Ask question

Ask question

All categories

3 years ago

5

A ___ is when too much water moves into an area

2 answers:

Marina86 [1]3 years ago

6 0

A. Flood.
Hope it helps!

sergey [27]3 years ago

3 0

The answer is A. Flood. I took the test and I got it right.

You might be interested in

A substance contains only one type of atom. The substance is a/an ____________. A. solution B. element C. mixture D. compound

Alex17521 [72]

The answer to this question is:B. Element

3 0

2 years ago

Read 2 more answers

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),

yuradex [85]

Answer:

$v=3.564\ m.s^{-1}$

$\Delta v =2.16\ m.s^{-1}$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

<u>Using the equation of motion:</u>

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3$

$v_J=5.94\ m.s^{-1}$

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30\times 5.94+0=(30+20)v$

$v=3.564\ m.s^{-1}$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

<u>Now we find the force along the slide due to the body weight:</u>

$F=m_J.g.sin\theta$

$F=30\times 9.8\times \frac{1.8}{3}$

$F=176.4\ N$

<em><u>Hence the net force along the slide:</u></em>

$F_R=71.4\ N$

<em>Now the acceleration of John:</em>

$a_j=\frac{F_R}{m_J}$

$a_j=\frac{71.4}{30}$

$a_j=2.38\ m.s^{-2}$

<u>Now the new velocity:</u>

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2\times 2.38\times 3$

$v_J_n=3.78\ m.s^{-1}$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^{-1}$

8 0

2 years ago

If we increase the force applied to an object and all other factors remain the same that amount of work will

worty [1.4K]

Hello there.

<span>If we increase the force applied to an object and all other factors remain the same that amount of work will

</span><span>C. Increase
</span>

5 0

2 years ago

Read 2 more answers

A taxi is travelling at 15m/s. Its driver accelerates with acceleration 3m/s^2 for 4 s. What would be the new velocity?..... pls

Aleksandr-060686 [28]

Answer:

27 m/s

Explanation:

Given:

v₀ = 15 m/s

a = 3 m/s²

t = 4 s

Find: v

v = at + v₀

v = (3 m/s²) (4 s) + (15 m/s)

v = 27 m/s

5 0

2 years ago

Read 2 more answers

In Millikan's oil drop experiment an oil drop is at rest between two large plates separated by 1.3 cm when the potential differe

sweet-ann [11.9K]

Answer:

Mass of the oil drop, $m=3.01\times 10^{-15}\ kg$

Explanation:

Potential difference between the plates, V = 400 V

Separation between plates, d = 1.3 cm = 0.013 m

If the charge carried by the oil drop is that of six electrons, we need to find the mass of the oil drop. It can be calculated by equation electric force and the gravitational force as :

$qE=mg$

$m=\dfrac{qE}{g}$

$q=6e$ , e is the charge on electron

E is the electric field, $E=\dfrac{V}{d}=\dfrac{400}{0.013}=30769.23\ V/m$

$m=\dfrac{6\times 1.6\times 10^{-19}\times 30769.23}{9.8}$

$m=3.01\times 10^{-15}\ kg$

So, the mass of the oil drop is $3.01\times 10^{-15}\ kg$ . Hence, this is the required solution.

5 0

3 years ago