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vlabodo
3 months ago
11

Suppose Sammy Sosa hits a home run which travels 361. ft (110. m). Leaving the bat at 50 degrees above the horizontal, how high

did it travel?

Physics
1 answer:
Yuliya22 [3.3K]3 months ago
5 0

Response:

The horizontal span of Sosa is 276.526 ft or 84.28 meters.

Explanation:

As illustrated in the diagram, let point O denote Sosa's starting position. She travels 361 ft at a 50-degree angle relative to the horizontal.

sin 50 = \frac{OM}{OP}

0.7660 = h / 361

h = 276.526 ft

h = 84.28 meters

The horizontal distance of Sosa is 276.526 ft or 84.28 meters.

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Three moles of an ideal gas with a molar heat capacity at constant volume of 4.9 cal/(mol∙K) and a molar heat capacity at consta
ValentinkaMS [3465]

Answer:

The amount of heat that enters the gas throughout this two-step process totals 120 cal.

Explanation:

Given that,

Moles present = 3

Heat capacity at volume held constant = 4.9 cal/mol.K

Heat capacity at pressure held constant = 6.9 cal/mol.K

Starting temperature = 300 K

Ending temperature = 320 K

We are tasked with determining the heat absorbed by the gas at constant pressure

Employing the heat formula

\Delta H_{1}=nC_{p}\times\Delta T

Substituting the values into the equation

\Delta H_{1}=3\times6.9\times(320-300)

\Delta H_{1}=414\ cal

Next, we calculate the heat absorbed by the gas at constant volume

Using the corresponding heat formula

\Delta H_{1}=nC_{v}\times\Delta T

Insert the values into the formula

\Delta H_{1}=3\times4.9\times(300-320)

\Delta H_{1}=-294\ cal

Now, it's necessary to evaluate the total heat flow into the gas during both steps

Using the total heat formula

\Delta H_{T}=\Delta H_{1}+\Delta H_{2}

\Delta H_{T}=414-294

\Delta H_{T}=120\ cal

Thus, the heat that transfers into the gas throughout this two-step process amounts to 120 cal.

7 0
3 months ago
In an amusement park rocket ride, cars are suspended from 3.40-m cables attached to rotating arms at a distance of 5.90 m from t
ValentinkaMS [3465]

Answer:

The rotational angular speed is measured at 1.34 rad/s.

Explanation:

Considering the following parameters,

Length = 3.40 m

Distance = 5.90 m

Angle = 45.0°

We are tasked with finding the angular speed of rotation

Using the balance equation

Horizontal component

T\cos\theta=mg

T=\dfrac{mg}{\cos\theta}

Vertical component

T\sin\theta=m\omega^2 r

Substituting the tension value

mg\tan\theta=m\omega^2(d+L\sin\theta)

\omega=\sqrt{\dfrac{g\tan\theta}{(d+L\sin\theta)}}

Substituting the value into the equation

\omega=\sqrt{\dfrac{9.8\tan45.0}{5.90+3.40\sin45.0}}

\omega=1.34\ rad/s

Thus, the angular speed of rotation computes to 1.34 rad/s.

7 0
4 months ago
An electron with a charge value of 1.6 x 10-19 C is moving in the presence of an electric field of 400 N/C. What force does the
Yuliya22 [3333]
The force experienced by the electron is 4.0×10⁻¹⁷ N.
4 0
3 months ago
A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then
serg [3582]

Response:

A) W_{ff} =-744.12J

B) F_f=-W_{ff}*sin\theta /hy = 112.75N

C) F_{f2}=207.58N

Clarification:

The question is not fully provided. The complete question was:

A skateboarder with a mass of ms = 54 kg is at the top of a ramp with a height of hy = 3.3 m. He then jumps on his skateboard and goes down the ramp. His speed at the base is vf = 6.2 m/s.  

Part (a) Formulate an expression for the work, Wf, done by the frictional force on the skateboarder in terms of the variables listed in the problem.

Part (b) The ramp is at an angle θ with the ground, where θ = 30°. Formulate an expression for the frictional force's magnitude, fr, between the skateboarder and the ramp.

Part (c) Upon reaching the bottom, the skateboarder continues with speed vf onto a grass-covered flat surface. The friction between the grass and the skateboarder brings him to a halt after 5.00 m. Determine the frictional force, Fgrass in newtons, between the skateboarder and the grass.

For part A), we assess the energy balance to determine the work done by the friction:

W_{ff}=\Delta E

W_{ff}=1/2*m*vf^2-m*g*hy

W_{ff}=-744.12J

For part B), we utilize the previously calculated work:

W_{ff}=-F_f*(hy/sin\theta)   Rearranging for friction force:

F_f=-W_{ff}*sin\theta /hy

F_f=112.75N

For part C), we first ascertain the acceleration through kinematic equations and subsequently find the frictional force using dynamic methods:

Vf^2=Vo^2+2*a*d

Rearranging for 'a':

a=-3.844m/s^2

Now, using dynamics:

|F_f|=|m*a|

|F_f|=207.58N

8 0
3 months ago
A boat is floating in a small pond. the boat then sinks so that it is completely submerged. what happens to the level of the pon
ValentinkaMS [3465]
When the boat submerges completely in the pond, the water level of the pond rises.
4 0
2 months ago
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