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Ronch [10]
3 years ago
5

In which city rusting is a problem delhi or mumbai

Physics
1 answer:
ss7ja [257]3 years ago
3 0

Answer:

Mumbai

Explanation:

Because Mumbai is a coastal city

due to high humidity,vigorous rusting damages all the iron structures

∴ rusting is a major problem in Mumbai

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Two cars, initially at rest and 5 km apart at t=0 , simultaneously move toward each other. Car A travels at a constant speed of
Anastasy [175]

Answer:

<em>d. 268 s</em>

Explanation:

<u>Constant Speed Motion</u>

An object is said to travel at constant speed if the ratio of the distance traveled by the time taken is constant.

Expressed in a simple equation, we have:

\displaystyle v=\frac{d}{t}

Where  

v = Speed of the object

d = Distance traveled

t  = Time taken to travel d.

From the equation above, we can solve for d:

d = v . t

And we can also solve it for t:

\displaystyle t=\frac{d}{v}

Two cars are initially separated by 5 km are approaching each other at relative speeds of 55 km/h and 12 km/h respectively. The total speed at which they are approaching is 55+12 = 67 km/h.

The time it will take for them to meet is:

\displaystyle t=\frac{5}{67}

t = 0.0746 hours

Converting to seconds: 0.0746*3600 = 268.56

The closest answer is d. 268 s

8 0
3 years ago
A copper telephone wire has essentially
Lunna [17]

Answer:

128.21 m

Explanation:

The following data were obtained from the question:

Initial temperature (θ₁) = 4 °C

Final temperature (θ₂) = 43 °C

Change in length (ΔL) = 8.5 cm

Coefficient of linear expansion (α) = 17×10¯⁶ K¯¹)

Original length (L₁) =.?

The original length can be obtained as follow:

α = ΔL / L₁(θ₂ – θ₁)

17×10¯⁶ = 8.5 / L₁(43 – 4)

17×10¯⁶ = 8.5 / L₁(39)

17×10¯⁶ = 8.5 / 39L₁

Cross multiply

17×10¯⁶ × 39L₁ = 8.5

6.63×10¯⁴ L₁ = 8.5

Divide both side by 6.63×10¯⁴

L₁ = 8.5 / 6.63×10¯⁴

L₁ = 12820.51 cm

Finally, we shall convert 12820.51 cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

12820.51 cm = 12820.51 cm × 1 m / 100 cm

12820.51 cm = 128.21 m

Thus, the original length of the wire is 128.21 m

5 0
3 years ago
A balloon filled with helium gas at 1.00 atm occupies 15.6 L. Will the volume of the balloon increase or decrease in the upper a
Mariana [72]

Answer:

The volume of the balloon increases in the upper atmosphere.

Explanation:

p1= 1 atm

p2= 0.15 atm

V1= 15.6 L

V2= ?

p1*V1= p2 * V2

V2= (p1/p2)*V1

V2= 104 L

6 0
3 years ago
Use the following half-life graph to answer the following question:
Temka [501]

Answer:

A 1.0 min

Explanation:

The half-life of a radioisotope is defined as the time it takes for the mass of the isotope to halve compared to the initial value.

From the graph in the problem, we see that the initial mass of the isotope at time t=0 is

m_0 = 50.0 g

The half-life of the isotope is the time it takes for half the mass of the sample to decay, so it is the time t at which the mass will be halved:

m'=\frac{50.0 g}{2}=25.0 g

We see that this occurs at t = 1.0 min, so the half-life of the isotope is exactly 1.0 min.

3 0
3 years ago
What is the pendulum length whose period is 2.0s ?
Mashutka [201]
Formula\ for\ period:\\\ T=2 \pi \sqrt{\frac{L}{g}}\\\ g-gravity=9,8 \frac{m}{s^2} ,\ L-pendulum \ length \\\\ \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\\\ \frac{T^2}{2 \pi } = \frac{L}{g} \\\\\ \frac{T^2}{2 \pi }*g=L\\\\ L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24mT=2 \pi   \sqrt{\frac{L}{g}} \\&#10; \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\&#10; \frac{T^2}{2 \pi }  = \frac{L}{g} \\&#10; \frac{T^2}{2 \pi }*g=L\\&#10;L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24m&#10;
7 0
3 years ago
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