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3 years ago

13

If the forces acting on a soccer ball are balanced , what should happen to the motion of the ball? A The ball will fall to the g

round 00 The ball wil bounce up and down. The ball will rise toward her head. The ball will not move .

2 answers:

lukranit [14]3 years ago

5 0

That’s what she said

m_a_m_a [10]3 years ago

5 0

Answer:

A

Explanation:

i took the test trust me:)

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A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA. What amount of vars must be added to bring the pf to 0.85

kvasek [131]

Answer:

$\mathtt{Q_{sh} = 600.75 \ vars}$

Explanation:

Given that:

A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA

Here:

the initial power factor i.e cos θ₁ = 0.7 lag

θ₁ = cos⁻¹ (0.7)

θ₁ = 45.573°

Active power P = 1500 watts

Apparent power S = 2100 VA

What amount of vars must be added to bring the pf to 0.85

i.e the required power factor here is cos θ₂ = 0.85 lag

θ₂ = cos⁻¹ (0.85)

θ₂ = 31.788°

However; the initial reactive power $Q_1$ = P×tanθ₁

the initial reactive power $Q_1$ = 1500 × tan(45.573)

the initial reactive power $Q_1$ = 1500 × 1.0202

the initial reactive power $Q_1$ = 1530.3 vars

The amount of vars that must therefore be added to bring the pf to 0.85

can be calculated as:

$Q_{sh} = P( tan \theta_1 - tan \theta_2)$

$Q_{sh} = 1500( tan \ 45.573 - tan \ 31.788)$

$Q_{sh} = 1500( 1.0202 - 0.6197)$

$Q_{sh} = 1500( 0.4005)$

$\mathtt{Q_{sh} = 600.75 \ vars}$

3 0

3 years ago

Suppose a small quantity of radon gas, which has a half-life of 3.8 days, is accidentally released into the air in a laboratory.

Daniel [21]

Answer:

1 day

Explanation:

Let the safe level = x

The current level = x + 0.2x = 1.2 x

Thus,

Half life = 3.8 days

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{3.8}\ days^{-1}$

The rate constant, k = 0.1824 days⁻¹

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

So,

$\frac {[A_t]}{[A_0]}$ = x / 1.2 x = 0.8333

t = ?

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.8333=e^{-0.1824\times t}$

t ≅ 1 day

<u>Lab must be vacated in 1 day.</u>

4 0

3 years ago

iris [78.8K]

Answer:

The answer, because i gotta

Explanation:

finish a challenge

7 0

3 years ago

A sample of water is heated at a constant pressure of one atmosphere. Initially, the sample is ice at 260 K, and at the end the

USPshnik [31]

In <u>370 K to 375 K </u>temperature intervals of 5 K, would be the greatest increase in the entropy of the sample.

Option: C

<u>Explanation</u>:

Because the largest difference in molar entropy occurs when a condensed phase (solid/liquid) transforms to the gas phase. Then change in entropy is equal to heat transfer divided by temperature: $\Delta \mathrm{S}=\frac{\Delta Q}{\mathrm{m} T}$ .

According to given ice sample at 260 K, when this solid sample start converting into liquid sample it will gain positive temperature and steam will take place near 373 K (273 K ice temperature + $100^{\circ} \mathrm{C}$ temperature of boiling water). Therefore it’s very obvious that greatest increase in entropy will occur during 370 K – 375 K.

5 0

3 years ago

Ice fishermen sit on top of frozen lakes in the winter and catch fish in the liquid water below through holes cut in the ice she

melomori [17]

Answer:

This is because below 4°c, water unlike other materials becomes less dense when it's temperature is further lowered.

Explanation:

Due to the unusual nature of water; at about 4°c, the behavior of the density of water in relation to its temperature reverses. This means that water becomes less dense as it becomes colder below 4°c. The colder parts therefore floats to the top of the water body while the warmer part sinks allowing the top to freeze and the remaining body below to remain in its liquid state.

The freezing of the top of the lake alone protects the remaining depth of water from freezing by acting as an insulator and preventing further heat loss from the water to the ambient space. If this had not been the case, and water froze all through, marine lives will freeze to death and it will be more difficult to melt the ice come the next summer.

This behavior is due to the hydrogen bonding of the water molecules.

8 0

3 years ago