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2 years ago

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A man runs at 7.5 m s-1 due east across the deck of a ship which moves at 10.0 m s-1 due north. Determine the resultant velocity

of the man( his velocity relative to the water)

1 answer:

german2 years ago

7 0

Answer:

12.5 m/s North/east

Explanation:

I thought of it as a graph. I basically just found the distance between the points (0,0) and (10, 7.5)!

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Lamar writes several equations trying to better understand potential energy. H = d with an arrow to the equation W = F d and P E

Debora [2.8K]

Answer:

The gravitational potential energy equals the work needed to lift the object.

Explanation:

here we know that

$H = \vec d$

work done is given as

$W = \vec F . \vec d$

Potential energy is given as

$PE_g = mgh$

force due to gravity is given as

$\vec F_g = mg$

now here if we plug in the value of distance and force in the formula of work done then we will have

$W = (mg)(h)$

so here we got

$W = PE_g$

so we can concluded that

The gravitational potential energy equals the work needed to lift the object.

3 0

2 years ago

Read 2 more answers

Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

kramer

Answer:

115 ⁰C

Explanation:

<u>Step 1:</u> The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_{1} +q_{2} =-q_{3}$ -----eqution 1

where,

$q_{1}$ is the heat absorbed by the solid at 0⁰C

$q_{2}$ is the heat absorbed by the liquid at 0⁰C

$q_{3}$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

<u>Step 2:</u> calculate how many moles of water you have in the 100.0-g sample

$=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O$

<u>Step 3: </u>calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_{2} = -q_{3}$

<u>Step 4:</u> calculate the final temperature of the water

$79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}$

Substitute in the values; we will have,

$79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})$

79.13 kJ + 990.66J* $(T_{f}-218})$ = -1463J* $(T_{f}-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_{f}-218})$ = -1.463KJ* $(T_{f}-100})$

$79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3$

collect like terms,

2.45366 $T_{f}$ = 283.133

∴ $T_{f} =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

6 0

2 years ago

An ocean liner is cruising at 10 meters/second and is about to approach a stationary ferryboat. A parcel is released from the oc

Afina-wow [57]

The parcel will undergo projectile motion, which means that it will have motion in both the horizontal and vertical direction.

First, we determine how long the parcel will fall using:

s = ut + 1/2 at²

where s will be the height, u is the initial vertical velocity of the parcel (0), t is the time of fall and a is the acceleration due to gravity.

5.5 = (0)(t) + 1/2 (9.81)(t)²
t = 1.06 seconds

Now, we may use this time to determine the horizontal distance covered by the parcel by using:
distance = velocity * time

The horizontal velocity of the parcel will be equal to the horizontal velocity of the cruise liner.

Distance = 10 * 1.06
Distance = 10.6 meters

The boat should be 10.6 meters away horizontally from the point of release.

4 0

2 years ago

A spring is stretched 6 in by a mass that weighs 8 lb. The mass is attached to a dashpot mechanism that has a damping constant o

olya-2409 [2.1K]

Answer:

y= 240/901 cos 2t+ 8/901 sin 2t

Explanation:

To find mass m=weighs/g

m=8/32=0.25

To find the spring constant

Kx=mg (given that c=6 in and mg=8 lb)

K(0.5)=8 (6 in=0.5 ft)

K=16 lb/ft

We know that equation for spring mass system

my''+Cy'+Ky=F

now by putting the values

0.25 y"+0.25 y'+16 y=4 cos 20 t ----(1) (given that C=0.25 lb.s/ft)

Lets assume that at steady state the equation of y will be

y=A cos 2t+ B sin 2t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -2A sin 2t+2B cos 2t

y"=-4A cos 2t-4B sin 2t

Now put the values of y" , y' and y in equation 1

0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t

So by comparing the coefficient both sides

30 A+ B=8

A-30 B=0

So we get

A=240/901 and B=8/901

So the steady state response

y= 240/901 cos 2t+ 8/901 sin 2t

6 0

2 years ago

An elephant's legs have a reasonably uniform cross section from top to bottom, and they are quite long, pivoting high on the ani

PilotLPTM [1.2K]

Answer:

t = 6,485 s , t_step = 25.94 s

the elephant gives 2.3 step very minute

Explanation:

Let's approximate this system to a simple pendulum that has angular velocity

w = √L / g

Angular velocity and period are related

w = 2π / T

T = 2π √g / L

Let's find the period

T = 2π √9.8 / 2.3

T = 12.97 s

Stride time is

t = T / 2

t = 12.97 / 2

t = 6,485 s

Frequency is inversely proportional to period

f = 1 / t

f = 1 / 6,485

f = 0.15 Hz

Since the elephant has 4 legs and each uses a time t, the total time for one step is

t_step = 4 t

t_step = 4 6.485

t_step = 25.94 s

f_step = 1/t_step =0.0385 s-1

Now let's use a proportion rule to find the number of steps in 60 s

#_step = 60 / t_step

#__step = 60 / 25.94

#_step = 2.3 steps

So the elephant gives 2.3 step very minute

4 0

2 years ago