if C is The vector sum of A and B C = A + B What must be true about The directions and magnitudes of A and B if C=A+B? What must be tre about the directions and magnitudes of A and B if C=0? ​

Answer:

D. It does not emit greenhouse gases.

Explanation:

D. It does not emit greenhouse gases.

Answer:

Here, m=9×10

−31

kg,

q=1.6×10

−19

C,v=3×10

7

ms

−1

,

b=6×10

−4

T

r=

qB

mv

=

(1.6×10

−19

)(6×10

−4

)

(9×10

−31

)×(3×10

7

)

=0.28m

v=

2πr

v

=

2πm

Bq

=

2×(22/7)×9×10

−31

(6×10

−4

)×(1.6×10

−19

)

=1.7×10

7

Hz

Ek=

2

1

mv

2

=

2

1

×(9×10

−31

)×(3×10

7

)

2

J

=40.5×10

−17

J=

1.6×10

−16

40.5×10

−17

keV

=2.53keV

Answer:

D newton third law

Explanation:

good luck

Answer:

2.5 km

Explanation:

Answer:

2.5 km

Explanation:

Distance = speed x time

So =5 x 0.5

Explanation:

CH3CH2OH

That is the answer I hope this helps

Answer:

Explanation :  B = μ₀i / 2πr

B1 = (2×10^-7) × 1.5/0.02

       = 15 μT

B2 = (2×10^-7 ) ×  1.5/0.04

       = 7.5 μT

total field = 15 μT + 7.5 μT

                  = 22.5 μT

Answer: 5.12x10∧-4N

Explanation:

Force = I B L

L = 6.4m

Let Current (I) I₁ = I₂= 14A

Distance of the wire = 42cm = 0.42m

BUT

B = μ₀I / 2πr

=(2X10∧-7 X 12) / 0.42

       B =5.714×10∧-6T

Force = I B L

Force = 14x [5.714×10-6]×6.4

Force = 5.12x10∧-4N

Explanation:

Recall that

[tex]K = \dfrac{v_0^2\sin2\theta}{g}\:\:\:\:\:\:\:\:\:(1)[/tex]

and

[tex]Q = \dfrac{v_0^2\sin^2\theta}{2g}\:\:\:\:\:\:\:\:\:(2)[/tex]

From Eqn(2), we can write

[tex]\sin\theta = \sqrt{\dfrac{2gQ}{v_0^2}}\:\:\:\:\:\:\:\:\:(3)[/tex]

Using the identity [tex]\sin\theta = 2\sin\theta \cos\theta[/tex], we can rewrite Eqn(1) as

[tex]\dfrac{gK}{2v_0^2} = \sin\theta \cos\theta[/tex]

Squaring the above equation, we get

[tex]\dfrac{g^2K^2}{4v_0^4} = \sin^2\theta \cos^2\theta[/tex]

[tex]\:\:\:\:\:\:\:\:\:=\sin^2\theta(1 - \sin^2\theta)\:\:\:\:\:\:\:(4)[/tex]

Use Eqn(3) on Eqn(4) and we will get the following:

[tex]\dfrac{g^2K^2}{4v_0^4} = \dfrac{2gQ}{v_0^2}(1 - \dfrac{2gQ}{v_0^2})[/tex]

This simplifies to

[tex]\dfrac{gK^2}{8v_0^2Q} = 1 - \dfrac{2gQ}{v_0^2}[/tex]

Rearranging this further, we get

[tex]1 = \dfrac{2gQ}{v_0^2} + \dfrac{gK^2}{8v_0^2Q}[/tex]

Putting [tex]v_0^2[/tex] to the left side, we get

[tex]v_0^2 = 2qQ + \dfrac{gK^2}{8Q}[/tex]

Finally, taking the square root of the equation above, we get the expression for the muzzle velocity [tex]v_0[/tex] as

[tex]v_0 = \sqrt{2gQ + \dfrac{gK^2}{8Q}}[/tex]

Answer:

whats the formula

Explanation:

Answer:

a) T = 0.5 s

b) v = 1.2π m/s ≈ 3.77 m/s

Explanation:

It makes two revolutions in one second so makes one revolution in ½ second

circumference of the circle is

C = 2πr = 0.6π m

which it traverses in one time period

0.6π m / 0.5 s = 1.2π m/s

To solve this, we must be knowing each and every concept related to speed and its calculations. Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.

Speed may be defined as the distance traveled by an item in the amount of time it requires to travel that distance. In other words, it measures how rapidly an item travels but does not provide direction.

