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3 years ago

PLEASE HELP ASAP...TIMED TEST. PLEASE ANSWER WITH AT LEAST A PARAGRAPH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Physics

1 answer:

charle [14.2K]3 years ago

8 0

Answer: I am pretty sure that you should pick radio waves.

Explanation: The scientist should use radio waves. I think this because you can use the radio waves to analyze the signals from outer space. This will work much better than anything there, to analyze it the best possible.

The best I could do.

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An automobile having a mass of 1,000 kg is driven into a brick wall in a safety test. The bumper behaves like a spring with cons

vlada-n [284]

Answer:

$v=2.02\frac{m}{s}$

Explanation:

Assuming no energy lost, according to the law of conservation of energy, the kinetic energy of the automobile becomes potential energy after the crash:

$K=U\\\frac{mv^2}{2}=\frac{kx^2}{2}$

Here m is the automobile's mass, v is the speed of the car before impact, k is the "bumper" constant and x is the compression of the bumper due to the collision. Solving for v:

$v=x\sqrt\frac{k}{m}\\v=2.63*10^{-2}m\sqrt{\frac{5.9*10^6\frac{N}{m}}{10^3kg}}\\v=2.02\frac{m}{s}$

8 0

3 years ago

A formula for the normal systolic blood pressure for a man age a , measured in mmhg, is given as p=0. 006a2−0. 02a 120. Find the

djyliett [7]

The age of a man whose normal blood pressure measures 123 mm of hg

9 years

<h3>What is Quadratic equation ?</h3>

A quadratic equation as an equation of degree 2, meaning that the highest exponent of this function is 2. The standard form of a quadratic equation is y = a $x^{2}$ + bx + c, where a, b, and c are numbers and a cannot be 0

P(A) = 0.006 $a^{2}$ - 0.02a + 120

123 = 0.006- 0.02a + 120

0=0.006 $a^{2}$ - 0.02a - 3

you can use the quadratic equation formula to solve for the man's age.

A = (-b ± ( $\sqrt{b^{2} - 4*a*c}$ ) ) / (2a)

A = (0.02 ± $\sqrt{(-0.02)^{2} - 4*0.006*(-3)}$ / (2*0.006)

A = (0.02 ± $\sqrt{0.0076}$ ) / 0.012

A = 9 , -5.67

Age of the man will be 9 years

To learn more about quadratic equation here

brainly.com/question/17177510?referrer=searchResults

#SPJ4

5 0

2 years ago

A student shakes a rope such that 20 complete vibrations are made in 4.00 seconds. Determine the vibrational frequency of the ro

fgiga [73]

Answer:

The vibrational frequency of the rope is 5 Hz.

Explanation:

Given;

number of complete oscillation of the rope, n = 20

time taken to make the oscillations, t = 4.00 s

The vibrational frequency of the rope is calculated as follows;

$Frequency = \frac{number \ of \ complete \ vibrations}{time \ taken} \\\\Frequency = \frac{20 }{4 \ s} \\\\Frequency = 5 \ Hz$

Therefore, the vibrational frequency of the rope is 5 Hz.

8 0

3 years ago

The howler monkey is the loudest land animal and, under some circumstances, can be heard up to a distance of 8.9 km. Assume the

Airida [17]

Answer:

$\frac{I}{I_0}=113.68$

Explanation:

P = Acoustic power = 63 µW

r = Distance to the sound source = 210 m

Acoustic power

$P=IA\\\Rightarrow I=\frac{P}{A}\\\Rightarrow I=\frac{63\times 10^{-6}}{4\times \pi \times 210^2}$

Threshold intensity = $I_0=1\times 10^{-12}\ W/m^2$

Ratio

$\frac{I}{I_0}=\frac{\frac{63\times 10^{-6}}{4\times \pi \times 210^2}}{1\times 10^{-12}}\\\Rightarrow \frac{I}{I_0}=113.68$

Ratio of the acoustic intensity produced by the juvenile howler to the reference intensity is 113.68

6 0

3 years ago

At one instant, the center of mass of a system of two particles is located on the x-axis at 2.0 cm and has a velocity of (5.0 m/

Nata [24]

Answer:

Explanation:

Given that,

At one instant,

Center of mass is at 2m

Xcm = 2m

And velocity =5•i m/s

One of the particle is at the origin

M1=? X1 =0

The other has a mass M2=0.1kg

And it is at rest at position X2= 8m

a. Center of mass is given as

Xcm = (M1•X1 + M2•X2) / (M1+M2)

2 = (M1×0 + 0.1×8) /(M1 + 0.1)

2 = (0+ 0.8) /(M1 + 0.1)

Cross multiply

2(M1+0.1) = 0.8

2M1 + 0.2 =0.8

2M1 = 0.8-0.2

2M1 = 0.6

M1 = 0.6/2

M1 = 0.3kg

b. Total momentum, this is an inelastic collision and it momentum after collision is given as

P= (M1+M2)V

P = (0.3+0.1)×5•i

P = 0.4 × 5•i

P = 2 •i kgm/s

c. Velocity of particle at origin

Using conversation of momentum

Momentum before collision is equal to momentum after collision

P(before) = M1 • V1 + M2 • V2

We are told that M2 is initially at rest, then, V2=0

So, P(before) = 0.3V1

We already got P(after) = 2 •i kgm/s in part b of the question

Then,

P(before) = P(after)

0.3V1 = 2 •i

V1 = 2/0.3 •i

V1 = 6 ⅔ •i m/s

V1 = 6.667 •i m/s

4 0

3 years ago