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natta225 [31]
3 years ago
7

Which of the following would NOT affect the level at which a canoe floats in a pond

Physics
1 answer:
nataly862011 [7]3 years ago
5 0
Sry i was knowing the answer i forgot ;(
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A 50-kg copper block initially at 140°c is dropped into an insulated tank that contains 90 l of water at 10°c. Determine the f
xxMikexx [17]

Answer:

T_f=24.71

Explanation:

From the question we are told that:

Mass of block m=50

Temperature of block T_b =140 \textdegree C

Volume of water V= 90L

Temperature of water T_w=10 \textdegree C

Density of water \rho=1000kg/m^3

Specific heat of water C_w=4.18KJ/kg-k

Specific heat of copper C_p=0.96KJ/kg-k

 

Generally the equation for equilibrium stage is mathematically given by

mC_p(T_b-T_f)=\rho*VV*c(T_f-T_w)

50*0.96(140-T_f)=1000*90*10^-3*c_w(T_f-10)

48(140-T_f)=376.2(T_f-10)

140-T_f=7.8375(T_f-10)

140-T_f=7.8375T_f-78.375

-8.8375T_f=-218.375

T_f=\frac{-218.375}{-8.8375}

T_f=\frac{-218.375}{-8.8375}

T_f=24.71

6 0
3 years ago
I need the solution to this
posledela

Answer:

He could jump 2.6 meters high.

Explanation:

Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:

v_0 = \sqrt{2gh}=\sqrt{2\cdot 9.8\frac{m}{s^2}\cdot 1.3m}=5.0\frac{m}{s}

With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.

Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:

v_0^2 = 2g_{1/2}h\implies \\h = \frac{v_0^2}{2g_{1/2}}=\frac{25\frac{m^2}{s^2}}{2\cdot 4.9\frac{m}{s^2}}=2.6m

This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.

6 0
3 years ago
a car with a mass of 2000 kilograms is moving around a circular curve at a uniform velocity of 25 meters per second. The curve h
melisa1 [442]
In the given problem, we say various information's that are going to help us reach the ultimate answer to the question. Let us first write the information's that have been presented in front of us.
Mass of the car = 2000 kg
Velocity of the car = 25 m/s^2
Radius of the circle = 80 m
Now we already know the equation for calculating the centripetal force and that is
Centripetal Force = [mass * (velocity)^2]/Radius
                            = [2000 * (25)^2]/80
                            = (2000 * 625)/80
                            = 1250000/80
                            = 15625
So the centripetal force on the car is 15625 Newtons
  
4 0
3 years ago
Read 2 more answers
Which of the following results when a crest and trough meet?
kifflom [539]
D. Destructive interference. An easy way to think about it is the waves are opposite each other, so they essentially cancel each other out, or make an effort to.
6 0
3 years ago
What would you get if you split a bar magnet in the middle?
ipn [44]

Answer:

I believe it's the first option

5 0
3 years ago
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