Ask question

Ask question

All categories

3 years ago

15

You drop a 30 g pebble down a well. You hear a splash 2.7 s later. What was the pebble’s speed just as it reaches the water? Ass

ume g = 9.8 m/s2

1 answer:

Flauer [41]3 years ago

7 0

vf=vi+at

No need for rearranging because it is already set up for Vf (final velocity)

a= -9.8m/s² (because it is falling)

vi= 0

t= 2.7

0 + -9.8(2.7) = vf

vf = -26.5 (-26.46)

You might be interested in

100 g of liquid nitrogen at its boiling point of 77 K is stirred into a beaker containing 500 g of 15°C water.

igomit [66]

Answer:

Option A: none

Explanation:

Specific heat capacity is the amount of heat energy required to raise the temperature of a substance per unit of mass.

From the question, the parameters given are; The heat of vaporization of nitrogen= 48 cal/g and that of water is 80 cal/g.

Using the formulae; Specific heat capacity,c= Q/ m× ∆T,----------–-----------------------------------------------------

STEP ONE: We will have to calculate all the energy numbers; 77k is approximately the boiling point of Nitrogen.

Energy required to decrease water from 15°C to 0°C = E(1).

0.1×10^3 g× 48 cal/gram..

= 4800 cal

Energy require to vaporize Nitrogen=E(2).

= 80 cal per gram×0.15(15-0)

= 180 cal

Energy required to decrease water from 15°C to 0°C is higher than that of the energy to vaporize Nitrogen, N2.

STEP 2: 180/4800× 15 = 0.5625

Therefore; 15-0.5625 =14.43

Answer= 14.43gC

8 0

4 years ago

Read 2 more answers

A child and her mother are at the park the child walked around a rectangular park while her mom waits on the bench the child ret

Genrish500 [490]

The distance is just the perimeter of the rectangle:

P = 2(411) + 2(475)
P = 822 + 950
P = 1772m

6 0

3 years ago

A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of

Lynna [10]

We have the relation

$\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}$

where $v_{A \mid B}$ denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

$v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}$

$v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}$

Let's assume the river flows South-to-North, so that

$\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath$

and let $-90^\circ < \theta < 90^\circ$ be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

$\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)$

Then the velocity of the boat relative to the Earth is

$\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath$

The crossing is 153.0 m wide, so that for some time $t$ we have

$153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s$

which is minimized when $\theta=0^\circ$ so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

$\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}$

The boat's position $\vec x$ at time $t$ is

$\vec x = \vec v_{B\mid E} t$

so that after 30.0 s, the boat's final position on the other side of the river is

$\vec x(30.0\,\mathrm s) = (153\,\mathrm m) \,\vec\imath + (111\,\mathrm m)\,\vec\jmath$

and the boat would have traveled a total distance of

$\|\vec x(30.0\,\mathrm s)\| = \sqrt{(153\,\mathrm m)^2 + (111\,\mathrm m)^2} \approx \boxed{189\,\mathrm m}$

3 0

2 years ago

Consider a simple but surprisingly accurate model for the hydrogen molecule: two positive point charges, each having charge e, a

lyudmila [28]

Answer:

a = R/2

Explanation:

Let R be the radius of the sphere and q = -2e be the charge in it. Let q₁ be the charge at radius a where the one of the point positive charges e is located. . Since the charge is uniformly distributed, the volume charge density is constant. So, q/4πR³ = q₁/4πa³

q₁ = q(a/R)³ = -2e(a/R)³. The electric force due to q₁ at r is F₁ = kq₁q/a² = kq²(a/R)³/a² = k(-2e)²a/R³ = 4ke²a/R³.

Let h be the distance between the two point charges. The electric force due to the one point charge on the other is F₂ = ke²/h²

If the net force on either charge is zero, then

F₁ = F₂

4ke²a/R³ = ke²/h²

a = R³/4h²

Since h = 2a, since the charges are equidistant from each other,

a = R³/4(2a)² = R³/8a²

a = R³/8a²

a³ = R³/8

a = ∛(R³/8)

a = R/2

7 0

3 years ago

Why does a catalyst cause a reaction to proceed faster??

prohojiy [21]

The activation energy is lowered only
There are more collisions per second and the collisions are of greater energy
There are more collisions per second only
The activation energy is lowered

7 0

4 years ago

Read 2 more answers