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sergiy2304 [10]
3 years ago
7

What is the term used when a ball is hit and the batter reaches the following bases safely (without being called out)?

Physics
2 answers:
trasher [3.6K]3 years ago
8 0
First base would be the term
andre [41]3 years ago
4 0
First base is the answer
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Two massless bags contain identical bricks, each brick having a mass M. Initially, each bag contains four bricks, and the bags m
stepladder [879]

Answer: F_{2}=\frac{3}{4}F_{1}

Explanation:

According to Newton's law of universal gravitation:

F=G\frac{m_{1}m_{2}}{r^2}

Where:

F is the module of the force exerted between both bodies

G is the universal gravitation constant.

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

In this case we have two situations:

1) Two bags with masses 4M and 4M mutually exerting a gravitational attraction F_{1} on each other:

F_{1}=G\frac{(4M)(4M)}{r^2}   (1)

F_{1}=G\frac{16M^2}{r^2}   (2)

F_{1}=16\frac{GM^2}{r^2}   (3)

2) Two bags with masses 2M and 6M mutually exerting a gravitational attraction F_{2} on each other (assuming the distance between both bags is the same as situation 1):

F_{2}=G\frac{(2M)(6M)}{r^2}   (4)

F_{2}=G\frac{12M^2}{r^2}   (5)

F_{2}=12\frac{GM^2}{r^2}   (6)

Now, if we isolate \frac{GM^2}{r^2} from (3):

\frac{F_{1}}{16}=\frac{GM^2}{r^2}   (7)

Substituting \frac{GM^2}{r^2}  found in (7) in (6):

F_{2}=12(\frac{F_{1}}{16})   (8)

F_{2}=\frac{12}{16}F_{1}   (9)

Simplifying, we finally get the expression for F_{2}  in terms of F_{1} :

F_{2}=\frac{3}{4}F_{1}  

5 0
3 years ago
A radio technician measures the frequency of an AM radio transmitter. The frequency is . What is the frequency in megahertz? Wri
sukhopar [10]

Complete Question

A radio technician measures the frequency of an AM radio transmitter. The frequency is 14603 kHz . What is the frequency in megahertz? Write your answer as a decimal.

Answer:

The value is  x =  14.6 \  MHz

Explanation:

From the question we are told that

  The  frequency is  f =  14603  \  kHz = 14603 *1000 = 14603000 \ Hz

Generally  

       1 Hz \to  1.0 *10^{-6} \  MHz

       14603000 \ Hz  \to x MHz

=>   x =  \frac{14603000 *  1.0*10^{-6}}{1 }

=>    x =  14.6 \  MHz

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