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3 years ago

Most of the Earth's surface is covered by what?

Physics

2 answers:

Fudgin [204]3 years ago

8 0

most of the earth is covered with oceans

Lera25 [3.4K]3 years ago

3 0

The earth surface is covered by water and 30% is taken on the seven continental landmasses . So i would say the answer is Oceans

Hope this helped.

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An object is 70 um long and 47.66um wide. how long and wide is the object in km?

Ganezh [65]

Answer:

length = 7*10^(-8)km

width = 4.666*10^(-8) km

Explanation:

We know that:

1 μm = 1*10^(-6) m

and

1km = 1*10^3 m

1m = 1*10^(-3) km

if we replace the meter in the first equation, we get:

1 μm = 1*10^(-6)*1*10^(-3) km

1 μm = 1*10^(-6 - 3)km

1 μm = 1*10^(-9)km

Now with this relationship we can transform our measures:

Length: 70 μm is 70 times 1*10^(-9)km, or:

L = 70*1*10^(-9)km = 7*10^(-8)km

And for width, we have 47.66um, this is 46.66 times 1*10^(-9)km, or:

W = 46.66*1*10^(-9)km = 4.666*10^(-8) km

7 0

3 years ago

Refer to Concept Simulation 4.4 for background relating to this problem. The drawing shows a large cube (mass = 28.9 kg) being a

Usimov [2.4K]

Answer:

smallest magnitud is P=33.3 N

Explanation:

We are analyze the situation as an external force is applied and there is a friction force. We have a problem with Newton's second law.

F = ma

As the two blocks go together they must have the same acceleration, so we can calculate this for the entire system

P = (m1 + m2) a

a = P / (m1 + m2)

In this case there is no friction force because the small block does not touch the ground.

In order to calculate the friction force, we must analyze each system component separately.

The large block on the X axis has an applied P force and as it moves feels a force from the small block. In the Y axis has the weight (W1) and the reaction to normal (N1)

For the small block on the X axis, the force it feels is the thrust of the large block, note that this is an action and reaction force between the two blocks, it is the same definition we have of the normal one, so we can call this force (N)

Y axis it has the weight (W2) down, the force of friction (fr) that opposes the movement, so it is directed upwards. we write these equations

N = m2 a

fr -W2 = 0

fr = W2

The definition of friction force is

fr = μ N

Let's replace and calculate

μ (m2 a) = m2 g

μ (P / (m1 + m2)) = g

P = g /μ (m1 + m2)

Let's calculate the value of this force

P = 9.8 / 0.710 (28.9 +4.4)

P = 13.80 (33.3)

P = 33.3 N

This is the minimum friction force that prevents the block from sliding down

6 0

3 years ago

Before 1960, people believed that the maximum attainable coefficient of static friction for an automobile tire on a roadway was

True [87]

This problem is looking for the minimum value of μs that is necessary to achieve the record time. To solve this problem:

Assuming the front wheels are off the ground for the entire ¼ mile = 402.3 m, the acceleration a = µs·9.8 m/s².

For a constant acceleration, distance = 402.3

m = 1/2at^2 = 804.6 m / (4.43 s)^2 = a = µs·9.8 m/s^2

µs = 804.6 m / (4.43s)^2 / 9.8 m/s^2 = 4.18

5 0

3 years ago

A rigid adiabatic container is divided into two parts containing n1 and n2 mole of ideal gases respectively, by a movable and th

kicyunya [14]

Answer:

Explanation:

Given

Pressure, Temperature, Volume of gases is

$P_1, V_1, T_1 & P_2, V_2, T_2$

Let P & T be the final Pressure and Temperature

as it is rigid adiabatic container therefore Q=0 as heat loss by one gas is equal to heat gain by another gas

$-Q=W+U_1----1$

$Q=-W+U_2-----2$

where Q=heat loss or gain (- heat loss,+heat gain)

W=work done by gas

$U_1 & U_2$ change in internal Energy of gas

Thus from 1 & 2 we can say that

$U_1+U_2=0$

$n_1c_v(T-T_1)+n_2c_v(T-T_2)=0$

$T(n_1+n_2)=n_1T_1+n_2T_2$

$T=\frac{n_1+T_1+n_2T_2}{n_1+n_2}$

where $n_1=\frac{P_1V_1}{RT_1}$

$n_2=\frac{P_2V_2}{RT_2}$

$T=\frac{\frac{P_1V_1}{RT_1}\times T_1+\frac{P_2V_2}{RT_2}\times T_2}{\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}}$

$T=\frac{P_1V_1+P_2V_2}{\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}}$

and $P=\frac{P_1V_1+P_2V_2}{V_1+V_2}$

6 0

3 years ago

The friction force between a bag and the floor is 4 N. what is the net force acting on the bag? What is the acceleration of the

tester [92]

Well, if the bag is moving across the floor at a constant velocity. Then I would assume that the Fnet force is equal to 0, and thus the value of acceleration would also be equal to 0.

6 0

3 years ago