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vazorg [7]
3 years ago
7

If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th

an the rest of the baseball field, at what height would the catcher need to hold his glove to catch the pitched ball? (Note: neglect air resistance, estimate the acceleration due to gravity as 10 m/s2, and assume the pitcher is only throwing the ball horizontally.)A. 2.0 m above the groundB. 1.8 m above the groundC. 0.5 m above the groundD. 0.2 m above the ground
Physics
1 answer:
neonofarm [45]3 years ago
7 0

Answer: A. 2.0m above ground

Explanation:

Height of the pitcher from the floor of his mound = 1.8m

Height of the mound above the rest of the field = 0.2m

Since the height difference between the mound and the ground the catcher stands on = 0.2m and the pitcher throws the ball horizontally, then the catcher has to hold his glove at (1.8 + 0.2)m

above ground.

= 2.0m above ground (option A)

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Which of the following is true about light waves and sound waves? A. Sound waves can move at different speeds, but light waves a
vivado [14]

Answer:

C

Explanation:

Sound waves speed up noticeably when moving through a solid or liquid, because all it is is just particles colliding; and particles are way closer together with those states of matter.

The speed of light can change when moving through different substances, but this is dependent on complicated factors such as frequency, polarization, intensity, et. cetera

The important part is that it does change speed, so your answer is C.

Hope this helps!

3 0
3 years ago
The brain mass of a human fetus during a particular trimester can be accurately estimated from the circumference of the head by
Paladinen [302]

Answer:

a. If c = 20 cm, then the mass of the brain is m = 5 g.

b. At c = 20 cm, the brain's mass is increasing at a rate of 15.75 g/cm.

Explanation:

From the equation

m\left(c\right) = \frac{c^3}{100}-\frac{1500}{c}

we have

a. for c = 20 cm

m\left(20\right)=\frac{20^3}{100}-\frac{1500}{20}=5,

then the mass is m(20) = 5 g.

b. In order to find the rate of change, first we derivate

\frac{dm}{dc}=\frac{3c^2}{100}+\frac{1500}{c^2}.

Evaluated at c = 20 cm, we have

\frac{dm}{dc}|_{c=20}=\frac{3\times 20^2}{100}+\frac{1500}{20^2}=15.75.

So, at c = 20 cm, the mass of the brain is increasing at a rate of 15.75 g/cm.

3 0
3 years ago
What is the energy (in evev) of a photon of visible light that has a wavelength of 500 nmnm?
lisabon 2012 [21]
<h3>Answer:</h3>
  • E≈2,5 eV
<h3>Explanation:</h3>

_______________

λ=500 nm = 500·10⁻⁹ m

c=3·10⁸ m/s

h=6,63·10⁻³⁴ J·s = 4,14·10⁻¹⁵ eV·s

_______________

E - ?

_______________

\displaystyle \boldsymbol{E}=h\nu =h \frac{c}{\lambda} =4,14\cdot 10^{-15} \; eV\cdot s\cdot \; \frac{3\cdot 10^8\; m/s}{500\cdot 10^{-9}\; m} =2,484\;  eV\approx \boldsymbol{2,5\; eV}

6 0
2 years ago
A block and tackle pulley system of 6 pulley uses 300N to overcome 1200N.find efficiency
Fittoniya [83]

Answer:

66.7%

Explanation:

There are 6 pulleys, so the mechanical advantage is 6.

The theoretical force needed to overcome 1200 N is therefore:

1200 N / 6 = 200 N

The actual force is 300 N.  So the efficiency is:

200 N / 300 N × 100% = 66.7%

4 0
3 years ago
Can someone solve this problem and explain to me how you got it​
Zarrin [17]

1) The electric force changes by a factor of 25

2) The electric force changes by a factor of 16/9

Explanation:

1)

The magnitude of the electrostatic force between two charges is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges

r is the separation between the two charges

In this problem, let's call F the initial force between the two charges when they are at a distance of r.

Later, the distance is changed by a factor of 5. Let's assume it has been increased to a factor of 5: so the new distance is

r' = 5r

Therefore, the new force between the charges is:

F' = k' \frac{q_1 q_2}{r'^2}=k' \frac{q_1 q_2}{(5r)^2}=\frac{1}{25}(k' \frac{q_1 q_2}{r'^2})=\frac{F}{25}

So, the force has changed by a factor of 25.

2)

The original force between the two charges is

F=k\frac{q_1 q_2}{r^2}

In this problem, we have:

- The distance between the charges is changed by a factor of 6:

r' = 6r

- The charges are both changed by a factor of 8:

q_1' = 8q_1

q_2' = 8q_2

Substituting into the equation, we find the new force:

F' = k' \frac{q_1' q_2'}{r'^2}=k' \frac{(8q_1) (8q_2)}{(6r)^2}=\frac{64}{36}(k' \frac{q_1 q_2}{r'^2})=\frac{16}{9}F

So, the force has changed by a factor of 16/9.

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0
3 years ago
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