A car driver spends 3hrs driving at an average speed of 80km/hr, stops for 30 minutes to
have some rest, and then drives at an average speed of 90km/hr for 2 hours.
Calculate the average speed during the whole journey​

Answer: 5.08 L.

Explation down below

Answer:

Ratio of kinetic energy of object A to B = 1:5

Explanation:

Given:

Mass of object A = 200 kg

Mass of object B = 1,000 kg

Find:

Ratio of kinetic energy of object A to B

Computation:

Kinetic energy = (1/2)(m)(v²)

Kinetic energy of object A = (1/2)(200)(v²)

Kinetic energy of object A = (100)(v²)

Kinetic energy of object B = (1/2)(1,000)(v²)

Kinetic energy of object B = (500)(v²)

Ratio of kinetic energy of object A to B = Kinetic energy of object A / Kinetic energy of object B

Ratio of kinetic energy of object A to B = (100)(v²) / (500)(v²)

Ratio of kinetic energy of object A to B = 100 / 500

Ratio of kinetic energy of object A to B = 1/5

Ratio of kinetic energy of object A to B = 1:5

Answer:

U=30cm

Explanation:

All you have to do is to put

Mirror formula , 1/f=1/u + 1/v

You should be careful in sign convention .

Virtual image is negative

we take focal length of convex lens negative even if its not given and so on...

No, because some scientist are bad and they must have a scientific attitude like honesty, curiosity, open-mindedness, optimistic and more. And they should follow others sometimes because for their own good.

Answer:

No

Explanation:

Science operates in the context of national and international laws, agreements and conventions. It requires scientists to conduct and communicate scientific work for the benefit of society, with excellence, integrity, respect, fairness, trustworthiness, clarity, and transparency.

Answer:

true it all changes

Explanation:

________________________

Answer:

The object is moving at constant speed.

Explanation:

The spaces between the dots are equal.

Answer:

B

Explanation:

The Lorentz force is the sum of the electric force and magnetic force.

F = qE + qvB

qE represents electric force and qvB represents magnetic force

Answer:

1)   P₁ = -2 D,   2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

          [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

          [tex]\frac{1}{f_1} = \frac{1}{q}[/tex]

           q = f₁

2) for an object located at p = 25 cm

            [tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}[/tex]

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

2)  

        [tex]\frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}[/tex]

        [tex]\frac{1}{f_2}[/tex] = 0.06

         f₂ = 16.67 cm

the expression for the power of the lenses is

          P = [tex]\frac{1}{f}[/tex]

where the focal length is in meters

1)       P₁ = 1/0.50

        P₁ = -2 D

2)     P₂ = 1 /0.16667

        P₂ = 6 D

Answer:

1/Re= 1/R1 + 1/R2

Explanation:

Two resistors are connected in parallel. If R1 and R2 represent the resistance in Ohms (Ω) of each resistor, then the total resistance R is given by  [tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex]. Thus, the rate of R changes when R₁ = 117 Ω and

R₂ = 112 Ω is 0.25 Ω/min

For a given resistor connected in parallel;

[tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex]

Making R from the left-hand side the subject of the formula, then:

[tex]\mathbf{R = \dfrac{R_1R_2}{R_1+R_2}}[/tex]

Given that:

Now, replacing the values in the above previous equation, we have:

[tex]\mathbf{R = \dfrac{13104}{229}}[/tex]

However, the differentiation of R with respect to time t will give us the rate at which R is changing when R1=117Ω and R2=112Ω.

So, by differentiating the given equation of the resistor in parallel with respect to time t;

[tex]\mathbf{\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}}[/tex],   we have:

[tex]\mathbf{\dfrac{1}{R^2}(\dfrac{dR}{dt})=\dfrac{1}{R_1^2}(\dfrac{dR_1}{dt})+\dfrac{1}{R_2^2}(\dfrac{dR_2}{dt})}[/tex]

[tex]\mathbf{(\dfrac{dR}{dt})=R^2 \Bigg[ \dfrac{1}{R_1^2}(\dfrac{dR_1}{dt})+\dfrac{1}{R_2^2}(\dfrac{dR_2}{dt})\Bigg]}[/tex]

[tex]\mathbf{\dfrac{dR}{dt}=(\dfrac{13104}{229})^2 \Bigg[ \dfrac{0.4}{117^2}+\dfrac{0.6}{112^2}\Bigg]}[/tex]

[tex]\mathbf{\dfrac{dR}{dt}=3274.44 \Bigg[ (7.7052 \times 10^{-5} )\Bigg]}[/tex]

[tex]\mathbf{\dfrac{dR}{dt}=0.25\ \Omega /min}[/tex]

Therefore, we can conclude that the rate at which R is changing R1=117Ω and R2=112Ω is 0.25 Ω/min

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Answer:

you may get bullied or teased for being a differrent race, ethnic.

