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3 years ago

What happens when an immovable object meets an unstoppable force?”

Physics

2 answers:

Alisiya [41]3 years ago

4 0

Answer:

Sorry to tell you this but there’s no answer for that

Explanation:

Gennadij [26K]3 years ago

3 0

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Explanation:

You might be interested in

If u add 1.362 and 25.2 what is the correct answer?

krek1111 [17]

Answer:

26.562

Explanation:

1+25= 26

0.362+0.2= 0.562

26+0.562= 26.562

or just use a calculator

6 0

2 years ago

Assume that speed = 10 and miles = 5. What is the value of each of the following expressions? a. speed + 12 – miles * 2 b. speed

Sergio [31]

Explanation:

Speed = 10 and miles = 5

(a) Speed + 12 – miles * 2

10 + 12 - 5 × 2

= 12

(b) speed + miles × 3

10 + 5 × 3

= 25

(10 + 5) × 3

= 45

(d) speed + speed × miles + miles

10 + 10 × 5 + 5

= 65

(e) (10 – speed) + miles / miles

(10 - 10) + 5/5

= 1

Therefore, this is the required solution.

6 0

4 years ago

Evgen [1.6K]

Answer:

mass

Explanation:

5 0

3 years ago

A piece of Nichrome wire has a radius of 7.1 x 10-4 m. It is used in a laboratory to make a heater that dissipates 3.20×102 W of

bekas [8.4K]

Answer:

the length of the Nichrome wire is 59.88 m

Explanation:

The resistance of a wire is determined by:

P = V² / R or R = V² / P

where

V is the voltage
R is the resistance
P is the power

We also know that

R = PL / A

where

A is the area and it is equal to πr²
P = 3.20 × 10² W

Therefore, the length of the Nichrome wire is

L = RA / ρ

L = (V²/P)(πr²) / ρ

L = V²πr² / ρ P

L = (110 V)²(π)(7.1 × 10⁻⁴ m)² / (3.2 × 10² W)(100 × 10⁻⁸ Ω·m)

L = 59.88 m

7 0

4 years ago

A point source at the origin emits sound of frequency 175 Hz uniformly in all directions. On the x-axis at x=100 m, the sound in

hammer [34]

Answer:

Multiple answers:

1. Power output P=17.59W

2.Intensity 160m I=17.6W/ $m^{2}$

3. dB = 77.3

4. f=178.5 Hz

Explanation:

First one comes from the expression

$I=\frac{P}{4\pi r^{2} }$

where I is the intensity, P is the power and r is the radio of the spherical wave, or in this case, the distance x. I solved for the Power by multiplying Intensity with the area (4 $\pi x^{2}$

Second one is done with:

$\frac{I_{2} }{I_{1} } =\frac{x^{2}_{1} }{x^{2} _{2}}$

Solving for Intensity 2, the result mentioned.

The third is simply computed with

$dB=10*log\frac{I}{10^{-12} }$

And finally the last one is done with doppler effect, taking into account the speed of the air as in 10ºC 337m/s.

$f=f_{initial} *(\frac{s+v_{receiver} }{s+v_{source} } )$

Where Finitial is the frequency emitted and s is the speed of the sound. The wind blowing in positive is, in principle, going away of the observer.

8 0

3 years ago