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3 years ago

Why is the current atomic model called the "electron could model"

1 answer:

8 0

The diagram is showing a 3d model of an atom, with all of the electrons demonstrated in a rounded shape, which resembles a cloud, thus being called an electron cloud.

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The electric force between two charged balloons is 0.12 newtons. If the distance between the two balloons is halved, what will b

marshall27 [118]

Answer:

The answer is D, I just took the test

Explanation:

5 0

3 years ago

The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the

kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is this case is 0 as the road is level

Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft$

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft$

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft$

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0

3 years ago

Sophie throws a tennis ball down from a height of 1.5 m at an angle of 450 with respect to vertical. She drops another tennis ba

Katena32 [7]

Given that,

Height =1.5 m

Angle = 45°

We need to find the greater speed of the ball

Using conservation of energy

$P.E_{i}+K.E_{f}=P.E_{f}+K.E_{f}$

$mgh+\dfrac{1}{2}mv_{i}^2=mgh+\dfrac{1}{2}mv_{f}^2$

Here, initial velocity and final potential energy is zero.

$mgh=\dfrac{1}{2}mv_{f}^2$

Put the value into the formula

$9.8\times1.5=\dfrac{1}{2}v_{f}^2$

$v_{f}^2=2\times9.8\times1.5$

$v_{f}=\sqrt{2\times9.8\times1.5}$

$v_{f}=5.42\ m/s$

Hence, the greater speed of the ball is 5.42 m/s.

5 0

3 years ago

n the railroad accident, a boxcar weighting 200 kN and traveling at 3 m/s on horizontal track slams into a stationary caboose we

USPshnik [31]

Answer:

ΔK = -6 10⁴ J

Explanation:

This is a crash problem, let's start by defining a system formed by the two trucks, so that the forces during the crash have been internal and the moment is preserved

initial instant. Before the crash

p₀ = m v₁ + M 0

final instant. Right after the crash

p_f = (m + M) v

p₀ = p_f

mv₁ = (m + M) v

v = $\frac{m}{m+M} \ v_1$

we substitute

v = $\frac{20}{20+40}$ 3

v = 1.0 m / s

having the initial and final velocities, let's find the kinetic energy

K₀ = ½ m v₁² + 0

K₀ = ½ 20 10³ 3²

K₀ = 9 10⁴ J

K_f = ½ (m + M) v²

K_f = ½ (20 +40) 10³ 1²

K_f = 3 10⁴ J

the change in energy is

ΔK = K_f - K₀

ΔK = (3 - 9) 10⁴

ΔK = -6 10⁴ J

The negative sign indicates that the energy is ranked in another type of energy

7 0

3 years ago

A pendulum of length =1.0 m is pulled to the side and released on the moon. It's period is measured to be 4.82 seconds. What is

Shkiper50 [21]

Answer:

Gravity on the moon, g = 1.69 m/s²

Explanation:

It is given that,

Length of pendulum, l = 1 m

Time period, T = 4.82 seconds

We have to find the gravity of the moon. The time period of the pendulum is given by :

$T=2\pi\sqrt{\dfrac{l}{g}}$

g = acceleration due to gravity on moon

$g=\dfrac{4\pi^2l}{T^2}$

$g=\dfrac{4\pi^2\times 1\ m}{(4.82\ s)^2}$

g = 1.69 m/s²

Hence, the gravity on the moon is 1.69 m/s².

7 0

3 years ago

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