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Lelechka [254]
3 years ago
15

A change in which property of light will have no effect on whether or not the photoelectric effect occurs??? Frequently energy i

ntensity and wavelength
Physics
2 answers:
defon3 years ago
8 0

Answer: intensity

Explanation:

MaRussiya [10]3 years ago
5 0
The intensity of the light has no connection with the photoelectric effect.

That's what was so baffling about it before the particle nature of light
was suspected ... a match with a blue flame might stimulate the
photoelectric effect, but a high-power red searchlight couldn't do it.
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Which direction will thermal energy flow if you pick up a snowball with your bare hand? Thermal energy will flow from the snowba
Lapatulllka [165]

Answer:

b. Thermal energy will flow from your hand to the snowball.

Explanation:

4 0
3 years ago
Read 2 more answers
Using a scale diagram, calculate the resultant force acting on a sailing boat when an easterly wind provides 2, point, 50, k, N,
EastWind [94]

Answer:

F = 3.6 kN, direction is 9.6º to the North - East

Explanation:

The force is a vector, so one method to find the solution is to work with the components of the vector as scalars and then construct the resulting vector.

Let's use trigonometry to find the component of the forces, let's use a reference frame where the x-axis coincides with the East and the y-axis coincides with the North.

Wind

X axis

          F₁ = 2.50 kN

Tide

         cos 30 = F₂ₓ / F₂

         sin 30 = F_{2y} / F₂

          F₂ₓ = F₂ cos 30

         F_{2y} = F₂ sin 30

         F₂ₓ = 1.20cos 30 = 1.039 kN

         F_{2y} = 1.20 sin 30 = 0.600 kN

the resultant force is

X axis

        Fₓ = F₁ₓ + F₂ₓ

        Fₓ = 2.50 +1.039

        Fₓ = 3,539 kN

        F_y = F_{2y}

        F_y = 0.600

to find the vector we use the Pythagorean theorem

         F = \sqrt{F_x^2 +F_y^2}

         F = \sqrt{ 3.539^2 + 0.600^2 }

         F = 3,589 kN

the address is

         tan θ = F_y / Fₓ

         θ = tan⁻¹ \frac{F_y}{F_x}

         θ = tan⁻¹  \frac{0.6}{3.539}0.6 / 3.539

         θ = 9.6º

the resultant force to two significant figures is

         F = 3.6 kN

the direction is 9.6º to the North - East

7 0
2 years ago
The movement of a magnetic pole away from the actual pole
Lunna [17]

The North Magnetic Pole is the point on the surface of Earth's Northern Hemisphere at which the planet's magnetic field points vertically downwards (in other words, if a magnetic compass needle is allowed to rotate about a horizontal axis, it will point straight down). There is only one location where this occurs, near (but distinct from) the Geographic North Pole and the Geomagnetic North Pole.
3 0
3 years ago
The total distance a group of movers move a refrigerator is 22 m. If they complete this task in 15 minutes, their efforts would
lianna [129]

Answer:

Their efforts would be expressed in units of Joules per second

Explanation:

The unit of their efforts can be derived from the formula of power which is given by the product of mass, acceleration and distance (the product is energy with unit joules) divided by time taken to complete the task (unit is seconds)

Therefore, the unit of their efforts would be joules per second

4 0
2 years ago
An insulated beaker with negligible mass contains 0.250 kg of water at 75.0C. How many kilograms of ice at -20.0C must be droppe
kkurt [141]

Answer:

The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg

Explanation:

Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water

Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C

To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.

Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C

Latent heat of ice = L = 334000 J/kg

Specific heat capacity of water = C = 4186 J/kg.°C

Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m

Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J

543600 m = 36627.5

m = 0.0674 kg = 67.4 g of ice.

3 0
3 years ago
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