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bulgar [2K]
3 years ago
10

Suppose for your cookout you need to make 100 hamburgers. You know that 2.00 pounds will make 9 hamburgers. How many pounds will

you need?
Physics
1 answer:
VMariaS [17]3 years ago
8 0
2 pounds = 9 burgers figure out ow many 9's you can get out of 100: 100/9=11 but that only makes 99 you need 100 so we would add another one making 12. now multiply 12 by 2: 12·2=24. You would need 24 pounds of meet :)

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baseball is hit into the air at an initial speed of 37.2 m/s and an angle of 49.3 ° above the horizontal. At the same time, the
Agata [3.3K]

Answer:

The average speed of the fielder is 5.24 m/s

Explanation:

The position vector of the ball after it was hit can be calculated using the following equation:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem.

When the ball is caught, its position vector will be (see r1 in the figure):

r1 = (r1x, 0.873 m)

Then, using the equation of the position vector written above:

r1x = x0 + v0 · t · cos α

0.873 m = y0 + v0 · t · sin α + 1/2 · g · t²

Since the frame of reference is located at the point where the ball was hit, x0 and y0 = 0. Then:

r1x = v0 · t · cos α

0.873 m = v0 · t · sin α + 1/2 · g · t²

Let´s use the equation of the y-component of r1 to obtain the time of flight of the ball:

0.873 m = 37.2 m/s · t · sin 49.3° - 1/2 · 9.8 m/s² · t²

0 = -0.873 m + 37.2 m/s · t · sin 49.3° - 4.9 m/s² · t²

Solving the quadratic equation:

t = 0.03 s and t = 5.72 s.

It would be impossible to catch the ball immediately after it is hit at t = 0.03 s. Besides, the problem says that the ball was caught on its way down. Then, the time of flight of the ball is 5.72 s.

With this time, we can calculate r1x which is the horizontal distance traveled by the ball from home:

r1x = v0 · t · cos α

r1x = 37.2 m/s · 5.72 s · cos 49.3°

r1x = 1.39 × 10² m

The distance traveled by the fielder is (1.39 × 10² m - 1.09 × 10² m) 30.0 m.

The average velocity is calculated as the traveled distance over time, then:

average velocity = treveled distance / elapsed time

average velocity = 30.0 m / 5.72 s = 5.24 m/s

8 0
3 years ago
The basic barometer can be used to measure the height of a building. If the barometric readings at the top and the bottom of a b
erma4kov [3.2K]

Answer:

h = 269.6 m

Explanation:

Pressure at the bottom of the building and at the top of the building must be related as

P_{top} = P_{bottom} - \rho g h

P_{top} = 730 mm Hg

P_{bottom} = 755 mm Hg

now we will have

(755 \times 10^{-3})(13.6 \times 10^3)(9.81) = P_{bottom}

P_{bottom} = 1.007 \times 10^5 Pa

P_{top} = (730\times 10^{-3})(13.6 \times 10^3)(9.81)

P_{top} = 0.974 \times 10^5

now we have

(1.007 - 0.974)\times 10^5 = 1.25 (9.81) h

h = 269.6 m

7 0
3 years ago
How many types of electrical charge are there in all materials and what are the charges
JulsSmile [24]

There are two types of electric charges; positive and negative

- If you need more info than this let me know

- hope this helps

4 0
3 years ago
What is a moment of a force
kirza4 [7]

Answer:

In physics and mechanics, torque is the rotational equivalent of linear force. It is also referred to as the moment, moment of force, rotational force or turning effect, depending on the field of study. The concept originated with the studies by Archimedes of the usage of levers

7 0
2 years ago
A football kicked in front of a goal post at an angle of 45 degree to the ground just clear the top by of the post 3m high. Calc
Firlakuza [10]

Answer:

A. 10.84 m/s

B. 1.56 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

Time (T) taken to hit the ground again =?

A. Determination of the velocity of projection.

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

H = u²Sine²θ / 2g

3 = u²(Sine 45)² / 2 × 9.8

3 = u²(0.7071)² / 19.6

Cross multiply

3 × 19.6 = u²(0.7071)²

58.8 = u²(0.7071)²

Divide both side by (0.7071)²

u² = 58.8 / (0.7071)²

u² = 117.60

Take the square root of both side

u = √117.60

u = 10.84 m/s

Therefore, the velocity of projection is 10.84 m/s.

B. Determination of the time taken to hit the ground again.

Angle of projection (θ) = 45°

Velocity of projection (u) = 10.84 m/s

Time (T) taken to hit the ground again =?

T = 2uSine θ /g

T = 2 × 10.84 × Sine 45 / 9.8

T = 21.68 × 0.7071 / 9.8

T = 1.56 s

Therefore, the time taken to hit the ground again is 1.56 s.

7 0
3 years ago
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