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Luden
1 month ago
11

A second baseman tosses the ball to the first baseman, who catches it at the same level from which it was thrown. The throw is m

ade with an initial speed of 18.0m/s at an angle of 37.5 above the horizontal. A) What is the horizontal component of the ball's velocity just before it is caught? B) How long is the ball in the air?. . Can someone please explain to me how this works not just the answer I'm somewhat confused.
Physics
1 answer:
Yuliya22 [3.3K]1 month ago
6 0
 (a) Considering the pitch angle of 37.5° along with the initial speed of the particle at 18.0 ms^-1, we calculate: 18.0*cos37.5 = v_x = 14.28 ms^-1, which indicates the horizontal component of the projectile. (b) We calculate the vertical component similarly: 18.0*sin37.5 = v_y = 10.96 ms^-1. This value is divided by acceleration, a_y, to find the time until maximum displacement, yielding 10.96/9.8 = 1.12 s. Finally, doubling this result provides the total time the particle remains with r_y > 0, which is 2.24 s. I hope this information is useful. Thank you for submitting your question.
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The convergence of light rays redirects them toward the focal point, resulting in a magnifying effect.

Explanation:

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2 months ago
A basketball player is running at a constant speed of 2.5 m/s when he tosses a basketball upward with a speed of 6.0 m/s. How fa
kicyunya [3294]
A basketball player maintains a steady pace of 2.5 m/s while throwing a basketball vertically at 6.0 m/s. How far does the player advance before getting the ball back? Air resistance is negligible. I was unsure which formula to apply to this scenario. Is there any relevance to an angle? First, we determine the duration to reach peak height. The total time for the flight will be double the ascent duration. According to Newton's equations of motion: v = u + at. At the highest point, v = 0, where u is 6 m/s. Thus, the equation becomes 0 = 6 - 9.81t, leading us to t = 0.61 seconds. Therefore, the total flight time equals 1.22 seconds as the player runs towards the ball at a horizontal speed of 2.5 m/s. The distance traveled can be calculated using distance = speed × time, resulting in distance = 2.5 m/s * 1.22, yielding a final distance of 6.11m.
3 0
2 months ago
To study torque experimentally, you apply a force to a beam. One end of the beam is attached to a pivot that allows the beam to
kicyunya [3294]

Answer:

a)  τ = i ^ (y F_{z} - z F_{y}) + j ^ (z Fₓ - x F_{z}) + k ^ (x F_{y} - y Fₓ)

b) τ = (-0.189i ^ -5.6 j ^ + 33.978k ^) N m

c) α = (-7.8 10⁻⁴ i ^ - 2.3 10⁻² j ^ + 1.4 10⁻¹ k ^) rad / s²

Explanation:

a) The torque can be expressed as

        τ = r x F

To tackle this equation, using the determinant approach is the most straightforward method

        \tau =\left[\begin{array}{ccc}i&j&k\\x&y&z\\F_{x}&F_{y} &F_{z}\end{array}\right]  

The resulting expression is

      τ = i ^ (y F_{z} - z F_{y}) + j ^ (z Fₓ - x F_{z}) + k ^ (x F_{y} - y Fₓ)

b) Now let's compute

     τ = i ^ (0.075 1.4 -0.035 8.4) + j ^ (0.035 2.8 - 4.07 1.4) + k ^ (4.07 8.4 - 0.075 2.8)

     τ = i ^ (- 0.189) + j ^ (-5.6) + k ^ (33,978)

     τ = (-0.189i ^ -5.6 j ^ + 33.978k ^) N m

c) To find angular acceleration, we use

       τ = I α

       α = τ / I

The moment of inertia being a scalar means that only the magnitude of each component changes, orientation remains constant.

           

     α = (-0.189i^  -5.6 j^  + 33.978k^) / 241

     α = (-7.8 10⁻⁴ i ^ - 2.3 10⁻² j ^ + 1.4 10⁻¹ k ^) rad / s²

8 0
2 months ago
A person who weighs 625 N is riding a 98-N mountain bike. Suppose that the entire weight of the rider and bike is supported equa
Keith_Richards [3271]

Answer:

A = 4.76 x 10⁻⁴ m²

Explanation:

Given data:

Person's weight = 625 N

Bike's weight = 98 N

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Find: Contact area per tire

Total system weight = 625 + 98 = 723 N

Let F represent the force supported by each tire

2F = 723 N

Therefore, F = 361.5 N

Using the formula F = P × A

A = \dfrac{F}{P}

A = \dfrac{361.5}{7.60 \times 10^5}

Contact area, A = 4.76 x 10⁻⁴ m²

7 0
3 months ago
If you double the mass of an object while leaving the net force unchanged what is the result
kicyunya [3294]

Answer:

If the net force acting on an object doubles, its acceleration also doubles. Conversely, if the mass is doubled, the acceleration will be halved. If both the net force and mass are doubled, the acceleration remains the same.

Explanation:

[[TAG_9]][[TAG_10]]
5 0
1 month ago
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