Speed may be calculated in Science. The speed equation is a scientific formula that is used to calculate various types of speed.

Mathematically, the formula for speed can be given as

speed= distance/time

Values that are given

Time period= 0.5 s

Circumference = 2πr = 0.6π m

substituting all the given values in the above equation, we get

speed     =0.6π m / 0.5 s

On calculations, we get

              = 1.2π m/s

              =3.77 m/s

Therefore, the angular speed of a model car moves round a circular path of radius 0.3m at 2 revolutions per secs is 3.77 m/.

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Answer:

Liquids evaporate faster as they heat up and more particles have enough energy to break away. The particles need energy to overcome the attractions between them. ... At this point the liquid is boiling and turning to gas. The particles in the gas are the same as they were in the liquid they just have more energy.

they offer fewer luxury services because they lack the money to provide them.

Answer:

Explanation:

Your body weighs more at the pole for two important reasons. Both have to do to the spin of the earth on its axis.

Because of its spin the earth is thicker around the equator than it is through the poles. This means that when you stand on the equator, you are farther away from the center of earth than you would be at the poles. As gravity decreases with the inverse of the square of distance, gravity will be weaker at the equator.

As you are also spinning with the earth, you will have a required centripetal acceleration and force to keep you attached to the ground, This force decreases the effect of gravity so again, you would weigh less at the equator.

Answer:

This is the answer

Explanation:

You can find the ans in the photo I attached.

Answer:

2 :π8 :3

The ratio between the most probable velocity,mean velocity and root mean square velocity is 2 :π8 :3. Vrms=M3RT.

Explanation:

pls mark brainliest

Answer:

A freely falling object has weight W=mg, where W-weight, m-mass of the object and g-acceleration produced due to the earth's gravity. ... This happens because the normal reaction force exerted on the object in the lift is equal to zero, and normal force equals to mg, which in turn equals the weight of the object

Explanation:

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Answer:

Gradually increases until the maximum weight reaches the surface of the earth

Explanation:

Answer:

A boat having stones floats in water. If stones are unloaded in water, what will happen to the level of water?​

Explanation:

Element is

[tex]\boxed{\sf {}^{27}Al_{13}}[/tex]

[tex]\\ \sf\longmapsto No\:of\:Protons=Atomic \:Number=13[/tex]

[tex]\\ \sf\longmapsto No\:of\:Neutrons=Mass\:number-Atomic\:Number[/tex]

[tex]\\ \sf\longmapsto No\:of\:Neutrons=27-13[/tex]

[tex]\\ \sf\longmapsto No\:of\:Neutrons=14[/tex]

And

[tex]\\ \sf\longmapsto No\:of\:electrons=No\:of\:Protons=13[/tex]

Answer:

tire advances to the right, the friction force must be directed to the left, that is, opposing the possible movement of the tire.

Explanation:

For the movement of the wheel to be composed of a rotating part and a translational part, it is necessary that there be a static friction force between the floor and the tire.

As the tire advances to the right, the friction force must be directed to the left, that is, opposing the possible movement of the tire.

Answer:

(A) Gravity is you're answer.

Explanation:

When an object or human is falling at an increased rate, The force of gravity is taking place.

We have that the spring constant is mathematically given as

[tex]k=2.37*10^{11}N/m[/tex]

Generally, the equation for angular velocity is mathematically given by

[tex]\omega=\sqrt{k}{m}[/tex]

Where

k=spring constant

And

[tex]\omega =\frac{2\pi}{T}[/tex]

Therefore

[tex]\frac{2\pi}{T}=\sqrt{k}{n}[/tex]

Hence giving spring constant k

[tex]k=m((\frac{2 \pi}{T})^2[/tex]

Generally

Mass of earth [tex]m=5.97*10^{24}[/tex]

Period for on complete resolution of Earth around the Sun

[tex]T=365 days[/tex]

[tex]T=365*24*3600[/tex]

Therefore

[tex]k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2[/tex]

[tex]k=2.37*10^{11}N/m[/tex]

In conclusion

The effective spring constant of this simple harmonic motion is

[tex]k=2.37*10^{11}N/m[/tex]

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Answer:

1) 65 m

2) 40 m/s downward

Explanation:

Using for both questions the kinematic equation

v² = u² + 2as

and ignoring air resistance.

1) h = 60 + √(20²/(2(9.8))) = 64.51753...

2) v = √(20² + 2(9.8)(60)) = 39.69886...

