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3 years ago

Electrical shock is more likely to be fatal when the path of current is through the

Physics

1 answer:

skad [1K]3 years ago

4 0

Electrical shock is most likely to be fatal when the path of the current is through the heart.

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archer shoots his arrow towards a target at a distance of 90 m, and hits ‘bullseye’ . Calculate the acceleration and time taken

mario62 [17]

Answer:

a =45 m/s2

t = 2 seconds

Explanation:

Hi, to answer this question we have to apply the next formula:

v^2 = u^2 +2 a d

Where:

v = final velocity = 90 m/s

u = initial velocity = 0 m/s (shots from rest)

a = acceleration (m/s2)

d = distance = 90m

90^2 = 0^2 + 2a(90)

Solving for a:

8,100= 180 a

8,100/180 = a

a = 45 m/s^2

For time:

v = u + at

90 = 0 + 45t

90/45=t

t =2 seconds

6 0

3 years ago

Which statement did Ernest Rutherford make about atoms?

PtichkaEL [24]

Answer:

Option A

Explanation:

Ernest Rutherford concluded that the atom has a small, dense center which constitutes the mass of the whole atom. He called it a "Nucleus". He also said that most of the space in the atom is empty.

6 0

4 years ago

Suppose Gabor, a scuba diver, is at a depth of 15m. Assume that: The air pressure in his air tract is the same as the net water

s2008m [1.1K]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at

the surface is $\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = 2.5$

The number of moles of gas that must be released is $n= 0.3538\ mols$

Explanation:

We are told from the question that the pressure at the surface is 1 atm and for each depth of 10m below the surface the pressure increase by 1 atm

This means that the pressure at the depth of the surface would be

$P_d = [\frac{15m}{10m} ] (1 atm) + 1 atm$

$= 2.5 atm$

The ideal gas equation is mathematically represented as

$PV = nRT$

Where P is pressure at the surface

V is the volume

R is the gas constant = 8.314 J/mol. K

making n the subject we have

$n = \frac{PV}{RT}$

Considering at the surface of the water the number of moles at the surface would be

$n_s = \frac{P_sV}{RT}$

Substituting $1 atm = 101325 N/m^2$ for $P_s$ , $6L = 6*10^{-3}m^3$ for volume , 8.314 J/mol. K for R , (37° +273) K for T into the equation

$n_s = \frac{(1atm)(6*10^{-3} m^3)}{(8.314J/mol \cdot K)(37 +273)K}$

$= 0.2359 mol$

To obtain the number of moles at the depth of the water we use

$n_d = \frac{P_d V}{RT}$

Where $P_d \ and \ n_d \$ are pressure and no of moles at the depth of the water

Substituting values we have

$n_d = \frac{(2.5)(101325 N/m^2)(6*10^{-3}m^3)}{(8.314 J/mol \cdot K)(37 + 273)K}$

$= 0.5897 mol$

Now to obtain the number of moles released we have

$n = n_d - n_s$

$= 0.5897mol - 0.2359mol$

$=0.3538 \ mol$

The molar concentration at the surface of water is

$[\frac{n}{V} ]_{surface} = \frac{0.2359mol}{6*10^-3m^3}$

$=39.31mol/m^3$

The molar concentration at the depth of water is

$[\frac{n}{V} ]_{15m} = \frac{0.5897}{6*10^{-3}}$

$= 98.28 mol/m^3$

Now the ratio of the gases in Gabor's lungs at the depth of 15m to that at the surface is

$\frac{(n/V)_{15\ m}}{(n/V)_{surface}} = \frac{98.28}{39.31} =2.5$

6 0

3 years ago

What are two differences between gravity and air resistance

Natalka [10]

1. Air resistance force usually upwards, but gravity doenwards.

2. Objects are affected by gravity the same, but air resistance can affect the speed of an object's descent.

Sorry if im wrong

4 0

3 years ago

An unbalanced force of 20 N is applied to a 10 kg mass. What is the acceleration of the mass?

NemiM [27]

Answer:

2 m/s2

Explanation:

Newton's 2nd law:

F=ma

20 N= (10 kg)a

a= 2 m/s2

8 0

3 years ago