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Anna007 [38]
3 years ago
5

A Force Play is when the defense...

Physics
1 answer:
ivolga24 [154]3 years ago
6 0
Hi i play softball so the answer is the letter:

A or D but i think D
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How does gravity affect balanced/unbalanced forces?
viva [34]

Answer:

The force of gravity pulling downward and the force of the table pushing upwards on the book are of equal magnitude and opposite directions. These two forces balance each other. Yet there is no force present to balance the force of friction. ... Unbalanced forces cause accelerations.

Explanation:

5 0
3 years ago
Read 2 more answers
The height of a triangle is 4 in. greater than twice its base. The area of the triangle is no more than 168 in.2. Which inequali
olga2289 [7]

<span>The height (h) of a triangle is 4 in. greater than twice its base (x): h = 2x + 4
</span>The area of the triangle is no more than 168 in.2: A ≤ 168 in²

The area of the triangle is A = x*h/2
Hence x*h/2 ≤ 168 in²

Substitute h in the formula for the area:
h = 2x + 4 = 2(x + 2)
x*h/2 ≤ 168 in²

x * 2(x + 2)/2 ≤ 168
x * (x + 2) ≤ 168
x² + 2x ≤ 168
x² + 2x - 168 ≤ 0

Using the formula for quadratic equation:
x_{1,2} = \frac{-b+/- \sqrt{ b^{2}-4ac } }{2a} = \frac{-2+/- \sqrt{ 2^{2}-4*1*(-168) } }{2*1}=  \frac{-2+/- \sqrt{ 4+672 } }{2}=  \\  \\ \frac{-2+/- \sqrt{ 676 } }{2}=\frac{-2+/- 26 }{2} \\  \\ &#10;x_1 =  \frac{-2+26}{2} = \frac{24}{2} =12 \\  \\ x_1 =  \frac{-2-26}{2} = \frac{-28}{2} =-14&#10;

6 0
3 years ago
Read 2 more answers
The speed of sound in air is 345 m/s. A tuning fork vibrates above the open end of a sound resonance tube. If sound waves have w
anzhelika [568]

Answer:

594.8 Hz

Explanation:

Parameters given:

Speed of sound, v = 345 m/s

Wavelength = 58 cm = 0.58 m

Speed of a wave is given as:

Speed = wavelength * frequency

Therefore:

Frequency = Speed/Wavelength

Frequency = 345/0.58

Frequency = 594.8 Hz

8 0
3 years ago
Help please I’ll mark as brainliest
kirza4 [7]

Answer:

yes

Explanation:

6 0
3 years ago
Phobos's Orbit. Phobos orbits Mars at a distance of 9,380 km from the center of the planet and has a period of 0.3189 days. Assu
Elenna [48]

Answer:

Explanation:

The relation between time period of moon in the orbit around a planet can be given by the following relation .

T² = 4 π² R³ / GM

G is gravitational constant , M is mass of the planet , R is radius of the orbit and T is time period of the moon .

Substituting the values in the equation

(.3189 x 24 x 60 x 60 s)²  = 4 x 3.14² x ( 9380 x 10³)³ / (6.67 x 10⁻¹¹ x M)

759.167 x 10⁶ = 8.25 x 10²⁰ x 39.43 / (6.67 x 10⁻¹¹ x M )

M = .06424  x 10²⁵

= 6.4 x 10²³ kg .

4 0
3 years ago
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