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alexgriva [62]
3 years ago
10

What might be done to prevent acid rain damage to objects made from metal and carbonate, such as limestone? Identify a metal obj

ect that could be damaged by acid rain. Then, describe what could be done to prevent acid rain damage to it.
Physics
1 answer:
serg [7]3 years ago
7 0

Answer:  Bronze is a metal that can be damaged in acid rain. It is harsh to metals by almost burning them and may create rust. Everbrite is a clear, protective coating that will seal metal and bronze.

Explanation:

I know this is correct i have done all the research

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A bobsled has a momentum of 3000 kg•m/s to the east. After the rider gives it a push, its momentum increases to 5500 kg•m/s to t
Archy [21]
Impulse, denoted as J, is defined by the change in momentum. Since we have our initial and our final, we can solve for the change in momentum.

J = p_{f} - p_{0} = 5500 - 3000 = 2500

J = 2500 kg*m/s
7 0
3 years ago
Read 2 more answers
A person sitting on a pier observes incoming waves that have a sinusoidal form with a distance of 2.5 m between the crests. Of a
Doss [256]

Answer:

Part(a): The frequency is \bf{0.2~Hz}.

Part(b): The speed of the wave is \bf{0.5~m/s}.

Explanation:

Given:

The distance between the crests of the wave, d = 2.5~m.

The time required for the wave to laps against the pier, t = 5.0~s

The distance between any two crests of a wave is known as the wavelength of the wave. So the wavelength of the wave is \lambda = 2.5~m.

Also, the time required for the wave for each laps is the time period of oscillation and it is given by T = 5.0~s.

Part(a):

The relation between the frequency and time period is given by

\nu = \dfrac{1}{T}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

Substituting the value of T in equation (1), we have

\nu &=& \dfrac{1}{5.0~s}\\~~~&=& 0.2~Hz

Part(b):

The relation between the velocity of a wave to its frequency is given by

v = \nu \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)

Substituting the value of \nu and \lambda in equation (2), we have

v &=& (0.2~Hz)(2.5~m)\\~~~&=& 0.5~m/s

5 0
3 years ago
Jasmine and her partner want to test to see if there is a relationship between how much of her solar panel is exposed to sunligh
katen-ka-za [31]
Any one trial might have been done incorrectly.
7 0
2 years ago
A tow truck exerts a force of 1850 N on a 840 kg car. What is the acceleration of the car during this time?
Harrizon [31]

Answer:

<h2>2.2 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{1850}{840}  =  \frac{185}{84} \\   = 2.20238095...

We have the final answer as

<h3>2.2 m/s²</h3>

Hope this helps you

4 0
3 years ago
A water discharge of 9 m3/s is to flow through this horizontal pipe, which is 0.98 m in diameter. If the head loss is given as 1
Mashutka [201]

Answer: The power required by the pump to produce a discharge of 9m³/s is 5990joules/secs.

Explanation: The given parameters from the questions are:

Flow rate Q = 9m³/s, Diameter D = 0.98m, acceleration a = 1.0, head loss(Pressure P) is given by the function 10v²/2g.

STEP 1. Find the velocity of water in the pipe from the equation:

Diameter D = (√4.Q/π.v), where v is the velocity, and Q is flow rate

Making v subject of the formula gives:

v = 4Q/π.√D =[ 4 × 9m³/s / 3.142 × (√0.98m)] = 11.69m/s.

STEP 2. Find the pressure from the relationship, P = 10v²/2g, NB. g = a

P = 10 × (11.69m/s)² / 2× 1.0m/s²

P = 683.25N/m² or Pascal.

STEP 3. Find force exerted by the pump;

Recall that Pressure P = Force/Area

But Area A = π.r², where r = D/2

Therefore, A = π.(D/2)²

A = 3.142 × [0.98m/2]² = 0.75m²

Therefore, Force = Pressure × Area

Force F = 683.25N/m² × 0.75m²

F = 512.44N.

STEP 4. Find work done

Work done W by the pump is = Force × distance d moved by the water

W = F . d

Also recall that flow rate Q = Velocity/time.

Q = v/t, we can write t = v/Q.

Time t = 11.69m/s / 9m³/s = 1.298s

Also recall that velocity v = distance d/time t, v = d/t, making d subject of formula gives v × t

Distance d = v × t = 11.69m/s × 1.298s = 15.17m.

Hence,

Work Done W = Force × distance

W = 512.44N × 15.17m = 7775.56Nm or joules.

Lastly, Power P = Work done/ time

P = 7775.56joules/1.298s

P = 5990.4joules/s.

8 0
3 years ago
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