Is the distance traveled during a specific unit of

A Scooter travelling at 10m/s speed up to 20m/s in 4 sec.find the acceleration of scooter

stira [4]

Answer:

2.5 m/s²

Explanation:

Given,

Initial speed ( u ) = 10 m/s

Final speed ( v ) = 20 m/s

Time ( t ) = 4 seconds

To find : Acceleration ( a ) = ?

Formula : -

a = ( v - u ) / t

a = ( 20 - 10 ) / 4

= 10 / 4

= 5 / 2

a = 2.5 m/s²

Therefore,

The acceleration of the scooter is 2.5 m/s²

7 0

2 years ago

A horizontal force of 92.7 N is applied to a 40.5 kg crate on a rough, level surface. If the crate accelerates at 1.13 m/s2, wha

lord [1]

Answer:

The value is $F_f = 46.935 \ N$

Explanation:

From the question we are told that

The magnitude of the horizontal force is $F = 92.7 \ N$

The mass of the crate is $m = 40.5 \ kg$

The acceleration of the crate is $a = 1.13 \ m/s$

Generally the net force acting on the crate is mathematically represented as

$F_{net} = F - F_f = ma$

Here $F_f$ is force of kinetic friction (in N) acting on the crate

So

$92.7 - F_f = 40.5 * 1.13$

=> $F_f = 46.935 \ N$

5 0

3 years ago

Which pair does NOT have an electric force between them?.

viva [34]

Two neutral objects will not have any electric force of attraction or repulsion between them.

<h3>What is the condition for the electric force between two objects?</h3>

As we know from the electrostatics that whenever there are two charges having a positive charge on one and a negative on the other will attract each other

similarly, if they are having like charges which are both of them having positive or both of them having a negative charge then there will be a force of repulsion between them.

But if both of them or even one of them is neutral then there will not be any electric force between them.

Thus neutral objects will not have any electric force of attraction or repulsion between them.

To know more about the nature of charged particles follow

brainly.com/question/22492496

5 0

2 years ago

200-grams of computer chips with a specific heat of 0.3 kJ/kg·K are initially at 25°C. These chips are cooled by placement in 0.

balu736 [363]

Answer:

a. -0.01324 kJ/K, b. = 0.03233 kJ/K , c. = 0.01909, Yes the process is possible

Explanation:

Heat transfer will occur between the chip and the surrounding fluid. Then, finally they will attain a common equilibrium temperature and heat transfer will stop. Now, if we assume that, after heat transfer, chip will attain the temperature of fluid, that is, -34 C,, So , to check whether this is possible

Amount of energy lost by the chip = m . c . (T(i) - T(f))

= 0.2 x 0.3 (25 + 34) = 3.54 KJ

Now, to evaluate the final state of the fluid, after the heat transfer completion,

Energy Gained = m(mew final – mew initial) = m[(μf+ x . μfg) - μf]

Note that heat transfer will change the internal energy of the fluid. Do not consider enthalpy change, as this is not a problem involving fluid flow in and out of the system

M[(μf+ x . μfg) - μf] = m(xμfg)

<u>Energy gained by the fluid will be equal to the energy lost by the chip (No energy loss to the surroundings)</u>

3.54 = 0.1 . X x 203.29

<u>x = 0.1741, which is the dryness fraction of fluid at the final state.</u>

Observe that the total energy lost by the chips is 3.45 kJ and fluid R-134a has got its value of mew fg at -34 C which is = 203.29 kJ/kg

So for 0.1kg of R-134a

0.1 x μfg= <u>20.329 kJ, which is much greater than 3.45 kJ</u>, therefore, it is certain that the state of fluid will be at -34 C only and at the saturation pressure of 69.56 KPa. So the chip will come to attain the temperature of -34 C.

a. Write the equation for the change of entropy in the chips

ΔSchips = mchips . c . ln(T2/T1), where mc is the mass of chips, c is the specific heat of chips, T2 is the temperature at state 2 and T1 is the temperature at state 1

Substitute mc = 0.2 kg, c = 0.3kJ/kg.K, T1 = 25 + 273, T2 = -34 + 273

delSchips = 0.2 x 0.3 x ln [(-34+273)/ (25+273)]

= -0.01324 kJ/K

There fore the change in entropy of the chips is -0.01324 kJ/K

b. Entropy change of fluid R- 134a

ΔS2 = m[Sfinal – S initial]

= m[Sf + x . Sfg - Sf]

= 0.2 x (0.1741 x 0.92859)

= 0.03233 kJ/K

c. Calculate the total change in the entropy of the entire system

delS = delSchips + delSR -134a

= -0.01324 + 0.03233

= 0.01909

<u>Since the total change in entropy of the entire system is positive that exactly explains that the actual processes are happening in the direction of increase of entropy therefore, the process is possible.</u>

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6 0

3 years ago

A parallel-plate capacitor is connected to a battery and becomes fully charged. The capacitor is then disconnected, and the sepa

AnnZ [28]

Answer:

Increases

Explanation:

The expression for the capacitance is as follows as;

$C=\frac{\epsilon _{0}A}{d}$

Here, C is the capacitance, $\epsilon _{0}$ is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.

It can be concluded from the above expression, the capacitance is inversely proportional to the distance. According to the given problem, the capacitor is disconnected from the battery and the distance between the plates is increased. Then, the capacitance of the given capacitor will decrease in this case.

The expression for the energy stored in the parallel plate capacitor is as follows;

$E=\frac{Q^{2}}{2C}$

Here, E is the energy stored in the capacitor, C is the capacitance and Q is the charge.

Energy stored in the given capacitor is inversely proportional to the capacitor. The charge on the capacitor is constant. In the given problem, as the distance between the parallel plates is being separated, the energy stored in this capacitor increases.

Therefore, the option (c) is correct.

8 0

3 years ago