Do acids or bases turn litmus paper red?

The minimum frequency of light needed to eject electrons from a metal is called the threshold frequency, ???? 0 ν0 . Find the mi

Allisa [31]

Answer:

<h2>E = 2.8028*10⁻¹⁹ Joules</h2>

Explanation:

The minimum energy needed to eject electrons from a metal with a threshold frequency fo is expressed as E = hfo

h = planck's constant

fo = threshold frequency

Given the threshold frequency fo = 4.23×10¹⁴ s⁻¹

h = 6.626× 10⁻³⁴ m² kg / s

Substituting this value into the formula to get the energy E

E = 4.23×10¹⁴ * 6.626 × 10⁻³⁴

E = 28.028*10¹⁴⁻³⁴

E = 28.028*10⁻²⁰

E = 2.8028*10⁻¹⁹ Joules

7 0

3 years ago

Joe is attempting to ramp his bicycle over a row of burning tires. If the ramp has a launch angle of 20 degrees, and Joe can ped

Luden [163]

Answer is 6 tires.

This is a projectile question.

First make sure units are consistent - express speed in m/s.

20 km/h = 20000m / 3600 s = 5.56 m/s

Assume the takeoff point of the ramp is at ground level (height, h, = 0m). We need to determine how long Joe is in the air, and use that time to calculate the horizontal distance he traveled.

Joe is traveling 5.56 m/s on a ramp angled at 20 degrees. There are vertical and horizontal components to his speed:

Vertical speed = 5.56sin20 = 1.90 m/s
Horizontal speed = 5.56cos20 = 5.22 m/s

An easy way to proceed is to calculate the time it takes for Joe’s vertical speed to reach 0m/s - this represents the time when Joe is at his maximum height and is therefore halfway through the trip. Double whatever time this is to find the total time of the trip. Remember he is decelerating due to gravity:

Time to peak:
a = Δv / Δt
-9.8 = -1.9 / Δt
Δt = 0.19s

Total trip time:
0.19 x 2 = 0.38s

Now that we have the total tome Joe is in the air, we can find the horizontal distance he traveled:

v = d / t
5.22 = d / 0.38
d = 1.98m

Now divide this total distance by the length of an individual tire to find the number of tires he will clear:

1.98 / 0.3 = 6.6 tires

Therefore he can jump 6 tires safely (he will land in the middle of the 7th tire).

Lots of steps I know but just try to think of the situation and keep track of the vertical and horizontal things!

4 0

3 years ago

How much heat is required to convert 2.55g of water at 28 degrees c to steam?

pantera1 [17]

There are two different processes here:
1) we must add heat in order to bring the temperature of the water from $28^{\circ}C$ to $100^{\circ}C$ (the temperature at which the water evaporates)
2) other heat must be added to make the water evaporates

1) The heat needed for process 1) is
$Q_1=m C_s \Delta T$
where
$m=2.55 g$ is the water mass
$C_s = 4.18 g/J^{\circ}C$ is the water specific heat
$\Delta T=100^{\circ}C-28^{\circ}C=72^{\circ}C$ is the variation of temperature of the water
If we plug the numbers into the equation, we find
$Q_1 = (2.55 g)(4.18 J/g^{\circ}C)(72^{\circ}C)=767.4 J$

2) The heat needed for process 2) is
$Q=m L_e$
where
$m=2.55 g$ is the water mass
$L_e = 2264.7 J/g$ is the latent heat of evaporation of water
If we plug the numbers into the equation, we find
$Q_2=(2.55 g)(2264.7 J/g)=5775.0 J$

So, the total heat needed for the whole process is
$Q=Q_1+Q_2=767.4 J+5775.0 J=6542.4 J$

4 0

2 years ago

Solve the problems. Show all work, be SI units for force for weight) .

Sav [38]

Answer:

1. 75 N

2. 6.8351 N

3. 700 N

Explanation:

1. Determination of the weight of the bowling ball:

Mass (m) = 7.5 kg

Acceleration due to gravity (g) = 10 m/s²

Weight (W) =?

Weight = mass × acceleration due to gravity

W = m × g

W = 7.5 × 10

W = 75 N

2. Determination of the weight of the object.

Mass (m) = 683.51 g

Weight (W) =?

We'll begin by converting 683.51 g to kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

683.51 g = 683.51 g × 1 Kg / 1000 g

683.51 g = 0.68351 Kg

Thus, 683.51 g is equivalent to 0.68351 Kg.

Finally, we shall determine the weight of the object as follow:

Mass (m) = 0.68351 Kg

Acceleration due to gravity (g) = 10 m/s²

Weight (W) =?

Weight = mass × acceleration due to gravity

W = m × g

W = 0.68351 × 10

W = 6.8351 N

3. Determination of the weight of the KHS student.

Mass (m) = 154 lb

Weight (W) =?

We'll begin by converting 154 lb to Kg. This can be obtained as follow:

2.2 lb = 1 Kg

Therefore,

154 lb = 154 lb × 1 Kg / 2.2 lb

154 lb = 70 kg

Thus, 154 lb is equivalent to 70 kg.

Finally, we shall determine the weight of the KHS student as follow:

Mass (m) = 70 Kg

Acceleration due to gravity (g) = 10 m/s²

Weight (W) =?

Weight = mass × acceleration due to gravity

W = m × g

W = 70 × 10

W = 700 N

8 0

2 years ago

Determine the gravitational force of attraction experienced by a two 5.00 kilogram masses separated by a distance of 2.5 meters.

drek231 [11]

The gravitational force between two objects may be calculated through the equation,
F = Gm₁m₂ / d²
where G is 6.67 x 10^-11 m³/s²kg
Substituting the known values to the equation,
F = (6.67 x 10^-11 m³/s²kg)(5.0 kg)(5.0 kg) / (2.5 m)²
F = 2.668 x 10^-10 N

4 0

3 years ago