Answer:

a

Explanation:

Answer:

what,     100.999m

Explanation:

convert 999 mm into meters, which is 0.999m and add that to a 100 m and that will make the total 100.999 m

The result of the addition of the two values is equal to 100.999 meters.

Given the following data:

To determine the result of the addition of the two values:

First of all, we would convert the value in millimeter (mm) to meter (m) as follows:

Conversion:

1 millimeter = 0.001 meter

999 millimeter = X meter

Cross-multiplying, we have:

[tex]X = 0.001 \times 999[/tex]

X = 0.999 meter.

For the result:

[tex]Result = 0.999 +100[/tex]

Result = 100.999 meters.

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Answer:

I don't get it rewrite please

Explanation:

[tex]v = \sqrt{2ax}[/tex]

Take the square of both sides:

[tex]v^2 = \left(\sqrt{2ax}\right)^2 = 2ax[/tex]

Divide both sides by 2a and you will get

[tex]x = \dfrac{v^2}{2a}[/tex]

The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.

In this case, we need to construct the Free Body Diagram of the sled-victim System in order to determine what Forces are doing Work. Then, we construct the respective Energy equation by Newton's Laws of Motion, Work-Energy Theorem and definition of Work.

Given that system experiments an uniform Acceleration, we must solve the resulting model for the work done by the Tension in the rope.

From the Free Body Diagram (see image attached), we see that both Weight of the sled and Friction between sled and snow are doing work in favor of gravity, whereas Tension forces is against gravity. Normal force is not doing work as its direction is perpendicular to the direction of motion. The energy equation of this system is:

[tex]-W_{T} + \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta = m\cdot a\cdot s[/tex] (1)

Where:

[tex]W_{T}[/tex] - Work done by tension, in joules.

[tex]m[/tex] - Mass of the sled-victim system, in kilograms.

[tex]\mu[/tex] - Coefficient of kinetic friction, no unit.

[tex]g[/tex] - Gravitational acceleration, in meters per square second.

[tex]s[/tex] - Travelled distance, in meters.

[tex]\theta[/tex] - Slope angle, in sexagesimal degrees.

[tex]a[/tex] - Net acceleration of the sled-victim system, in meters per square second.

If we know that [tex]\mu = 0.100[/tex], [tex]m = 55.3\,kg[/tex], [tex]g = 10\,\frac{m}{s^{2}}[/tex], [tex]s = 2.1\,m[/tex], [tex]\theta = 79.6^{\circ}[/tex] and [tex]a = -4.3\,\frac{m}{s^{2}}[/tex], then the work done by the tension in the rope is:

[tex]-W_{T} + \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta = m\cdot a\cdot s[/tex]

[tex]W_{T} = \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta -m\cdot a\cdot s[/tex]

[tex]W_{T} = (0.100)\cdot \left(55.3\,kg\right)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (2.1\,m)\cdot \cos 79.6^{\circ} + \left(55.3\,kg\right)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (2.1\,m)\cdot \sin 79.6^{\circ} - (55.3\,kg)\cdot \left(-4.3\,\frac{m}{s^{2}} \right) \cdot (2.1\,m)[/tex]

[tex]W_{T} = 1662.544\,J[/tex]

The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.

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Answer:

A solenoid is a type of electromagnet, the purpose of which is to generate a controlled magnetic field through a coil wound into a tightly packed helix. The coil can be arranged to produce a uniform magnetic field in a volume of space when an electric current is passed through it.

The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.

Centripetal force is described as the force applied to a body that is travelling in a circular motion and is pointed in the direction towards the center of the circular path.

Here,

Mass of the car, m = 0.5 kg

Velocity of the car, v = 6 m/s

Radial friction between the car and the track, f = 7 N

The necessary centripetal force for the car to execute the circular motion is provided by the maximum radial frictional force between the car and the track.

So, the condition that the car cannot continue to travel in the circular path is that the centripetal force required is greater than the maximum radial friction.

So,

mv²/r > f

0.5 x 6²/r > 7

Therefore, the radius of the circular track,

r < 18/7

r < 2.57 m

Hence,

The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.

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Answer:

q 1 = q 2 = Q

E = k * Q / r²

The combined electric field:

E = 2 * k * Q / r²

k = 9 · 10^9 Nm²/C²

r = 0.075 : 2 = 0.0375 m

45 N/C = 2 · 9 · 10^9 · Q / 0.0375²

45 = 18 · 10 ^9 Q / 0.0014062

Q = 45 · 0.0014062 / 18 · 10^9

Q = 0.003515 · 10^(-9) C

Q = 3.5155 · 10 ^ (-12) C = 3.5155 pC