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alexandr402 [8]
3 years ago
15

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

Physics
1 answer:
alina1380 [7]3 years ago
6 0

Answer:

Shoukd be a base hit

Explanation:

......

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A road sign says "Tucson, Arizona - 120 miles." How many kilometers is this?
Irina-Kira [14]
Slightly less than 200 (193)
3 0
2 years ago
Read 2 more answers
PLEASE HELP ME ASAP!!!!!!! <br> giving brainliest for correct answers
Vera_Pavlovna [14]

Answer:

b

Explanation:

6 0
3 years ago
A person travelled 350 m east from his home and returns back home an hour has displacement of_?​
Svetradugi [14.3K]

Answer:

vector of zero magnitude

Explanation:

The displacement is a vector magnitude, therefore, in addition to being a module, it has direction and sense.

In this case it moved 350 m and then returned the same 350 m, so the total displacement is zero.

If we draw the vector, one has a directional direction to the right and the other direction to the left, therefore when adding the two vectors gives a vector of zero magnitude

7 0
3 years ago
In moving out of a dormitory at the end of the semester, a student does 1.20 x 104 J of work. In the process, his internal energ
Zielflug [23.3K]

Answer:

(a) W=1.20×10⁴J

(b) U= -5.46×10⁴J

(c) Q= -4.26×10⁴J

Explanation:

Given that student does 1.20×10⁴J work

(a) W=1.20×10⁴J

Work done by student,so positive sign

During the process, his internal energy decreases by 5.46×10⁴J.

(b) U= -5.46×10⁴J.

As the Energy decreases therefore negative sign

For (c) Q

We know the formula

Q=W+U\\Q=1.2*10^{4}+(-5.46*10^{4} )\\ Q=-4.26*10^{4}J

8 0
3 years ago
Find the fundamental frequency and the next three frequencies that could cause standing-wave patterns on a string that is 30.0 m
maksim [4K]

Answer:

0.786 Hz, 1.572 Hz, 2.358 Hz, 3.144 Hz

Explanation:

The fundamental frequency of a standing wave on a string is given by

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

where

L is the length of the string

T is the tension in the string

\mu is the mass per unit length

For the string in the problem,

L = 30.0 m

\mu=9.00\cdot 10^{-3} kg/m

T = 20.0 N

Substituting into the equation, we find the fundamental frequency:

f=\frac{1}{2(30.0)}\sqrt{\frac{20.0}{(9.00\cdot 10^{-3}}}=0.786 Hz

The next frequencies (harmonics) are given by

f_n = nf

with n being an integer number and f being the fundamental frequency.

So we get:

f_2 = 2 (0.786 Hz)=1.572 Hz

f_3 = 3 (0.786 Hz)=2.358 Hz

f_4 = 4 (0.786 Hz)=3.144 Hz

6 0
3 years ago
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