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3 years ago

A graduated cylinder contains water to the 15-cm'level. A stone was placed in the

Physics

1 answer:

qaws [65]3 years ago

6 0

Answer:

The anwser is 6

Explanation:

You subtract the 21-15

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A ball is thrown straight up from the ground with speed v0. At the same instant, a second ball is dropped from rest from a heigh

dlinn [17]

Answer:

a) t = H/v0

b) H = -(v0)²/g

Explanation:

Hi there!

a)The position of the balls can be calculated using the following equation:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the ball at time t.

y0 = initial height.

v0 = initial velocity.

g = acceleration due to gravity.

t = time.

For the ball that is thrown upwards, the initial height is zero, then, the equation can be written as follows:

y = v0 · t + 1/2 · g · t²

The second ball is initially at a height H and the initial velocity is zero. The equation of height for the second ball will be:

y = H + 1/2 · g · t²

When the two balls collide, their height is the same. Then, equalizing both equations we can obtain the time at which they collide:

v0 · t + 1/2 · g · t² = H + 1/2 · g · t²

v0 · t = H

t = H/v0

b) When the first ball is at the highest point its velocity is zero. Using the equation of velocity we can find the time at which the ball is at that point. The equation of velocity is the following:

v = v0 + g · t

At the highest point v = 0.

0 = v0 + g · t

Solving for t:

-v0/g = t

The time at which the first ball is at the highest point is t = -v0/g

The time at which both balls collide was calculated above:

t = H/v0

Then, equalizing both times and solving for H:

H/v0 = -v0/g

H = -v0/g · v0

H = -(v0)²/g

3 0

4 years ago

Write about Archimedes principle

Schach [20]

Answer:

Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially, is equal to the weight of the fluid that the body displaces. Archimedes' principle is a law of physics fundamental to fluid mechanics.

8 0

2 years ago

Read 2 more answers

If the bond enthalpy for a C-H bond is 413 kJ, what will happen when the C-H bond is broken?

pentagon [3]

Explanation:

Bond Enthalpy : It is defined as amount of energy required to break a the particular bond in there gaseous state. It is also known as bond energy. It units are kJ/mol.

Breaking of a bond is an Endothermic process (energy absorbed from the surroundings).

Formation of bond is an Exothermic process (energy is released to the surroundings).

If the average bond enthalpy for a C-H bond is 413 kJ/mol, When the C-H bond breaks in which energy will be required ,which will be an endothermic reaction.

8 0

3 years ago

the nucleus and the subatomic particles in the nucleus, what is the charge of the nucleus of an atom?

pychu [463]

The Nucleus contains Protons and Neutrons.

The Neutrons does not have a charge.

The Protons are positively charge.

Hence the charge on the Nucleus, would be the charge of the proton, which is positive.

Hence Nucleus is Positively Charged.

4 0

3 years ago

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

6 0

3 years ago