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3 years ago

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Select the correct answer. x y 2.5 6.25 9.4 88.36 15.6 243.63 19.5 380.25 25.8 665.64 The table lists the values for two paramet

ers, x and y, of an experiment. What is the approximate value of y for x = 4? A. 11 B. 16 C. 24 D. 43

1 answer:

In-s [12.5K]3 years ago

6 0

Answer: B. 16

Explanation:

The given table:

x y

2.5 6.25

9.4 88.36

15.6 243.63

19.5 380.25

25.8 665.64

Here, we observed that value of y is the square value of x.

i.e. $y=x^2$ [For example 2.5²=6.25, 9.4²=88.36, 15.6²=243.63]

Put x=4, we get

$y=4^2=16$

Hence, the approximate value of y for x = 4 is 16.

So, the correct option is B. 16 .

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1. A 59 kg person is in a vehicle travelling at 41 m/s. The vehicle runs into a telephone pole. At impact, it

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Answer:

a. 12,600 N

b. 1290 kg

Explanation:

a. Impulse = change in momentum

F Δt = m Δv

F (0.192 s) = (59 kg) (0 m/s − 41 m/s)

F = -12,600 N

b. F = mg

12,600 N = m (9.8 m/s²)

m = 1290 kg (or 2,830 lbs)

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Which special effects technique is being used in television weather reports in which meteorologists stand in front of moving map

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A wooden block of mass M resting on a frictionless, horizontal surface is attached to a rigid rod of length ℓ and of negligible

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Answer:

a)

$mv l$

b)

$\frac{M }{(M + m)}$

Explanation:

Complete question statement is as follows :

A wooden block of mass M resting on a friction less, horizontal surface is attached to a rigid rod of length ℓ and of negligible mass. The rod is pivoted at the other end. A bullet of mass m traveling parallel to the horizontal surface and perpendicular to the rod with speed v hits the block and becomes embedded in it.

(a) What is the angular momentum of the bullet–block system about a vertical axis through the pivot? (Use any variable or symbol stated above as necessary.)

(b) What fraction of the original kinetic energy of the bullet is converted into internal energy in the system during the collision? (Use any variable or symbol stated above as necessary.)

a)

$m$ = mass of the bullet

$v$ = velocity of the bullet before collision

$r$ = distance of the line of motion of bullet from pivot = $l$

$L$ = Angular momentum of the bullet-block system

Angular momentum of the bullet-block system is given as

$L = m v r$

$L = mv l$

b)

$V$ = final velocity of bullet block combination

Using conservation of momentum

Angular momentum of bullet block combination = Angular momentum of bullet

$(M + m) V l = m v l\\V =\frac{mv}{(M + m)}$

$K_{o}$ = Initial kinetic energy of the bullet

Initial kinetic energy of the bullet is given as

$K_{o} = (0.5) m v^{2}$

$K_{f}$ = Final kinetic energy of bullet block combination

Final kinetic energy of bullet block combination is given as

$K_{f} = (0.5) (M + m) V^{2}$

Fraction of original kinetic energylost is given as

Fraction = $\frac{(K_{o} - K_{f})}{K_{o}} = \frac{((0.5) m v^{2} - (0.5) (M + m) V^{2})}{(0.5) m v^{2}}$

Fraction = $\frac{(m v^{2} - (M + m) (\frac{mv}{(M + m)})^{2})}{m v^{2}} = \frac{(Mm v^{2} + m^{2} v^{2} - m^{2} v^{2})}{(M + m) m v^{2}}$

Fraction = $\frac{(Mm v^{2} + m^{2} v^{2} - m^{2} v^{2})}{(M + m) m v^{2}}\\ \frac{M }{(M + m)}$

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3 years ago

At a baseball game, the batter hit a fly ball at time t = 0 s. The outfielder caught the ball at t = 5.8 s. When was the ball at

Agata [3.3K]

We have the following equation for height:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
Where,
a: acceleration
vo: initial speed
h0: initial height.
The value of the acceleration is:
a = -g = -9.8 m / s ^ 2
For t = 0 we have:
h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
h (0) = h0
h0 = 0 (reference system equal to zero when the ball is hit).
For t = 5.8 we have:
h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
(1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
vo = (1/2) * (9.8) * (5.8)
vo = 28.42
Substituting values we have:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
Rewriting:
h (t) = -4.9 * t ^ 2 + 28.42 * t
The maximum height occurs when:
h '(t) = -9.8 * t + 28.42
-9.8 * t + 28.42 = 0
t = 28.42 / 9.8
t = 2.9 seconds.
Answer:
The ball was at maximum elevation when:
t = 2.9 seconds.

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A single insulated duct flow experiment using air operating at steady-state is performed in a lab. One measurement location (Sta

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Answer:

a) -0.0934 kJ/kg. K

b) The direction of flow is from right to left.

Explanation:

A free flow diagram of the horizontal insulated duct is as shown below.

NOW,

Let assume that the direction of flow is from left to right and consider the following relation for the entropy rate balance equation for a control volume as:

$\frac{\sigma_{cv}}{m}= (s_2-s_1) \geq 0 \ \ \ -------> \ \ \ 1$

Now; if the value for this relation is greater than zero; then we conclude that our assumption is correct.

If the value is less than zero; then we conclude that the assumption is wrong.

Then, the flow is said to be in the opposite direction

Formula for the change in specific entropy can be calculated as:

$s_2-s_1 = s^0(T_2) - s^0(T_1)-R \ In ( \frac{P_2}{P-1}) \ \ \ -------> \ \ \ 2$

where;

$s_1, s_2 , s^0(T_2), s^0(T1)$ are specific entropies

R = universal gas constant

$P_1$ = pressure at location 1

$P_2$ = pressure at location 2

We obtain the specific properties of air at temperature at $T_1$ = (67°C + 273)K = 340 K from the table A-22 ( Ideal gas properties of air)

$s^0(T1)$ = 1.8279 kJ/kg.K

We also obtain the specific properties of air at temperature $T_2$ = 22°C + 273) K = 295 K

From the table A- 22

$s^0(T_2)$ = 1.68515 kJ/kg . K

R = $\frac{8.314 kJ}{28.97 kg.K}$

$P_1 =$ 0.95 bar

$P_2 =$ 0.8 bar

Now replacing our values into equation (2) from above; we have;

$s_2-s_1 = s^0(T_2) -s^0(T_1)-R \ In (\frac{P_2}{P_1} )$

$s_2-s_1 = 1.68515 -1.8279-\frac{8.314}{28.97} \ In (\frac{0.8}{0.95} )$

$s_2-s_1 = 1.68515 -1.8279+ 0.0493$

$s_2-s_1 =-0.0934 \ kJ/kg.K$

Equating our result to equation (1)

$s_2-s_1 \geq 0\\-0.0934 \leq 0$

Therefore , our assumption is wrong and the direction of flow is said to be from right to left.

We therefore conclude that the direction of flow is from right to left.

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3 years ago