All categories

3 years ago

By the time she managed/had managed to open the door, the postman already went/had already gone.

Physics

1 answer:

leva [86]3 years ago

8 0

Answer:

<h3>Answer Sentence: </h3><h2><em><u>By the time she had managed to open the door, the postman had already gone</u></em>.</h2>

You might be interested in

A box with mass (m) it's sliding along on a friction-free surface at 9.87 m/s at a height of 1.81 meters. It travels down the hi

Rus_ich [418]

A) The answer is 11.53 m/s

The final kinetic energy (KEf) is the sum of initial kinetic energy (KEi) and initial potential energy (PEi).
KEf = KEi + PEi

Kinetic energy depends on mass (m) and velocity (v)
KEf = 1/2 m * vf²
KEi = 1/2 m * vi²

Potential energy depends on mass (m), acceleration (a), and height (h):
PEi = m * a * h

So:
KEf = KEi + <span>PEi
</span>1/2 m * vf² = 1/2 m * vi² + m * a * h
..
Divide all sides by m:
1/2 vf² = 1/2 vi² + a * h

We know:
vi = 9.87 m/s
a = 9.8 m/s²
h = 1.81 m

1/2 vf² = 1/2 * 9.87² + 9.8 * 1.81
1/2 vf² = 48.71 + 17.74
1/2 vf² = 66.45
vf² = 66.45 * 2
vf² = 132.9
vf = √132.9
vf = 11.53 m/s

b) The answer is 6.78 m

The kinetic energy at the bottom (KE) is equal to the potential energy at the highest point (PE)
KE = PE

Kinetic energy depends on mass (m) and velocity (v)
KE = 1/2 m * v²

Potential energy depends on mass (m), acceleration (a), and height (h):
PE = m * a * h

KE = PE
1/2 m * v² = m * a * h

Divide both sides by m:
1/2 * v² = a * h
v = 11.53 m/s
a = 9.8 m/s²
h = ?

1/2 * 11.53² = 9.8 * h
1/2 * 132.94 = 9.8 * h
66.47 = 9.8 * h
h = 66.47 / 9.8
h = 6.78 m

3 0

3 years ago

The nonreflective coating on a camera lens with an index of refraction of 1.29 is designed to minimize the reflection of 634-nm

eimsori [14]

Answer:

minimum thickness of the coating = 122.868 nm

Explanation:

Given data

lens index of refraction = 1.29

wavelength = 634 nm

glass index of refraction = 1.53

to find out

minimum thickness of the coating

solution

we have given non reflective coating

we know that minimum thickness of the coating formula

minimum thickness of the coating = Wavelength / 4n

here n is coating index of refraction

so put here both value to get thickness

minimum thickness of the coating = Wavelength / 4n

minimum thickness of the coating = 634 / 4 ( 1.29 )

so minimum thickness of the coating = 122.868 nm

5 0

3 years ago

If a 1 kg book has 46 Joules of gravitational potential energy how high is the shelf it is on?

Mashcka [7]

Answer:

4.7m

Explanation:

Given parameters:

Mass of the book = 1kg

Gravitational potential energy = 46J

Unknown:

Height of the shelf = ?

Solution:

The potential energy is due to the position of a body above the ground.

Gravitational potential energy = mgh

m is the mass,

g is the acceleration due gravity = 9.8m/s²

h is the height which is unknown

46 = 1 x 9.8 x h

h = 4.7m

4 0

3 years ago

Question 1 of 10

Novay_Z [31]

Answer:

C. 5.6 × 10^11 N/C

Explanation:

The electric field $E$ at a distance $R$ from a charge $Q$ is given by

$E = k\dfrac{Q}{R^2}$

where $k = 9*10^9Nm/C$ is the coulomb's constant.

Now, in our case

$R = 0.0075m$

$Q = 0.0035C$ ;

therefore,

$E = (9*10^9)\dfrac{0.0035C}{(0.0075m)^2}$

$\boxed{E = 5.6*10^{11}N/C.}$

which is choice C from the options given<em> (at least it resembles it).</em>

6 0

3 years ago

A uniform marble rolls down a symmetrical bowl, start- ing from rest at the top of the left side. The top of each side is a dist

Paha777 [63]

Answer:

Part a)

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

Explanation:

As we know by energy conservation the total energy at the bottom of the bowl is given as

$\frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy

now on the right side of the bowl there is no friction

so its rotational kinetic energy will not change and remains the same

so it will have

$\frac{1}{2}mv^2 = mgh'$

now we know that

$I = \frac{2}{5}mr^2$

$\omega = \frac{v}{r}$

so we have

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh$

$\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh$

$\frac{7}{10}mv^2 = mgh$

$\frac{1}{2}mv^2 = \frac{10}{14}mgh$

so the height on the smooth side is given as

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

6 0

3 years ago

Read 2 more answers