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2 years ago

What is the kinetic energy of a 1.40 kg discus with a speed of 22.5 m/s?

1 answer:

8 0

Kinetic energy = (1/2) (mass) (speed)²

   = (1/2) (1.4 kg) (22.5 m/s)²

   = (0.7 kg) (506.25 m²/s² )

   = 354.375 kg-m²/s² = 354.375 joules .

This is just the kinetic energy associated with a 1.4-kg glob of
mass sailing through space at 22.5 m/s. In the case of a frisbee,
it's also spinning, and there's some additional kinetic energy stored
in the spin.

You might be interested in

The maximum wavelength For photoelectric emissions in tungsten is 230 nm. What wavelength of light must be use in order for elec

notka56 [123]

Answer:

λ = 1.8 x 10⁻⁷ m = 180 nm

Explanation:

First we find the work function of tungsten by using the following formula:

∅ = hc/λmax

where,

∅ = work function = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λmax = maximum wavelength for photoelectric emission = 230 nm

λmax = 2.3 x 10⁻⁷ m

Therefore,

∅ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.3 x 10⁻⁷ m)

∅ = 8.64 x 10⁻¹⁹ J

Now we convert Kinetic Energy of electron into Joules:

K.E = (1.5 eV)(1.6 x 10⁻¹⁹ J/1 eV)

K.E = 2.4 x 10⁻¹⁹ J

Now, we use Einstein's Photoelectric Equation:

Energy of Photon = ∅ + K.E

Therefore,

Energy of Photon = 8.64 x 10⁻¹⁹ J + 2.4 x 10⁻¹⁹ J

Energy of Photon = 11.04 x 10⁻¹⁹ J

but,

Energy of Photon = hc/λ

where,

λ = wavelength of light = ?

Therefore,

11.04 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ

λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(11.04 x 10⁻¹⁹ J)

5 0

2 years ago

An initially motionless test car is accelerated to 115 km/h in 8.58 s before striking a simulated deer. The car is in contact wi

hoa [83]

Answer:

a) a = 3.72 m / s², b) a = -18.75 m / s²

Explanation:

a) Let's use kinematics to find the acceleration before the collision

v = v₀ + at

as part of rest the v₀ = 0

a = v / t

Let's reduce the magnitudes to the SI system

v = 115 km / h (1000 m / 1km) (1h / 3600s)

v = 31.94 m / s

v₂ = 60 km / h = 16.66 m / s

et's calculate

a = 31.94 / 8.58

a = 3.72 m / s²

b) For the operational average during the collision let's use the relationship between momentum and momentum

I = Δp

F Δt = m v_f - m v₀

F = $\frac{m ( v_f - v_o)}{t}$

F = m [16.66 - 31.94] / 0.815

F = m (-18.75)

Having the force let's use Newton's second law

F = m a

-18.75 m = m a

a = -18.75 m / s²

4 0

2 years ago

Two horizontal forces,225N and 165N,are exerted on s canoe in the same direction. Find the net horizontal foce

Readme [11.4K]

Since they are in the same direction, you would add them together. Let’s also assume said direction is positive. 225 N + 165 N = 390 N

5 0

3 years ago

Which of the following is a potential result of preparing appropriately before starting an experiment in a lab?

oksano4ka [1.4K]

Answer:

Your project goes well.

Explanation:

Because that's how it works.

6 0

2 years ago

Which of the following is the smallest conceivable amount of time that could pass between a lunar eclipse and a solar eclipse .

pickupchik [31]

The time required for a moon to orbit around the earth is about 27-28 days

In order for lunar eclipse to occur the line that should be formed is:
Sun-Earth-Moon
because earth is making shade on moon

in order for solar eclipse to occur the line is now:
Sun-Moon-Earth
because moon is making a shade on earth (blocking sun = solar eclipse)

Therefore moon needs to make half of its orbit to go from behind the earth to in front of the earth.

28/2 = 14

Answer is 14

5 0

2 years ago