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3 years ago

A 1kg block of ice at 0o C falls into alake whose

Physics

1 answer:

Readme [11.4K]3 years ago

3 0

Answer:

340.01019 m

Explanation:

m = Mass of ice = 1 kg

$L_f$ = Latent heat of fusion = $333.55\times 10^3\ J/kg$

g = Acceleration due to gravity = 9.81 m/s²

M = Mass of water = 0.01 kg

The potential energy will balance the heat released

$mgh=ML_f\\\Rightarrow h=\dfrac{ML_f}{mg}\\\Rightarrow h=\dfrac{0.01\times 333.55\times 10^3}{1\times 9.81}\\\Rightarrow h=340.01019\ m$

The minimum altitude is 340.01019 m

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A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein

RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of $8m^{3}/s$ . As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi r^{2}h$

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

$\frac {r}{h}=\frac{4}{5}$

When solving for r, we get:

$r=\frac{4}{5}h$

so we can substitute this into our volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h$

which simplifies to:

$V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h$

$V_{cone}=\frac{16}{75} \pi h^{3}$

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

$\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}$

Which simplifies to:

$\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}$

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

$\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}$

Now we can substitute the provided values into our equation. So we get:

$\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}$

so:

$\frac{dh}{dt}=0.25m/s$

3 0

3 years ago

A car slams on its breaks,producing friction between the tires and the road.Into which type of energy is the mechanical energy o

Step2247 [10]

The answer is Heat Energy

4 0

3 years ago

Read 2 more answers

The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 440g . The vibra

ANEK [815]

Answer:

Tension, T = 2038.09 N

Explanation:

Given that,

Frequency of the lowest note on a grand piano, f = 27.5 Hz

Length of the string, l = 2 m

Mass of the string, m = 440 g = 0.44 kg

Length of the vibrating section of the string is, L = 1.75 m

The frequency of the vibrating string in terms of tension is given by :

$f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$

$\mu=\dfrac{m}{l}$

$\mu=\dfrac{0.44}{2}=0.22\ kg/m$

$T=4L^2f\mu$

$T=4\times (1.75)^2\times (27.5)^2 \times 0.22$

T = 2038.09 N

So, the tension in the string is 2038.09 N. Hence, this is the required solution.

6 0

3 years ago

A player cathces a ball. Consider the action force to be the impact of the ball against the players glove. What is the reaction

Anuta_ua [19.1K]

Force the glove exerts on the ball

Explanation:

The reaction to this force is the force the glove exerts on the ball.

According to Newton's third ;aw of motion "Action and reaction forces are equal and opposite in direction".

The action force is the impact of the ball against the players glove.
The reactive force is the force the glove exerts on the ball.

This reactive force is directed in the opposite direction and it is the reason why the motion of the incoming ball is halted.

Learn more;

Newton's law brainly.com/question/11411375

#learnwithBrainly

6 0

3 years ago

An object moves in a straight path. Its position x as a function of time t is presented by the equation x(t) = at – bt2+c, where

ankoles [38]

Answer:

The answer is below

Explanation:

Given that:

x(t) = at – bt2+c

a) x(t) = at – bt2+c

Substituting a = 1.4 m/s, b = 0.06 m/s2 and c =50 m gives:

x(t) = 1.4t - 0.06t² + 50

At t = 5, x(5) = 1.4(5) - 0.06(5)² + 50 = 55.5 m

At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m

The average velocity (v) is given as:

$v=\frac{x(5)-x(0)}{5-0}\\ \\v=\frac{55.5-50}{5-0}=1.1\\ \\v=1.1\ m/s$

b) x(t) = 1.4t - 0.06t² + 50

At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m

At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m

The average velocity (v) is given as:

$v=\frac{x(10)-x(0)}{10-0}\\ \\v=\frac{58-50}{10-0}=0.8\\ \\v=0.8\ m/s$

c) x(t) = 1.4t - 0.06t² + 50

At t = 15, x(5) = 1.4(15) - 0.06(15)² + 50 = 57.5 m

At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m

The average velocity (v) is given as:

$v=\frac{x(15)-x(10)}{15-10}\\ \\v=\frac{57.5-58}{15-10}=0.1\\ \\v=0.1\ m/s$

6 0

3 years ago