Putting the ball back into play by the offense is called a ____ ___

A rescue pilot drops a survival kit while her plane is flying at an altitude of 2000.0 m with a forward velocity of 100.0 m/s.

jeyben [28]

I believe it’s 2020m? I’m sorry if it’s wrong

8 0

4 years ago

Increasing the mass attached to a spring will decrease the angular frequency of its vibrations. True or False

Scrat [10]

Answer:

decreases

Explanation:

The angular frequency of the mass is given by

$\omega =\sqrt{\frac{k}{m}}$

where, k is the spring constant and m be the mass of the body which is doing oscillations.

if the mass of the body increases, so the value of angular frequency decreases, as the angular frequency is inversely proportional to the square root of the mass of the body.

4 0

4 years ago

0.001225 kg/L x 720 000 000L =?

I am Lyosha [343]

Answer:

0.001225 kg/L × 720 000 000 L = 882000 kg

Explanation:

Given:

The equation to solve is given as:

0.001225 kg/L × 720 000 000 L = ?

Let us write each term of the product in terms of power of 10.

As 0.001225 has 6 digits after the decimal place, therefore, we use the exponent 6 for 10 and the sign is negative. This gives,

$0.001225\ kg/L = 1225\times 10^{-6}\ kg/L$

Now, for 720000000 L there are 6 zeros after 720. So, we use exponent 6 but with a positive sign. This gives,

$720000000\ L=720\times 10^{6}\ L$

Now, finding the product, we get:

$0.001225\ kg/L\times 720000000\ L\\\\=1225\times 10^{-6}\ kg/L\times 720\times 10^6\ L\\\\=(1225\times 720)\times (10^6\times 10^{-6})\ (\frac{kg}{L}\times L)\\\\=882000\times 10^{6-6}\ kg\\\\=882000\times 10^0\ kg\\\\=882000\times 1\ kg\\\\=882000\ kg$

Therefore, the product is equal to:

0.001225 kg/L × 720 000 000 L = 882000 kg

6 0

4 years ago

Which force can affect an object without direct physical contact?

denis-greek [22]

Gravity affet everything and it touches nothing.
Hope this helps!

8 0

4 years ago

Read 2 more answers

While tuning a string to the note C at 523 Hz, a piano tuner hears 2.00 beats/s between a reference oscillator and the string.

lara31 [8.8K]

Answer:

a)the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b) 526hz

c)0.989 or a 1.14% decrease in tension

Explanation:

a) While tuning a string at 523 Hz,piano tuner hears 2.00 beats/s between a reference oscillator and the string.

The possible frequencies of the string can be calculated by

fl=f' - B

where

fl= lower limit of the possible frequency

f'= frequency of the string

B= beat heard by the tuner

fl= 523hz + Or - (2beats/secs * 1hz/1beat per sc)

fl= 521hz or 525hz

So the possible frequencies are 521hz ,522hz, 523, 524hz,525hz

b)fl=f' - B

523hz= f' - 3

f'= 523 + 3= 526hz

c) The tension is directly proportional to the square of the frequencies

T1/T2 =f1^2/f2^2

523^2 / 526^2 = 0.989 or a 1.14% decrease in tensio

6 0

3 years ago

Putting the ball back into play by the offense is called a ____ ____

Putting the ball back into play by the offense is called a