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koban [17]
3 years ago
15

Acid rain is an example of which type of chemical weathering?

Physics
2 answers:
tino4ka555 [31]3 years ago
6 0
The answer is Carbonic acid
Virty [35]3 years ago
4 0
An example of a chemical weathering is Carbonic acid
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7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh
pentagon [3]

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

F= \mu R

where \mu is the coefficient of friction = 0.02

R = Normal reaction of the load = mgcos\theta = 25 \times 9.81 \times cos 10 = 241.52N

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

F=0.02 \times 241.52N

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

4 0
3 years ago
If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle
ad-work [718]

This question is incomplete, the complete question is;

The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10⁻⁹ N⋅m on an electric dipole oriented at 30° from the direction of the external field. The dipole moment of the dipole is 7.5 × 10⁻¹² C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Answer:

- the magnitude of the external electric field is 5333.3 N/C

- the magnitude of the charge on each particle is 3.0 × 10⁻¹² C  ≈ 3 nC

Explanation:

Given that;

Torque = 20.0 × 10⁻⁹ N⋅m

dipole moment = 7.5 × 10⁻¹²

∅ = 30°

The moment T of restoring couple is;

T = PEsin∅

E = T/Psin∅

we substitute

E = 20.0 × 10⁻⁹ N⋅m / (7.5 × 10⁻¹²) sin(30°)

E = 20.0 × 10⁻⁹ / 3.75 × 10⁻¹²

E =  5333.3 N/C

Therefore, the magnitude of the external electric field is 5333.3 N/C

The dipole moment is given by the expression;

p = ql

q = p / l

given that l = 2.5 mm = 0.0025 m

we substitute

q = 7.5 × 10⁻¹² / 0.0025

q = 3.0 × 10⁻¹² C ≈ 3 nC

Therefore, the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

7 0
2 years ago
Simple Pendulum: A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light
SIZIF [17.4K]

Answer:

The correct answer is "4.443 sec".

Explanation:

Given:

Mass of child,

= 34 kg

Mass of swing,

= 18 kg

Length,

= 4.9 m

The time period of pendulum will be:

T = 2 \pi \sqrt{4g}

  = 2 \pi \sqrt{\frac{4.9}{9.8} }

  = 4.443 \ sec  

5 0
3 years ago
Read 2 more answers
Remy wonders if the height of the mountain has anything to do with the eventual size of the tsunami wave. How should Remy test t
Anna71 [15]

Answer:

<em>The answer is B</em>

Explanation:

<em>I got this from study island</em>

5 0
3 years ago
How much heat is needed to boil 120 kg of water ?
nekit [7.7K]
Q = mc<span>∆t, where:
q = energy flow
m = mass, 120 000 g
c = specific heat capacity, 4.81 J/gC
</span><span>∆t = change in temperature, ~75 (100 - 25, which is room temperature)

Substituting in the values, we get:
q = 120000 x 4.81 x 75 = 43290000 Joules = 43.29 MJ

Hope I helped!! xx


</span>
7 0
3 years ago
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