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Simora [160]
3 years ago
5

A 98.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.69 r

ad/s . A monkey drops a 9.29 kg bunch of bananas vertically onto the platform. They hit the platform at 45 of its radius from the center, adhere to it there, and continue to rotate with it. Then the monkey, with a mass of 20.3 kg , drops vertically to the edge of the platform, grasps it, and continues to rotate with the platform. Find the angular velocity of the platform with its load. Model the platform as a disk of radius 1.51 m .
Physics
1 answer:
sertanlavr [38]3 years ago
6 0

Answer:

Explanation:

The problem is based on conservation of angular momentum.

Moment of inertia of the disc = 1/2 m R² , m is mass of the disc and R is its radius.

= 1/2 x 98.1 x 1.51²

= 111.84 kg m²

Moment of inertia of disc + moment of inertia of bananas + monkey

= 1/2 x 98.1 x 1.51² + 9.29 x .45 x 1.51 + 20.3 x 1.51² ( moment of inertia of banana and monkey will be equal to mass x radial distance from axis² )

= 111.84 + 6.31 +46.28

= 164.43 kg m²

Now applying law of conservation of angular momentum

= I₁ ω₁ = I₂ω₂

111.84 x 1.69 = 164.43 x ω₂

ω₂ = 1.15 rad / s

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Alecsey [184]

Answer:

The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

Explanation:

Given that,

First charge q_{1}= 20\mu C

second charge q_{2}= 8\mu C

Distance = 20 cm

We need to calculate the electric field

For first charge,

Using formula of electric field

E_{1}= \dfrac{kq_{1}}{r^2}

Put the valueinto the formula

E_{1}=\dfrac{9\times10^{9}\times20\times10^{-6}}{10\times10^{-2}}

E_{1}=18\times10^{5}\ N/C

Direction of electric field along AB

We need to calculate the electric field

For second charge,

Using formula of electric field

E_{2}= \dfrac{kq_{2}}{r^2}

Put the valueinto the formula

E_{2}=\dfrac{9\times10^{9}\times8\times10^{-6}}{10\times10^{-2}}

E_{2}=7.2\times10^{5}\ N/C

Direction of electric field along AO

We need to calculate the net electric field at midpoint

E_{net}=E_{1}-E_{2}

E_{net}=(18-7.2)\times10^{5}\ N/C

E_{net}=10.8\times10^{5}\ N/C

Direction of net electric field along AB

Hence, The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

8 0
3 years ago
A 5 meter long ladder leans against a wall. The bottom of the ladder slides away from the wall at the constant rate of 1 3 m/s.
Oksana_A [137]

Answer:9.75 m/s

Explanation:

Given

Length of ladder (L)=5 m

Foot the ladder is moving away with speed of \frac{\mathrm{d} x}{\mathrm{d} t}=13 m/s

From diagram

x^2+y^2=L^2------1

at x=3

y^2=25-9=16

y=4 m

Now differentiating equation 1 w.r.t time

2x\frac{\mathrm{d} x}{\mathrm{d} t}+2y\frac{\mathrm{d} y}{\mathrm{d} t}=0

x\frac{\mathrm{d} x}{\mathrm{d} t}=-y\frac{\mathrm{d} y}{\mathrm{d} t}

3\times 13=-4\times \frac{\mathrm{d} y}{\mathrm{d} t}

\frac{\mathrm{d} y}{\mathrm{d} t}=-\frac{3\times 13}{4}=-9.75 m/s

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5 0
3 years ago
Answer all these questions
Kazeer [188]

Answer:

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The higher the frequency, the shorter the wavelength.

Explanation:

5 0
3 years ago
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If you walk 3 kilometers in 30 minutes , what is the average speed in kilometers per hour?
Pepsi [2]

Answer:

6 km/h

Explanation:

V avg = ∆x/∆t = 3km / 30 min ×(60min/1h) = 3 km× 2 /h = 6 km/h

4 0
2 years ago
A comet is cruising through the solar system at a speed of 50,000 kilometers per hour foe 4 hours time. What is the total distan
Triss [41]

Answer:

<em>The distance covered by comet is </em>200,000 km

Explanation:

Speed is defined as the rate of change of distance with time. It is given by the equation speed= \bold{\frac{distance}{time}}

Thus distance= speed*time

In this problem it is given that speed of comet= \frac{50,000km}{hr}

time travelled by the comet= 4 hours

Thus distance= speed*time

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5 0
3 years ago
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