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amid [387]
3 years ago
12

The rope can support a maximum tension of 2700 n . is this rope strong enough to do the job? choose the correct answer and expla

nation.
Physics
1 answer:
USPshnik [31]3 years ago
5 0
Need more info.
It depends on what the job is.
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Where does the
GREYUIT [131]

Answer:

it comes from your knowledge and the information you have to get the reason why that is the answer so you are putting together things that you already know what the new information you have

5 0
3 years ago
A tuning fork vibrates at a frequency of 512 hertz
Naily [24]

Answer: The correct answer is " longitudinal wave with air molecules

vibrating parallel to the direction of travel".

Explanation:

In longitudinal wave, the particles vibrate parallel to the direction to the propagation of the wave. For example, sound wave is a longitudinal wave.

It needs a medium for its propagation. It can travel in solid, liquid and gas. It travel faster in solid in comparison to the liquid.

In the given problem, a tuning fork vibrates at a frequency of 512 hertz  when struck with a rubber hammer. The sound  produced by the tuning fork will travel through air. Here, a longitudinal wave with air molecules vibrating parallel to the direction of travel.

Therefore, the correct option is (1).

4 0
3 years ago
Read 2 more answers
g If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resi
Anettt [7]
<h2>Question:</h2>

In this circuit the resistance R1 is 3Ω, R2 is 7Ω, and R3 is 7Ω. If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resistance be?

Answer:

9.1Ω

Explanation:

The circuit diagram has been attached to this response.

(i) From the diagram, resistors R1 and R2 are connected in parallel to each other. The reciprocal of their equivalent resistance, say Rₓ, is the sum of the reciprocals of the resistances of each of them. i.e

\frac{1}{R_X} = \frac{1}{R_1} + \frac{1}{R_2}

=> R_{X} = \frac{R_1 * R_2}{R_1 + R_2}             ------------(i)

From the question;

R1 = 3Ω,

R2 = 7Ω

Substitute these values into equation (i) as follows;

R_{X} = \frac{3 * 7}{3 + 7}

R_{X} = \frac{21}{10}

R_{X} = 2.1Ω

(ii) Now, since we have found the equivalent resistance (Rₓ) of R1 and R2, this resistance (Rₓ) is in series with the third resistor. i.e Rₓ and R3 are connected in series. This is shown in the second image attached to this response.

Because these resistors are connected in series, they can be replaced by a single resistor with an equivalent resistance R. Where R is the sum of the resistances of the two resistors: Rₓ and R3. i.e

R = Rₓ + R3

Rₓ = 2.1Ω

R3 = 7Ω

=> R = 2.1Ω + 7Ω = 9.1Ω

Therefore, the combination of the resistors R1, R2 and R3 can be replaced with a single resistor with an equivalent resistance of 9.1Ω

4 0
3 years ago
N4M.6 A board has one end wedged under a rock having a mass of 380 kg and is supported by another rock that touches the bottom s
Hitman42 [59]

Answer:

it is safe to stand at the end of the table

Explanation:

For this exercise we use the rotational equilibrium condition

         Στ = 0

         W x₁ - w x₂ - w_table x₃ = 0

         M x₁ - m x₂ - m_table x₃ = 0

where the mass of the large rock is M = 380 kg and its distance to the pivot point x₁ = 850 cm = 0.85m

the mass of the man is 62 kg and the distance

            x₂ = 4.5 - 0.85

            x₂ = 3.65 m

the mass of the table (m_table = 22 kg) is at its geometric center

            x_{cm} = L/2 = 2.25 m

            x₃ = 2.25 -0.85

            x₃ = 1.4 m

let's look for the maximum mass of man

            m_{maximum} = \frac{ M x_1 -m_{table} x_3}{ x_2}

let's calculate

             m_{maximum} = \frac{ 380 \ 0.85 - 22 \ 1.4}{3.65}(380 0.85 - 22 1.4) / 3.65

             m_{maximum} = 80 kg

we can see that the maximum mass that the board supports without turning is greater than the mass of man

             m_{maximum}> m

consequently it is safe to stand at the end of the table

5 0
3 years ago
A crate is sliding down a ramp that is inclined at an angle 30.6 ° above the horizontal. The coefficient of kinetic friction bet
Vlada [557]

Answer:1.95 m/s^2

Explanation:

Given

inclination \theta =30.6^{\circ}

coefficient of kinetic friction \mu =0.36

As crate is moving Down therefore friction will oppose the motion

using FBD

mg\sin \theta -f_r=ma

f_r=\mu N

f_r=\mu mg\cos \theta

mg\sin \theta -\mu mg\cos \theta =ma

a=g\sin \theta -\mu g\cos \theta

a=g(\sin (30.6)-0.36\cdot \cos (30.6))

a=9.8\times 0.199

a=1.95 m/s^2          

6 0
3 years ago
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