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g100num [7]
3 years ago
8

During a baseball game, a baseball is struck at ground level by a batter. The ball leaves the baseball bat with an initial speed

v0 = 38 m/s at an angle θ = 34° above horizontal. Let the origin of the Cartesian coordinate system be the ball's position the instant it leaves the bat. Air resistance may be ignored throughout this problem.
Physics
1 answer:
slava [35]3 years ago
4 0

Answer:

The horizontal distance is 136.6 m.

Explanation:

Given that,

Initial speed = 38 m/s

Angle = 34°

Suppose, Calculate the horizontal distance x_{max} in meters the ball has traveled when it returns to ground level.

We need to calculate the horizontal distance

Using formula of range

R=\dfrac{v_{0}^2\sin(2\theta)}{g}

Where, v= speed

g = acceleration due to gravity

Put the value into the formula

R=\dfrac{(38)^2\sin(2\times34)}{9.8}

R=136.6\ m

Hence, The horizontal distance is 136.6 m.

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2 years ago
A 20-cm long spring is attached to the wall. When pulled horizontally with a force of 100N, the spring stretches to a length of
suter [353]

Answer:

Explanation:

Part 0

All the spring moves is 2 cm

x = 2 cm * [1 m / 100 cm ]

x = 0.020 meters

F = k*d

100N = k * 0.02 m

100 N / 0.02 = k

5000 N / m

Part A

The spring feels a force of 100 N - - 100N = 200 N because each person is pulling in the opposite direction.

F = k * x

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200 / 5000 = d

d = 0.04 meters.

Part B

10.2 kg must be converted to a force as experienced here on earth.

F = m * g

g = 9.81

m = 10.2

F = 10.2 * 9.81

F = 100.06 N

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100.06 = 5000 * d

d = 100.06 / 5000

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3 0
3 years ago
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4 0
3 years ago
gas has a volume of 185 ml and pressure of 310 mm hg. The desiered volume is 74.0 ml. What is the required new pressure
Mamont248 [21]

Answer:

The required new pressure is 775 mm hg.

Explanation:

We are given that gas has a volume of 185 ml and a pressure of 310 mm hg. The desired volume is 74.0 ml.

We have to find the required new pressure.

Let the required new pressure be '\text{P}_2'.

As we know that Boyle's law formula states that;

                    P_1 \times V_1 = P_2 \times V_2

where, P_1 = original pressure of gas in the container = 310 mm hg

           P_2 = required new pressure

            V_1 = volume of gas in the container = 185 ml

            V_2 = desired new volume of the gas = 74 ml

So,  P_2 = \frac{P_1 \times V_1}{V_2}  

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            =  775 mm hg

Hence, the required new pressure is 775 mm hg.

7 0
3 years ago
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