While investigating Kirchhoff's Laws, you begin observing a blackbody, such as a star, from Earth using advanced technology that can analyze spectra. While pointing it at the star with nothing between you and the star, you observe a full spectrum. You come back and repeat this same experiment a year later using the same star, except this time you observe an absorption spectrum. What is the most likely explanation for this

Static friction (magnitude Fs) keeps the car on the road, and is the only force acting on it parallel to the road. By Newton's second law,

Fs = m a = W a / g

(a = centripetal acceleration, m = mass, g = acceleration due to gravity)

We have

a = v ² / R

(v = tangential speed, R = radius of the curve)

so that

Fs = W v ² / (g R)

Solving for v gives

v = √(Fs g R / W)

Perpendicular to the road, the car is in equilibrium, so Newton's second law gives

N - W = 0

(N = normal force, W = weight)

so that

N = W

We're given a coefficient of static friction µ = 0.4, so

Fs = µ N = 0.4 W

Substitute this into the equation for v. The factors of W cancel, so we get

v = √((0.4 W) g R / W) = √(0.4 g R) = √(0.4 (9.80 m/s²) (105 m)) ≈ 20.3 m/s

An arrow pointing from the bottom of the ramp to the top, I assume it would be friction.

Answer:

[tex]Vm=0.894m/s[/tex]

Explanation:

From the question we are told that

Velocity if travel [tex]v=4m/s[/tex]

Diameter of  prototype [tex]d_1=20m[/tex] and [tex]d_2=110m[/tex]

Scale ratio=[tex]\frac{1}{20}[/tex]

Generally Velocity of of the model using Froud's model is mathematically given as

[tex]Fm=Fp[/tex]

[tex]\frac{Vm}{\sqrt{Lmg}} =\frac{Vp}{\sqrt{Lpg}}[/tex]

[tex]Vm=Vp*\frac{Vp}{\sqrt{Lpg} }[/tex]

[tex]Vm=4*\frac{1}{\sqrt{20}}[/tex]

[tex]Vm=0.894m/s[/tex]

Answer:

Force = 2240 Newton.

Explanation:

Given the following data;

Mass A= 65kg

Mass B = 215kg

Acceleration = 8m/s²

To find the force;

Force is given by the multiplication of mass and acceleration.

Mathematically, Force is;

[tex] F = ma[/tex]

Where;

F represents force.

m represents the mass of an object.

a represents acceleration.

First of all, we would have to find the total mass.

Total mass = Mass A + Mass B

Total mass = 65 + 215

Total mass = 280kg

Substituting into the equation, we have

[tex] Force = 280 * 8 [/tex]

Force = 2240 Newton.

Answer:

Explanation:

equation of wave is given by the following equation

y = (2.6 cm) sin[1.8 - (5.8 s-1)t].

Comparing it with standard form of wave

y = A sin ( ωt - kx )

we get

ω = 5.8

2πn = 5.8

n = .92 per second

kx = 1.8

k x 6 = 1.8

k = 0.3

[tex]\frac{2\pi}{\lambda}[/tex] = 0.3

λ = 20.9 cm

Answer:

it will option option A hope it helps

Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ([tex]\dot Q[/tex]), measured in watts, that is, joules per second, by the following formula:

[tex]\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}[/tex] (1)

Where:

[tex]m[/tex] - Mass of the sphere, measured in kilograms.

[tex]c[/tex] - Specific heat of the material, measured in joules per kilogram-degree Celsius.

[tex]T_{o}[/tex], [tex]T_{f}[/tex] - Initial and final temperatures of the sphere, measured in degrees Celsius.

[tex]\Delta t[/tex] - Time, measured in seconds.

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

[tex]\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}[/tex] (2)

Where:

[tex]m_{I}[/tex], [tex]m_{X}[/tex] - Masses of the iron and unknown spheres, measured in kilograms.

[tex]\Delta t_{I}[/tex], [tex]\Delta t_{X}[/tex] - Times of the iron and unknown spheres, measured in seconds.

