Has a man he has married many women but has never been married before who is he

750 kg car zooms away from a red light with an acceleration of 7.8 m/s squared . What is the average net force in Newtons that t

natta225 [31]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 750 × 7.8

We have the final answer as

Hope this helps you

5 0

3 years ago

One gallon of paint (volume = 3.79 10-3 m3) covers an area of 17.8 m2. What is the thickness of the fresh paint on the wall?

AURORKA [14]

Answer:

0.2129 mm

Explanation:

We have given volume of the paint = 1 gallon $=3.79\times 10^{-3}m^3$

Area that covers the paint $=17.8 m^2$

We have to find the thickness of the fresh paint

So $thickness=\frac{volume}{area}=\frac{3.79\times 10^{-3}}{17.8}=0.2129\times 10^{-3}m=0.2129mm$

So the thickness of fresh paint on the wall is 0.2129 mm

6 0

3 years ago

If a force of 10 N acts on an object and an additional force of 6 N acts on the object in

PIT_PIT [208]

Answer:

D

Explanation:

For this kind of problem, forces add. F = F1 + F2

F1 = 6 N

F2 = 10 N

F = 6N + 10N

F = 16N

6 0

2 years ago

A mass free to vibrate on a level, frictionless surface at the end of a horizontal spring is pulled 35 cm from its equilibrium p

saul85 [17]

Answer:

0.67 s

Explanation:

This is a simple harmonic motion (SHM).

The displacement, $x$ , of an SHM is given by

$x = A\cos(\omega t)$

A is the amplitude and $\omega$ is the angular frequency.

We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or $\frac{\pi}{4}$ radian.

From trigonometry, $\sin A =\cos B$ if A and B are complementary.

At $t = 0$ , $x = 3.5$

$3.5 = A\cos(\omega \times0)$

$A =3.5$

So

$x = 3.5\cos(\omega t)$

At $t = 0.12$ , $x = 1.5$

$1.5 = 3.5\cos(0.12\omega)$

$\cos(0.12\omega)=\dfrac{1.5}{3.5}=0.4286$

$0.12\omega =\cos^{-1}0.4286$

$0.12\omega = 1.13$

$\omega = 9.4$

The period, $T$ , is related to $\omega$ by

$T = \dfrac{2\pi}{\omega} = \dfrac{2\times3.14}{9.4}=0.67$

5 0

3 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>

<u />

6 0

3 years ago

Has a man he has married many women but has never been married before who is he​

Has a man he has married many women but has never been married before who is he