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3 years ago

Q2) The

Physics

2 answers:

Sindrei [870]3 years ago

8 0

Smallerrrrrrrrrrrrrrrrrrrrrrrro

kobusy [5.1K]3 years ago

4 0

Answer:

greater.

Explanation:

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A coin is dropped into a wishing well. It takes 1.1 seconds for a splash to be heard. Calculate the depth of the wishing well

Zigmanuir [339]

Answer:

If the wishing well was in a vacuum, then s=ut + 0.5 a t^2 (s=distance, ... wishing well if you drop a coin into it and hear the splash 10 seconds

Explanation:

8 0

3 years ago

a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its

lilavasa [31]

Answer:

D/H =15

Explanation:

We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
At this point, we can find the value of H, applying the following kinematic equation:

$v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)$

If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

$H = \frac{v_{0} ^{2} }{2*g} (2)$

We can use the same equation, to find the value of D, as follows:

$v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)$

In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
When we replace these values in (1), we find that v₁ = -v₀.
Replacing in (3), we have:

$(4*v_{0})^{2} - (-v_{0}) ^{2} = 2* g* D\\ \\ 15*v_{0}^{2} = 2*g*D$

Solving for D:

$D = \frac{15*v_{0} ^{2} }{2*g}$

From (2) we know that H can be expressed as follows:

$H = \frac{v_{0} ^{2} }{2*g}$

⇒ D = 15 * H

$\frac{D}{H} = 15$

3 0

3 years ago

Which statement best describes the law of conservation of energy?

shtirl [24]

Answer:

b!!!! did it help? :)))

7 0

3 years ago

Read 2 more answers

Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y

WITCHER [35]

Question:

A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity $3.65\times 10^5N/C$

Explanation:

A point charge ,q = $-2.14\times 10^{-6} C$ is located in the center of a spherical cavity of radius , $r =6.55\times 10^{-2}$ m inside an insulating spherical charged solid.