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andriy [413]
3 years ago
11

Why does ice float on water? A. The temperature of ice is lower than the temperature of water. B. The temperature of water and i

ce are the same. C. The density of ice is less than the density of water. D. The density of ice is greater than the density of water.
Physics
2 answers:
Allushta [10]3 years ago
4 0

It is C, ice is about 9% less dense than water! This is true info

Vesnalui [34]3 years ago
4 0

Answer:

C. The density of ice is less than the density of water.

Explanation:

As we know that when ice floats on the water then we can say that

Buoyancy force on ice will be counterbalanced by the weight of ice block

So here we have

F_b = mg

now we say that for flotation of ice cube the buoyant force must be greater than the force of gravity

\rho_w Vg > \rho_{ice} Vg

so here we can say

\rho_w > \rho_{ice}

so here density of ice will be less than the density of water

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A bus moving in a straight line at speed of 25m/s. what time does the bus take to cover 5km?<br>​
wlad13 [49]

Answer:

20 seconds

Explanation:

we know,

speed=distance by time

therefore time= distance by speed

so,

time=5000m by 25m|s

=20 seconds

6 0
3 years ago
Have you ever written a bio-data or an application letter? Share your experience in the
Georgia [21]

Answer:

I found the experience tasking

Explanation:

I wouldn't say it was hard, neither was it easy. I'd rather go for something like it being tasking. It's worthy of note that it was my first time, and I think it's very normal especially when one hasn't been doing something of that nature previously. Of course I did my draft, which unsurprisingly happened to be not good enough, and I had to look for templates to guide me through the acceptable way.

I still did it in my own way, but in the right way. Ever since then though, I have never stuttered when writing application letters, as it had since then seem inborn

8 0
3 years ago
A balsa wood raft is 1.50 m long, 1.00 m wide, and 0.120 m high. the density of balsa is 380 kg/m^3. what is the mass of the raf
topjm [15]

Answer:

the mass of the raft is 68.4 kg

Explanation:

Since Mass is defined as Volume times Density, start by calculating the volume of the raft:

Volume = length x width x high = 1.5 m x 1.0 m x 0.12 m = 0.18 m^3

and now multiply it times the given density in order to find its mass:

Mass = Volume x Density = 0.18 m^3 x 380 kg/m^3 = 68.4 kg.

Notice that the m^3 units cancel out (they are in numerator and in denominator) leaving just the kg (a unit of mass) in the answer.

Therefore, the mass of the raft is 68.4 kg

4 0
4 years ago
A superball with a mass m = 61.6 g is dropped from a height h = [02]____________________ m. It hits the floor and then rebounds
aleksklad [387]

<em>There is not enough data to solve the problem, but I'm assuming the initial height as h = 10 m for the question to have a valid answer and the student can have a reference to solve their own problem</em>

Answer:

(a) \Delta P=1.67 \ kg.m/s

(b) \Delta P=0.86\ m/s

Explanation:

<u>Change of Momentum</u>

The momentum of a given particle of mass m traveling at a speed v is given by

P=m.v

When this particle changes its speed to a value v', the new momentum is

P'=m.v'

The change of momentum is

\Delta P=m.v'-m.v

\Delta P=m.(v'-v)

Defining upward as the positive direction, we'll compute the change of momentum in two separate cases.

(a) The initial height of the superball of m=61.6 gr = 0.0616 Kg is set to h= 10 m. This information leads us to have the initial potential energy of the ball just after it's dropped to the floor:

U=m.g.h=0.0616\cdot 9.8\cdot 10 =6.0368\ J

This potential energy is transformed into kinetic energy just before the collision occurs, thus

\displaystyle \frac{1}{2}mv^2=6.0368

Solving for v

\displaystyle v=\sqrt{\frac{6.0368\cdot 2}{0.0616}}

v=-14\ m/s

This is the speed of the ball just before the collision with the floor. It's negative because it goes downward. Now we'll compute the speed it has after the collision. We'll use the new height and proceed similarly as above. The new height is

h'=88.5\% (10)=8.85\ m

The potential energy reached by the ball at its rebound is

U'=m.g.h'=0.0616\cdot 9.8\cdot 8.85 =5.342568\ J

Thus the speed after the collision is

\displaystyle v'=\sqrt{\frac{5.342568\cdot 2}{0.0616}}

v'=13.17\ m/s

The change of momentum is

\Delta P=0.0616\cdot (13.17+14)

\Delta P=1.67 \ kg.m/s

(b) If the putty sticks to the floor, then v'=0

\Delta P=0.0616\cdot (0+14)

\Delta P=0.86\ m/s

3 0
4 years ago
By what factor does the force required to stop a car increase if the initial speed is doubled and the stopping distance remains
mojhsa [17]
 <span>Since the equation is F = ma the acceleration would change by a factor of 2, since a = v/2. The force would double.</span>
4 0
4 years ago
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