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3 years ago

A rock dropped from a 5 m height accelerates at 10 m/s and strikes the ground 1 s later. If the rock is dropped

Physics

1 answer:

antoniya [11.8K]3 years ago

7 0

Answer:

10 m/s²

Explanation:

The above question simply indicates motion under gravity.

The acceleration due to gravity (i.e acceleration of free fall) has a constant value of 10 m/s².

Whether the rock is dropped from a height of 5 m or 2.5 m, it will accelerate at 10 m/s² before striking the ground. The only thing that will be different is the time taken for the rock to strike the ground when released from both 5 m and 2.5 m.

Thus, the rock will have a constant acceleration of 10 m/s² irrespective of the height to which it was released.

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if the radius of the capillary tube is doubled, what changes will take place in the hieght of rise of liquid with capacity tube

tangare [24]

Explanation:

The height of the rise of liquid with capillary tube is given by the formula as follows :

$h=\dfrac{2S\cos\theta}{r\rho g}$

Where

r is radius

It is clear that the height of the rise of liquid is inversely proportional to the radius of the capillary tube.

If the radius of the capillary tube is doubled, it means the height of rise of liquid with capillary tube become half.

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3 years ago

a crane lifts a 75 kg mass at a height of 8m . calculate the gravitational potential energy gained by the mass (g= 10N)

Bogdan [553]

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3 years ago

(b) Show that for certain incident energies there is 100 percent transmission. Suppose that we model an atom as a one-dimensiona

bixtya [17]

Answer:

Explanation:

100 persent transmission implies that the T=1

Therefore using the previous result we have

$1+\frac{\sin^2\sqrt{\frac{2m}{\hbar^2}(E+V_0)}a}{4\frac{E}{V_0}\frac{(E+V_0)}{V_0}}=1
$

$\sin\sqrt{\frac{2m}{\hbar^2}(E+V_0)}a=0\Rightarrow \sqrt{\frac{2m}{\hbar^2}(E+V_0)}=0\Rightarrow E=-V_0
$

The depth of the well for 100% transmission should be

$V_0=-0.7~{\rm{eV}}$

7 0

3 years ago

A 34.9-kg child starting from rest slides down a water slide with a vertical height of 16.0 m. (Neglect friction.)

aalyn [17]

Answer:

a) 12.528 m/s

b) 15.344 m/s

Explanation:

Given:

Mass of the child, m = 34.9 kg

Height of the water slide, h = 16.0 m

Now,

a) By the conservation of energy,

loss in potential energy = gain in kinetic energy

mgh = $\frac{\textup{1}}{\textup{2}}\textup{m}\times\textup{v}^2$

where,

g is the acceleration due to the gravity

v is the velocity of the child

thus,

at halfway down, h = $\frac{\textup{16}}{\textup{2}}$ = 8 m

therefore,

34.9 × 9.81 × 8 = $\frac{\textup{1}}{\textup{2}}\textup{34.9}\times\textup{v}^2$

v = 12.528 m/s

at three-fourth way down

height = $\frac{\textup{3}}{\textup{4}}\times16$ = 12 m

thus,

loss in potential energy = gain in kinetic energy

34.9 × 9.81 × 12 = $\frac{\textup{1}}{\textup{2}}\textup{34.9}\times\textup{v}^2$

v = 15.344 m/s

5 0

3 years ago

Read 2 more answers

An airplane flying parallel to the ground undergoes two consecutive dis- placements. The first is 75 km 30.0° west of north, and

torisob [31]

Answer: displacement of airplane is 172 km in direction 34.2 degrees East of North

Explanation:

In constructing the two displacements it is noticed that the angle between the 75 km vector and the 155 km vector is a right angle (90 degrees).

Hence if the plane starts out at A, it travels to B, 75 km away, then turns 90 degrees to the right (clockwise) and travels to C, 155 km away from B. Angle ABC is 90 degrees, hence we can use Pythagoras theorem to solve for AC

AC2 = AB2 + BC2 ; AC^2 = 752 + 1552 ; from this we get AC = 172 km (3 significant figures)

Angle BAC = Tan-1(155/75) ; giving angle BAC = 64.2 degrees

Hence AC is in a direction (64.2 - 30) = 34.2 degrees East of North

Therefore the displacement of the airplane is 172 km in a direction 34.2 degrees East of North

5 0

3 years ago