/tex]

[tex]|F_R|=\sqrt{10.34^2+9.13^2}=13.8\ \text{N}[/tex]

The magnitude of the resultant is 13.8 N

Direction is given by

[tex]\tan^{-1}=\dfrac{y}{x}=\tan^{-1}\dfrac{9.13}{10.34}=41.44^{\circ}[/tex]

The angle of the resultant is [tex]41.44^{\circ}[/tex]

Answer:

The total distance of the entire trip is 16 Km

Explanation:

Distance and Displacement

A moving object constantly travels for some distance at certain periods of time. The total distance is the sum of each individual distance the object traveled. It can be written as:

dtotal = d1 + d2 + d3 + ... + dn

This sum is obtained independently of the direction the object moves.

The displacement only takes into consideration the initial and final positions of the path defined by the object in its moving. The displacement, unlike distance, is a vectorial magnitude and can be even zero if the object starts and ends the movement at the same point.

The described movements are 6 Km North and 10 Km West. Regardless of the speed or time taken, the total distance is:

d = 6 Km + 10 Km = 16 Km

The total distance of the entire trip is 16 Km

Answer: c. incident ray

Explanation:

Answer:

30 m

__________________________________________________________

Explanation:

We are given:

Initial velocity (u) = 10 m/s

Final velocity (v) = 20 m/s

Time interval (t) = 2 seconds

Distance travelled by the bus (s) = s meters

Solving for the distance travelled:

Solving for the acceleration:

v = u + at                                                            [first equation of motion]

20 = 10 + a(2)                                                    [replacing the given values]

2a = 10                                                               [subtracting 10 from both sides]

a = 5 m/s²                                                         [dividing both sides by 2]

Solving for the distance:

s = ut + 1/2 (at²)                                                  [second equation of motion]

s = 10(2) + 1/2(5)(2)²                                           [replacing the given values]

s = 20 + 10

s = 30 m

Therefore, the bus travelled 30 m in the given time interval

Answer:

10778292789403987593790

Explanation:

I am a Cow'

Answer:Average Velocity of the trip=98.77km/h

Explanation:

Given that:

Black Panther started his journey with

Distance ran towards east  = 567.5km

Time ran towards east  = 2.3 hours

and then  turned back,

Distance ran towards west,  = 2218 km

Time taken towards west,  = 1.2 hours

Let us treat the westward movement as negative movement of the east

So total displacement =567.5 km east + (-2218km east )= 1,650.5km

But it seems the value of the distance ran west is wrong , because i do not think he can run  2218km under 1.2hours considering he ran 567.5m due east at 2.3hours .

So Let we use 221.8 km as distance ran due west.

So that Total Displacement becomes = 567.5 km east + (-221.8km east )= 345.7km

Total time = 2.3 + 1.2 = 3.5hrs.

Average velocity = Total displacement / total time  

.345.7km / 3.5hr =  98.77km/h

Answer:

The correct answer is B

An echo underwater and an echo in the air will return at different times. The echo underwater will return more quickly than the echo in the air.

Explanation:

The physics of this is simple.

Water and air are both made up of particles. The particles for water are more closely or densely arranged that those of the air molecules. Hence sound travels faster in water than in air. When measured, the speed actually differs by as much as 5 times with water being the fastest medium.

Think of it like this. Assume you have two stacks of dominoes, one closely packed than the other but exactly the same amount of dominos, you'd notice that the stack that is more tightly arranged will be the first to topple over because it takes less time for the kinetic energy from the first domino to reach the next and on and on like that until the last domino.

Cheers

Answer:

b. at different times, the echo under water returning more quickly.

Answer:

Explanation:

The weight of the object can be found by using the formula

[tex]f = \frac{w}{d} \\ [/tex]

w is the workdone

d is the distance

From the question we have

[tex]f = \frac{62500}{25} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

x = 1.26 R

Explanation:

For this exercise let's find the magnetic field using the Biot-Savart law

            B = μ₀ I/4π ∫ ds x r^ / r²

In the case of a loop or loop, the quantity ds is perpendicular to the distance r, therefore the vector product reduces to the algebraic product and the direction of the field is perpendicular to the current loop

suppose that the spiral eta in the yz plane, therefore the axis is in the x axis

         B = μ₀ I/4π ∫ ds / (R² + x²)

The total magnetic field has two components, one parallel to the x axis and another perpendicular, this component is annual when integrating the entire loop, so the total field is

            B = Bₓ i^

using trigonometry

            Bₓ = B cos θ

we substitute

            Bₓ = μ₀ I/4π ∫ ds cos θ / (x² + R²)

the cosine function is

            cos θ = R /√(x² + R²)

