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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

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What kind of surface is suitable for infared radiation

nlexa [21]

Answer:

Different surfaces

<h3>You can see that dull surfaces are good absorbers and emitters of infrared radiation. Shiny surfaces are poor absorbers and emitters (but they are good reflectors of infrared radiation</h3>

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3 years ago

The forklift exerts a 1,500.0 N force on the box and moves it 3.00 m forward to the stack. How much work does the forklift do ag

Deffense [45]

The answer is D using the work formula
W= F•d but if it was against gravity, it would be 0 if gravity is exerting the same amount, I would pick D using the formula, but I'm not so sure sorry

7 0

3 years ago

Guys please help me

Likurg_2 [28]

Answer:

1) $t=2.26\: s$

2) $S=33.9\: m$

3) $v=26.77\: m/s$

4) $\alpha=55.92$

Explanation:

1)

We can use the following equation:

$y_{f}=y_{0}+v_{iy}t-0.5*g*t^{2}$

Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.

$0=25-0.5*9.81*t^{2}$

$t=2.26\: s$

2)

The equation of the motion in the x-direction is:

$v_{ix}=\frac{S}{t}$

$15=\frac{S}{2.26}$

$S=33.9\: m$

3)

The velocity in the y-direction of the stone will be:

$v_{fy}=v_{iy}-gt$

$v_{fy}=0-(9.81*2.26)$

$v_{fy}=-22.17\: m/s$

Now, the velocity in the x-direction is 15 m/s then the velocity will be:

$v=\sqrt{v_{x}^{2}+v_{fy}^{2}}=\sqrt{15^{2}+(-22.17)^{2}}$

$v=26.77\: m/s$

4)

The angle of this velocity is:

$tan(\alpha)=\frac{22.17}{15}$

$\alpha=tan^{-1}(\frac{22.17}{15})$

$\alpha=55.92$

Then α=55.92° negative from the x-direction.

I hope it helps you!

6 0

3 years ago

Every football field has two 300kg field goal posts separated by 110m of football field (that includes the endzones). What is th

Mama L [17]

Answer:

4.96×10¯¹⁰ N

Explanation:

The following data were obtained from the question:

Mass 1 (M1) = 300 Kg

Mass 2 (M2) = 300 Kg

Separating distance (r) = 110 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Gravitational force (F) =?

The gravitational force between the two goal posts can be obtained as follow:

F = GM1M2 / r²

F = 6.67×10¯¹¹ × 300 × 300 / 110²

F = 6.003×10¯⁶ / 12100

F = 4.96×10¯¹⁰ N

Therefore the gravitational force between the two goal posts is 4.96×10¯¹⁰ N

3 0

3 years ago

Diagnostic ultrasound of frequency 3.82 MHz is used to examine tumors in soft tissue. (a) What is the wavelength in air of such

maxonik [38]

Explanation:

It is given that,

Frequency of diagnostic ultrasound, f = 3.82 MHz = 3820 Hz

The speed of the sound in air, v = 343 m/s

(a) We need to find the wavelength in air of such a sound wave. Let it is given by λ₁

i.e. $\lambda=\dfrac{v}{\nu}$

$\lambda_1=\dfrac{343\ m/s}{3820\ Hz}$

$\lambda_1=0.089\ m$

(b) If the speed of sound in tissue is 1650 m/s .

$\lambda_2=\dfrac{v}{\nu}$

$\lambda_2=\dfrac{1650\ m/s}{3820\ Hz}$

$\lambda_2=0.43\ m$

Hence, this is the required solution.

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3 years ago