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kirza4 [7]
2 years ago
8

A baseball player wants to hit a home run over the wall of a stadium. He hits the ball 0.751 m above the ground with at an angle

of 30 degrees above the horizontal and a speed of 37.5 m/s. What is the tallest wall the player can clear if the wall is 150 m away?
Physics
2 answers:
tekilochka [14]2 years ago
7 0

Answer:

The baseball does not reach the wall, because the ball falls at 125.33 meters and the wall is at 150 meters

Explanation:

V= 37.5 m/s

α= 30º

g= 9.8 m/s²

Vx= V * cos(30º) = 32.47 m/s

Vy= V * sin(30º) = 18.75 m/s

flytime of the baseball:

t= 2 * Vy/g

t= 3.86 sec

distance of baseball fall:

d= Vx * t

d= 125.33 m

Trava [24]2 years ago
5 0

Answer: 18.67 m wall

Explanation:this is projectile.

For Max height,

H = v²sin²©/2g

H= (37.5)²sin²(30)/(2*9.81)

H = 17.919 m

But he is standing 0.751 m above ground, so total height = 17.919+0.751= 18.67m

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Points a and b lie in a region where the y-component of the electric field is Ey=α+β/y2. The constants in this expression have t
Drupady [299]

Answer:

V_{a} - V_{b} = 89.3

Explanation:

The electric potential is defined by

         V_{b} - V_{a} = - ∫ E .ds

In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.

         V_{b} - V_{a} = - ∫ E ds

We substitute

         V_{b} - V_{a} = - ∫ (α + β/ y²) dy

We integrate

          V_{b} - V_{a} = - α y + β / y

We evaluate between the lower limit A  2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m

           V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)

            V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33

            V_{b} - V_{a} = - 89.3 V

As they ask us the reverse case

             V_{b} - V_{a} = - V_{b} - V_{a}

             V_{a} - V_{b} = 89.3

3 0
2 years ago
a proton is moving horizontally at 7.9 x 10^5 m/s. It passes through a vertical magnetic field that points downward. In which di
blondinia [14]
There are four key characteristics of the magnetic force on a moving charge.

First, Its magnitude is proportional to the magnitude of the charge. 
Second, the magnitude of the force is also proportional
Third, the magnetic force depends on the particle’s
velocity. 
Fourth, we find by experiment that the magnetic force \vec{F} does not have the same direction as the magnetic field \vec{F} but instead is always perpendicular to both \vec{B} and the velocity \vec{v} 

So, in a mathematical language this is given by:

\vec{F}=q\vec{v} \times \vec{B}

So, this is a cross product. Therefore, applying the Right-hand rule:

The answer is:

<span>D. to the left of its velocity.

Explaining this in other words:

"The force </span>\vec{F} is directed into the plane of the paper, so this is the direction of the proton"

7 0
2 years ago
A plane traveling north at 100.0 km/h through the air gets caught in a 40.0 km/h crosswind blowing west. This turbulence caused
mojhsa [17]

Answer:

Explanation:

If I assume that the wind did not cause the plane to chage its velocity.

The plane will have a velocity of vp =  (0*i + 100*j) km/h relative to ground

The cart has a velocity of vc = (0*i - 20*j) km/h relative to the plane

vc' = vc + vp

vc' = (0*i + 100*j) + (0*i - 20*j) = (0*i + 80*j) km/h relative to the ground.

If I assume that the wind move the plane:

The plane will have a velocity of vp =  (-40*i + 100*j) km/h relative to ground

The cart has a velocity of vc = (0*i - 20*j) km/h relative to the plane

vc' = vc + vp

vc' = (-40*i + 100*j) + (0*i - 20*j) = (-40*i + 80*j) km/h relative to the ground.

In reality the wind would move the plane a little, not to the full speed of the wind, somewhere between these two values, but without more data it cannot be calculated.

6 0
2 years ago
You are wallpapering two walls of a room. One wall measures 15 ft by 12 ft and the other measures 9 ft by 12 ft. The wall paper
enyata [817]

Answer:

5.76 round off to 6

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wall 1 = 15 × 12 = 180

wall 2 = 9 × 12 = 108

now 1 roll covers 50 square feet

formula = wall 1 + wall 2 / 50

= 180 + 108 / 50

= 288÷ 50

= 5.76

4 0
2 years ago
A 0.500 kg aluminum pan on a stove is used to heat 0.250 liters of water from 20.0ºC to 80.0ºC. (a) How much heatis required? Wh
Jet001 [13]

Answer:

heat used to rise temperature pan =  30.1%

heat used to rise temperature water =  69.9%

Explanation:

Given data

mass of water = 0.250 liter = 0.250 kg

aluminum pan mass = 0.500 kg

initial temperature = 20.0ºC

final temperature =  80.0ºC

to find out

heat used to rise temperature of  pan and water

solution

we find here heat transferred to the water that is

heat transferred to the water = mass of water × specific heat of water × change in temperature    ...........1

specific heat of water is 4186 J/kgºC

so

heat transferred to the water = 0.250 × 4186 × (80-20) kJ

heat transferred to the water = 62.8 kJ

and

heat transferred to the aluminum that is

heat transferred to the aluminum = mass of aluminum × specific heat of aluminum × change in temperature    ...........2

here specific heat of aluminum is 900 J/kgºC

heat transferred to the aluminum = 0.500 × 900 × (80-20) kJ

heat transferred to the aluminum = 27 kJ

so

total heat = 62.8 + 27 = 89.8 kJ

so

heat used to rise temperature pan = 27/89.8 ×100% = 30.1%

heat used to rise temperature water = 62.8 / 89.8 ×100% = 69.9%

8 0
2 years ago
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