The differential is

            ds = R dθ

we substitute

             Bₓ = μ₀ I/4π ∫ (R dθ)  R /√( (x² + R²)³ )

we integrate from 0 to 2π

              Bₓ =μ₀ I/4π R² / √(x² + R²)³   2pi

therefore the final expression is

            B = μ₀ I R²/ 2√(x² + R²)³   i^

In our case the distance is requested where B is half of B in the center of the bone loop x = 0

Spire center field   x=0

              B₀ = μ₀ I/2R

Field at the desired point (x)

              B = B₀ / 2

we substitute

              R² /√(x² + R²)³ = ½  1 /R

              2R³ =√(x² + R²)³

              (x² + R²)³ = 4 (R²)³

              (x²/R² + 1)³ = 4

The exact result is the solution of this equation, but it is quite laborious, we can find an approximate result assuming that the distance x is much greater than R (x »R)

           B = μ₀ I/2x³ 

we substitute

            R² / x³ = 1/2   1 / R

            2R³ = x³

          x = ∛2  R

            x = 1.2599 R

The distance at which the magnetic field strength is half the strength of the field at the center of the loop in terms of  R is 0.766 R

Suppose we consider a magnetic field located at point z on the axis of the current loop with radius R carrying a current (I), then the magnetic field can be represented as:

[tex]\mathbf{B = \dfrac{\mu_o}{2} \dfrac{IR^2}{(z^2+R^2)^{^{\dfrac{3}{2}}}}}[/tex]

And the field situated at the center of the loop is:

[tex]\mathbf{B_{z=0} =\dfrac{\mu_o}{2} \dfrac{I}{R} }[/tex]

Let consider a distance (z) on the axis of the loop, in which the magnetic field as a result of the loop is equal to half the strength of the magnetic field at the center of the loop;

Then;

[tex]\mathbf{B(z) = \dfrac{1}{2}B_{z=0}}[/tex]

[tex]\mathbf{\dfrac{\mu_o}{2} \dfrac{IR^2}{(z^2+R^2)^{\frac{3}{2}}} =\dfrac{1}{2} \Big (\dfrac{\mu_o}{2} \dfrac{I}{R} \Big )}[/tex]

Multiply both sides by (2);

[tex]\mathbf{\dfrac{R^2}{(z^2+R^2)^{\frac{3}{2}}} = \dfrac{1}{2R}}[/tex]

Cross multiply;

[tex]\mathbf{2R^3 = (z^2 +R^2)^{\dfrac{3}{2}}}[/tex]

[tex]\mathbf{4R^6 = (z^2 +R^2)^3}[/tex]

[tex]\mathbf{z = \sqrt{4^{1/3} R^2 -R^2 }}[/tex]

[tex]\mathbf{z = R\sqrt{4^{1/3} -1 }}[/tex]

z = 0.766 R

Therefore, we can conclude that the distance at which the magnetic field strength is half the strength of the field at the center of the loop in terms of  R is 0.766 R

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Answer:

a. 77.9 nT

Explanation:

Given;

average value of the electromagnetic wave, E = 0.724 W/m^2

The average value of a Poynting vector is given by;

[tex]S = \frac{1}{2\mu_o} *E_oB_o\\\\But, E_o = cB_o\\\\S = \frac{1}{2\mu_o} *cB_o^2\\\\cB_o ^2= 2\mu_o S\\\\B_o^2 = \frac{2\mu_o S}{c}\\\\B_o = \sqrt{ \frac{2\mu_o S}{c}} \\\\B_o = \sqrt{ \frac{2(4\pi*10^{-7}) (0.724)}{3*10^8}}\\\\B_o = 7.79*10^{-8} \ T\\\\B_o = 77.9*10^{-9} \ T\\\\B_o = 77.9 \ nT[/tex]

Therefore, the maximum value of the magnetic field in the wave is 77.9 nT.

Answer:

A I think : )

Explanation:

Answer:

Explanation:

Step one:

given data

mass of cable= 4lb/ft

mass of coal= 1000lb

dept of mine= 700ft

Step two:

Required

the work-done to lift the coal and the rope combined

Work-done to lift coal

Wc=1000*700= 700,000 lb-ft

Work-done to lift rope

[tex]Wr=\int\limits^{700} _0 {4(700-y)} \, dx \\\\Wr=4(700y-\frac{1}{2}y^2 )\limits^{700}_0[/tex]

substitute y=700 we have, since y=0 will result to 0

[tex]Wr=4(700*700-\frac{1}{2}*700^2 )\\\\Wr=4(490000-245000)\\\\Wr=4(245000)\\\\Wr=980000ft-lbs[/tex]

Answer:

e

Explanation:

cause you have any attachments are not sure if you can get it

When a block slides down an inclined plane, normal force does zero work. Hence, option (a) is correct.

