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zysi [14]
3 years ago
14

Lacie kicks a football from ground level at a velocity of 13.9 m/s and at an angle of 25.0° to the ground. You have determined t

hat the football would travel 15.1 m before landing. How would this value change if the football was kicked at an angle of 35.0°?
Physics
2 answers:
Tems11 [23]3 years ago
5 0

Accttuallyyy its

This is for the other one sowwy

1.2 s and 15.1 m

poizon [28]3 years ago
3 0
Based on the information given: 
The vertical velocity is 13.9 * sin (35°) = 7.97 m/s
The horizontal velocity is 13.9 * cos (35°) = 11.39 m/s

  First, let's see how long the ball will be in the air, given its initial vertical velocity of 7.97 m/s. To do this, let's see when the vertical velocity equals zero (when the ball reaches the vertical peak of its trajectory, or the moment just before the ball starts to come down). Remember that acceleration due to gravity is -9.8 /s²
v(t) = vi - at
0 = 7.97 + -9.8 * t     Subtract 7.97 from both sides
-7.97 = -9.8 * t          Divide both sides by -9.8
t = .8135
The height at this point was:
h(t) = vi (t) + (1/2) a *t²
h(t) = 7.97 (.8135) + (1/2) (-9.8) (.8135²)
h(t) = 6.48 - 3.24
h(t) = 3.24m

How long until the ball came down? How long does it take a ball to fall 3.24 meters, with an initial vertical velocity of 0?
3.24 = 0 - (1/2) (9.8) (t²)           Divide both sides by (9.8/2)
t² = .66                                     Take the square root of both sides
t = .835
/
The ball took about the same amount of time, .8 seconds, to come down as it did to come up. Total time ball was in air: .813 + .835 = 1.648

The balled traveled for 1.648 seconds at a constant horizontal speed (11.39 m/s) before coming down. How far did it travel?
1.648 * 11.39 = 18.77 m

What is the change given this new angle, compared to the 25° angle?
18.77 - 15.1 = 3.67

How do the two angles differ in terms of the general path of the ball? In the 25° angle, the ball was kicked at a shallower angle (kind of like a line drive hit in baseball), and traveled with a higher horizontal velocity and lower initial vertical velocity. This means the ball was in the air for less time. the 35° angle had a higher initial vertical and lower horizontal velocity, but was in the air for longer. This extra hang time means that the ball was able to travel farther before hitting the ground. 
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A proton orbits a long charged wire, making 1.80 ×106 revolutions per second. The radius of the orbit is 1.20 cm What is the wir
Fantom [35]

Answer:

linear charge density = -9.495 × 10^{-34} C/m

Explanation:

given data

revolutions per second = 1.80 × 10^{6}

radius = 1.20 cm

solution

we know that when proton to revolve around charge wire then centripetal force is require to be in orbit of radius around provide by electric force

so

- q × E = m × w² × r     ..................1

- 9 × 10^{9}  × \frac{2*linear\ charge\ density}r} q =  m × w² × r   ............2

and w = \frac{2*\pi}{T}  

w = \frac{d\theta }{dt}

w = 1.80 × 10^{6} × \frac{2*\pi}{1}

w = 11304000 rad/s

so here from equation 2

- 9 × 10^{9}  × \frac{2*linear\ charge\ density}{0.012} 1.80 × 10^{6} =  1.672 × 10^{-27} × 11304000² × 0.0120  

linear charge density = -9.495 × 10^{-34} C/m

8 0
3 years ago
An engine causes a car to move 10 meters with a force of 100 N. The engine produces 10,000 J of energy. What is the efficiency o
kodGreya [7K]

Answer:

Part A

The efficiency of the engine is 10%

Part B

The change in internal energy is 300 J

Part C

The change in volume is 1 m³ which is one cubic meter

Explanation:

Part A

Efficiency is defined as the ratio of energy output to energy input;

The given parameters are;

The distance the car is moved, d = 10 meters

The force which moves the car, F = 100 N

The amount of energy produced by the engine = 1,000 J

Therefore, we have;

The energy output to the car = The work done on the car = Force applied to the car, F × The distance the car moves, d;

∴ The energy output to the car by the engine = F × d = 100 N × 10 m = 1,000 J

The energy input from the engine = The energy produced by the engine = 10,000 J

The efficiency of the engine = (The energy output)/(The energy input)= 1,000J/10,00J = 0.1

The efficiency in percentage = 0.1 × 100 = 10 %

The efficiency of the engine = 10%

Part B

The amount of heat added to the substance, ΔQ = 1,000J

The amount of work the substance does on the atmosphere, W = 700 J

The change in internal energy, ΔU is given as follows;

ΔQ = W + ΔU

∴ ΔU = ΔQ - W

For the substance, we have;

The change in internal energy, ΔU = 1,000 J - 700 J = 300 J

Part C

The work done by the piston, W = 1,000 J

The pressure, in the piston, P = 1,000 Pa = constant

The work done by the piston in a constant pressure process, W = P × ΔV

Where;

W = The work done

P = The constant pressure applied

ΔV = The change in volume = V₂ - V₁

V₂ = The final volume

V₁ = The initial volume

∴The change in volume ΔV = W/P = 1,000 J/(1,000Pa) = 1 m³

The change in volume ΔV = 1 m³

3 0
3 years ago
How quickly do muscles become fatigued?​
agasfer [191]

Answer:

Fatigue is usually defined as the reversible decline of performance during activity, and most recovery occurs within the first hour. However, there is also a slowly reversible component that can take several days to reverse (155). Muscle injury also causes a decline in performance that reverses only very slowly.

5 0
3 years ago
Two cylinders each contain 0.30 mol of a diatomic gas at 320 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cy
Svetllana [295]

Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is 7.86\times10^{-3}\ m^3

(d). The final volume of the gas in the cylinder B is 5.7\times10^{-3}\ m^3

Explanation :

Given that,

Number of mole n = 0.30 mol

Initial temperature = 320 K

Pressure = 3.0 atm

Final pressure = 1.0 atm

We need to calculate the initial volume

Using formula of ideal gas

P_{1}V_{1}=nRT

V_{1}=\dfrac{nRT}{P_{1}}

Put the value into the formula

V_{1}=\dfrac{0.30\times8.314\times320}{3.039\times10^{5}}

V_{1}=2.62\times10^{-3}\ m^3

(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}

Put the value into the formula

T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}

T_{2}=233.7\ K

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

P_{1}V_{1}=P_{2}V_{2}

V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}

Put the value into the formula

V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}

V_{2}=0.00786\ m^3

V_{2}=7.86\times10^{-3}\ m^3

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}

V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}

V_{2}=0.0057\ m^3

V_{2}=5.7\times10^{-3}\ m^3

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is 7.86\times10^{-3}\ m^3

(d). The final volume of the gas in the cylinder B is 5.7\times10^{-3}\ m^3

6 0
3 years ago
Explain why sound energy is not a type of potential energy.
Natasha2012 [34]

Answer:

A

Explanation:

3 0
4 years ago
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