Ask question

Ask question

All categories

pishuonlain [190]

3 years ago

11

A cyclist is travelling eastwards at a velocity of 40 ms and rain is falling vertically at a speed of 10 ms. Find the velocity o

f the rain relative to the cyclist.using vectors to solve it

1 answer:

Genrish500 [490]3 years ago

6 0

41.23

v(B,E)=40 GOING EAST

V(R,E)=10 COMING DOWN

then 41.23

You might be interested in

During each cycle of operation a refrigerator absorbs 56 cal from the freezer compartment and expels 81 cal tothe room. If one c

Alexandra [31]

Answer:

138.18 minutes

Explanation:

$L_v$ = Latent heat of water at 0°C = 80 cal/g

m = Mass of water = 570 g

Heat removed for freezing

$Q=mL_v\\\Rightarrow Q=570\times 80\\\Rightarrow Q=45600\ cal$

Let N be the number of cycles and each cycle removes 56 cal from the freezer.

So,

$55\times N=45600\\\Rightarrow N=\frac{45600}{55}$

Each cycle takes 10 seconds so the total time would be

$\frac{45600}{55}\times \frac{10}{60}=138.18\ minutes$

The total time taken to freeze 138.18 minutes

6 0

3 years ago

Please solve the Problem.

prisoha [69]

The scalar reading during the process is 170. 5 Newton

<h3>How to determine the scalar reading</h3>

force = mass × acceleration

Given the mass = 55kg

In finding the acceleration, use the first equation of motion

v = u + at

u = 0

t = 2s

v = 6.5 m/s

Substitute the values

6.5 = 0 + 2a, make 'a' the subject

a = 6. 5÷ 2 = 3. 1 m/s²

Substitute the value of 'm' and 'a' in the original equation

F = 55× 3.1 = 170. 5 Newton

Hence, the scalar reading during the process is 170. 5 Newton

Learn more about force here:

brainly.com/question/13164598

#SPJ1

7 0

2 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:

a

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

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

b

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

Work occurs when

defon

I believe the answer is the second option.

7 0

4 years ago

A spherical snowball is melting in such a way that its radius is decreasing at a rate of 0.1 cm/min. At what rate is the volume

Ne4ueva [31]

Answer: 80.384 cubic cm /min

Explanation:

Let V denote the volume and r denotes the radius of the spherical snowball .

Given : $\dfrac{dr}{dt}=-0.1\text{cm/min}$

We know that the volume of a sphere is given by :-

$V=\dfrac{4}{3}\pi r^3$

Differentiating on the both sides w.r.t. t (time) ,w e get

$\dfrac{dV}{dt}=\dfrac{4}{3}\pi(3r^2)\dfrac{dr}{dt}\\\\\Rightarrow\ \dfrac{dV}{dt}=4\pi r^2 (-0.1)=-0.4\pi r^2$

When r= 8 cm

$\dfrac{dV}{dt}=-0.4(3.14)(8)^2=-80.384$

Hence, the volume of the snowball decreasing at the rate of 80.384 cubic cm /min.

6 0

3 years ago