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2 years ago

14

iron ball weight 400 gram inside water when it is completely impressed in water 53 gram water is displaced what will be the weig

ht of iron ball in air?

1 answer:

Alika [10]2 years ago

8 0

Answer:

453 gm

Explanation:

<u>Immersed </u>objects are buoyed up by force equal to mass of displaced liquid

400 + 53 = 453 gm in air

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Read the question below.

Oxana [17]

Answer:

Experimental

Explanation:

Here the dependency of pressure on the temperature is to be determined. For finding this factor we need to calculate the pressure of the object at different temperature.

The process of calculating pressure for different temperature is experimental investigation because we need to perform this experiment for finding these value. On the basis of observed values the we can conclude the dependency.

Thus, this investigation is experimental.

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3 years ago

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What must be applied to move a load?

Misha Larkins [42]

An applied force must be applied to move a load. This applied force must be large enough to overcome any opposing forces in order to move the load.

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4 years ago

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Find the useful power output (in W) of an elevator motor that lifts a 2600 kg load a height of 30.0 m in 12.0 s, if it also incr

Annette [7]

Answer:

P = 251,916.667 W

Cost = 2,267.25 cents

Explanation:

To solve this question we will use the Work Energy Theorem, which is

$W = dP + dK\\$

Where

$dP$ = Change in Potential Energy

$dK$ = Change in Kinetic Energy

Change in Potential Energy

$P_{i} = mgh_{i}\\ P_{f} = mgh_{f}$

Where

$P_{i}$ = Initial Potential Energy

$P_{f}$ = Final Potential Energy

$m$ = Mass of System = 10,000 kg

$g$ = Acceleration due to gravity = 9.81 m/s

$h_{i}$ = Initial Height = 0

$h_{f}$ = Final Height = 30 m

Inputting the values we get the answer for $dP$

$dP = P_{f} - P_{i}\\dP= mgh_{f} - mgh_{i}\\ dP= 10000(9.81)(30) - 0\\ dP= 2943000$

Change in Kinetic Energy

$K_{i} = \frac{1}{2} mv_{i} ^2\\ K_{f} = \frac{1}{2} mv_{f} ^2$

Where

$K_{i}$ = Initial Kinetic Energy

$K_{f}$ = Final Kinetic Energy

$m$ = Mass of System = 10,000 kg

$g$ = Acceleration due to gravity = 9.81 m/s

$v_{i}$ = Initial Velocity = 0 m/2

$v_{f}$ = Final Velocity = 4 m/s

Inputting the values we get the answer for $dK$

$dK = K_{f} - K_{i}\\ dK = \frac{1}{2} mv_{f} ^2 - \frac{1}{2} mv_{i} ^2\\ dK = \frac{1}{2} (10000)(4)^2 - 0 \\ dK = 80000$

Total Work

$W = dP + dK\\$

Inputting the values

$W = 2943000 + 80000$

$W = 3,023,000$

a) Finding the useful Power Output

$P = \frac{W}{t}$

Where

$P$ = Power Output

$W$ = Work Done = 3,023,000J

$t$ = Time = 12s

Inputting the values

$P = \frac{3,023,000}{12}\\ P = 251,916.667$

P = 251,916.667 W

b) Finding the Total Cost

Cost = $0.0900 x P/1000

Cost = $0.0900 x (251,916.667/1000)

Cost = $22.67 or 2,267.25 cents

4 0

4 years ago

A ball of mass 0.250 kg and a velocity of + 5.00 m/s collides head-on with a ball of mass 0.800 kg that is initially at rest. No

Mademuasel [1]

Answer:

$v_1=-2.616\ m/s$

Explanation:

Given that,

Mass of ball 1, $m_1=0.25\ kg$

Initial speed of ball 1, $u_1=5\ m/s$

Mass of ball 2, $m_2=0.8\ kg$

Initial speed of ball 2, $u_2=0$ (at rest)

After the collision,

Final speed of ball 2, $v_2=2.38\ m/s$

Let $v_2$ is the final speed of ball 1.

Initial momentum of the system is :

$p_i=m_1u_1+m_2u_2$

$p_i=0.25\times 5+0$

$p_i=1.25\ m/s$

Final momentum of the system is :

$p_f=m_1v_1+m_2v_2$

$p_f=0.25\times v_1+0.8\times 2.38$

$p_f=0.25 v_1+1.904$

According the law of conservation of linear momentum :

initial momentum = final momentum

$1.25=0.25 v_1+1.904$

$v_1=-2.616\ m/s$

So, the final velocity of ball 1 is (-2.616)m/s.

8 0

3 years ago

If the rise and fall of your lungs is considered to be simple harmonic motion, how would you relate the period of the motion to

AURORKA [14]

Answer:

Breaths per minute is a frequency. The period is its reciprocal.

Explanation:

In simple harmonic motion, a period (T) is the time taken for one point to start in a position and reach that position again, in other words to complete a cycle or lapse. In this case, a period is the time one takes from starting to inspire the air to releasing all of it from the lungs.

In simple harmonic motion, the frequency (f) is how many times a point completes a cycle or lapse in one unity of time (could be one second, one minute, one hour, etc). In this case, the frequency is how many times one breathes in one minute. This is the breathing rate, since it is breathings per minute. Breaths per minute is a frequency.

Period (T) and frequency (f) relate to each other in the following formulae: $T=\frac{1}{f}$ or $f=\frac{1}{T}$ .

Therefore, breaths per minute is a frequency, and since it is related to the period, we say the period is reciprocal to it.

8 0

4 years ago