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ivolga24 [154]
3 years ago
10

A soccer ball is kicked horizontally at 15.8 m/s off the top of a field house and lands 33.9 meters from the base of the field h

ouse. Determine the height of the field house
Physics
1 answer:
Leviafan [203]3 years ago
5 0
First figure out the time by using the equation (you know v and x):
 v_{x} = \frac{x}{t} 

Then, use the following equation to figure out what y (height of the field house is):
y= v_{y} t+ \frac{1}{2}at^{2}

Remember that you can always analyze half of the scenario by dividing the time in half, and setting vy = 0 if at all it helps.

You might be interested in
Which of the following is most likely to be an observation made by a physiologist
Naya [18.7K]
Do you have the answer choices ?
3 0
3 years ago
An object is thrown straight up with an initial velocity of 10 m/s, and there is an air resistance force causing an acceleration
lana [24]

Answer:

Vf= 7.29 m/s

Explanation:

Two force act on the object:

1) Gravity

2) Air resistance

Upward motion:

Initial velocity = Vi= 10 m/s

Final velocity = Vf= 0 m/s

Gravity acting downward =  g = -9.8 m/s²

Air resistance acting downward = a₁ = - 3 m/s²

Net acceleration = a = -(g + a₁ ) = - ( 9.8 + 3 ) = - 12.8 m/s²

( Acceleration is consider negative if it is in opposite direction of velocity )

Now

2as = Vf² - Vi²

⇒ 2 * (-12.8) *s = 0 - 10²

⇒-25.6 *s = -100

⇒ s = 100/ 25.6

⇒ s = 3.9 m

Downward motion:

Vi= 0 m/s

s = 3.9 m

Gravity acting downward =  g = 9.8 m/s²

Air resistance acting upward = a₁ = - 3 m/s²

Net acceleration = a = g - a₁  =  9.8 - 3  = 6.8 m/s²

Now

2as = Vf² - Vi²

⇒ 2 * 6.8 * 3.9 = Vf² - 0

⇒ Vf² = 53. 125

⇒ Vf= 7.29 m/s

8 0
3 years ago
Water of density 1000 kg/m3 falls without splashing at a rate of 0.373 L/s from a height of 40.5 m into a 0.64 kg bucket on a sc
Sphinxa [80]

Answer:

       F_scale = 20.18 N

Explanation:

The scale reading corresponds to two factors, the first the weight of the water in the container and the second the force of the liquid that is falling at the moment of reading.

* Let's find the amount of liquid in the container for a time of t = 2.93 s

Let's use a direct proportion rule. If 0.373 l falls in one second at t = 2.93 s, how many liters are there

        V_{water} = 2.93 s (0.373 l / 1s) = 1.09 l

        V_{water} = 1.09 10⁻³ m³

the amount of water is

       ρ = m / V

       m = ρ V

       m = 1000 1.09 10⁻³

       m = 1.09 kg

so the weight of the liquid in the container for this time is

       W = mg

       W = 1.09 9.8

       W = 10.68 N

* Let's look for the force of the falling jet

Let's use Bernoulli's equation, where the subscript 1 is for the container and the subscript 2 is for the water at a height h

        P₁ + 1/2 ρ g v₁² + ρ g y₁ = P₂ + 1/2  ρ g v₂² + ρ g y₂

In this case, the water falls freely, so the external pressure is atmospheric.

         P₂ = P_{atm}

since they indicate that the water falls, we assume that its initial velocity is zero v₂ = 0

let's use kinematics to find the speed of a drop when it reaches the container y = 0

         v² = v₀² - 2 g (y-y₀)

         v = \sqrt{0 -2 g ( 0-y_o)}

let's calculate

         v = √(2 9.8 40.5)

         v = 28.17 m / s

this is the speed in the container v₁ = 28.17 m / s

the height from where it falls is y₂ = 40.5 and reaches the container y₁ = 0

we substitute in Bernoulli's equation

         P₁ +1/2 ρ g v₁² + 0 = P_{atm} + 0 + ρ g y₂

         P₁ + ½ ρ g v₁² = P_{atm} + ρ g y₂

         P₁ = P_{atm} + ρ g y₂ - ½ ρ g v₁²

         P₁ = 1 10⁵ + 1000 9.8 40.5 - ½ 1000 28.17²

         P₁ = 1 10⁵ + 3.97 10⁵ - 3.69 10⁵

         P₁ = 1.28 10⁵ Pa

The definition of Pressure is

         P = F / A

         F = P A

We must suppose a time to carry out the reading suppose an average time of the modern equipment t = 0.1 s, in this time how much is now arriving

          m₂ = 0.373 0.2 = 0.0746 l = 0.0746 10⁻³ m³

the volume is V = A l

if the length of l = 1 m

A = 0.0746 10⁻³ m³ = 7.45 10⁻⁵ m²

the force of this jet is

            F = P A

            F = 1.28 10⁵  7.46 10⁻⁵

            F = 9.5 N

with these data let's use the equilibrium equation

           F_ scale -W - F = 0

           F_scale = W + F

           F_scale = 10.682 + 9.5

           F_scale = 20.18 N

4 0
3 years ago
When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the
MaRussiya [10]

Answer:

(a) \alpha=-111.26rad/s

(b) s=4450.6in

(c) 8.66in

Explanation:

First change the units of the velocity, using these equivalents 1rev=2\pi rad and 1 min =60s

4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s

The angular acceleration \alpha the time rate of change of the angular speed \omega according to:

\alpha=\frac{\Delta \omega}{\Delta t}

\Delta  \omega=\omega_i-\omega_f

Where \omega_i is the original velocity, in the case the velocity before starting the deceleration, and \omega_f is the final velocity, equal to zero because it has stopped.

\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s

b) To find the distance traveled in radians use the formula:

\theta = \omega_i t + \frac{1}{2} \alpha t^2

\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad

To change this result to inches, solve the angular displacement \theta for the distance traveled s (r is the radius).

\theta=\frac{s}{r} \\s=\theta r

s=890.12(5)=4450.6in

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

\frac{890.12}{2\pi}=141.6667

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance  between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle \gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120 is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which  is also the net displacement):

c^2=a^2+b^2-2abcos(\gamma)

c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in

4 0
3 years ago
All of the following equations are statements of the ideal gas law except
elena55 [62]

Answer:

a. P = nRTV

Explanation:

The question is incomplete. Here is the complete question.

"All of the following equations are statements of the ideal gas law except a. P = nRTV b. PV/T = nR c. P/n = RT/v d. R = PV/nT"

Ideal gas equation is an equation that describes the nature of an ideal gas. The molecule of an ideal gas moves at a particular velocity depending on the temperature. This gases collides with one another elastically. The collision that an ideal gas experience is a perfectly elastic collision.

The ideal gas equation is expressed as shown:

PV = nRT where:

P is the pressure of the gas

V is the volume

n is the number of moles

R is the ideal gas constant

T is the temperature.

Based on the formula given for an ideal gas, it can be inferred that the equation. P = nRTV is not a statement of an ideal gas equation.

The remaining option will results to an ideal gas equation if they are cross multipled.

7 0
3 years ago
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