[tex]c_{I}[/tex], [tex]c_{X}[/tex] - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

[tex]c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}[/tex]

If we know that [tex]\Delta t_{I} = 6.35\,s[/tex], [tex]\Delta t_{X} = 4.59\,s[/tex], [tex]m_{I} = 0.515\,kg[/tex], [tex]m_{X} = 1.263\,kg[/tex] and [tex]c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}[/tex], then the specific heat of the unknown material is:

[tex]c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)[/tex]

[tex]c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}[/tex]

Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Answer:

LAW 1 :  For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation.  

---------------------------------------------

LAW 2: For a given metal, there exists a certain frequency below which the photoelectric emission does not take place. This frequency is called threshold frequency.

-----------------------------------------------

LAW 3: For a frequency greater than the threshold frequency, the kinetic energy of photoelectrons is dependent upon frequency or wavelength but not on the intensity of light.

-----------------------------------------------

LAW 4: Photoelectric emission is an instantaneous process. The time lag between incidence of radiations and emission of electron is 10^-9 seconds.

Explanation:

Answer:

LAW 1 : For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation. ... LAW 4: Photoelectric emission is an instantaneous process.

Answer:

6N

Explanation:

Given parameters:

Mass of object  = 6kg

Initial velocity  = 5m/s

Final velocity  = 25m/s

Time  = 30s

Unknown:

Net force acting on the object  = ?

Solution:

From Newton's second law of motion:

   Force  = mass x acceleration

Acceleration is the rate of change of velocity with time

  Acceleration  = [tex]\frac{Final velocity - Initial velocity }{time}[/tex]  

  Force  = mass x  [tex]\frac{Final velocity - Initial velocity }{time}[/tex]  

So;

 Force  = 6 x [tex]\frac{25 - 5}{30}[/tex]    = 6N

Answer:

B

Explanation:

Answer:

The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.

Explanation:

Let suppose that RC car-slingshot-steelpellet is a conservative system, that is, that non-conservative forces (i.e. friction, air viscosity) can be neglected. The velocity of the steel pellet can be found by means of the Principle of Energy Conservation and under the consideration that change in gravitational potential energy is negligible and that the RC car travels at constant velocity:

[tex]\frac{1}{2}\cdot (m_{C}+m_{P})\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{C}\cdot v_{o}^{2} + \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]

[tex]\frac{1}{2}\cdot m_{P}\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]

[tex]m_{P}\cdot v_{o}^{2} + k\cdot x^{2} = m_{P}\cdot v^{2}[/tex]

[tex]v^{2} = v_{o}^{2} + \frac{k}{m_{P}}\cdot x^{2}[/tex]

[tex]v = \sqrt{v_{o}^{2}+\frac{k}{m_{P}}\cdot x^{2} }[/tex] (1)

Where:

[tex]v_{o}[/tex] - Initial velocity of the steel pellet, measured in meters per second.

[tex]v[/tex] - Final velocity of the steel pellet, measured in meters per second.

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]m_{P}[/tex] - Mass of the steel pellet, measured in kilograms.

[tex]m_{C}[/tex] - Mass of the RC car, measured in kilograms.

[tex]x[/tex] - Initial deformation of the spring, measured in meters.

If we know that [tex]v_{o} = 5.6\,\frac{m}{s}[/tex], [tex]k = 85\,\frac{N}{m}[/tex], [tex]m_{P} = 0.025\,kg[/tex] and [tex]x = 0.035\,m[/tex], then the velocity of the steel pellet relative to the ground when it leaves the sling shot is:

[tex]v = \sqrt{\left(5.6\,\frac{m}{s} \right)^{2}+\frac{\left(85\,\frac{N}{m} \right)\cdot (0.035\,m)^{2}}{0.025\,kg} }[/tex]

[tex]v \approx 5.960\,\frac{m}{s}[/tex]

The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.

