PHYSICS COLLEGE

A car traveling at 45 km/h starts to brake, and comes to a stop over a distance of 18 m. Calculate the accelerationof the braking car.

Answers

Answer 1

Answer:

Answer:

Acceleration, $a=8.68\ m/s^2$

Explanation:

Given that,

Initial speed of a car, u = 45 km/h = 12.5 m/s

Final speed, v = 0 (as they comes to rest)

Distance, d = 18 m

We need to find the acceleration of the breaking car. Using third equation of motion as follows :

$v^2-u^2=2ad\n\n\text{Where a is acceleration of the car}\n\na=(v^2-u^2)/(d)\n\na=((12.5)^2)/(18)\n\na=8.68\ m/s^2$

So, the acceleration of the braking car is $8.68\ m/s^2$ .

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A person is making homemade ice cream. She exerts a force of magnitude 26 N on the free end of the crank handle on the ice-cream maker, and this end moves on a circular path of radius 0.26 m. The force is always applied parallel to the motion of the handle. If the handle is turned once every 2.0 s, what is the average power being expended?
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A pilot performs an evasive maneuver by diving vertically at 270. If he can withstand an acceleration of 9.0 's without blacking out, at what altitude must he begin to pull out of the dive to avoid crashing into the sea? y= ?m

Answers

Answer:

The pilot must be began at an altitude of 826.53 m to avoid crash into the sea.

Explanation:

Given that,

Velocity = 270 m/s

Acceleration = 9.0g s

We need to calculate the altitude

Using formula of centripetal acceleration

$a_(c)=(v^2)/(r)$

$r=(v^2)/(a_(c))$

Where, v = velocity

r = altitude

a = acceleration

Put the value into the formula

$r=(270^2)/(9.0*9.8)$

$r=826.53\ m$

Hence, The pilot must be began at an altitude of 826.53 m to avoid crash into the sea.

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.55 m and the acceleration achieved during the time the jumper is extending their legs is 1.2 times the acceleration due to gravity, g .How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Answers

Answer:

1.32 m.

Explanation:

Below is an attachment containing the solution.

A rocket with a mass of 62,000 kg (including fuel) is burning fuel at the rate of 150 kg/s and the speed of the exhaust gases is 6,000 m/s. If the rocket is fired vertically upward from the surface of the Earth, determine its height after 744 kg of its total fuel load has been consumed. Since the mass of fuel consumed is small compared to the total mass of the rocket, you can consider the mass of the rocket to be constant for the time interval of interest.

Answers

Answer:

h≅ 58 m

Explanation:

GIVEN:

mass of rocket M= 62,000 kg

fuel consumption rate = 150 kg/s

velocity of exhaust gases v= 6000 m/s

Now thrust = rate of fuel consumption×velocity of exhaust gases

=6000 × 150 = 900000 N

now to need calculate time t = amount of fuel consumed÷ rate

= 744/150= 4.96 sec

applying newton's law

M×a= thrust - Mg

62000 a=900000- 62000×9.8

acceleration a= 4.71 m/s^2

its height after 744 kg of its total fuel load has been consumed

$h= (1)/(2)at^2$

$h= (1)/(2)4.71*4.96^2$

h= 58.012 m

h≅ 58 m

wo charged spheres are 1.5 m apart and are exerting an electrostatic force (Fo) on each other. If the charge on each sphere decreases by a factor of 9, determine (in terms of Fo) how much electrostatic force each sphere will exert on the other.

Answers

Answer:

F0 / 81

Explanation:

Let the two charges by Q and q which are separated by d.

By use of coulomb's law

F0 = k Q q / d^2 ......(1)

Now the charges are decreased by factor of 9.

Q' = Q / 9

q' = q / 9 ......(2)

Now the Force is

F' = k Q' q' / d^2

F' = k (Q /9) (q / 9) / d^2

F' = k Q q / 81d^2

F' = F0 / 81

A spectroscope breaks light up into its colors, allowing scientists to analyze light from the solar system and universe. By studying the spectral line patterns, widths, strengths and positions, scientists can determine the speed, position, and _____ of celestial bodies.A) age
B) origin
C) rotation
D) temperature

Answers

It is D - temperature

Final answer:

A spectroscope analyses light to determine various parameters of celestial bodies. The missing parameter in this context is the 'temperature' of the celestial body (option D). The spectral lines, based on their pattern and strengths helps in determining this.

Explanation:

A spectroscope decomposes or breaks white light into its spectrum of colors, allowing scientists to study them and understand various aspects of celestial bodies. When scientists analyze the spectral line patterns, widths, strengths, and positions, they can discern essential parameters. These parameters include the speed and position of the celestial body, and more importantly, the correct answer to your question, its temperature (option D). This is because every element when heated, absorbs or emits light at characteristic wavelengths, that give us the 'spectral lines'. By studying these we can determine the temperature of the celestial body.

Learn more about Spectroscope here:

brainly.com/question/31240398

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What is the wavelength of the photons emitted by hydrogen atoms when they undergo n =5 to n =3 transitions?

Answers

Answer:

$\lambda=1282nm$

Explanation:

The wavelength of the photons emitted due to an atomic electron transition in a hydrogen atom, is given by the Rydberg formula:

$(1)/(\lambda)=R_H((1)/(n_1^2)-(1)/(n_2^2)})$

Here $R_H$ is the Rydberg constant for hydrogen and $n_1,n_2$ are the lower and higher quantum number for the energy levels of the atomic electron transition, respectively. Replacing the given values and solving for $\lambda$

$(1)/(\lambda)=1.097*10^7m^(-1)((1)/(3^2)-(1)/(5^2)})\n(1)/(\lambda)=7.81*10^5m^(-1)\n\lambda=(1)/(7.81*10^5m^(-1))\n\lambda=1.282*10^(-6)m\n\lambda=1.282*10^(-6)m*(1nm)/(10^(-9)m)\n\lambda=1282nm$

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