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2 years ago

In the formula vy=a t what is the value of a for an object in projectile motion

1 answer:

8 0

I THINK that formula gives the vertical speed of a projectile at the time
't' seconds after it's launched.

(I can't really be sure of anything, because whenever you show a formula
to other people, you're supposed to tell them what each variable in it means,
and you didn't do that.)

If I'm on the right track, then the 'a' is the acceleration of gravity wherever
the projectile is flying. If that happens to be on Earth, then the acceleration
is 9.8 meters per second² .

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Please help answer question

nika2105 [10]

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, $v_t=60\ m/s$

Area of cross section, $A=0.33\ m^2$

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

R = W

$\dfrac{1}{2}\rho CAv_t^2=mg$

Where

$\rho$ is the density of air = 1.225 kg/m³

C is drag coefficient

So,

$C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01$

So, the drag coefficient is 1.01.

4 0

2 years ago

An auditorium measures 40.0 m 3 20.0 m 3 12.0 m. The density of air is 1.20 kg/m3. What are (a) the volume of the room in cubic

Burka [1]

We use the formula,

m = V\rho

Here, m is the mass, V is the volume and $\rho$ density

Also

$V = l w h$

Here l is length, w is width and h is height.

(a) The volume of the room,

$V=40.0 \ m \times 3 20.0 \ m \times 3 12.0\ m = 3.99 \times 10^{6} m^3$

The volume of the room in cubic feet,

$V = 3.99 \times 10^{6} m^3 \times(\frac{3.281 \ ft}{m} )^3 = 4.3 \times 10^7 \ ft^3$

(b) Now the mass of the air in room,

$m= (3.99 \times 10^{6} \ m^3) (1.20 \ kg/m^3) = 4.8 \times 10^6 kg$ .

Therefore, the weight of the air in room,

$W = mg= 4.8 \times 10^6 kg \times 9.8 m/s^2 = 46.9 \times 10^6 \ N \\\\ W = 4.69 \times 10^7 \ N$ .

The weight of air in the room in pounds,

$W = 4.69 \times 10^7 \ N ( \frac{1 \ lb}{4.448 \ N} ) = 1.1 \times 10^7 \ lb$

3 0

2 years ago

How does upwelling affect the weather of a coastal region?

SIZIF [17.4K]

c.The warm surface water results in moist air and more rainfall.

Explanation:

During upwelling, cold water in the ocean is stirred up and brought to the surface.
The warmer surface water is then taken into deeper parts of the ocean.
Upwelling allows for nutrient mixing in the ocean and allows for useful gases to circulate well.
The warm surface water causes the air to be moisty.
When the air is carried landward towards the coast, it leads to rainfall when the saturated air releases the water.
The air then becomes cold and dry and it rises up.
Therefore, warm surface water results in moist air and more rainfall.

Learn more:

Ocean current brainly.com/question/4117397

#learnwithBrainly

4 0

2 years ago

Read 2 more answers

A plate in a parallel-plate capacitor has an area of 0.03 m2 and is 0.5 × 10–3 m from the other plate. The space between the pla

Tems11 [23]

Answer:

A. 4 × 10–9 F

4 0

2 years ago

A syringe of volume 16 cm3 is filled with air to a pressure of 1.03 atm. If the piston of the syringe is pushed to change the vo

denpristay [2]

<h3>Answer:</h3>

189.07 kPa

<h3>Explanation:</h3>

Concept tested: Boyle's law

<u>We are given;</u>

Initial volume of the syringe, V1 is 16 cm³
Initial pressure of the syringe, P1 is 1.03 atm
New volume of the syringe, V2 is 8.83 cm³

We are required to calculate the new pressure of the syringe;

We are going to use the concept on Boyle's law of gases.
According to the Boyle's law, for a fixed mass of a gas, the pressure is inversely proportional to its volume at constant temperature.

That is; P α 1/V

At varying pressure and volume, k(constant) = PV and P1V1=P2V2

Therefore, to get the new pressure, P2, we rearrange the formula;

P2 = P1V1 ÷ V2

= ( 16 cm³ × 1.03 atm) ÷ 8.83 cm³

= 1.866 atm.

Thus, the new pressure is 1.866 atm
But, we need to convert pressure to Kpa
Conversion factor is 101.325 kPa/atm

Thus;

Pressure = 1.866 atm × 101.325 kPa/atm

= 189.07 kPa

Hence, the new pressure of the air in the syringe is 189.07 kPa

3 0

3 years ago