Which of the following would NOT affect the level at which a canoe floats in a pond

What crop is least likely to do well when the temperatures are very hot?

torisob [31]

a. Sweet corn and possibly d. okra.

3 0

4 years ago

A beam of ultraviolet radiation, with frequency

andreyandreev [35.5K]

To develop this problem, we will apply Einstein's relationship which is in charge of the work done with the kinetic energy of the body versus the total energy of the system.

The energy can be calculated as

$E = hf$

Here,

h = Planck's Constant

f = Frequency

Our values are given as,

$f = 2.78*10^{15} Hz ,$

$W = 3.80 eV$

Therefore the Energy is

$E = hf$

$E = 6.625*10^{-34}J\cdot s(2.78*10^{15}Hz)$

$E = 1.84*10^{-18}J \rightarrow 1 J = 6.242*10^{18}eV$

Then,

$E = 11.48eV$

Applying the Einstein Relation we have that

$E = W+KE$

$KE = E -W$

$KE = 11.48eV-3.80 eV$

$KE = 7.68eV$

Therefore the maximum kinetic energy for an electron dislodged fromthe surface by the radiation is 7.68eV

6 0

3 years ago

1. A person of mass 55 kg is in a bumper car that has a mass of 75 kg. They are together at a velocity of 8 m/s.

ExtremeBDS [4]

The momentum of the bumper car after the collision is 1,040 kgm/s.

<h3>Momentum of bumper</h3>

The change in momentum of the bumper is calculated as follows;

P = v(m1 + m2)

P = 8(55 + 75)

P = 1,040 kgm/s

The momentum of the bumper car after the collision is 1,040 kgm/s.

The direction is still the same.

Learn more about momentum here: brainly.com/question/7538238

#SPJ1

7 0

2 years ago

Determine the rate at which the electric field changes between the round plates of a capacitor, 5.8 cm in diameter, if the plate

Natalka [10]

Answer:

The rate at which the electric field changes between the round plates of a capacitor is $125\times 10^{3}Vs^{-1}$ .

Explanation:

It is given in the problem that the round plates of a capacitor are spaced some distance apart and the voltage across them is changing.

The expression for the electric field in terms of voltage is as follows;

$E=\frac{V}{d}$

Here, E is the electric field, V is the voltage and d is the distance of separation.

Differentiate expression of the electric field with respect to time, t.

$\frac{dE}{dt}=\frac{1}{d}\frac{dV}{dt}$

Convert the distance of separation from mm to m.

d= 1.2 mm

$d=1.2\times 10^{-3}m$

Calculate the rate at which the electric field changes.

$\frac{dE}{dt}=\frac{1}{d}\frac{dV}{dt}$

Put $\frac{dV}{dt}=150 Vs^{-1}$ and $d=1.2\times 10^{-3}m$

$\frac{dE}{dt}=\frac{1}{1.2\times 10^{-3}}(150)$

$\frac{dE}{dt}=125\times 10^{3}Vs^{-1}$

Therefore, the rate at which the electric field changes is $125\times 10^{3}Vs^{-1}$ .

4 0

3 years ago

A particle moves along the x-axis so that its velocity at anytime is t greater than or equal to 0 is given by v(t)=1-sin(2pi t)

UkoKoshka [18]

A) We differentiate the expression for velocity to obtain an expression for acceleration:
v(t) = 1 - sin(2πt)
dv/dt = -2πcos(2πt)
a = -2πcos(2πt)

b) Any value of t can be plugged in as long as it is greater than or equal to 0.

c) we integrate the expression of velocity to find an expression for displacement:
∫v(t) dt = ∫ 1 - sin(2πt) dt
x(t) = t + cos(2πt)/2π + c
x(0) = 0
0 = = + cos(0)/2π + c
c = -1/2π
x(t) = t + cos(2πt)/2π -1/2π

5 0

4 years ago