The definition of force in physics is: The push or pull on a massed object changes its velocity.

An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude. A spring balance can be used to calculate the Force. The Newton is the SI unit of force.

The component of a contact force in mechanics known as the normal force is perpendicular to the surface that an item encounters.

As normal force always acts perpendicular to the motion, this force does zero work.

But friction force the weight acts in opposite direction of the motion, this force does work against the motion.

In an inclined plane, the parallel component of  weight does work along the motion.

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Answer:

F = 1.8 KN

Explanation:

Given that the spacecraft is 2.00 Earth radii above the Earth's surface, the force of Earth's gravity on it can be determined by;

F = [tex]\frac{GMm}{(R + h)^{2} }[/tex]

Where: F is the force, G is the universal gravitation constant, M is the mass of the earth, m is the mass of the spacecraft, and R is the radius of the earth and h is the distance of the spacecraft to the surface of the earth.

G = 6.67 x [tex]10^{-11}[/tex] N[tex]m^{2}[/tex][tex]kg^{-2}[/tex], M = 5.972 x [tex]10^{24}[/tex] kg, m = 1650 kg, R = 6371 km, h = (2.0 x 6371 km) = 12742 km.

Thus,

F = [tex]\frac{6.67*10^{-11}*5.972*10^{24} *1650 }{(6371*10^{3}+12742*10^{3}) ^{2} }[/tex]

  = [tex]\frac{6.5725*10^{17} }{(19113000)^{2} }[/tex]

  = 1799.17

F = 1.8 KN

The force of the Earth's gravity on the spacecraft is 1.8 KN.

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Actually Welcome to the Concept of the Kinematics in real world.

So, as given here, we have to find the Mass of the bus from the given momentum, so we get as,

P = m * V

momentum = mass * velocity

here, P= 152625 kgm/s and v= 11.1 m/s

so substituting we get as,

m = 152625 ÷ 11.1 => 13,750 kg

hence,the mass of the bus is 13,750 kg.

Answer:

   y= y'  0.67

Explanation:

This is an exercise in refraction of light,

         n₁ sin θ₁ = n₂ sin θ₂

where subscript 1 is used for the incident medium and subscript 2 for the refracted medium

          sin θ₁ = n2 /n1   sin θ₂

the incident medium is air with refractive index n1 = 1 and the medium where the ray is refracted is water with n = 1.33

        let's calculate

        sin θ₁ = 1.33 / 1 sin  25

        θ₁ = sin⁻¹ (0.56208)

        θ₁ = 34.2º

when the ray is refracted we can assume that the adjacent leg (water surface) is the same for the two media

let's use the trigonometry relationship

          tan θ₁ = x / y

          tan θ₂ = x / y '

         tan θ₁ = y’ tan θ₂

         y = y ’ tanθ₂ / tan θ₁

to finish exercise you must know the depth of the object

         y =y'  tan 25/ tan 34.2

         y= y'  0.67

Answer:

TO POWER ELECTRIC GADGETS AND SAVE EARTH FROM POLLUTION

Explanation:

Answer:

D. The carbon dioxide levels in the atmosphere began increasing rapidly around 1950.

Explanation:

Study Island

The carbon dioxide levels in the atmosphere began increasing rapidly around 1950. Therefore, the correct option is option D.

One carbon atom is covalently doubly connected to the two oxygen atoms in each of the molecules that make up carbon dioxide. At room temperature, it exists as a gas.

In the atmosphere, carbon dioxide serves as a greenhouse gas because it absorbs infrared radiation despite being transparent to visible light. It has increased from which was before values of 280 ppm to be a gas in the atmosphere in the Stratosphere at 431 parts per thousand, or roughly 0.04% by volume. The carbon dioxide levels in the atmosphere began increasing rapidly around 1950. This statement is true.

Therefore, the correct option is option D.

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Answer:

Explanation:

Step one:

given data

mass of bullet m1=25g= 0.025kg

initial velocity of bullet u1=240m/s

mass of block m2=?

initial velocity of bullet u2 of block=0m/s

final velocity of the system after impact v= 27m/s

Required; the mass of the block m2

Step two:

applying the conservation of linear moment/elastic collision we have

m1u1+m2u2=v(m1+m2)

substituting we have

0.025*240+m2*0=27(0.025+m2)

6=0.675+27m2

6-0.675=27m2

5.325=27m2

divide both sides by 27

m2=5.325/27

m2=0.197

m2=0.2kg approx.

