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3 years ago

What is one of the most precious natural resources we have is A) crops B) soil C) trees

Physics

1 answer:

brilliants [131]3 years ago

8 0

Soil i believe
......................

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How much work is required to pull a sled 5 meters if you use 60N of force?

almond37 [142]

Answer:

300Nm

Explanation:

work = force x distance = 60x5=300

5 0

3 years ago

Read 2 more answers

assuming that the seafloor is spreading at a rate of 1.3cm/yr, calculate how long it took africa's west coast to move 2400 kn aw

Verdich [7]

Answer: 184,615,384.6 years

Explanation:

This problem can be solve by the following equation:

$V=\frac{d}{t}$ (1)

Where:

$V=1.3 \frac{cm}{year} \frac{1 km}{100000 cm}=0.000013 \frac{km}{year}$ is the spreading rate of the seafloor, its velocity

$d=2400 km$ is the distance the Africa's west coast moved at this rate

$t$ is the time it took to the coast to move the descibed distance

Isolating $t$ from (1):

$t=\frac{d}{V}$ (2)

$t=\frac{2400 km}{0.000013 \frac{km}{year}}$ (3)

Finally:

$t=184,615,384.6 years$ This is the time it took to the Africa's west coastto move away from the Mid atlantic ridge.

3 0

3 years ago

A 0.415-kg mass suspended from a spring undergoes simpleharmonic oscillations with a period of 1.4 s. How much mass, inkilograms

koban [17]

Answer:

m=0.893kg

Explanation:

time period of oscillations is given by= 2π√(m/k)

m: mass of the object

k: spring constant

when T=1.5 and m=0.415

1.5= 2π√(0.415/k)

k= 7.27 N/m

when T= 2.2s

2.2= 2π√(m/7.27)

m=0.893kg

6 0

3 years ago

The drawings show three charges that have the same magnitude but may have different signs. In all cases the distance d between t

Radda [10]

Answer:

a) $F = 6816.5680 N$

b) $F' = 0 N$

c) $F''=28195.5N$

Explanation:

Magnitude of charges $M=1.6\muC$

Distance $d=2.6$

Generally the equation for Net Force is mathematically given by

For First Drawing

$F =\frac{ k q^2}{d^2}+\frac{k q^2}{d^2}$

$F =\frac{2* 9*10^9* (1.6*10^-6)^2}{ (2.6*10^-3^2}$

$F = 6816.5680 N$

For second Drawing

$F' =\frac{ k q^2}{d^2 }-\frac{ k q^2}{ d^2}$

$F' = 0 N$

For Third Drawing

$F'' =\frac{ k q^2}{d^2} * \sqrt (2)$

$F'' = 9*10^9* (6.4*10^-6)^2 * \frac{\sqrt(2)}{(4.3*10^-3)^2}$

$F''=28195.5N$

4 0

3 years ago

An atom that changes so that it has an<br> electrical charge is a(n)

MrMuchimi

Answer:

An ion.

Explanation:

Atoms can gain or lose electrons and become ions, which are atoms that have a positive or negative charge, because they have unequal numbers of protons and electrons.

6 0

3 years ago