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4 years ago

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A brick is released with no initial speed from the roof of a building and strikes the ground in 1.80 s , encountering no appreci

able air drag. how tall, in meters, is the building?

1 answer:

Mars2501 [29]4 years ago

8 0

<span>Kinematics is used in this problem. The mass does not matter here because the question is mass independent. vi = 0 vf = x d = ? d = vi + 1/2 a t^2 d = 0 + 1/2 (9.8) (1.8)^2 d = 15.9 m (counting sig figs)</span>

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A recent study found that electrons that have energies between 3.45 eV and 19.9 eV can cause breaks in a DNA molecule even thoug

vlada-n [284]

Answer:

The Minimum wavelength is $\lambda_{min}= 382.2nm$

The Maximum wavelength is $\lambda_{max}= 624.2nm$

Explanation:

From the question we are told that

The energy range is $E_r = 3.25eV \ and \ 19.9eV$

Considering $E = 19.9eV$

When a single photon is transferred to to an electron the energy obtained can be calculated as follows

$E = 19.9eV = 19.9 *1.6 *10^{-19}J$

This energy is mathematically represented as

$E = \frac{hc}{\lambda_{max}}$

Here h is the Planck's constant with value of $h= 6.625*10^{-34}J\cdot s$

c is the speed of light with value of $c = 3*10^8 m/s$

Substituting values and making $\lambda$ the subject of the formula

$\lambda_{max} = \frac{hc}{E}$

$= \frac{6.625*10^{-34} * 3.0*10^{8}}{19.9*1.6*10^{-19}}$

$\lambda_{max}= 624.2nm$

Considering $E = 3.25eV$

When a single photon is transferred to to an electron the energy obtained can be calculated as follows

$E = 19.9eV = 3.25 *1.6 *10^{-19}J$

This energy is mathematically represented as

$E = \frac{hc}{\lambda_{min}}$

Substituting values and making $\lambda$ the subject of the formula

$\lambda_{min} = \frac{hc}{E}$

$= \frac{6.625*10^{-34} * 3.0*10^{8}}{3.25*1.6*10^{-19}}$

$\lambda_{min}= 382.2nm$

3 0

3 years ago

Three objects lie in the x, y plane. Each rotates about the z axis with an angular speed of 5.58 rad/s. The mass m of each objec

QveST [7]

Answer:

a) V1=11.05m/s V2=92.07m/s V3=17.24m/s

b) KE = 16238.26J

Explanation:

For tangential speeds:

$V1 = \omega*R1=5.58*1.98=11.05m/s$

$V2 = \omega*R2=5.58*16.5=92.07m/s$

$V3 = \omega*R3=5.58*3.09=17.24m/s$

For the kinetic energy, it can be calculated as:

$KE=1/2*\omega^2*(I1+I2+I3)$

Where:

$I1 = m1*R1^2=5.46*1.98^2=21.4kg.m^2$

$I2 = m2*R2^2=3.64*16.5^2=990.99kg.m^2$

$I3 = m3*R3^2=3.21*3.09^2=30.65kg.m^2$

So,

$KE=1/2*5.58^2*(21.4+990.99+30.65)$

KE=16238.26J

4 0

3 years ago

Radiation, conduction, and convection are methods of energy transfer. Of these methods, radiation is the only one in which energ

8_murik_8 [283]

C. Space.

Radiation is a process by which heat energy is transferred through space.

4 0

3 years ago

Read 2 more answers

A motor must lift a 1500-kg elevator cab. The cab's maximum occupant capacity is 400 kg, and its constant "cruising" speed is 1.

natka813 [3]

Answer:

12900 W

24200 W

Explanation:

Given:

v₀ = 0 m/s

v = 1.3 m/s

t = 2.0 s

Find: a and Δx

v = at + v₀

(1.3 m/s) = a (2.0 s) + (0 m/s)

a = 0.65 m/s²

Δx = ½ (v + v₀) t

Δx = ½ (1.3 m/s + 0 m/s) (2.0 s)

Δx = 1.3 m

While accelerating:

Newton's second law:

∑F = ma

F − mg = ma

F = m (g + a)

F = (1500 kg + 400 kg) (9.8 m/s² + 0.65 m/s²)

F = 19855 N

Power = work / time

P = W / t

P = Fd / t

P = (19855 N) (1.3 m) / (2.0 s)

P ≈ 12900 W

At constant speed:

Newton's second law:

∑F = ma

F − mg = 0

F = mg

F = (1500 kg + 400 kg) (9.8 m/s²)

F = 18620 N

Power = work / time

P = W / t

P = Fd / t

P = Fv

P = (18620 N) (1.3 m/s)

P ≈ 24200 W

5 0

4 years ago

A brick is moving at a speed of 3 m/s and a pebble is moving at a speed of 5 m/s. If both objects have the same kinetic energy,

Cerrena [4.2K]

Answer:

let M be the mass of brick

let m be the mass of the pebble

.5M(3)^2 =KE

.5m(5)^2= KE

.5M9 = .5m25

9M = 25m

(9/25)M = m

7 0

3 years ago