Answer:

POSITIVE

Explanation:

The acceleration of an object is given by the rate of change of velocity.

u and v are initial and final velocities

If the final velocity is more than that of the initial velocity, the acceleration of the object is positive. It means positive acceleration occurs when an object speeds up. Hence, the correct option is (a) "positive".

Answer:

Positive

Explanation:

Answer:

This refers when a spiker quickly strides towards the net before they jump in the air for an attack.

hi ilove you jk it stoo stsfad

Answer:

c) 1.0 kg

Explanation:

The mass of the stick will be located at the centre of the metre rule. Since the rock is located 0.25m from the pivot, the mass of the meter rule is also 0.25m to the Right of the support

According to law of moment

Sum of clockwise moment = sum of anti clockwise moments

Clockwise moment = M×0.25(mass of metre rule is M)

CW moment = 0.25M

Anti clockwise moment = 0.25×1

ACW moments = 0.25kgm

Equate;

0.25M = 0.25

M = 0.25/0.25

M = 1.0kg

Hence the mass of the metre rule is 1.0kg

Answer:

-7/19

Explanation:

Answer:

The options are

A. Older rocks are commonly remitted over huge regions

B. Older rocks have been uplifted and eroded away

C. Large parts of the continent are subducted deep within the mantle

D. Parts of the continent have been added by the accretion of tectonic terraces.

The answer is D. Parts of the continent have been added by the accretion of tectonic terraces.

The major reason why the age of the oldest rocks can vary from one part of the continent to another is that parts of continent have been added by the accretion of tectonic terraces.

Answer:

36 m/s

Explanation:

t = 3.6s

u = 0m/s

a = +g = 10m/s²

v = ?

using,

v = u + at

v = 0 + 10(3.6)

v = 36 m/s

Answer:

v = 120 m/s

Explanation:

We are given;

earth's radius; r = 6.37 × 10^(6) m

Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s

Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.

The angle will be;

θ = ¾ × 90

θ = 67.5

¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.

The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:

v = r(cos θ) × ω

v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)

v = 117.22 m/s

Approximation to 2 sig. figures gives;

v = 120 m/s

Answer:

i. F =  1.3 x [tex]10^{-7}[/tex] N

ii. The direction of the force of attraction exerted by the proton on the electron is towards the itself (i.e a pull).

Explanation:

Since the given charges are opposite, then the force of attraction is experienced. The force of attraction between the two charges can be determined by:

F = [tex]\frac{kq_{1} q_{2} }{d^{2} }[/tex]

where F is the force, k is the constant, [tex]q_{1}[/tex] is the charge of the electron, [tex]q_{2}[/tex] is the charge on the proton, and d is the distance between them.

So that; k = 9.0 x [tex]10^{9}[/tex] N[tex]m^{2}[/tex][tex]C^{-2}[/tex] , [tex]q_{1}[/tex] = 1.6 x [tex]10^{-19}[/tex] C, [tex]q_{2}[/tex] = 1.6 x

Thus,

F = [tex]\frac{9.0*10^{9}*1.6*10^{-19}*1.6*10^{-19} }{(4.2*10^{-11}) ^{2} }[/tex]

  = [tex]\frac{2.304*10^{-28} }{1.764*10^{-21} }[/tex]

  = 1.3061 x [tex]10^{-7}[/tex]

F = 1.3 x [tex]10^{-7}[/tex] N

The force between the charges is 1.3 x [tex]10^{-7}[/tex] N.

ii. The direction of the force of attraction exerted by the proton on the electron is towards the itself.

Answer:

v = 5.24 m/s

Explanation:

Given that,

No of spokes of a Ferris wheel = 60

The diameter of a wheel, d = 100 m

Radius, r = 50 m

It makes one rotation every 60 seconds.

We need to find the speed of the passengers when the Ferris wheel is rotating at this rate. Let it is v. It can be calculated as follows :

[tex]v=r\omega\\\\=\dfrac{2\pi r }{T}\\\\=\dfrac{2\pi \times 50 }{60}\\\\v=5.235\ m/s[/tex]

or

v = 5.24 m/s

So, the speed of the passengers is 5.24 m/s. Hence, the correct option is (A).

Answer:

50,000N

Explanation:

According to Newton's second law of motion;

Net Force = Mass * acceleration

Given

Mass = 5000kg

Let the acceleration = 10m/s²

Net force = 5000 * 10

Net force = 50,000N

Hence the net force acting on the bus is 50000N