What season is the Southern hemisphere in

Answer:

Current in a parallel circuit = 0.61 amps (Approx)

Explanation:

Given:

Voltage V = 6 volt

Two resistors = 17.2 , 22.4 in parallel circuit

Find:

Current in a parallel circuit

Computation:

1/R = 1/r1 + 1 / r2

1/R = 1/17.2 + 1 / 22.4

R = 9.73 ohms (Approx)

Current in a parallel circuit = V / R

Current in a parallel circuit = 6 / 9.73

Current in a parallel circuit = 0.61 amps (Approx)

Explanation:

angular velocity = velocity/radius

= 5/2

= 2.5 rad/s

Answer:

Jumping battle slams - just move the rope up and down

Alternating jump wave - jump and move the rope side to side

Alternating wide circles - move the rope in a circle position

Jumping jacks

Squat to sholder

Explanation:

The guy above me is correct give him Brainliest

Answer:

Explanation:

Force of friction at car B ( break was applied by car B ) =μ mg = .65 x  2100 X 9.8  = 13377 N .

work done by friction = 13377 x 7.30 = 97652.1 J

If v be the common velocity of both the cars after collision

kinetic energy of both the cars = 1/2 ( 2100 + 1500 ) x v²

= 1800 v²

so , applying work - energy theory ,

1800 v² = 97652.1

v² = 54.25

v = 7.365 m /s

This is the common velocity of both the cars .

To know the speed of car A , we shall apply law of conservation of momentum  .Let the speed of car A before collision be v₁ .

So , momentum before collision = momentum after collision of both the cars

1500 x v₁ = ( 1500 + 2100 ) x 7.365

v₁ = 17.676 m /s

= 63.63 mph .

( b )

yes Car A was crossing speed limit by a difference of

63.63 - 35 = 28.63 mph.

(a) The speed of car A just before the collision is 51.58 mph.

(b) With the given speed limit of 35 miles per hour, car A was crossing the speed limit by 16.58 mph.

What is collision?

The event when two objects strike each other from either direction, then such event is known as a collision. During the collision, the speed of colliding objects may vary according to the direction of the approach.

Given data -

The mass of car A is,  mA = 1500 kg.

The mass of car B is, mB = 2100 kg.

The length of the skid mark is, d = 7.30 m.

The coefficient of kinetic friction between tires and road is, [tex]\mu = 0.65[/tex].

(a)

The combined kinetic energy of both cars is,

[tex]KE_{T}=\dfrac{1}{2} (mA+mB)v^{2}\\\\KE_{T}=\dfrac{1}{2} (1500+2100)v^{2}\\\\KE_{T}=1800v^{2}[/tex]

Applying the work-energy principle as,

Work done due to kinetic friction = Combined kinetic energy of cars

[tex]F \times d = KE_{T}\\\\(\mu \times (mA+mB)\times g) \times d = KE_{T}\\\\(0.65 \times (1500+2100)\times 9.8) \times 7.30 = 1800v^{2}\\\\v = 9.64 \;\rm m/s[/tex]

Converting into mph as,

[tex]v = 9.64 \times 2.23\\\\v = 21.49 \;\rm mph[/tex]

To know the speed of car A , we shall apply the law of conservation of momentum. Let the speed of car A before collision be v₁.

So , momentum before collision = momentum after collision of both the cars

1500 x v₁ = ( 1500 + 2100 ) x 21.49

v₁ = 51.58 mph

Thus, we can conclude that the speed of car A just before the collision is 51.58 mph.

(b)

With the given speed limit of 35 mph, the obtained speed of car A before the collision is 51.58 mph. Clearly, car A is crossing the speed limit. And the difference is,

= 51.58 - 35 = 28.63 mph.

= 16.58 mph

Thus, we can conclude that car A was crossing the speed limit by 16.58 mph.

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Answer:

Height of its bounce

Explanation:

The dependent variable is always what is being measured or the data collected.

Answer:

The final velocity of the 2kg ball is 1.270 m/s

Explanation:

According to Newton's second and third laws of motion

Newton's second law state that "the rate of change of momentum is proportional to the applied force and takes place in the direction of that force".

Newton's third law state that "for every action, there must be an equal and opposite reaction".

