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Veronika [31]
3 years ago
12

How do you think the temperature difference between the beakers relates to the rate of heat transfer?

Physics
1 answer:
Bogdan [553]3 years ago
8 0

Answer and Explanation:

The rate (in W) at which heat energy moves from the more blazing article to the colder item increments with the temperature distinction between the items.  

The more noteworthy the temperature contrast, the more prominent the rate at which exchange of heat energy takes place.

Thus the temperature difference between the beakers relates in the same manner as described above.

The exchange of heat energy continues until a thermal equilibrium is attained.

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a physics student throws a stone horizontally off a cliff. one second later, he throws a second identical stone in exactly the s
Virty [35]

The second stone hits the ground exactly one second after the first.

The distance traveled by each stone down the cliff is calculated using second kinematic equation;

h = v_0_yt + \frac{1}{2} gt^2

where;

  • <em>t is the time of motion </em>
  • <em />v_0_y<em> is the initial vertical velocity of the stone = 0</em>

h = \frac{1}{2} gt^2

The time taken by the first stone to hit the ground is calculated as;

t_1 = \sqrt{\frac{2h}{g} }

When compared to the first stone, the time taken by the second stone to hit the ground after 1 second it was released is calculated as

t_2 = \sqrt{\frac{2h}{g} } + 1

t_2 = t_1 + 1

Thus, we can conclude that the second stone hits the ground exactly one second after the first.

"<em>Your question is not complete, it seems be missing the following information;"</em>

A. The second stone hits the ground exactly one second after the first.

B. The second stone hits the ground less than one second after the first

C. The second stone hits the ground more than one second after the first.

D. The second stone hits the ground at the same time as the first.

Learn more here:brainly.com/question/16793944

8 0
3 years ago
A concrete piling of 50 kg is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. How much will the wire stretch?
pentagon [3]

Explanation:

It is given that,

Mass of concrete pilling, m = 50 kg

Diameter of wire, d = 1 mm

Radius of wire, r = 0.0005 m

Length of wire, L = 11.2

Young modulus of steel, Y=20\times 10^{10}\ N/m^2

The young modulus of a wire is given by :

Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}

Y=\dfrac{F.L}{A\Delta L}

\Delta L=\dfrac{F.L}{A.Y}

\Delta L=\dfrac{50\ kg\times 9.8\ m/s^2\times 11.2\ m}{\pi (0.0005\ m)^2\times 20\times 10^{10}\ N/m^2}

\Delta L=0.034\ m

So, the wire will stretch 0.034 meters. Hence, this is the required solution.

8 0
3 years ago
What are the formulas to calculate acceleration
IRINA_888 [86]

In middle school, the formula you'll use most often when you're
working with acceleration is . . .

Acceleration = (change in speed during some time) / (time for the change)


5 0
3 years ago
Read 2 more answers
Use the impulse-momentum theorem to find how long a stone falling straight down takes to increase its speed from 3.7 m/s to 9.90
11111nata11111 [884]
<h2>Time taken is 0.632 seconds</h2>

Explanation:

Impulse momentum theorem is change in momentum is impulse.

Change in momentum = Impulse

Final momentum - Initial momentum = Impulse

Mass x Final velocity - Mass x Initial Velocity = Force x Time

Mass x Final velocity - Mass x Initial Velocity =Mass x Acceleration x Time

Final velocity - Initial Velocity = Acceleration x Time

Final velocity = 9.9 m/s

Initial Velocity = 3.7 m/s

Acceleration = 9.81 m/s²

Substituting

9.9 - 3.7 = 9.81 x Time

Time = 0.632 seconds

Time taken is 0.632 seconds

8 0
2 years ago
A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio
Alekssandra [29.7K]

Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

\sum F_{x}=0

we can see there are only two forces in x, which are the weight on x and the static friction, so:

-W_{x}+f_{s}=0

when solving for the static friction we get:

f_{s}=W_{x}

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

W_{x}=mg sin \alpha

we can substitute this on our sum of forces equation:

f_{s}=mg sin \alpha

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

f_{s}=N\mu_{s}

so we can substitute this on our equation:

N\mu_{s}=mg sin \alpha

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

\sum F_{y}=0

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

N-W_{y}=0

when solving for N we get:

N=W_{y}

When seeing the free-body diagram we can determine that:

W_{y}=mg cos \alpha

so we can substitute that in the sum of y-forces equation, so we get:

N=mg cos \alpha

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

mg cos \alpha \mu_{s}=mg sin \alpha

we can divide both sides of the equation into mg so we get:

cos \alpha \mu_{s}=sin \alpha

as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for \mu_{s} we get:

\mu_{s}=\frac{sin \alpha}{cos \alpha}

when simplifying this we get that:

\mu_{s}=tan \alpha

now we can solve for the angle so we get:

\alpha= tan^{-1}(\mu_{s})

and we can substitute the given value so we get:

\alpha= tan^{-1}(0.350)

which yields:

α=19.29°

which rounds to:

α=19°

8 0
3 years ago
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