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Ne4ueva [31]
3 years ago
6

The long term weather patterns that are typical in a location are the locations ___

Physics
2 answers:
11Alexandr11 [23.1K]3 years ago
8 0
Climate is correct....
Dmitry [639]3 years ago
5 0

<em>The correct answer would be Climate.</em>

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An object weighing 49 N is pushed across a floor by a force of 12 N. What is the acceleration of the object?
NISA [10]

Answer:

Explanation:

Given parameters:

Weight of object  = 49N

Force applied = 12N

Unknown:

Acceleration of object  = ?

Solution:

The acceleration of the object is found by dividing the force by the weight;

 Acceleration  = \frac{12}{49}   = 0.25m/s²

3 0
3 years ago
How many species go extinct every day??
Law Incorporation [45]
Approximately 150-200 species.
5 0
3 years ago
Explanation for question i &amp; ii. Thank you.
suter [353]
Do u know Chinese ?if yes I can explain to u easily
6 0
3 years ago
The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371
arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

q=m\times c\times (T_2-T_1)

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = 4.18J/g.K

T_1 = initial temperature of water = 293.0 K

T_2 = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

q=250g\times 4.18J/g.K\times (371.2-293.0)K

q=81719J

Now we have to calculate the enthalpy of combustion of octane.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

\Delta H=\frac{-81719J}{0.015mole}

\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0
4 years ago
A child bounces a 48 g superball on the sidewalk. the velocity change of the superball is from 26 m/s downward to 17 m/s upward.
Nataly_w [17]
By definition we have the momentum is:
 P = m * v
 Where,
 m = mass
 v = speed
 Before the impact:
 P1 = (0.048) * (26) = 1.248 kg * m / s
 After the impact:
 P2 = (0.048) * (- 17) = -0.816 Kg * m / s.
 Then we have that deltaP is:
 deltaP = P2-P1
 deltaP = (- 0.816) - (1,248)
 deltaP = -2,064 kg * m / s.
 Then, by definition:
 deltaP = F * delta t
 Clearing F:
 F = (deltaP) / (delta t)
 Substituting the values
 F = (- 2.064) / (1/800) = - 1651.2N
 answer:
 the magnitude of the average force exerted on the superball by the sidewalk is 1651.2N
3 0
4 years ago
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