The combinations of these two laws resulted in an elastic collision

Given that:

m1 = 2kg

u1 = 2.20m/s

m2 = 4.00kg

u2 = 0m/s

An Elastic collision is when kinetic energy before = kinetic energy after

E.K before = [tex]1/2mv^{2}[/tex]

E.K before = 1/2 * 2 * (2.20)^2

E.K = 1/2 * 2 * 4.84

E.K before = 4.84j

E.K after = 1/2 x (4 + 2)v^2

E.K after = 1/2(6v^2)

E.K after = 3v^2

Since E.K before = E.K after

4.84 = 3v^2

Divide through by 3

4.84/3 = 3v^2/3

1.6133 = v^2

[tex]V = \sqrt{1.6133} \\V = 1.270 m/s[/tex]

Explanation:

What impulse occurs when a cart that is originally at rest experiences an average force of

N for 2.5 s? *

(10 Points)

25 N

25 Nm

25 Ns

25 kg m/s

Answer:

the correct one is 2. the equipotential lines must be closer together where the field has more intensity

Explanation:

The equipotential line concept is a line or surface where a test charge can move without doing work, therefore the potential in this line is constant and they are perpendicular to the electric field lines.

In this exercise we have a charge and a series of equipotential lines, if this is a point charge the lines are circles around the charge, where the potential is given by

           V = k q / r

also the electric field and the electuary potential are related

           E =  [tex]- \frac{dV}{dr}[/tex]

therefore the equipotential lines must be closer together where the field has more intensity

When checking the answers, the correct one is 2

Answer:

Aerobics I think.

Explanation:

Answer:

The circuit is missing attached below is the required circuit

answer :

a) Ic = 1.944 mA

  Rp = 288.66 kΩ

b) The Emitter-base Junction of the BJT is forward biased while its collector-base junction is reverse biased

Explanation:

Rc = 3.6 kΩ

VBE = 0.8 v

1) predict Ic and specify Rp to establish Vce at 5 V

we will apply Kirchhoff's voltage law to resolve this

solution attached below

b ) The BJT is said to be in Forward reverse bias because The Emitter-base Junction of the BJT is forward biased while its collector-base junction is reverse biased

2 people

Explanation:

Answer:

Cayla ? please whats going on ?

Explanation:

Answer:

terminal velocity is;

v = 117.54 m/s

v = 423.144 km/hr

Explanation:

Given the data in the question;

we know that, the force on a body due to gravity is;

[tex]F_g[/tex] = mg

where m is mass and g is acceleration due to gravity

Force of drag is;

[tex]F_d[/tex] = [tex]\frac{1}{2}[/tex]pCAv²

where p is the density of fluid, C is the drag coefficient, A is the area and v is the terminal velocity.

Terminal velocity is reach when the force of gravity is equal to the force of drag.

[tex]F_g = F_d[/tex]

mg =  [tex]\frac{1}{2}[/tex]pCAv²

we solve for v

v = √( 2mg / pCA )

so we substitute in our values

v = √( [2×(86 kg)×9.8 m/s² ] / [ 1.21 kg/m³ × 0.7 × 0.145 m²] )

v = √( 1685.6 / 0.122015 )

v = √( 13814.6949 )

v  = 117.54 m/s

v = ( 117.54 m/s × 3.6 ) = 423.144 km/hr

Therefore terminal velocity is;

v = 117.54 m/s

v = 423.144 km/hr

Answer:

Height of the ramp

Explanation:

The independent variable is always what the person doing the experiment can change or modify.  So if the question is about whether the height of the ramp with  effect the car, that is indeed what they are changing or modifying in the experiment.

Answer:

Explanation:

Thermal energy or internal energy gain or loss = mass x specific heat x temperature  

specific heat of water = 4.2 kJ / kg degree Celsius

mass of water lost in 10 second = rate of loss x time = .5 x 10 = 5 kg .

heat energy associated with lost water = 5 x 4.2 x ( 60 - 0 ) = 1260 kJ .

Heat energy lost = 1260 kJ .

Answer: 0.52N

Explanation:

Formula

F=ke

Givens

e= 4.0cm>>>0.04m

k=13N/m

Plug givens into formula

F= (13N/m)(0.04m)

F=0.52N

The force on the ball at the moment the spring is released will be F =0.52N

When a spring is stretched or compressed , it displaces from equilibrium position . As a result , a restoring force will act ( which act opposite to the direction of force applied on the string ) and tends to retract the spring back to its original position . The force is called spring force

Given

x = 4cm  = 0.04 m

spring constant is(k) =  13 N/m

F( spring force on the ball ) = ?

