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allsm [11]
3 years ago
5

Coal is a major fossil fuel which is mined in many countries around the world. Which of these is most likely to happen if coal i

s stacked near mines?
a. groundwater will get heated up due to coal
b. atmospheric temperature will increase due to coal
c. rain water will evaporate rapidly when it falls on coal
d. rain water will carry contaminants in coal into surface water
Physics
1 answer:
Troyanec [42]3 years ago
6 0
<span>If coal is stacked near mines, the most likley thing that will happen is that rain watter will carry contaminants in coal into durface water. When rain falls on coal, it erodes it just like any other rock. So amounts of coal can be washed out of its holding area and into surface *and* groundwater. Correct answer:D</span>
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An airplane accelerates from a speed of 88m/s to a speed of 132 m/s during a 15 second time interval. How far did the airplane t
Gelneren [198K]

Answer:

1650\:\mathrm{m}

Explanation:

We can use the following kinematics equations to solve this problem:

v_f=v_i+at,\\{v_f}^2={v_i}^2+2a\Delta x.

Using the first one to solve for acceleration:

132=88+a(15),\\15a=44,\\a=\frac{44}{15}=2.9\bar{3}\:\mathrm{m/s^2}.

Now we can use the second equation to solve for the distance travelled by the airplane:

132^2=88^2+2\cdot2.9\bar{3}\cdot \Delta x,\\\Delta x= \frac{9680}{2\cdot2.9\bar{3}},\\\Delta x =\fbox{$ 1650\:\mathrm{m}$}(three significant figures).

6 0
3 years ago
At the start of a basketball game, a referee tosses a basketball straight into the air by giving it some initial speed. After be
max2010maxim [7]

Answer:

0.32 m.

Explanation:

To solve this problem, we must recognise that:

1. At the maximum height, the velocity of the ball is zero.

2. When the velocity of the ball is 2.5 m/s above the ground, it is assumed that the potential energy and kinetic energy of the ball are the same.

With the above information in mind, we shall determine the height of the ball when it has a speed of 2.5 m/s. This can be obtained as follow:

Mass (m) = constant

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) = 2.5 m/s

Height (h) =?

PE = KE

Recall:

PE = mgh

KE = ½mv²

Thus,

PE = KE

mgh = ½mv²

Cancel m from both side

gh = ½v²

9.8 × h = ½ × 2.5²

9.8 × h = ½ × 6.25

9.8 × h = 3.125

Divide both side by 9.8

h = 3.125 / 9.8

h = 0.32 m

Thus, the height of the ball when it has a speed of 2.5 m/s is 0.32 m.

3 0
3 years ago
what is the energy (in eV units) carried by one photon violet light that has a wavelength of 4.5e-7?
DaniilM [7]
The energy of a photon is given by
E=hf
where h is the Planck constant and f is the photon frequency.

We can find the photon's frequency by using the following relationship:
f= \frac{c}{\lambda}
where c is the speed of light and \lambda is the photon's wavelength. By plugging numbers into the equation, we find
f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{4.5 \cdot 10^{-7} m}=6.67 \cdot 10^{14}Hz

And so now we can find the photon energy
E=hf=(6.6 \cdot 10^{-34} Js)(6.67 \cdot 10^{14}Hz )=4.4 \cdot 10^{-19} J

We know that 1 Joule corresponds to
1 J = 1.6 \cdot 10^{-19} eV
So we can convert the photon's energy into electronvolts:
E= \frac{4.4 \cdot 10^{-19} J }{1.6 \cdot 10^{-19} J/eV}=2.75 eV
4 0
4 years ago
Two astronauts on opposite ends of a spaceship are comparing lunches. One has an apple, the other has an orange. They decide to
german

Answer: v= 1.23 m/s θ = 75.3º

Explanation:

First of all, we define the direction in which both fruits are tossed as the x axis, so all initial momenta have horizontal components only.

Now, if no external forces act during collision (due to the infinitesimal time during which collision takes place) momentum must be conserved.

As momentum is a vector, both components must be conserved, so we can write the following equations:

p₁ₓ = p₂ₓ ⇒ -m₁ . vi₁ +m₂. vi₂ = m₁ . vf₁ . cos θ  (1)

p₁y = p₂y ⇒ 0 =m₂ . vf₂ - m₁. vf₁. sin θ (2)

Replacing by the values of m1, m₂, vi₁, vi₂, and vf₂, we can calculate the value of the angle θ, that the apple forms with the horizontal, as follows:

(1) -0.13 Kg. 1.05 m/s + 0.15 Kg. 1.18 m/s = 0.13. vf . cos θ  

(2) 0.15 Kg. 1.03 m/s = 0.13 vf. sin θ

sin θ / cos θ = 3.82 ⇒ tg θ = 3.82 ⇒ θ = arc tg (3.82) = 75.3º

Replacing this value of θ in (2), we get:

0.15 kg. 1.03 m/s = 0.13 vf . sin 75.3º = 0.13 . vf . 0.967

Solving for vf, we get:

vf = 0.15 kg. 1.03 m/s / 0.13. 0.967 = 1.23 m/s

5 0
3 years ago
Suppose a piece of food is on the edge of a rotating microwave oven plate. Does it experience nonzero tangential acceleration, c
Rus_ich [418]

Answer:

Explanation:

(a) When the plate starts to spin:

Its angular velocity increases, so the angular acceleration is non zero. As the direction of velocity keeps on changing every instant so the linear acceleration is also non zero.

(b) When the plate rotates at constant angular velocity:

Its angular velocity is constant so the angular acceleration is zero. As the direction of velocity keeps on changing every instant so the linear acceleration is also non zero.

(c) When the plate sows to halt:

Its angular velocity decreases, so the angular acceleration is non zero( but negative). As the direction of velocity keeps on changing every instant so the linear acceleration is also non zero.

4 0
3 years ago
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