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77julia77 [94]
3 years ago
14

Pulling up on a rope, you lift a 4.25 kg bucket of water from a well with an acceleration of 1.80 m/s2 . What is the tension in

the rope?
Physics
1 answer:
sattari [20]3 years ago
8 0

Answer:

49.3 N

Explanation:

Given that Pulling up on a rope, you lift a 4.25 kg bucket of water from a well with an acceleration of 1.80 m/s2 . What is the tension in the rope?

The weight of the bucket of water = mg.

Weight = 4.25 × 9.8

Weight = 41.65 N

The tension and the weight will be opposite in direction.

Total force = ma

T - mg = ma

Make tension T the subject of formula

T = ma + mg

T = m ( a + g )

Substitutes all the parameters into the formula

T = 4.25 ( 1.8 + 9.8 )

T = 4.25 ( 11.6 )

T = 49.3 N

Therefore, the tension in the rope is 49.3 N approximately.

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Temperature change time base problem. Suppose V = 24 V, I = 0.1 A, for water: mw = 51 gm, cw = 4.18 J/gm ∘K-1, for resistor: mr
nirvana33 [79]

Answer:

t = 444.125 sec

Explanation:

Given data:

V = 24 volt

I  = 0.1 ampere

mass of water mw = 51 gm

cr = 4.18 J/gm degree K^-1

mass of resistor = 8 gm

cr = 3.7 J/gm degree K^-1

we know that power is given as

Power P = VI

But P =E/t

so equating both side we have

\frac{E}{t} = VI

solving for t

t = \frac{E}{VI}

t = \frac{m_w C_w \Delta T}{VI}

t = \frac{51 \times 4.18 \times (5 -0)}{24\times 0.1}

t = 444.125 sec

5 0
3 years ago
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You make a straight line when an object is
swat32

Answer:

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Explanation:

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Suppose the moon rotated on its axis just as quickly as Earth. Would we still always see the same side of the moon from Earth?
jasenka [17]

Answer:

No, The Moon, on the other hand, rotates once around its every 28 days, and once around the Earth in that same 28 days. The result of this combination is that the same side of the Moon is always facing the Earth.

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What is the focal length (in meters) of a lens whose radius of curvature is 9. 2 m and has a refractive index 1. 2?
Fed [463]

The focal length (in meters) of a lens whose radius of curvature is 9. 2 m and has a refractive index  1.2 will be 18.4 m

The focal length of a lens is determined when the lens is focused at infinity. Lens focal length tells us the angle of view—how much of the scene will be captured—and the magnification—how large individual elements will be.

Focal length of a lens is the distance between center of lens and focal point . Focal point is a point on principal axis , at which light rays parallel to principal axis meet after refraction through lens or seem to meet after refraction .

The radius of curvature is the radius of sphere formed by the convex or concave mirror. It is also equal to the distance between the pole and center of curvature. The sign convention for focal length and radius of curvature is the same.

focal length = 2 * radius of curvature

given

radius of curvature = 9.2 m

focal length = 2 * 9.2

                    = 18.4 m

To learn more about focal length here

brainly.com/question/16188698

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4 0
2 years ago
In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.42 m. The mug
telo118 [61]

Answer:

a) V_{x}=3.72m/s, b) ∠=-54.83°

Explanation:

In order to solve this problem, we must start with a drawing of the situation, this will help us visualize the problem better. (See picture attached).

a)

Now, the idea is that the beer mug has a horizontal speed and no vertical speed at initial conditions. So knowing this, we can start finding the initial velocity of the mug.

In order to do so, we need to find the time it takes for the mug to reach the ground. We can find it by using the following equation:

y=y_{0}+V_{y0}t+\frac{1}{2}a_{y}t^{2}

We can see from the drawing that y and the initial velocity in y are zero, so we can simplify our formula:

0=y_{0}+\frac{1}{2}a_{y}t^{2}

so we can solve for t, so we get:

t=\sqrt{\frac{-(2)y_{0}}{a}}

so now we can substitute the known values, so we get:

t=\sqrt{\frac{-(2)(1.42)}{-9.8}}

which yields:

t=0.538s

So we can use this value to find the velocity in x:

V_{x}=\frac{x}{t}

When substituting we get:

V_{x}=\frac{2m}{0.538s}

which yields:

V_{x}=3.72m/s

b)

In order to solve part b, we need to find the y-component of the velocity, for which we can use the following formula:

\Delta y=\frac{V_{f}^{2}-V_{0}^{2}}{2a}

We know that V_{0} is zero, so we can simplify the expression:

\Delta y=\frac{V_{yf}^{2}}{2a}

So we can solve the equation for V_{yf}^{2} so we get:

V_{yf}=\sqrt{2\Delta y a}

and when substituting the known values we get:

V_{yf}=\sqrt{2(-1.42m)(-9.8m/s^{2})}

which yields:

V_{yf}=-5.28m/s

Once we got the final velocity in y, we can use it together with the velocity in x to find the angle.

So we can use the following formula:

tan \theta =\frac{V_{y}}{V_{x}}

when solving for theta we get:

\theta = tan^{-1}(\frac{V_{y}}{V_{x}})

We can substitute so we get:

\theta = tan^{-1}(\frac{-5.28m/s}{3.72m/s})

which yields:

\theta = -54.83^{o}

7 0
3 years ago
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