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2 years ago

How do you find the mechanical advantage of a wheel and axle

1 answer:

alexandr1967 [171]2 years ago

7 0

Answer: <em>In order to calculate the mechanical advantage of a wheel and axel, you can divide the radius of the wheel by the radius of the axle. If a wheel's radius is 60cm and its axle is 30cm, what is the mechanical advantage of the wheel and axle? The output force of the wheel and axle is 2 times greater than the input force.</em>

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What are the 3 tools used to collect weather data

sergey [27]

Anemometer, Psychrometer, <span>Barometer</span>

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2 years ago

A box has a mass of 5.8kg. The box is lifted from the garage floor and placed on a shelf 2.5m off the ground. How much gravitati

Damm [24]

Gravitational potential energy =

(mass) x (gravity) x (height)

= (5.8 kg) x (9.8 m/s²) x (2.5 m)

= 142.1 Joules (C)

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2 years ago

A student places her 490 g physics book on a frictionless table. She pushes the book against a spring, compressing the spring by

Paul [167]

Answer:

book speed is 3.99 m/s

Explanation:

given data

mass m = 490 g = 0.490 kg

compressing x = 7.10 cm = 0.0710 m

spring constant k = 1550 N/m

to find out

book speed

solution

we know energy is conserve so

we can say

loss in spring energy is equal to gain in kinetic energy

$\frac{1}{2}*k*x^2 = \frac{1}{2}*m*v^2$ ..................1

put here value

$\frac{1}{2}*1550*0.071^2 = \frac{1}{2}*0.490*v^2$

v = 3.99 m/s

so book speed is 3.99 m/s

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2 years ago

One end of a 34-m unstretchable rope is tied to a tree; the other end is tied to a car stuck in the mud. The motorist pulls side

anyanavicka [17]

Answer:

Fc = 89.67N

Explanation:

Since the rope is unstretchable, the total length will always be 34m.

From the attached diagram, you can see that we can calculate the new separation distance from the tree and the stucked car H as follows:

L1+L2=34m

$L1^2=L2^2=L^2=2^2+(H/2)^2$ Replacing this value in the previous equation:

$\sqrt{2^2+H^2/4}+ \sqrt{2^2+H^2/4}=34$ Solving for H:

$H=\sqrt{52}$

We can now, calculate the angle between L1 and the 2m segment:

$\alpha = atan(\frac{H/2}{2})=60.98°$

If we make a sum of forces in the midpoint of the rope we get:

$-2*T*cos(\alpha ) + F = 0$ where T is the tension on the rope and F is the exerted force of 87N.

Solving for T, we get the tension on the rope which is equal to the force exerted on the car:

$T=Fc=\frac{F}{2*cos(\alpha) } = 89.67N$

7 0

2 years ago

What is the most significant challenge for organisms that live in estuaries?

marysya [2.9K]

"Changing water salinity" is the most significant challenge for organisms that live in estuaries.

<u>Answer:</u> Option D

<u>Explanation:</u>

For estuaries, alkalinity levels are usually the maximum at a river's mouth where the ocean water falls for, and the minimum upstream where freshwater falls in. Although salinity vary throughout the tidal cycle. In estuaries, salinity rates usually decrease in spring as snow melt and rain raises the freshwater flow from streams and groundwater.

It influences the chemical environments within the estuary, especially the dissolved oxygen (DO) levels in the water. The level of oxygen that would get dissolved in water or its solubility get declined when the alkalinity rises.

8 0

2 years ago