Answer:

15km/h

Explanation:

→ Speed = Distance ÷ Time

30 ÷ 2 = 15km/h

Answer:

If the force goes in the direction of movement, the speed must increase and if the net force goes in the opposite direction, the speed must decrease.

Explanation:

The principle of inertia or Newton's first law states that every body remains static or with constant velocity if there is no net force acting on it.

Based on this principle, if we have a net force, the velocity of the body changes by having an unbalanced force acting.

If the force goes in the direction of movement, the speed must increase and if the net force goes in the opposite direction, the speed must decrease.

Answer:

9.3m/s

Explanation:

Based on the law of conservation of momentum

Sum of momentum before collision = sum of momentum after collision

m1u1 +m2u2 = m1v1+m2v2

m1 = 8kg

u1 = 15.4m/s

m2 = 10kg

u2 = 0m/s(at rest)

v1 = 3.9m/s

Required

v2.

Substitute

8(15.4)+10(0) = 8(3.9)+10v2

123.2=31.2+10v2

123.2-31.2 = 10v2

92 = 10v2

v2 = 92/10

v2 = 9.2m/s

Hence the velocity of the 10.0 kg object after the collision is 9.2m/s

Solution :

1. Rayleigh scattering takes place when the particle size is smaller than the wavelength (λ).

2. Mie scattering takes place when particle size is nearly equal to the wavelength (λ).

3. Non-selective scatter takes place when particle size in greater than the wavelength  (λ).

We have the sizes of different particles :

[tex]$O_2 \rightarrow 10^{10} \ m $[/tex]

Smoke particles [tex]$\rightarrow 3 \times 10^{-7} \ m$[/tex]

Cloud droplets [tex]$\rightarrow 2 \times 10^{-5} \ m$[/tex]

Rain droplets [tex]$\rightarrow 3 \times 10^{-3} \ m$[/tex]

Wavelength           [tex]$ O_2 $[/tex]         Smoke particles    Cloud droplets     Rain droplets

                            [tex]$10^{-10} \ m$[/tex]        [tex]$ 3 \times 10^{-7} \ m$[/tex]           [tex]$ 2 \times 10^{-5} \ m$[/tex]              [tex]$ 3 \times 10^{-3} \ m$[/tex]

[tex]$5500 \times 10^{-4} \ m$[/tex]      Rayleigh  Non-selective      Non-selective     Non-selective

[tex]$11 \times 10^{-6} \ m $[/tex]         Rayleigh    Rayleigh            Non-selective      Non-selective

[tex]$1600 \times 10^{-10} \ m $[/tex]    Rayleigh  Non-selective      Non-selective     Non-selective

[tex]$10^{-2} \ m $[/tex]                 Rayleigh      Rayleigh               Rayleigh          Mie

Answer:

C. Is traveling on a curve in the road

    Hope this helps :3

The pick up truck has a changing velocity because, it is travelling on a curve in the road. A change in direction results in its change in velocity because, velocity is a vector quantity.

Velocity is a physical quantity that measures the distance covered by an object per unit time. It is a vector quantity, thus having magnitude as well direction.

The rate of change in velocity is called acceleration of the object. Like velocity, acceleration also is a vector quantity. Thus, a change in magnitude or direction or change in both for velocity make the object to accelerate.

Here, all the three vehicles  are travelling with the same velocity. But, the truck is moving to a curve on the road. The curvature in the path will make a change in its velocity.

Find more on velocity:

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The image related with this question is attached below:

Answer:

A. 72000 C

B. 1100 W

C. 26.4 cents.

Explanation:

From the question given above, the following data were obtained:

Current (I) = 10 A

Voltage (V) = 110 V

Time (t) = 2 h

A. Determination of the charge.

We'll begin by converting 2 h to seconds. This can be obtained as follow:

1 h = 3600 s

Therefore,

2 h = 2 h × 3600 s / 1 h

2 h = 7200 s

Thus, 2 h is equivalent to 7200 s.

