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padilas [110]
2 years ago
12

PLEASE ANSWER

Physics
1 answer:
Irina-Kira [14]2 years ago
8 0

Answer:

Farm = 98.1 [N]

Explanation:

To solve this problem we must draw the respective free body diagram, with the forces acting on the monkey. An analysis of the sums on the y-axis must be performed, in this axis the weight is acting down and the forces of both arms pulling up.

Weight is defined as the product of mass by gravitational acceleration.

W = m*g

where:

m = mass = 20 [kg]

g = gravity acceleration = 9.81 [m/s²]

W = 196.2 [N] (units of Newtons)

As this force points down, the force of both arms must go up, therefore each arm exerts a force of:

Farm = 196.2 / 2

Farm = 98.1 [N]

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If the distance from a light source triples, how does light intensity change? The intensity will be 3x greater. The intensity wi
Tcecarenko [31]

Answer:

The intensity will be 1/9 as much.

Explanation:

The intensity of the light or any source is inversely related to the square of the distance.

I\alpha \frac{1}{r^{2} }

Now according to the question the distance is increased by three times than,

\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{r_{2}^{2} }

Therefore,

\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{(3r_{1})^{2} }\\\frac{I_{2} }{I_{1} }=\frac{1}{9} \\{I_{2}=\frac{1}{9}{I_{1} }

Therefore the intensity will become 1/9 times to the initial intensity.

3 0
3 years ago
Which of the following is related to a sound's pitch?
FromTheMoon [43]
A bell or a siren or a ring in somewhere
3 0
3 years ago
A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500
defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

M = \frac{\mu_0 N_1 N_2A_1}{l_1}

Where,

\mu =permeability of free space

N_1 = Number of turns in solenoid 1

N_2 = Number of turns in solenoid 2

A_1= Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

\mu_0 = 4\pi *10^{-7}H/m

N_1 = 500

N_2 = 40

A = 7.5*10^{-4}m^2

l = 0.5m

Substituting,

M = \frac{\mu_0 N_1 N_2A_2}{l_1}

M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}

M = 3.77*10^{-4}H

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0
3 years ago
A circle graph shows that one-fourth of the students in a particular school ride the bus every day. What is this equivalent meas
Aleonysh [2.5K]

Answer:

it's 1.57... radians and 90°

3 0
2 years ago
A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are
Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, \Sigma F_y = 0

The vertical component of the tension, T_y, in each half of the rope is given as follows;

T_y = T × sin(θ)

∴ \Sigma F_y = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

4 0
2 years ago
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