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3 years ago

How are the land area and the oceans affected during an ice age

2 answers:

34kurt3 years ago

8 0

Well, if the ice falls on the land and water which can damage a lot of things. Also that when it melts, it makes the ocean a bit bigger and the land smaller. And if a huge chunk of ice breaks off, it can cause the ocean waters to rise and form huge waves.

amm18123 years ago

3 0

The land was frozen everywhere and a part of the ocean froze over to create a bridge between two countries. Some people may disagree with me over the ice bridge, though.

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This is about the magnet fields. thanks in advance.

Wewaii [24]

The answer is A. The outer lines change as it moves

8 0

2 years ago

PLZZ HELP WILL MARK BRAINLY The function of the respiratory system is to ___________.

pickupchik [31]

Breathe and now I’m just filling in more letters so it’ll go thru

3 0

4 years ago

A certain corner of a room is selected as the origin of a rectangular coordinate system. If a fly is crawling on an adjacent wal

Helga [31]

Answer:

2.59 m

Explanation:

Coordinates of origin = (0, 0)

Coordinates of Point p where the fly reach = (2.3 m, 1.2 m)

Use the distance formula of coordinates to find the distance between the origin and the point P.

$d=\sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$

$d=\sqrt{\left ( 2.3- 0 \right )^{2}+\left ( 1.2-0 \right )^{2}}$

d = 2.59 m

Thus, the distance between the origin and the point P is 2.59 m.

5 0

3 years ago

A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretch

ANTONII [103]

Answer:

A. A = 0.913 m

B. amax = 132.24m/s^2

C. Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

$v_{max}=A\omega$ (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

$\omega=\sqrt{\frac{k}{m}}$ (2)

k: spring constant = 355 N/m

m: mass of the object = 2.54 kg

$\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}$

Next, you solve the equation (1) for A and replace the values of vmax and w:

$A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m$

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

$a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}$

The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

$F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N$

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

4 0

3 years ago

Suppose that in a lightning flash the potential difference between a cloud and the ground is 0.96×109 V and the quantity of char

Dvinal [7]

(a) $2.98\cdot 10^{10} J$

The change in energy of the transferred charge is given by:

$\Delta U = q \Delta V$

where

q is the charge transferred

$\Delta V$ is the potential difference between the ground and the clouds

Here we have

$q=31 C$

$\Delta V = 0.96\cdot 10^9 V$

So the change in energy is

$\Delta U = (31 C)(0.96\cdot 10^9 V)=2.98\cdot 10^{10} J$

(b) 7921 m/s

If the energy released is used to accelerate the car from rest, than its final kinetic energy would be

$K=\frac{1}{2}mv^2$

where

m = 950 kg is the mass of the car

v is the final speed of the car

Here the energy given to the car is

$K=2.98\cdot 10^{10} J$

Therefore by re-arranging the equation, we find the final speed of the car:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2.98\cdot 10^{10})}{950}}=7921 m/s$

5 0

4 years ago