Answer:

Workload

Explanation:

You may feel stressed about a heavier workload or more too do. You will also be paid as an employee unlike, a student.

Answer:

I think its d

Explanation:

I'm not sure I'm sorry if I'm wrong

Answer:

B) It appears to be SHM (simple harmonic motion) which is also the projection of an object moving in a circular path on either the x or y axes.

Answer:

Both sound and waves

Explanation:

Thank me later

Answer:

2

Explanation:

To catch a 14.715N ball traveling at a velocity of 37.5m/s, required mechanical work is 1056.10 joule.

Physics' definition of work makes clear how it is related to energy: anytime work is performed, energy is transferred.

In a scientific sense, a work requires the application of a force and a displacement in the force's direction. Given this, we can state that

Work is the product of the component of the force acting in the displacement's direction and its magnitude.

Weight of the ball = 14.715 N.

Mass of the ball = 14.715 N ÷ (9.8 m/s²) = 1.502 kg.

Velocity of the ball = 37.5 m/s

Kinetic energy of the ball = 1/2 × 1.502 × 37.5² Joule = 1056.10 Joule.

Hence, to catch a 14.715N ball traveling at a velocity of 37.5m/s, required work is 1056.10 joule.

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Answer:

0.1 kg/s.

Explanation:

The density of water, d = 1000 kg/m³

Volume, V = 2 L

Time, t = 20 s

We need to find the mass flow rate from the faucet. We know that the density of an object is given by :

[tex]d=\dfrac{m}{V}\\\\m=d\times V\\\\\dfrac{m}{t}=\dfrac{dV}{t}\\\\\dfrac{m}{t}=\dfrac{1000\times 0.002}{20}\\\\=0.1\ kg/s[/tex]

So, the mass flow rate is equal to 0.1 kg/s.

Answer:

P = 4.5 watts

Explanation:

Given that,

EMF of the circuit, E = 3 volt

The resistance  of the resistors, R = 2 ohms

We need to find the power of this circuit. The relation between power, emf and resistance is given by the formula as follows :

[tex]P=\dfrac{V^2}{R}[/tex]

Substitute all the values,

[tex]P=\dfrac{3^2}{2}\\\\P=4.5\ W[/tex]

So, the power of this circuit is equal to 4.5 watts.

Answer:

[tex]0.886[/tex] N buoyant force is exerted by air

Explanation:

My weight is [tex]75[/tex] Kg

Weight = mass * gravity

As we know

Buoyant Force is equal to the product of density * acceleration due to gravity and volume of the body

Assuming weight density is a bit less than that of water or equal to water i.e [tex]997.77[/tex] kg/m3

Volume is equal to mass / density

[tex]= 75[/tex] Kg * g/[tex]997.777[/tex]

[tex]= 0.0751[/tex] * g

Buoyant Force

= Volume * g * density

[tex]= 0.0751 * 9.8 * 1.2041[/tex]kg/m3

[tex]= 0.886[/tex] N

Answer:

the answer is 86400seconds.

Explanation:

1day= 24 hours

24hours to seconds =

24×60×60= 86400 seconds.

( 60 second = 1 minutes)

( 60 minutes = 1 hour)

Answer: 0.516 ft/s

Explanation:

Given

Length of ladder L=20 ft

The speed at which the ladder moving away is v=2 ft/s

after 1 sec, the ladder is 5 ft away from the wall

So, the other end of the ladder is at

[tex]\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft[/tex]

Also, at any instant t

[tex]\Rightarrow l^2=x^2+y^2[/tex]

differentiate w.r.t.

[tex]\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s[/tex]

Answer:

The answer is "15.56 cm".