Finally, we shall determine the charge. This can be obtained as follow:

Current (I) = 10 A

Time (t) = 7200 s

Charge (Q) =?

Q = It

Q = 10 × 7200

Q = 72000 C

B. Determination of the power.

Current (I) = 10 A

Voltage (V) = 110 V

Power (P) =?

P = IV

P = 10 × 110

P = 1100 W

C. Determination of the cost of operation.

We'll begin by converting 1100 W to KW. This can be obtained as follow:

1000 W = 1 KW

Therefore,

1100 W = 1100 W × 1 KW / 1000 W

1100 W = 1.1 KW

Thus, 1100 W is equivalent to 1.1 KW

Next, we shall determine the energy consumption of the range. This can be obtained as follow:

Power (P) = 1.1 KW

Time (t) = 2 h

Energy (E) =?

E = Pt

E = 1.1 × 2

E = 2.2 KWh

Finally, we shall determine the cost of operation. This can be obtained as follow:

1 KWh cost 12 cents.

Therefore, 2.2 KWh will cost = 2.2 × 12

= 26.4 cents.

Thus, the cost of operating the range for 2 h is 26.4 cents.

Answer:

1/9

Explanation:

Let A denote the bigger piece and let B denote the smaller piece.

We are told that one with three times the mass of the other.

Therefore, we have;

M_a = 3M_b

Firecracker is placed in the block and it explodes and thus, momentum is conserved.

Thus;

V_ai = V_bi = 0

Where V_ai is initial velocity of piece A and V_bi is initial velocity of piece B.

Since initial momentum equals final momentum, we have;

P_i = P_f

Thus;

0 = (M_a × V_af) + (M_b × V_bf)

Since M_a = 3M_b, we have;

(3M_b × V_af) + (M_b × Vbf) = 0

Making V_af the subject, we have;

V_af = -⅓V_bf

The kinetic energy gained by each block during the explosion will later be lost due to the negative work done by friction. Thus;

W_f = -½M_b•(v_bf)²

Now, let's express the work is in terms of the force and the distance.

Thus;

W_f = F_f × Δx × cos 180°

Frictional force is also expressed as μmg

Thus;

W_f = -μM_b × g × Δx

Earlier, we saw that;

W_f = -½M_b•(v_bf)²

Thus;

-½M_b•(v_bf)²= -μM_b × g × Δx

Δx = (v_bf)²/2μg

Let the distance travelled by block A be Δx_a and that travelled by B be Δx_b

Thus;

Δx_a/Δx_b = ((v_ba)²/2μg)/((v_bf)²/2μg)

Δx_a/Δx_b = ((v_af)²/((v_bf)²)

Δx_a/Δx_b = (-⅓V_bf)²/(V_bf)²

Δx_a/Δx_b = 1/9

Answer:I think it’s self monitoring sorry if wrong

Explanation:

Answer:

It self monitoring

Explanation:

I took the test

Answer:

average speed is 1 m/s and average velocity is 0 m/s.

Explanation:

Given that :

Length of round trip = 50 m

Time taken = 100 seconds

The average speed :

Total distance / total time taken

Length of complete round trip :

(50 + 50) m, total. Distance = 100 m

100 / 100 = 1m/s

The average velocity :

Total Displacement / total time taken

Total Displacement of round trip = end point - start point = 0

0 / 100 = 0

Average speed is 1 m/s and average velocity is 0 m/s.

The average speed is defined as the ratio of distance to time. Speed is a scalar quantity hence it does not take direction into account while velocity is a vector quantity hence it takes direction into account.

The speed is obtained from;

Speed = Distance/time = 2(50 m)/100 s = 1 m/s.

The velocity is 0 m/s since it is complete round-trip lap.

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Answer:

force = mass * acceleration

therefore the SI unit is kg*m/s2 or newton's

it's a vector quantity because it has both direction(acceleration) and size (mass)

Answer:

It's true I took the test on Edge.

