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bezimeni [28]
3 years ago
12

A golfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. What is the horizontal velocity compo

nent of the ball? Round the answer to the nearest tenth of a m/s.
Physics
2 answers:
Grace [21]3 years ago
7 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
<span>The vertical velocity (vy) is -569 meter/second.</span>
ololo11 [35]3 years ago
5 0

Answer:

25.4 m/s

Explanation:

When doing a question such as this, always assume the object traveling (a golf ball in this case) travels as a particle, where there is no air resistance.

The golf ball has an initial velocity or 31 m/s, and is hit at an angel of 35°.

When breaking down its velocity into its x and y components, we use different formulas. To acquire the x component of velocity, the following formula is used:

Vₓ = v cos(θ)  (where v is the overall velocity, and theta-θ is the angel at which it is hit)

Using this formula, we can calculate the x component of the velocity:

Vₓ = v cos(θ)

Vₓ = (31 m/s) cos(35)

Vₓ = 25.3937 m/s

Since the question asks to round to the nearest tenth, the final answer is:

Vₓ = 25.4 m/s

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4 years ago
The unit for momentum is which of the following?
Alika [10]

(any unit of mass) x (any unit of speed)  is a unit of momentum.

The only choice on the list with the dimensions of (mass) x (speed)
is 'c'.

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6 0
3 years ago
A decorative plastic film on a copper sphere of 10 mm diameter in an oven at 750C. Upon removal from the oven, the sphere is sub
AlladinOne [14]

Answer:

3.26 secs

Explanation:

Diameter of sphere ( D )= 10 mm

T1 = 75°C

P = 1 atm

T∞ = 23°C

T2 = 35°c

Velocity = 10 m/s

<u>Determine how long it will take to cool the sphere to 35°C</u>

<em>Using the properties of copper and air given in the question</em>

Nu = 2 + (Re)^0.8 (Pr)^0.33

hd / k = 2 + ( vd/v )^0.8 (Pr)^0.33

∴ h ≈  2594.7 W/m^2k

Given that :

(T2 - T∞) / ( T1 - T∞ )  = exp [ ( -hA / pv CP ) t ]  

( 35 - 23 ) / ( 75 - 23 ) = exp [  - 2594.7 * 6 * t / 8933 * 387 * 10 * 10^-3 ]

=  ln ( 12/52 ) = -1.466337069  =     - 0.45032919 *  t

∴ t ≈ 3.26 secs        ( -1.466337069 / -0.45032919 )

6 0
3 years ago
A pressure vessel that has a volume of 10m3 is used to store high-pressure air for operating a supersonic wind tunnel. If the ai
Anna71 [15]

Answer:

The mass of air stored in the vessel is 235.34 kilograms.

Explanation:

Let supossed that air inside pressure vessel is an ideal gas, The density of the air (\rho), measured in kilograms per cubic meter, is defined by following equation:

\rho = \frac{P\cdot M}{R_{u}\cdot T} (1)

Where:

P - Pressure, measured in kilopascals.

M - Molar mass, measured in kilomoles per kilogram.

R_{u} - Ideal gas constant, measured in kilopascal-cubic meters per kilomole-Kelvin.

T - Temperature, measured in Kelvin.

If we know that P = 2026.5\,kPa, M = 28.965\,\frac{kg}{kmol}, R_{u} = 8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} and T = 300\,K, then the density of air is:

\rho = \frac{(2026.5\,kPa)\cdot \left(28.965\,\frac{kg}{kmol} \right)}{\left(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} \right)\cdot (300\,K)}

\rho = 23.534\,\frac{kg}{m^{3}}

The mass of air stored in the vessel is derived from definition of density. That is:

m = \rho \cdot V (2)

Where m is the mass, measured in kilograms.

If we know that \rho = 23.534\,\frac{kg}{m^{3}} and V = 10\,m^{3}, then the mass of air stored in the vessel is:

m = \left(23.534\,\frac{kg}{m^{3}} \right)\cdot (10\,m^{3})

m = 235.34\,kg

The mass of air stored in the vessel is 235.34 kilograms.

7 0
3 years ago
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5 0
4 years ago
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