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2 years ago

9

What is the value of a in the function’s equation?

1 answer:

Dennis_Churaev [7]2 years ago

6 0

Answer:

Explanation:

What is the value of a in the function's equation?

A. -2

B. -3

C. 3

D. 2

ANSWER: c.

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A mxiture of n2 and H2 has mole fraction of 0.4 and 0.6 respectively. Determine the density of the mixture at one bar and 0 c.

SOVA2 [1]

Answer:

The density of the mixture is 0.55kg/m^3

Explanation:

P = 1bar = 100kN/m^2, T = 0°C = 273K, n = 0.4+0.6 = 1mole

PV = nRT

V = nRT/P = 1×8.314×273/100 = 22.70m^3

Mass of N2 = 0.4×28 = 11.2kg

Mass of H2 = 0.6×2 = 1.2kg

Mass of mixture = 11.2 + 1.2 = 12.4kg

Density of mixture = mass/volume = 12.4/22.7 = 0.55kg/m^3

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4 years ago

22. A 2.0 kg ball, traveling at 1.0 m/s, rolls off of a cliff of height 3.0 m. If the

RoseWind [281]

GOOD MORNING HAVE A NICE DAY

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3 years ago

A 3.42 kg mass hanging vertically from a spring on the Earth (where g = 9.8 m/s2) undergoes simple oscillatory motion. If the sp

sineoko [7]

Answer:

Time period of oscillation on moon will be equal to 3.347 sec

Explanation:

We have given mass which is attached to the spring m = 3.42 kg

Spring constant K = 12 N/m

We have to find the period of oscillation

Period of oscillation is equal to $T=2\pi \sqrt{\frac{m}{K}}$ , here m is mass and K is spring constant

So period of oscillation $T=2\times 3.14\times \sqrt{\frac{3.42}{12}}$

$T=2\times 3.14\times 0.533=3.347sec$

So time period of oscillation will be equal to 3.347 sec

As it is a spring mass system and from the relation we can see that time period is independent of g

So time period will be same on earth and moon

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3 years ago

A ball of mass 2.0 kg falls vertically and hits the ground with speed 7.0 ms-1 as shown below.

Sauron [17]

Answer:

8.0 Ns

Explanation:

Change in momentum is given as:

Final momentum - Initial momentum

= m*v - m*u

Where m = mass of ball

v = final velocity

u = initial velocity

Change in momentum = (2.0 * 3.0) - (2.0 * 7.0)

= 6.0 - 14.0 = -8.0 Ns

The magnitude will be |-8.0| = 8.0 Ns

7 0

4 years ago

Read 2 more answers

A raft of mass 199 kg carries two swimmers of mass 52 kg and 70 kg. The raft is initially floating at rest. The two swimmers sim

Minchanka [31]

To solve this problem we will apply the concept related to the conservation of the Momentum. We will then start considering that the amount of initial momentum must be equal to the amount of final momentum. Considering that all the objects at the initial moment have the same initial velocity (Zero, since they start from rest) the final moment will be equivalent to the multiplication of the mass of each object by the velocity of each object, so

Initial Momentum = Final Momentum

$(m_B+m_1+m_2)v_i = m_1v_1+m_2v_2+m_Bv_B$

Here,

$m_B$ = mass of Raft

$m_1$ = Mass of swimmers 1

$m_2$ = Mass of swimmers 2

$v_i$ = Initial velocity (of the three objects)

$v_B$ = Velocity of Raft

Replacing,

$(199+52+70)*0 = (52)(4)+(70)(-4)+199v_B$

Solving for $v_B$

$vB = \frac{72}{199}$

$v_B = 0.3618m/s$

Therefore the velocity the rarft start to move is 0.3618m/s

5 0

4 years ago