F = -k x

F  = -(13)(.04)

F ( spring force) = - 0.52N

The force on the ball at the moment the spring is released will be F =0.52N

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#SPJ3

Answer:

d

Explanation:

Answer:

the laws of physics are the same for all non-accelerating observers

Explanation:

Answer:

In the Both time

Explanation:

A plane takes off at St.Louis, flies straight to Denver, and then returns the same way. The plane flies at the same speed with respect to the ground during ...

Depending on whether the plane flies against the earth's rotation or with it. Is this expectation is true statement.

Physical quantities such as work, temperature, and distance can all be completely represented in daily life by their magnitude. The laws of arithmetic can, however, be used to explain how these physical values relate to one another.

Motion in two dimensions is another name for motion in a plane. For instance, a projectile moving in a circle. The origin, along with the two coordinate axes X and Y, will serve as the reference point for the investigation of this kind of motion.

Therefore,  Depending on whether the plane flies against the earth's rotation or with it. Is this expectation is true statement.

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Answer:

354 m/s

Explanation:

For the second overtune (Third harmonic) of an open pipe,

λ = 2L/3................................ Equation 1

Where L = Length of the open pipe, λ = Wave length.

Given: L = 1.75 m.

Substitute into equation 1

λ = 2(1.75)/3

λ = 1.17 m.

From the question,

V = λf.......................... Equation 2

V = speed of sound in the room, f = frequency

Given: f = 303 Hz.

Substitute into equation 2

V = 1.17(303)

V = 353.5

V ≈ 354 m/s

Hence the right answer is 354 m/s

q1 = -5.00 x 10-6 C

q2 = -5.00 x 10-6 C

q3 = -5.00 x 10-6 C

E1 = kq/r^2 = ( 9 x 10^9)( 5 x 10^-6)/(0.5^2) = 180000 N/C to the left

E2 = kq/r^2 = ( 9 x 10^9)( 5 x 10^-6)/(0.25^2) = 720000 N/C to the right

E net = 720000 - 180000 = 540000 N/C to the right

F = qE

F = (-5 x 10^6 C)(540000 N/C) = - 2.7 N

The force on q2 is 2.7 N to the left.

The net electrostatic force on the q2 is 2.7N owards left

The equation for electrostatic force is

        [tex]F= k\frac{q_{1}q_{2} }{r^{2} }[/tex]

where k = [tex]9*10^{9} Nm^{2}/C^{2}[/tex] and r is the distance separating charges q1 and q2.

the force has to be calculated on a charge q2 = -5.0 ×[tex]10^{-6}[/tex] C by the charges q1=  -5.0 ×[tex]10^{-6}[/tex] C and q3=  -5.0 ×[tex]10^{-6}[/tex] C

distance between q1 and q2 is 0.5 m = 5×[tex]10^{-1}[/tex]m

distance between q2 and q3 is 0.25 m = 25×[tex]10^{-2}[/tex]m

force due to charge q1

           [tex]F_{1}[/tex] = 9×[tex]10^{9}[/tex]×(-5)×(-5)×[tex]10^{-12}[/tex]/25×[tex]10^{-2}[/tex] N = +0.9N =  0.9N towards right

           [tex]F_{2}[/tex] = 9×[tex]10^{9}[/tex]×(-50)×(-4)×[tex]10^{-12}[/tex]/625×[tex]10^{-4}[/tex] N = -3.6N = 3.6N towards left

hence net force F = [tex]F_{1}+F_{2}[/tex]

                              = 0.9N - 3.6N = -2.7N

                           F = 2.7 N towards left

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Answer:

conduction (the heat is transferring to the air)

Answer:

6km

Explanation:

Answer:

its the third one

Answer:

Bitter Magnet inside a superconducting magnet

Explanation:

Since there are no options available, generally, the electromagnet that is considered the strongest is the Bitter Magnet inside a superconducting magnet.

This electromagnet produces 45 Tesla units which is a result of bitter magnet producing 33.5 Tesla and the superconducting coil produces the additional 11.5 Tesla.

Hence, justifying that the greater the current in the coil the stronger the electromagnet. ​