Ask question

Ask question

All categories

lisabon 2012 [21]

4 years ago

5

Four penguins that are being playfully pulled along very slippery (frictionless) ice by a curator. The masses of three penguins

and the tension in two of the cords are 1 12 m kg = , 3 15 m kg = , 4 20 m kg = , 2 111TN= , 4 222TN= . Find the penguin mass 2m that is not given.

1 answer:

Sati [7]4 years ago

7 0

Answer:

$m_2 = 23 kg$

Explanation:

As we know that the tension in two strings are

$T_2 = 111 N$

$T_4 = 222 N$

now we have

$F_{net} = ma$

so we can say

$(m_1 + m_2)a = T_2$

$(12 + m_2)a = 111$

also we have

$(m_1 + m_2 + m_3 + m_4) a = T_4$

$(12 + m_2 + 15 + 20) a = 222$

now divide two equations

$\frac{222}{111} = \frac{47 + m_2}{12 + m_2}$

$2m_2 + 24 = 47 + m_2$

$m_2 = 23 kg$

You might be interested in

Why is the rotation system used in volleyball? Physics was the closest thing to PE

tankabanditka [31]

Answer:

A rotation occurs after every side out, which is when the receiving team gains the right to serve by winning a rally. ... The new serving team will rotate clockwise one spot. The purpose of this is to rotate all the players through the serving position. If you continue winning points, you stay in position.

3 0

3 years ago

Read 2 more answers

Fill in the blank. Consider the inverse square law: When light leaves a light bulb, it spreads out over more and more space as i

aliya0001 [1]

Answer:

Explanation:

Intensity of light is inversely proportional to distance from source

I ∝ 1 /r² where I is intensity and r is distance from source . If I₁ and I₂ be intensity at distance r₁ and r₂ .

I₁ /I₂ = r₂² /r₁²

If r₂ = 4r₁ ( given )

I₁ / I₂ = (4r₁ )² / r₁²

= 16 r₁² / r₁²

I₁ / I₂ = 16

I₂ = I₁ / 16

So intensity will become 16 times less bright .

"16 times " is the answer .

8 0

3 years ago

A 60 kg man jumps down from a 0.8 m table. What is the speed when he

Rom4ik [11]

Answer: Speed = 4 m/s

Explanation:

The parameters given are

Mass M = 60 kg

Height h = 0.8 m

Acceleration due to gravity g= 10 m/s2

Before the man jumps, he will be experiencing potential energy at the top of the table.

P.E = mgh

Substitute all the parameters into the formula

P.E = 60 × 9.8 × 0.8

P.E = 470.4 J

As he jumped from the table and hit the ground, the whole P.E will be converted to kinetic energy according to conservative of energy.

When hitting the ground,

K.E = P.E

Where K.E = 1/2mv^2

Substitute m and 470.4 into the formula

470.4 = 1/2 × 60 × V^2

V^2 = 470.4/30

V^2 = 15.68

V = square root (15.68)

V = 3.959 m/s

Therefore, the speed of the man when hitting the ground is approximately 4 m/s

4 0

3 years ago

This picture represents<br> A: Reflection<br> B: Refraction<br> C: Interference<br> D: Diffraction

-BARSIC- [3]

Answer:

I think it's a because it goes thru it and reflects

5 0

2 years ago

Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc

Crazy boy [7]

Answer:

Temperature at the exit = $267.3 C$

Explanation:

For the steady energy flow through a control volume, the power output is given as

$W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

Inlet area of the turbine = $60cm^{2}= 0.006m^{2}$

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume $v_{1}$

as

$v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg$

$m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec$

for Ideal gasses, the enthalpy change can be calculated using the formula

$h_{2}-h_{1}=C_{p}(T_{2}-T_{1})$

hence we have

$W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

$250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})$

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have $T_{2}=267.3C$

Hence, the temperature at the exit = $267.3 C$

5 0

3 years ago