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VladimirAG [237]

3 years ago

11

Rank them from fastest to slowest in how quickly each warms up: (a) steel, 450 j/kgââc ; (b) aluminum, 910 j/kgââc ; (c) copper,

390 j/kgââc .

1 answer:

s2008m [1.1K]3 years ago

5 0

<span>The rate of conductivity is different among different substances, like aluminum, steel, and copper. Aluminum conducts heat the fastest at 910 j/kgaac. Steel is next and conducts heat at 450 j/kgaac. Copper conducts heat the slowest at 390 j/kgaac.</span>

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A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed

liq [111]

Answer:

A) total time = 55.5 seconds

B) average velocity = 25.27 m/s

Explanation:

It starts from rest, so initial velocity, u = 0 m/s

We are given;

acceleration; a = 2 m/s²

Final velocity; v = 31 m/s

From Newton's first law of motion,

v = u + at

So, 31 = 0 + 2t

t = 31/2

t = 15.5 sec

We are told that, after this time of 15.5 sec, the car travels 35 sec at a constant speed and after that it takes 5 sec additional time to stop. Thus;

(a) Total time in which car is in motion = 15.5 + 35 +5 = 55.5 seconds

b)Total distance traveled during first 15.5 sec would be gotten from Newton's second equation of motion which is;

S = ut + ½at²

S1 = 0 + ½(2 * 15.5²)

S1 = 240.25 m

Distance traveled in 35 sec with with velocity of 31 m/sec is;

S2 = velocity x time

S2 = 35 × 31 = 1085 m

Now, for the final stage, final velocity (v) will now be 0 since the car comes to rest while initial velocity(u) will be 31 m/s.

From the first equation of motion,

a = (v - u)/t

a = (0 - 31)/5

a = -6.2 m/s²

So, distance travelled is;

S3 = ut + ½at²

S3 = (31 × 5) + ½(-6.2 × 5²)

S3 = 155 - 77.5

S3 = 77.5 m

So overall total distance = S1 + S2 + S3

Overall total distance = 240.25 + 1085 + 77.5 = 1402.75 m

Average velocity = total distance/total time

Average velocity = 1402.75/55.5 = 25.27 m/s

6 0

2 years ago

An object of mass 6 kg. is resting on a horizontal surface. A horizontal force

son4ous [18]

Answer:

a) The work done by the applied force is 1500 joules.

b) The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) 300 joules of energy are lost during motion.

Explanation:

a) Since the object has a constant mass, on which a constant horizontal force is exerted. The work done by the force ( $W$ ), measured in joules, is defined by the following expression:

$W = F\cdot \Delta x$ (1)

Where:

$F$ - Force, measured in newtons.

$\Delta x$ - Distance, measured in meters.

If we know that $F = 15\,N$ and $\Delta x = 100\,m$ , then the work done by the force exerted on the object is:

$W = (15\,N)\cdot (100\,m)$

$W = 1500\,J$

The work done by the applied force is 1500 joules.

b) At first we need to calculate the net acceleration of the object ( $a$ ), measured in meters per square second. By assuming a constant acceleration, we use the following kinematic formula:

$\Delta x = v_{o}\cdot t +\frac{1}{2}\cdot a\cdot t^{2}$ (2)

Where $v_{o}$ is the initial velocity of the object, measured in meters per second.

We clear the acceleration within the equation above:

$\frac{1}{2}\cdot a \cdot t^{2} = \Delta x-v_{o}\cdot t$

$a = \frac{2\cdot (\Delta x - v_{o}\cdot t)}{t^{2}}$

If we know that $\Delta x = 100\,m$ , $v_{o} = 0\,\frac{m}{s}$ and $t = 10\,s$ , then the net acceleration experimented by the object is:

$a = \frac{2\cdot \left[100\,m-\left(0\,\frac{m}{s} \right)\cdot (10\,s)\right]}{(10\,s)^{2}}$

$a = 2\,\frac{m}{s^{2}}$

By the 2nd Newton's Law, we construct the following equation of equilibrium under the consideration of a friction force acting against the motion of the object:

$\Sigma F = F - f = m\cdot a$ (3)

Where:

$F$ - External force exerted on the object, measured in newtons.

$f$ - Kinetic friction force, measured in newtons.

If we know that $F = 15\,N$ , $m = 6\,kg$ and $a = 2\,\frac{m}{s^{2}}$ , the kinetic friction force is:

$f = F-m\cdot a$

$f = 15\,N-(6\,kg)\cdot \left(2\,\frac{m}{s^{2}} \right)$

$f = 3\,N$

The work done by friction ( $W'$ ), measured in joules, is:

$W' = f\cdot \Delta x$ (4)

$W' = (3\,N) \cdot (100\,m)$

$W' = 300\,J$

And the net work experimented by the object is:

$\Delta W = 1500\,J - 300\,J$

$\Delta W = 1200\,J$

By the Work-Energy Theorem we understand that change in translational kinetic energy ( $\Delta K$ ), measured in joules, is equal to the change in net work. That is:

$\Delta K = \Delta W$ (5)

If we know that $\Delta W = 1200\,J$ , then the change in translational kinetic energy is:

$\Delta K = 1200\,J$

The kinetic energy of the block after 10 seconds is 1200 joules.

c) The magnitude of the force of friction is 3 newtons and its direction is against motion.

d) The energy lost by the object is equal to the work done by the force of friction. Therefore, 300 joules of energy are lost during motion.

7 0

3 years ago

On his honeymoon, James Joule attempted to explore the relationships between various forms of energy by measuring the rise of te

Marizza181 [45]

Answer:

The maximum temperature rise = 0.047 °C

Explanation:

Potential Energy, P = mgh

Energy transfered, Q=mcΔT

Potential energy = Energy transfered

mgh = mcΔT

gh = cΔT

ΔT = gh/c

ΔT = (9.81 * 20) / 4186

ΔT = 0.047 °C

8 0

3 years ago

Which nucleus completes the following equation?

IgorLugansk [536]

Answer: $^{50}_{25}Mn\rightarrow ^{0}_1e+ ^{50}_{24}Cr$

The chromium nucleus will complete the reaction.

Explanation:

The given reaction is a type of radioactive nuclei decay:

Positron emission: It is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. This is also known as $\beta ^+$ -decay. In this the mass number remains same.

$_A^Z\textrm{X}\rightarrow _{Z-1}^A\textrm{Y}+_{+1}^0e$

Now according to the given reaction:

$^{50}_{25}Mn\rightarrow ^{0}_1e+ ^{50}_{24}Cr$

The chromium nucleus will complete the reaction.

4 0

3 years ago

Read 2 more answers

A bicycle tire is spinning counterclockwise at 2.70 rad/s. During a time period Δt = 1.50 s, the tire is stopped and spun in the

Andru [333]

Answer:

Δω = -5.4 rad/s

αav = -3.6 rad/s²

Explanation:

<u>Given</u>:

Initial angular velocity = ωi = 2.70 rad/s

Final angular velocity = ωf = -2.70 rad/s (negative sign is

due to the movement in opposite direction)

Change in time period = Δt = 1.50 s

<u>Required</u>:

Change in angular velocity = Δω = ?

Average angular acceleration = αav = ?

<u>Solution</u>:

<u>Angular velocity (Δω):</u>

Δω = ωf - ωi

Δω = -2.70 - 2.70

Δω = -5.4 rad/s.

<u> Average angular acceleration (αav):</u>

αav = Δω/Δt

αav = -5.4/1.50

αav = -3.6 rad/s²

Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.

7 0

2 years ago

Read 2 more answers