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Mkey [24]
3 years ago
12

A 150 kg uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. a) Calculate the m

agnitude of the vertical component of the force that the wall exerts on the left end of the beam if the angle between the cable and horizontal is θ = 47°.
Physics
1 answer:
Rus_ich [418]3 years ago
4 0

Answer:

1004.277N

Explanation:

Given that g = 9.817 ms^-2

If the angle between the cable and horizontal is 47°.

The angle between the cable and vertical is 90-47= 43°.

The weight of the beam is mg

The weight of the beam is

= 150*9.817

The weight of the beam is

= 1472.55N

magnitude of the vertical component of the force is mgsin(tita)

= 1472.55*sin43

= 1004.277N

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Answer:

A motion diagram represents the motion of an object by displaying its location at various equally spaced times on the same diagram. Motion diagrams are a pictorial description of an object's motion. They show an object's position and velocity initially, and present several spots in the center of the diagram.

Explanation:

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Desde lo alto de un acantilado de 140 m, se lanza verticalmente un objeto hacia abajo con velocidad de 3m/s. Entonces la magnitu
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Answer:

54

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7 0
3 years ago
A garden hose with a diameter of 0.64 in has water flowing in it with a speed of 0.46 m/s and a pressure of 1.9 atmospheres. At
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Answer:

(a).The speed of the water in the nozzle is 3.014 m/s.

(b). The pressure in the nozzle is 1.86 atm.

Explanation:

Given that,

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Hose pipe diameter = 0.64 in = 0.016256 m

Pressure = 1.9 atm =192518 Pa

(a). We need to calculate the speed of the water in the nozzle

Flow Speed at the inlet pipe will be given by using Continuity Equation

Q_{1}=Q_{2}

v_{1}A_{1}=v_{2}A_{2}

v_{1}=v_{2}\times(\dfrac{A_{2}}{A_{1}})

Where, A = area of pipe

A=\pi\times \dfrac{d^2}{4}

v_{1}=v_{2}\times(\dfrac{d_{2}^2}{d_{1}^2})

Put the value into the formula

v_{1}=0.46\times\dfrac{(0.016256)^2}{(0.00635)^2}

v_{1}=3.014\ m/s

The speed of the water in the nozzle is 3.014 m/s.

(b). We need to calculate the pressure in the nozzle

Using Bernoulli's Theorem,

P_{1}+\dfrac{1}{2}\rho\times v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho\times v_{2}^2+\rho gh_{2}

Where, h_{1}=h_{2}

P_{1}+\dfrac{1}{2}\rho\times v_{1}^2=P_{2}+\dfrac{1}{2}\rho\times v_{2}^2

P_{1}=P_{2}+\dfrac{1}{2}\rho(v_{2}^2-v_{1}^2)

Put the value into the formula

P_{1}=192518 +\dfrac{1}{2}\times1000\times((0.46)^2-(3.014)^2)

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P=1.86\ atm

Hence, (a).The speed of the water in the nozzle is 3.014 m/s.

(b). The pressure in the nozzle is 1.86 atm.

7 0
3 years ago
Determine the ratio of the resistivity of pure water to silver?
shutvik [7]
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r= \frac{10^3 \Omega m}{10^{-8} \Omega m}  \sim 10^{11}

Therefore, we can say that the correct answer is
3 \cdot 10^{11} : 1
3 0
3 years ago
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