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3 years ago

9

wo ships, one 200200 metres in length and the other 100100 metres in length, travel at constant but different speeds. When trave

lling in opposite directions, it takes 1010 seconds for them to completely pass each other. When travelling in the same direction, it takes 2525 seconds for them to completely pass each other. The speed of the faster ship, in metres per second, is

1 answer:

shutvik [7]3 years ago

7 0

Answer:

The speed of the faster ship is 22.5 m/s

Explanation:

The length of the ships are;

Ship 1 = 200 m

Ship 2 = 100 m

The time it takes for the ships to completely cross each other when travelling in opposite directions = 10 seconds

The time it takes both ships to cross each other when travelling in the same direction = 25 seconds

Let x represent the speed of the first ship, ship 1, and y represent the speed of the second ship, ship, 2, we have;

(x + y) × 10 = 200 + 100 = 300

10·x + 10·y = 300...(1)

(x - y) × 20 = 200 + 100 = 300

20·x - 20·y = 300...(2)

Multiply equation (1) by 2, to get;

(x + y) × 10 × 2 = 300 × 2

20·x + 20·y = 600...(3)

Adding equation (1) to equation (3) gives;

20·x + 20·y + 20·x - 20·y = 600 + 300

40·x = 900

x = 900/40 = 22.5

x = 22.5

The speed of the first ship, ship 1 = x = 22.5 m/s

From equation (1), we have;

10·x + 10·y = 300

y = (300 - 10·x)/10 = (300 - 10×22.5)/10 = 7.5

y = 7.5

The speed of the second ship, ship 2 = y = 7.5 m/s

Therefore, the faster ship is ship 1 with a speed of 22.5 m/s

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3 years ago

A child on a sled slides (starting from rest) down an icy slope that makes an angle of 15◦ with the horizontal. After sliding 20

statuscvo [17]

Answer:

A) v₁ = 10.1 m/s t₁= 4.0 s

B) x₂= 17.2 m

C) v₂=7.1 m/s

D) x₂=7.5 m

Explanation:

A)

Assuming no friction, total mechanical energy must keep constant, so the following is always true:

$\Delta K + \Delta U = (K_{f} - K_{o}) +( U_{f} - U_{o}) = 0 (1)$

Choosing the ground level as our zero reference level, Uf =0.
Since the child starts from rest, K₀ = 0.
From (1), ΔU becomes:
$\Delta U = 0- m*g*h = -m*g*h (2)$
In the same way, ΔK becomes:
$\Delta K = \frac{1}{2}*m*v_{1}^{2} (3)$
Replacing (2) and (3) in (1), and simplifying, we get:

$\frac{1}{2}*v_{1}^{2} = g*h (4)$

In order to find v₁, we need first to find h, the height of the slide.
From the definition of sine of an angle, taking the slide as a right triangle, we can find the height h, knowing the distance that the child slides down the slope, x₁, as follows:

$h = x_{1} * sin \theta_{1} = 20.0 m * sin 15 = 5.2 m (5)$

Replacing (5) in (4) and solving for v₁, we get:

$v_{1} = \sqrt{2*g*h} = \sqrt{2*9.8m/s2*5.2m} = 10.1 m/s (6)$

As this speed is achieved when all the energy is kinetic, i.e. at the bottom of the first slide, this is the answer we were looking for.
Now, in order to finish A) we need to find the time that the child used to reach to that point, since she started to slide at the its top.
We can do this in more than one way, but a very simple one is using kinematic equations.
If we assume that the acceleration is constant (which is true due the child is only accelerated by gravity), we can use the following equation:

$v_{1}^{2} - v_{o}^{2} = 2*a* x_{1} (7)$

Since v₀ = 0 (the child starts from rest) we can solve for a:

$a = \frac{v_{1}^{2}}{2*x_{1} } = \frac{(10.1m/s)^{2}}{2* 20.0m} = 2.6 m/s2 (8)$

Since v₀ = 0, applying the definition of acceleration, if we choose t₀=0, we can find t as follows:

$t_{1} =\frac{v_{1} }{a} =\frac{10.1m/s}{2.6m/s2} = 4.0 s (9)$

B)

Since we know the initial speed for this part, the acceleration, and the time, we can use the kinematic equation for displacement, as follows:

$x_{2} = v_{1} * t_{2} + \frac{1}{2} *a_{2}*t_{2}^{2} (10)$

Replacing the values of v₁ = 10.1 m/s, t₂= 2.0s and a₂=-1.5m/s2 in (10):

$x_{2} = 10.1m/s * 2.0s + \frac{1}{2} *(-1.5m/s2)*(2.0s)^{2} = 17.2 m (11)$

C)

From (6) and (8), applying the definition for acceleration, we can find the speed of the child whem she started up the second slope, as follows:

$v_{2} = v_{1} + a_{2} *t_{2} = 10.1m/s - 1.5m/s2*2.0s = 7.1 m/s (12)$

D)

Assuming no friction, all the kinetic energy when she started to go up the second slope, becomes gravitational potential energy when she reaches to the maximum height (her speed becomes zero at that point), so we can write the following equation:

$\frac{1}{2}*v_{2}^{2} = g*h_{2} (13)$

Replacing from (12) in (13), we can solve for h₂:

$h_{2} =\frac{v_{2} ^{2}}{2*g} = \frac{(7.1m/s) ^{2}}{2*9.8m/s2} = 2.57 m (14)$

Since we know that the slide makes an angle of 20º with the horizontal, we can find the distance traveled up the slope applying the definition of sine of an angle, as follows:

$x_{3} = \frac{h_{2} }{sin 20} = \frac{2.57m}{0.342} = 7.5 m (15)$

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3 years ago

Ideally, when a thermometer is used to measure the temperature of an object, the temperature of the object itself should not cha

nordsb [41]

Answer : The temperature of water is, $57.7^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of thermometer = $804J/kg.^oC=0.804J/g^oC$

$c_2$ = specific heat of water = $4.18J/g.^oC$

$m_1$ = mass of thermometer = 33.0 g

$m_2$ = mass of water = 149 g

$T_f$ = final temperature = $56.0^oC$

$T_1$ = initial temperature of thermometer = $15.4^oC$

$T_2$ = initial temperature of water = ?

Now put all the given values in the above formula, we get:

$33.0g\times 0.804J/g^oC\times (56.0-15.4)^oC=-149g\times 4.18J/g.^oC\times (56.0-T_2)^oC$

$T_2=57.7^oC$

Therefore, the temperature of water is, $57.7^oC$

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