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2 years ago

What factors affect the ability of a substance to transfer thermal energy to heat or be heated by it's surroundings?

1 answer:

4 0

The prime factors that affect the ability of substances to transfer the thermal energy to heat are the temperature difference between the two objects, area of cross-section, time, and distance travelled by the thermal energy.

Explanation: 

The process of heat conduction takes place through contact between two or more objects. But this conduction depends on multiple factors that are responsible for thermal conduction. They are-

Temperature Difference( $\Delta T$ ) - The two objects must have a temperature difference else there will be no thermal conduction between them. The more the difference in their temperatures, the more thermal energy flows from one object to the other.
Area of Cross-section (A) - Larger areas of contact provide as better medium of thermal conduction.
Time (t) - The more time we give for the thermal conduction, the more energy is transmitted from one system to the other.
Distance Travelled (l) - The longer the distance, lesser the conduction. Means, the distance should be minimized in order to achieve the optimum thermal conduction between two objects.

Consider metal pot and its handle, it is being boiled for 15 m. The molecules present near the source of heat, showing fast vibration and bounce off. It actually indicates the heats of substance. That’s why, handle remains hot as heat conduction takes place. It can be estimated by,

$Q=\frac{k A \Delta T t}{l}$

k - Thermal conductivity of the material, measured in J/s.m. $^{\circ} \mathrm{C}$

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A drag racer accelerates from rest at an average rate of +13.2 mls for a distance of 100. m. The driver coasts for 0.5 then uses

gtnhenbr [62]

Complete Question:

A drag racer accelerates from rest at an average rate of +13.2 m/s² for a distance of 100. m. The driver coasts for 0.5 s then uses the brakes and parachute to decelerate until the end of the track. If the total length of the track is 180 m, what minimum deceleration rate must the racer have in order to stop prior to the the end of the track?

Answer:

-31.92 m/s²

Explanation:

The drag races do a retiling uniform variated movement. There are 3 steps in the movement, first, it accelerates from rest until 100 m, second it coasts to 0.5 s, and then it decelerates. So, let's analyze each one of the steps:

Step 1

The initial velocity is v0 = 0 (because it was at rest), the acceleration is +13.2 m/s², and the distance ΔS = 100.0 m, so the final velocity, v, is:

v² = v0² + 2aΔS

v² = 2*13.2*100

v² = 2640

v = √2640

v = 51.38 m/s

Step 2

Know it's initial velocity is 51.38 m/s, it take 0.5s, and has the same acceleration, so, after 0.5 s, the velocity will be:

v = v0 + at

v = 51.38 + 13.2*0.5

v = 57.98 m/s

Thus, the distance it travels is:

v² = v0² + 2aΔS

57.98² = 51.38² + 2*13.2*ΔS

3361.6804 = 2639.9044 + 26.4ΔS

26.4ΔS = 721.776

ΔS = 27.34 m

Step 3

The initial velocity of the drag racer is 57.98 m/s, and it travels the final distance of the track: 180 - 100 - 27.34 = 52.66 m. So, when it stops, its final velocity will be 0. The minimum deceleration must be the one that it would stop at the end of the track (less than that it would cross the final track):

v² = v0² + 2aΔS

0 = 57.98² + 2a*52.66

-105.32a = 3361.6804

a = - 31.92 m/s²

4 0

2 years ago

A cannon is mounted on a tower above a wide, level field. The barrel of the cannon is 20 m above the ground below. A cannonball

OLga [1]

Answer:

Cannonball will be in flight before it hits the ground for 2.02 seconds

Explanation:

Initial height from ground = 20 meter.

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate time when s = 20 meter.

Substituting

$20=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.02 seconds$

So it will take 2.02 seconds to reach ground.

5 0

2 years ago

a 75 kg man is standing at rest on ice while holding a 4kg ball. if the man throws the ball at a velocity of 3.50 m/s forward, w

AysviL [449]

Answer:

His resulting velocity will be 0.187 m/s backwards.

Explanation:

Given:

Mass of the man is, $M=75\ kg$

Mass of the ball is, $m=4\ kg$

Initial velocity of the man is, $u_m=0\ m/s(rest)$

Initial velocity of the ball is, $u_b=0\ m/s(rest)$

Final velocity of the ball is, $v_b=3.50\ m/s$

Final velocity of the man is, $v_m=?\ m/s$

In order to solve this problem, we apply law of conservation of momentum.

It states that sum of initial momentum is equal to the sum of final momentum.

Momentum is the product of mass and velocity.

Initial momentum = Initial momentum of man and ball

Initial momentum = $Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm$

Final momentum = Final momentum of man and ball

Final momentum = $Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14$

Now, initial momentum = final momentum

$0=75v_m+14\\\\75v_m=-14\\\\v_m=\frac{-14}{75}\\\\v_m=-0.187\ m/s$

The negative sign implies backward motion of the man.

Therefore, his resulting velocity is 0.187 m/s backwards.

3 0

2 years ago

Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The object

Tatiana [17]

Answer:

C. the area of the rectangle plus the area of the triangle under the line

Explanation:

Based on the information provided, the velocity vs. time graph is a line with a positive slope and a y-intercept of (0, 3). The displacement is the area under this line. This area can be divided into a triangle and a rectangle. So of the options available, C is the correct one.

6 0

2 years ago

Read 2 more answers

A small rock is thrown vertically upward with a speed of 22.0 m/s from the edge of the roof of a 30.0-m-tall building. The rock

RSB [31]

To solve this problem we will apply the kinematic equations of linear movement. For this purpose we will begin to define the final speed of the body before hitting the street. The first equation will begin using the difference in velocities as a function of acceleration (gravity) and position. And the second will use the concept of acceleration, time and speed, to find the time variable.

PART A) Equation of motion is

$v^2-u^2 = 2as$