Explanation:

[tex]v= 2.5 \ cm\\\\f= 2.154 \ cm[/tex]

Calculating object of length is x so:

[tex]u= -x[/tex]

Using formula:

[tex]\to \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\\\\to \frac{1}{2.5}-\frac{1}{-x}=\frac{1}{2.154}\\\\\to \frac{1}{x}=\frac{1}{2.154}-\frac{1}{2.5}\\\\\to \frac{1}{2.5}-\frac{1}{-x}=\frac{1}{2.154}\\\\\to x= 15.56 \ cm[/tex]

Answer:

[tex]4.25\ \text{m/s}[/tex]

[tex]3391.22\ \text{N}[/tex]

Explanation:

y = Height of compression = 0.38 m

m = Mass of basketball player = 101 kg

h = Height of center of gravity after jump = 0.92 m

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

Energy balance of the system is given by

[tex]mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.92}\\\Rightarrow v=4.25\ \text{m/s}[/tex]

The velocity of the player when he leaves the floor is [tex]4.25\ \text{m/s}[/tex]

[tex]Fy=mgy+\dfrac{1}{2}mv^2\\\Rightarrow F=\dfrac{mgy+\dfrac{1}{2}mv^2}{y}\\\Rightarrow F=\dfrac{101\times 9.81\times 0.38+\dfrac{1}{2}\times 101\times 4.25^2}{0.38}\\\Rightarrow F=3391.22\ \text{N}[/tex]

The force exerted on the floor is [tex]3391.22\ \text{N}[/tex].

Answer:

Some plains form as ice and water erodes, or wears away, the dirt and rock on higher land. Water and ice carry the bits of dirt, rock, and other material, called sediment, down hillsides to be deposited elsewhere. As layer upon layer of this sediment is laid down, plains form. Volcanic activity can also form plains.

Answer:

[tex]23^{\circ}[/tex]

Explanation:

n = Refractive index of air = 1

[tex]n_1[/tex] = Refractive index of contact lens = 1.6

[tex]n_2[/tex] = Refractive index of cornea = 1.4

[tex]n_3[/tex] = Refractive index of fluid = 1.3

From Snell's law

[tex]n\sin30^{\circ}=n_1\sin\theta\\\Rightarrow \theta=\sin^{-1}\dfrac{1\sin30^{\circ}}{1.6}\\\Rightarrow \theta=18.21^{\circ}[/tex]

[tex]n_1\sin\theta=n_2\sin\theta_1\\\Rightarrow \theta_{1}=\sin^{-1}\dfrac{1.6\times \sin18.21^{\circ}}{1.4}\\\Rightarrow \theta_1=20.92^{\circ}[/tex]

[tex]n_2\sin\theta_1=n_3\sin\theta_3\\\Rightarrow \theta_3=\sin^{-1}\dfrac{1.4\sin20.92^{\circ}}{1.3}\\\Rightarrow \theta_3=22.62^{\circ}\approx 23^{\circ}[/tex]

The angle is the light traveling in the fluid behind her cornea is [tex]23^{\circ}[/tex].

The angle is the light traveling in the fluid will be 23⁰. Light is traveling in a particular direction with an angle.

"The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given color and for a given set of media,

The given data in the problem is;

n is the refractive index of air = 1

n₁ is the refractive index of contact lens = 1.6

n₂ is the refractive index of cornea = 1.4

n₃ is the refractive index of fluid = 1.3

According to Snell's law. The formula for Snell's law is

[tex]\rm n sin30^0 = n_1 sin\theta \\\\ \theta = sin^{- 1}(\frac{1sin30^0}{1.6} )\\\\ \theta = 18.21 ^0[/tex]

For contact lenses;

[tex]\rm n_1sin\theta = n_2 sin\theta_1 \\\\ \theta_1 = sin^{-1}\frac{1.6 \times sin 18.21^0}{1.4} \\\\ \theta_1 =20.92 ^0[/tex]

For fluid;

[tex]n_2 sin\theta_1 = n_2 sin \theta_3\\\\ \theta_3 = sin^{-1}\frac{1.4 sin 20.92^0}{1.3} \\\\ \theta_3 = 22.62 ^ 0 =23^0[/tex]

Hence the angle is the light traveling in the fluid will be 23⁰.

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