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s344n2d4d5 [400]

3 years ago

14

A lump of steel of mass 10kg at 627 degree Celsius is dropped in 100kg oil at 30 degree Celsius . the specific heat of steel And

oil are 0.5kj/kg.k and 3.5kj/kg.k calculate the entropy change in steel,oil and in the universe.

1 answer:

Naily [24]3 years ago

8 0

Answer:

700J

Explanation:

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A bat emits a sound at a frequency of 30.0 kHz as it approaches a wall. The bat detects beats such that the frequency of the ech

Alex

Answer:

The speed of the bat is 5.02 m/s.

Explanation:

Given that,

Frequency = 30.0 kHz

Frequency of echo = 900 Hz

We need to calculate the frequency

Using formula of beat frequency

$f_{b}=f_{1}-f_{2}$

Put the value into the formula

$900=f_{1}-30\times10^{3}$

$f_{1}=900+30000$

$f_{1}=30900\ Hz$

We need to calculate the speed of the bat

Using Doppler equation

$f_{apr}=f\times(\dfrac{v_{sound}+v_{observer}}{v_{s}-v_{source}})$

Put the value into the formula

$30900=30000\times(\dfrac{340+v_{bat}}{340-v_{bat}})$

$\dfrac{30900}{30000}=\dfrac{340+v_{bat}}{340-v_{bat}}$

$1.03=\dfrac{340+v_{bat}}{340-v_{bat}}$

$340\times1.03-340=v_{b}+1.03v_{b}$

$10.2=2.03v_{bat}$

$v_{bat}=\dfrac{10.2}{2.03}$

$v_{bat}=5.02\ m/s$

Hence, The speed of the bat is 5.02 m/s.

6 0

3 years ago

. A new game has been invented where the player has to combine a dive from a board with shooting of a basketball through a hoop

skad [1K]

Explanation:

Not being able to see the picture attached and supposing the ground level is considered to be the water level so to be able to find out the velocity you need to score into the hoop you need to use the combined force ( resulting force) to throw the ball at 6.50 m distance, 1 m below to where you are standing

5 0

3 years ago

Read 2 more answers

Two cars A and B are 100m apart moving towards each other with

maxonik [38]

Let car A's starting position be the origin, so that its position at time <em>t</em> is

A: <em>x</em> = (40 m/s) <em>t</em>

and car B has position at time <em>t</em> of

B: <em>x</em> = 100 m - (60 m/s) <em>t</em>

<em />

They meet when their positions are equal:

(40 m/s) <em>t</em> = 100 m - (60 m/s) <em>t</em>

(100 m/s) <em>t</em> = 100 m

<em>t</em> = (100 m) / (100 m/s) = 1 s

so the cars meet 1 second after they start moving.

They are 100 m apart when the difference in their positions is equal to 100 m:

(40 m/s) <em>t</em> - (100 m - (60 m/s) <em>t</em>) = 100 m

(subtract car B's position from car A's position because we take car A's direction to be positive)

(100 m/s) <em>t</em> = 200 m

<em>t</em> = (200 m) / (100 m/s) = 2 s

so the cars are 100 m apart after 2 seconds.

3 0

3 years ago

A 1.0 kg cart moving right at 5.0 m/s on a frictionless track collides with a second cart moving at 2.0m/s. The 1.0 kg cart has

____ [38]

Answer:

m₂ = 3kg

Explanation:

The question wasn't clear about what direction the initial velocity of the second cart was, so I'll assume it was going left at 2.0m/s.

Anyway, this is a conservation of momentum problem. The equation you need to use is the one written in blue. They want you to solve for the mass of the second cart, so do some algebra and rearrange that blue equation in term of m₂.

Now that you have the equation for m₂, plug in all the values given from the question and you'll get 3kg.

3 0

4 years ago

A racing car travels with a constant tangential speed of 80.8 m/s around a circular track of radius 602 m. (a) Find the magnitud

KatRina [158]

Answer:

10.84 m/s2 radially inward

Explanation:

As the car is traveling an a constant tangential speed of 80.8 m/s, the total acceleration only consists of the centripetal acceleration and no linear acceleration. The formula for centripetal acceleration with respect to tangential speed v = 80.8 m/s and radius r =602 m is

$a = \frac{v^2}{r} = \frac{80.8^2}{602} = 10.84 m/s^2$

b) The direction of this centripetal acceleration is radially inward

6 0

4 years ago