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4 years ago

Why are clouds considered to be a part of the hydrosphere?

1 answer:

7 0

It is considered parts of the Hydrosphere because water condenses in clouds and falls as precipitation so it is a vital part of the hydrosphere.

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A boy is pulling a cart by a force of 100N. The frictional force experienced by the cart is 20N. The force causing the motion of

Serga [27]

Answer:

3 is right i guss look : 100N-20N=80N

4 0

3 years ago

How many carbon-14 atoms are left after 1 half-life

Marizza181 [45]

Answer:

The time it takes for 14C to radioactively decay is described by its half-life. C has a half-life of 5,730 years. In other words, after 5,730 years, only half of the original amount of 14C remains in a sample of organic material. After an additional 5,730 years–or 11,460 years total–only a quarter of the 14C remains.

Explanation:

Hope this helps

7 0

3 years ago

In traveling a distance of 2.5 km between points A and D, a car is driven at 99 km/h from A to B for t seconds and 48 km/h from

Andreyy89

Answer:

d = 1.954 Km

Explanation:

given,

total distance, D = 2.5 Km

in stretch A to B =

speed = 99 Km/h = 99 x 0.278 = 27.22 m/s time =t

in stretch B to C

time = 3.4 s

In stretch C to D

speed = 48 Km/h = 48 x 0.278 = 13.34 m/s time =t

we know,

distance = speed x time

distance of BC

using equation of motion

v = u + a t

27.22 = 13.34 - a x 3.4

a = 4.08 m/s²

uniform deceleration is equal to 4.08 m/s²

distance traveled in BC

$s = ut + \dfrac{1}{2}at^2$

$s = 13.34\times 3.4 + \dfrac{1}{2}\times 4.08 \times 3.4^2$

s = 68.94 m

$3000 = 99 \times \dfrac{1000\ t}{3600}+ 68.94 + 48\times \dfrac{1000\ t}{3600}$

3000 = 27.5 t + 68.94 + 13.33 t

40.83 t = 2931.06

t = 71.79 s

distance travel in AB

distance = s x t

d = 27.22 x 71.79

d = 1954 m

d = 1.954 Km

distance between A and B is equal to 1.954 Km.

4 0

4 years ago

An electron in the n = 7 level of the hydrogen atom relaxes to a lower energy level, emitting light of 397 nm. what is the value

Dimas [21]

Answer:

$n_f=2$

Explanation:

It is given that,

Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})$