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2 years ago

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The table below shows the measurements you took in an experiment. Trial Length ( miles) 1.9 4.2 N 3 5.9 4 What is the longest me

asurement you made in units 1mi = 1.61 km . O A. 5.2 km B. 8.7 km C. 3.7 km D. 9.5 km

1 answer:

spin [16.1K]2 years ago

7 0

Answer:

9.5km

Explanation: just got it right

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A light string is wrapped around the edge of the smaller disk, and a 1.50 kg block is suspended from the free end of the string.

LenaWriter [7]

Answer: 7.41 m/s

Explanation: By using the law of of energy, kinetic energy of the brick as it falls equals the potential energy before falling.

Kinetic energy = mv²/2, potential energy = mgh

mv²/2 = mgh

v²/2 = gh

v² = 2gh

v = √2gh

Where g = 9.8 m/s², h = 2.80m

v = √2×9.8×2.8 = 7.41 m/s

7 0

3 years ago

In a carrom game, a striker weighs three times the mass of the other pieces, the carrom men and the queen, which each have a mas

Mila [183]

Answer:

- The final velocity of the queen is (3/2) of the initial velocity of the striker. That is, (3V/2)

- The final velocity of the striker is (1/2) of the initial velocity of the striker. That is, (V/2)

Hence, the relative velocity of the queen with respect to the striker after collision

= (3V/2) - (V/2)

= V m/s.

Explanation:

This is a conservation of Momentum problem.

Momentum before collision = Momentum after collision.

The mass of the striker = M

Initial Velocity of the striker = V (+x-axis)

Let the final velocity of the striker be u

Mass of the queen = (M/3)

Initial velocity of the queen = 0 (since the queen was initially at rest)

Final velocity of the queen be v

Collision is elastic, So, momentum and kinetic energy are conserved.

Momentum before collision = (M)(V) + 0 = (MV) kgm/s

Momentum after collision = (M)(u) + (M/3)(v) = Mu + (Mv/3)

Momentum before collision = Momentum after collision.

MV = Mu + (Mv/3)

V = u + (v/3)

u = V - (v/3) (eqn 1)

Kinetic energy balance

Kinetic energy before collision = (1/2)(M)(V²) = (MV²/2)

Kinetic energy after collision = (1/2)(M)(u²) + (1/2)(M/3)(v²) = (Mu²/2) + (Mv²/6)

Kinetic energy before collision = Kinetic energy after collision

(MV²/2) = (Mu²/2) + (Mv²/6)

V² = u² + (v²/3) (eqn 2)

Recall eqn 1, u = V - (v/3); eqn 2 becomes

V² = [V - (v/3)]² + (v²/3)

V² = V² - (2Vv/3) + (v²/9) + (v²/3)

(4v²/9) = (2Vv/3)

v² = (2Vv/3) × (9/4)

v² = (3Vv/2)

v = (3V/2)

Hence, the final velocity of the queen is (3/2) of the initial velocity of the striker and is in the same direction.

The final velocity of the striker after collision

= u = V - (v/3) = V - (V/2) = (V/2)

The relative velocity of the queen withrespect to the striker after collision

= (velocity of queen after collision) - (velocity of striker after collision)

= v - u

= (3V/2) - (V/2) = V m/s.

Hope this Helps!!!!

3 0

3 years ago

Read 2 more answers

The ratio of output power to input power, in percent, is called

kow [346]

That ratio is called"efficiency". It doesn't need to be a percent.
It can just as well be a fraction or a decimal number.

3 0

3 years ago

Explain the importance of measurement<br>Physics.<br><br>

KiRa [710]

Answer:

Measurements are an important part of comparing things, as they provide the basis on comparing objects to other objects. Measurements allow us to recognize three hours and see how it's shorter than five hours, without having to observe the hours passing by themselves.

7 0

3 years ago

Which of the following metals require ultraviolet light to exhibit the photoelectric effect?The options available: a. Cs, work f

Eddi Din [679]

Answer:

b. AG, work function=4.74eV

Explanation:

Ultraviolet light starts at the end of the visible light spectrum, where violet light ends:

$\lambda=380 nm =3.8\cdot 10^{-7}m$ (wavelength of lowest-energy ultraviolet light)

So, the lowest energy of ultraviolet light can be found by using the formula

$E=\frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

Substituting,

$E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.8\cdot 10^{-7} m}=5.23\cdot 10^{-19}J$

And keeping in mind that

$1 eV = 1.6\cdot 10^{-19}J$

This energy converted into electronvolts is

$E=\frac{5.23\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=3.27 eV$

The work function of a metal is the minimum energy needed to extract a photoelectron from the surface of the metal. Therefore, the metals that exhibit photoelectric effect are the ones whose work function is larger than the energy we found previously, so:

b. AG, work function=4.74eV

Because for all the other metals, visible light will be enough to extract photoelectrons.

7 0

3 years ago