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victus00 [196]
3 years ago
13

A Porsche challenges a Honda to a 500 m race

Physics
1 answer:
zlopas [31]3 years ago
8 0
A Porsche will always win no matter what
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If a scuba diver fills his lungs to full capacity of 5.5 L when 10 m below the surface, to what volume would his lungs expand if
umka21 [38]

Answer:

The volume at the surface is 10.97 L.

Explanation:

Given that,

Volume = 5.5 L

Height = 10 m

Density of sea water= 1025 kg/m³

We need to calculate the pressure at that point

Using formula of pressure

P'=P+\rho gh

Put the value into the formula

P'=1.01\times10^{5}+1025\times9.8\times10

P'=201450\ Pa

We need to calculate the volume at the surface

Using equation of ideal gas

PV= RT

So, for both condition

PV=P'V'

Put the value into the formula

V=\dfrac{201450\times5.5}{1.01\times10^{5}}

V=10.97\ L

Hence, The volume at the surface is 10.97 L.

3 0
3 years ago
Read 2 more answers
The robot HooRU is lost in space, floating around aimlessly, and radiates heat into the depths of the cosmos at the rate of 13.1
ahrayia [7]

Answer:

The temperature is  T  = 168.44 \ K

Explanation:

From the question ewe are told that

   The rate of heat transferred is    P  = 13.1 \ W

     The surface area is  A = 1.55 \ m^2

      The emissivity of its surface is  e = 0.287

Generally, the rate of heat transfer is mathematically represented as

           H  =  A e \sigma  T^{4}

=>         T  =  \sqrt[4]{\frac{P}{e* \sigma } }

where  \sigma is the Boltzmann constant with value  \sigma  = 5.67*10^{-8} \ W\cdot  m^{-2} \cdot  K^{-4}.

substituting value  

             T  =  \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }

            T  = 168.44 \ K

7 0
3 years ago
A(n) 131 g ball is dropped from a height
larisa [96]

Answer:

26.59 N/m

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 131 g

Extention (e) = 4.82755 cm

Acceleration due to gravity (g) = 9.8 m/s²

Spring constant (K) =?

Next, we shall convert 131 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

131 g = 131 g × 1 Kg / 1000 g

131 g = 0.131 Kg

Thus, 131 g is equivalent to 0.131 Kg.

Next, we shall the force exerted by the ball on the spring. This can be obtained as follow:

Mass (m) = 0.131 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = ma

F = 0.131 × 9.8

F = 1.2838 N

Next, we shall convert 4.82755 cm to metre (m)

This can be obtained as follow:

100 cm = 1 m

Therefore,

4.82755 cm = 4.82755 cm × 1 m / 100 cm

4.82755 cm = 0.0482755 m

Thus, 4.82755 cm is equivalent to 0.0482755 m

Finally, we shall determine the spring constant as follow:

Force (F) = 1.2838 N

Extention (e) = 0.0482755 m

Spring constant (K) =?

F = Ke

1.2838 = K × 0.0482755

Divide both side by 0.0482755

K = 1.2838 / 0.0482755

K = 26.59 N/m

Thus the spring constant is 26.59 N/m

7 0
3 years ago
what is the approximate weight of a 20-kg cannonball on the moon if the acceleration due to gravity is 1.6m/s^2
monitta
On Earth, a cannonball with a mass of 20 kg would weigh 196 Newtons.
With the formula F=mg, where F is the weight in Newtons, m is the mass, and g is the acceleration due to gravity on the Earth which is 9.8m/s^2.
F=20kg x 9.8m/s^2= 196 Newtons

BUT on the moon, acceleration due to gravity is 1.6 m/s^2,
so F=mg=20kgx1.6m/s^2= 32 N
5 0
3 years ago
What is an easy way to balance chemical equations?
Lorico [155]

write out what you have on both sides, then just use basic multiplication to try and even out both sides. I can help if you need me to balance some for you!!

8 0
3 years ago
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