There is T__ as much voltage. This<br> produces twice the current

3 PHYSICAL SCIENCE: In order to figure out what is in a sealed box, you perform some tests. The box seems heavy for its size and

rjkz [21]

Some sort of magnetic metal

Metals are heavier per cubic unit than other materials such as air or water, and also are much more magnetic than other materials

7 0

3 years ago

An astronaut on Pluto attaches a small brass ball to a 1.00-m length of string and makes a simple pendulum. She times 10 complet

pickupchik [31]

Answer:

The acceleration due to gravity at Pluto is 0.0597 m/s^2.

Explanation:

Length, L = 1 m

10 oscillations in 257 seconds

Time period, T = 257/10 = 25.7 s

Let the acceleration due to gravity is g.

Use the formula of time period of simple pendulum

$T = 2\pi\sqrt{\frac{L}{g}}\\\\25.7 = 2 \times 31.4\sqrt{\frac{1}{g}}\\\\g = 0.0597 m/s^2$

7 0

3 years ago

A pressure wave is what type of wave

Alex777 [14]

propagated disturbance is a variation

5 0

2 years ago

Water at room temperature is discharged from a pipe at a rate of 1000 gallons per minute (gpm). Express this flow rate in cubic

marshall27 [118]

Answer

given,

discharge rate from pipe = 1000 gallons/minutes

now,

flow rate in cubic meters per second

1 gallon = 0.00378541 m³

1 min = 60 s

Q = $1000\times \dfrac{0.00378541\ m^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}$

Q = 0.063 m³/s

flow rate in liters per minute

1 gallon = 3.78541 L

Q = $1000\times \dfrac{3.78541\ m^3}{1\ gallon}$

Q = 3785.41 m³/min

flow rate in cubic feet per second

1 gallon = 0.133681 ft³

1 min = 60 s

Q = $1000\times \dfrac{0.133681\ ft^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}$

Q = 2.23 ft³/s

4 0

3 years ago

You are trying to overhear a juicy conversation, but from your distance of 20.0 m , it sounds like only an average whisper of 30

12345 [234]

Answer:

r₂ = 0.2 m

Explanation:

given,

distance = 20 m

sound of average whisper = 30 dB

distance moved closer = ?

new frequency = 80 dB

using formula

$\beta = 10 log(\dfrac{I_1}{I_0})$

I₀ = 10⁻¹² W/m²

now,

$30 = 10 log(\dfrac{I_1}{10^{-12}})$

$\dfrac{I_1}{10^{-12}}= 10^3$

$I_1= 10^{-8}\ W/m^2$

to hear the whisper sound = 80 dB

$80 = 10 log(\dfrac{I_2}{10^{-12}})$

$\dfrac{I_2}{10^{-12}}= 10^8$

$I_2= 10^{-4}\ W/m^2$

we know intensity of sound is inversely proportional to square of distances

$\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}$

$\dfrac{10^{-8}}{10^{-4}}=\dfrac{r_2^2}{20^2}$

$10^{-4}=\dfrac{r_2^2}{20^2}$

r₂ = 0.2 m

6 0

3 years ago

There is T__ as much voltage. This produces twice the current