Explanation:

Answer:

True

Explanation:

Got it right on edg

Distance = (speed) x (time)

Car A: Distance = (8 m/s) x (43 s)  =  344 meters

Car B: Distance = (7 m/s) x (50 s)  =  350 meters

350 meters is a longer distance than 344 meters.

Car-B traveled a longer distance than Car-A did.

Answer:

[tex]\boxed {\boxed {\sf Car \ B : 350 \ meters }}[/tex]

Explanation:

Distance is equal to the product of speed and time.

[tex]d=s*t[/tex]

1. Car A

Car A has a speed of 8 meters per second and travels for 43 seconds.

[tex]s= 8 \ m/s \\t= 43 \ s[/tex]

Substitute the values into the formula.

[tex]d= 8 \ m/s *43 \ s[/tex]

Multiply and note that the seconds will cancel out.

[tex]d= 8 \ m*43= 344 \ m[/tex]

2. Car B

Car B has a speed of 7 meters per second and travels for 50 seconds.

[tex]s= 7 \ m/s \\t= 50 \ s[/tex]

Substitute the values in and multiply.

[tex]d= 7 \ m/s * 50 \ s[/tex]

[tex]d= 7 \ m * 50 = 350 \ m[/tex]

350 meters is a longer distance than 344 meters, so Car B traveled the longer distance.

Answer:

58.8J

Explanation:

Given parameters;

Mass of ball  = 4kg

Height above the floor  = 1.5m

g  = 9.8n/kg

Unknown:

Potential energy  = ?

Solution:

The potential energy of a body is the energy due to the position of the body.

It is mathematically expressed as:

  Potential energy = mass x acceleration due to gravity x height

  Potential energy  = 4 x 9.8 x 1.5  = 58.8J

Answer:

T = 10010 N

Explanation:

To solve this problem we must use the translational equilibrium relation, let's set a reference frame

X axis

       Fₓ-Fₓ = 0

       Fₓ = Fₓ

whereby the horizontal components of the tension in the cable cancel

Y Axis  

        [tex]F_{y} + F_{y} - W =0[/tex]

        2[tex]F_{y}[/tex] = W

let's use trigonometry to find the angles

        tan θ = y / x

        θ = tan⁻¹ (0.30 / 0.50 L)

        θ = tan⁻¹ (0.30 / 0.50 15)

        θ = 2.29º

the components of stress are

         F_{y} = T sin θ

we substitute

       2 T sin θ = W

       T = W / 2sin θ

        T = [tex]\frac{ 800}{ 2sin 2.29}[/tex]

        T = 10010 N

Answer:

form 1 question??????????

Answer:

A

Explanation:

Ke = 1/2 MV^2

Answer:

the charge on the object is 71.043×10^-20 C and the number of electron is 4.44

Explanation:

from coulumbs law, The force that is acting over both charge can be computed as

F=( kq1q2)/r^2..............eqn(1)

Where

F=electrostatic force= 3.89 x 10-21 N

k= column constant= 9 x 10^9 Nm^2/C^2

q1 and q2= magnitude of the charges

r= distance between the charges= 1.08 x 10-3 m.

Since both charges are experiencing the same force, eqn(1) can be written as

F=( kq^2)/r^2.

We can make q subject of the formula

q= √(Fr^2)/k

= √[(3.89 x 10^-21× (1.08 x 10^-3)^2]/8.99 x 10^9

q= 71.043×10^-20 C

Hence, the charge is 71.043×10^-20 C

From quantization law, the number of electron can be computed as

N=q/e

Where

N= number of electron

q= charges

=1.6×10^-19C

N=71.043×10^-20/1.6×10^-19

=4.44

Hence, the charge on the object is 71.043×10^-20 C and the number of electron is 4.44

Answer:

49 Ns

Explanation:

Given data

Force= 14N

time = 3.5seconds

Applying the expression for impulse

P= Ft

substitute

P=14*3.5

P=49 Ns

Hence the impulse is 49 Ns