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velikii [3]
3 years ago
11

A golfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. What is the horizontal velocity compo

nent of the ball? Round the answer to the nearest tenth of a m/s
Physics
2 answers:
Sever21 [200]3 years ago
8 0
The horizontal component of the velocity of the ball is calculated by multiplying the speed by the cosine of the given angle. 
                        x-component of speed = (31 m/s)(cos 35°) 
                                                               = 25.39 m/s
Thus, the horizontal velocity component of the ball is 25.39 m/s. 
lawyer [7]3 years ago
5 0

Answer:

Horizontal component of the speed is given as

v_x = 25.4 m/s

Explanation:

As we know that if a vector is inclined at some angle with the horizontal then the component of the vector along and perpendicular to horizontal is given as

A_x = Acos\theta

A_y = Asin\theta

now here we need to find the component of velocity

so we know that

v = 31 m/s

\theta = 35^0

now we will have

v_x = 31cos35

v_x = 25.4 m/s

so horizontal component of the speed will be 25.4 m/s

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Naturally occurring element X exists in three isotopic forms: X-28 (27.977 amu, 92.23% abundance), X-29 (28.976 amu, 4.67% abund
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Answer:

C. 28.09 amu

Explanation:

The natural occurring element exist in 3 isotopic forms: namely X-28 (27.977 amu, 92.23% abundance),  X-29 (28.976 amu, 4.67% abundance) and  X-30 (29.974 amu, 3.10% abundance).

The atomic weight of elements depends on the isotopic abundance. If you know the fractional abundance and the mass of the isotopes the atomic weight can be computed.

The atomic weight is computed as follows:

atomic weight = mass of X-28 × fractional abundance + mass of X-29 × fractional abundance + mass of  X-30 × fractional abundance

atomic weight = 27.977 × 0.9223 + 28.976 × 0.0467 + 29.974 ×  0.0310

atomic weight = 25.8031871 + 1.3531792 + 0.929194

atomic weight = 28.0855603 amu

To 2 decimal place atomic weight = 28.09 amu

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Three metal spheres are placed as shown in the image. Sphere A weighs 5 kg, B weighs 8 kg, and C weighs 3 kg.
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B

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The following equation, N2 + 3 H2 —>2 NH3 ,describes a
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Within which type of water body does water move the slowest?
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An aquarium open at the top has 30-cm-deep water in it. You shine a laser pointer into the top opening so it is incident on the
Setler [38]

Answer:

You must add 8cm of water to the tank

Explanation:

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This law relates the index of refraction of the water (n2), the index of refraction of the air (n1), the incidence angle relative to the vertical (theta1) and the refraction angle relative to the vertical (theta2) by using the next equation:

(n1)*(sin(theta1))=(n2)*(sin(theta2))

Then we will find the refraction angle relative to the vertical this way:

(n1/n2)*(sin(theta1))=sin(theta2)

(1/1.33)*(sin(45))=sin(theta2)

Then, theta2=32.12°

Now that we have this information we can imagine a triangle with a 30cm height and a 32.12° angle. This way we can find how much X is, this X will be the distance between the vertical line and the spot the beam hits the bottom, so we can use some trigonometry to find it, this way:

tan(32.12)=(X/30cm)

X=(tan(32.12))*(30cm)

Then, X=18.8cm, we can approximate it to 19cm

Once we have X we will add 5cm to it which is how much the beam needs to be moved, then the new X will be 24cm

Now, with the new horizontal distance we will find the new vertical distance, let´s call it Y, this way we will know how much water we must add to move the beam, then we will have a triangle with a vertical distance called Y, the same 32.12° angle will be used as we are still working with the air-water interface and a 19cm horizontal distance, then:

tan(32.12)=(24cm/Y)

Y=(24cm/tan(32.12))

Then, Y=38cm

In this case, you must add 8cm of water to the tank to move the beam on the bottom 5cm

5 0
3 years ago
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