The change in velocity of the baseball is 39.72 m/s
<h3>What is change in velocity?</h3>
This is the difference between the final and the initial velocity of a body.
To calculate the change in velocity of the ball, we use the formula below.
<h3>Formula:</h3>
- Ft = mΔv.............. Equation 1
Where:
- F = Force exerted by the person
- t = time
- m = mass of the baseball
- Δv = Change in velocity.
Make Δv the subject of the equation
- Δv = Ft/m........... Equation 2
From the question,
Given:
- F = 12 N
- t = 0.480 s
- m = 0.145 kg
Substitute these values into equation 2
- Δv = (12×0.48)/0.145
- Δv = 39.72 m/s.
Hence, The change in velocity of the baseball is 39.72 m/s
Learn more about change in velocity here: brainly.com/question/112886
Answer:
v = 4.25 m / s
Explanation:
To solve this exercise we must use Newton's second law, let's set a reference system that is horizontal and vertical
Axis y
- W = 0
X axis
Tₓ = m a
The acceleration is centripetal
a = v² / r
Tₓ = m v² / r
v² = Tₓ r / m (1)
Let's use trigonometry to find the tension components,
sin θ = Tₓ / T
cos θ = / T
= T cos θ
Tₓ = T sin θ
Let's look angle for the maximum tension 103 N
= T cos θ = W
θ = cos⁻¹ W / T
θ = cos⁻¹ (5.77 9.8 / 103)
θ = 56.7°
Now let's find the radius of the circle
sin 56.7 = r / L
r = L sin 56.7
We substitute in the speed equation (1)
v² = T sin 56.7 L sin 56.7 / m
v = √ T L sin² 56.7 / m
Let's calculate
v = √ (103 1.45 sin² 56.7 /5.77)
v = 4. 25 m / s
Answer:
B
Explanation:
The more negative the colder the higher the celcius the hotter it is
Answer:
The amount of moisture air can hold is dependent on its temperature and pressure. The warmer the air the greater the quantity of water vapour it can contain. The air temperature is measured with a normal thermometer this is the Dry-Bulb reading. The actual amount of moisture known as the mixing ratio is measured in grams of water per kilogram of dry air. When air at a certain temperature is saturated it cannot hold any more moisture. The relative humidity of the air is the ratio of the actual amount of moisture in the air to the fully saturated amount.
We can use evaporation to measure the amount of moisture in the air. A wet cloth is placed over the bulb of a thermometer and then air is blown over the cloth causing the water to evaporate. Since evaporation takes up heat, the thermometer will cool to a lower temperature than a thermometer with a dry bulb at the same time and place. The depression in Wet-Bulb temperature allows the humidity to be calculated. If the air is fully saturated (100% relative humidity) the water cannot evaporate, so both the wet and dry bulb temperatures are the same. If the dry and wet bulb temperatures are set to the same value the calculator will show the saturation mixing ratio the amount of water in saturated air at that temperature.
If partly saturated air is cooled without changing its pressure or amount of water vapour, a point is reached when it becomes saturated. The moisture will be given up as dew or ice crystals. This temperature is the Dew Point. This is why condensation will form on a bottle of cold beer, as the air in the vicinity of the bottle is cooled below its dew point. Meteorological reports usually quote the temperature and dew point as well as the station pressure. Moisture content and relative humidity can be calculated from these figures.
The Psychrometer is the name of a device containing both a wet and dry bulb thermometer. This can be a fixed device for meteorology or a handheld sling psychrometer that is often used in air conditioning applications.
Explanation:
Answer:
Same magnitude of the 10 nc charge cause the electric field is external.
Explanation:
To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.
As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.
F = qE
If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.
