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3 years ago

7

A ship travels with velocity given by 12, with current flowing in the direction given by 11 with respect to some co-ordinate axe

s.
What is the velocity of the ship in the direction of the current?

1 answer:

nataly862011 [7]3 years ago

7 0

Answer:

$v_x = 11.78 m/s$

Explanation:

Velocity of the ship is given as

$v = 12 units$

the direction of the velocity of the ship is making an angle of 11 degree with the current

so we will have two components of the velocity

1) along the direction of the current

2) perpendicular to the direction of the current

so here we know that the component of the ship velocity along the direction of the current is given as

$v_x = v cos\theta$

$v_x = 12 cos11$

$v_x = 11.78 m/s$

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A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of wate

vivado [14]

Answer: The molar heat capacity of aluminum is $25.3J/mol^0C$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of water = 130.0 g

$m_2$ = mass of aluminiunm = 23.5 g

$T_{final}$ = final temperature = $26.0^oC=(273+26)K=299K$

$T_1$ = temperature of water = $23^oC=(273+23)K=296K$

$T_2$ = temperature of aluminium = $100^oC=273+100=373K$

$c_1$ = specific heat of water= $4.184J/g^0C$

$c_2$ = specific heat of aluminium= ?

Now put all the given values in equation (1), we get

$130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]$

$c_2=0.938J/g^0C$

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity = $0.938J/g^0C\times 27g/mol=25.3J/mol^0C$

5 0

3 years ago

A "moving sidewalk" in an airport terminal moves at 1.0 m/s and is 35.0 m long. If a woman steps on at one end and walks at 1.5

pishuonlain [190]

Answer:

a.14 s

b.70 s

Explanation:

a.Let the sidewalk moving in positive x- direction.

Speed of sidewalk relative to ground= $v_s=1m/s$

Speed of women relative to sidewalk=v=1.5m/s

The speed of women relative to the ground

$v_w=v_s+v=1+1.5=2.5m/s$

Distance=35 m

Time= $\frac{distance}{speed}$

Using the formula

Time taken by women to reach the opposite end if she walks in the same direction the sidewalk is moving= $\frac{35}{v_w}=\frac{35}{2.5}=14s$

b.If she gets on at the end opposite the end in part (a)

Then, we take displacement negative.

Speed of sidewalk relative to ground= $v_s=1m/s$

Speed of women relative to sidewalk=v=-1.5 m/s

The speed of women relative to the ground= $v_w=v_s+v=1-1.5=-0.5m/s$

Time= $\frac{-35}{-0.5}=70 s$

Hence, the women takes 70 s to reach the opposite end if she walks in the opposite direction the sidewalk is moving.

3 0

3 years ago

Which of the following is a high wave that crashes down onto the

andrew-mc [135]

A breaker is a high wave that crashes onto the ocean floor.

4 0

3 years ago

Read 2 more answers

A force F with arrow= (5xî+ 4yĵ), where F with arrow is in newtons and x and y are in meters, acts on an object as the object mo

harkovskaia [24]

Answer:

w= 62.75 J

Explanation:

Given that

Force vector F= 5 x i + 4 y j

Space vector or displacement vector d= 5.01 i

We know that work (w)

w=∫ F.ds

w= ∫(5 x i + 4 y j) .dx ( only object is moving in x- direction)

$w=\int_{0}^{5.01}5xi.dxi$

$w=\int_{0}^{5.01}5x.dx$

$w=\left [\dfrac{5}{2}x^2\right ]_0^{5.01}$

w= 2.5 x 5.01² J

w= 62.75 J

6 0

3 years ago

How much work would have to be done to bring a 1150kg automobile traveling at 86km/h to a stop?

krok68 [10]

Explanation:

We have,

Mass of an automobile is 1150 kg

The automobile traveling at 86 km/h and then it comes to stop.

86 km/h = 23.88 m/s

It is required to find work done by the automobile.

Concept used : Work energy theorem

Th change in kinetic energy of an object is equal to the work done by it. The work done is then given by :

$W=\dfrac{1}{2}m(v^2-u^2)$

Here, v = 0

$W=-\dfrac{1}{2}mu^2\\\\W=-\dfrac{1}{2}\times 1150\times (23.88)^2\\\\W=-327896.28\ J$

or

$W=3.27\times 10^5\ J$

Therefore, the work done by the automobile is $-3.27\times 10^5\ J$ .

7 0

3 years ago