Ask question

Ask question

All categories

2 years ago

15

when a bucket full of water is pulled of from a well,in which condition: either inside or outside the water is it difficult to l

ift up and why?

2 answers:

Alexeev081 [22]2 years ago

7 0

Upthrust due to water is greater than that of air. Upthrust exerted by water decreases weight of body. Hence, it is easier to pull a bucket of water from the well until it is inside the water but difficult when it is out of water.

julsineya [31]2 years ago

6 0

Answer:

Well there is gravitational force acting on the bucket ,gravitational force is a downward force so it will be acting on the bucket making it harder to lift it

You might be interested in

Saturn moves in an orbit around the Sun with radius 10 AU. How many degrees does it move on the Celestial in one year? (Hint: Ca

Lana71 [14]

Answer:

B. About 12 degrees

Explanation:

The orbital period is calculated using the following expression:

T = 2π*( $\sqrt{\frac{r^3}{Gm}}$ )

Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.

Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.

Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:

$\frac{Tsaturn}{Tearth}$ = $\sqrt{\frac{rSaturn^3}{rEarth^3} } = \sqrt{10^3}}$

Knowing that the orbital period of the earth is 1 year, the orbital period of Saturn will be $\sqrt{10^3}}$ years, or 31.62 years.

We find the amount of degrees it moves in 1 year:

$1year * \frac{360degrees}{31.62years} = 11.38 degrees$

or about 12 degrees.

6 0

3 years ago

A ball is dropped from a rooftop 60m high. <br> How long is the ball in the air?

alina1380 [7]

Answer: 3.49 s

Explanation:

We can solve this problem with the following equation of motion:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0 m$ is the final height of the ball

$y_{o}=60 m$ is the initial height of the ball

$V_{o}=0 m/s$ is the initial velocity (the ball was dropped)

$g=9.8 m/s^{2}$ is the acceleratio due gravity

$t$ is the time

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (2)

$t=\sqrt{\frac{2 (60 m)}{9.8 m/s^{2}}}$ (3)

Finally we find the time the ball is in the air:

$t=3.49 s$ (4)

7 0

3 years ago

An intravenous (IV) system is supplying saline solution to a patient at the rate of 0.06 cm3/s through a needle of radius 0.2 mm

horsena [70]

Answer:

Pressure applied to the needle is 7528 Pa

Explanation:

As we know by poiseuille's law of flow of liquid through a cylindrical pipe

the rate of flow through the pipe is given as

$Q = \frac{\Delta P \pi r^4}{8\eta L}$

now we know that

$Q = 0.06 \times 10^{-6} m^3/s$

radius = 0.2 mm

Length = 6.32 cm

$\eta = 1\times 10^{-3} Pa s$

now we have

$6 \times 10^{-8} = \frac{\Delta P \pi (0.2 \times 10^{-3})^4}{8(1 \times 10^{-3})6.32 \times 10^{-2}}$

$3.03 \times 10^{-11} = \Delta P 5.02 \times 10^{-15}$

$\Delta P = 6028 Pa$

now we have

$P - 1500 = 6028 Pa$

$P = 7528 Pa$

8 0

2 years ago

How much heat is required to heat 2 kg of water from 25°C to 40°C?

Dominik [7]

Answer:

126000 J

Explanation:

Applying,

Q = cm(t₂-t₁).................. Equation 1

Where Q = Amount of heat, c = specifc heat capacity of water, m = mass of water, t₁ = Initial temperature, t₂ = Final temperature.

From the question,

Given: m = 2 kg, t₁ = 25°C, t₂ = 40°C

Constant: c = 4200 J/kg.°C

Substitute these value into equation 1

Q = 2×4200(40-25)

Q = 2×4200×15

Q = 126000 J

5 0

2 years ago

Explain the following defects of a simple electric cell:

eimsori [14]

Answer:

Explanation:

The two major defects of simple electric cells causes current supplied to be for short time. These defects are: polarization and local action.

a. Polarization: This is a defect caused by an accumulation of hydrogen bubbles at the positive electrode of the cell. It can be prevented by the use of vent, using a hydrogen absorbing material or the use of a depolarizer.

b. Local Action: This is the gradual wearing away of the electrode due to impurities in the zinc plate. It can be controlled by the amalgamation of the zinc plate before it is used.

4 0

2 years ago