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Alina [70]
4 years ago
11

If a photon has a frequency of 5.20 × 10ÜÊ hertz, what is the energy of the photon?

Physics
1 answer:
ASHA 777 [7]4 years ago
5 0
Is think is B.) -7.91*10^-16


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PlS help <br>I really need help.<br>very urgent<br><br>PLS help
horsena [70]
Draw a freebody diagram, it will explain it really well

the boat is floating on top of the water, which means that the net acceleration in the y direction must be zero

the boat is not sinking (dominant downwards acceleration/force)
the boat is not flying (dominant upward acceleration/force)

that measn
F_{net_y} =0

now, if you drew the FBD, you only have 2 forces acting on the boat.
the upward bouyancy force on the boat and the downward force due to weight

F_{net_y}= F_{bouyancy} - F_{weight}

since the net force is equal to zero

F_{bouyancy} - F_{weight} = 0

and thus

F_{bouyancy} = F_{weight}
4 0
3 years ago
PLEASE HELP MEE THIS IS DUE IN 45 MINS
guajiro [1.7K]

Answer:

The distance travelled does not depend on the mass of the vehicle. Therefore, s = d

Explanation:

This deceleration situation can be analyzed by means of Work-Energy Theorem, where change in translational kinetic energy is equal to the work done by friction:

\frac{1}{2}\cdot m\cdot v^{2}-\mu\cdot m\cdot g \cdot s = 0 (1)

Where:

m - Mass of the car, in kilogram.

v - Initial velocity, in meters per second.

\mu - Coefficient of friction, no unit.

s - Travelled distance, in meters.

Then we derive an expression for the distance travelled by the vehicle:

\frac{1}{2}\cdot v^{2} = \mu \cdot g \cdot s

s = \frac{v^{2}}{\mu\cdot g}

As we notice, the distance travelled does not depend on the mass of the vehicle. Therefore, s = d

3 0
3 years ago
A 2000-kg sailboat experiences an eastward force of 2000 N by the ocean tide and a wind force against its sails with magnitude o
BaLLatris [955]

Answer:

The force exerted by the ocean tide is directed to the right (east).

The force exerted by the wind is directed to the northwest (45° N of W).

We should separate the x- and y- components of the wind force, and evaluate each component separately.

F_x = 6000\cos(\pi/4)(-\^{x})\\F_y = 6000\sin(\pi/4)(+\^{y})

We denote the direction to the right as the positive direction, so the x-component of the wind force is in the negative direction.

The resultant force is as follows:

F_R = [2000 - 4242.64](\^{x}) + [4242.64](\^{y})\\F_R = -2242.64\^{x} + 4242.64\^{y}

The resultant acceleration can be found by Newton's Second Law:

F = ma\\\\a_R = F_R/m = -1.12\^{x} + 2.12\^{y}

The magnitude of the resultant acceleration is

|a_R| = \sqrt{(1.12)^2 + (2.12)^2} = 2.4 ~m/s^2

7 0
3 years ago
I need help with one through six please
dybincka [34]

Answer:

1] 8500000 = <u>8.5 × 10⁶</u>

2] .000072 = <u>7.2 × 10⁻⁵</u>

3] 5.3 × 10⁴ = <u>53000</u>

4] 2.8 × 10⁻³ = <u>0.0028</u>

5] Velocity = \frac{distance}{time}

 V = \frac{50}{10}

 <u>V = 5 m/s</u>

6] Acceleration = \frac{V1-V2}{time}

 A = \frac{30-15}{3}

 A = \frac{15}{3}

 <u>A = 3 m/s²</u>

7 0
2 years ago
Solids are usually more dense than: gases liquids plasmas all of the above
vichka [17]
Yes, this is because particles in a solid there are more particles which are touching. They can only vibrate. But particles in a gas are far apart. This is the same for liquids to.
7 0
4 years ago
Read 2 more answers
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