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Rzqust [24]
3 years ago
12

What are the conditions necessary for a terrestrial planet to have a strong magnetic field?.

Physics
1 answer:
Stells [14]3 years ago
6 0
Both a molten metallic core and reasonably fast rotation.
You might be interested in
An alligator swims to the left with a constant velocity of 5 \,\dfrac{\text{m}}{\text s}5
dexar [7]

Answer:

The alligator will take t = 10 s to reach the final speed of 35 m/s

Explanation:

As we know that the initial speed of the alligator is 5 m/s

then it accelerate by given acceleration to reach the final speed of 35 m/s

so we will have

v_i = 5 m/s

v_f = 35 m/s

a = 3m/s^2

now we have

v_f = v_i + at

35 = 5 + 3 t

t = 10 s

5 0
3 years ago
A 50-cm3 block of wood is floating on water, and a 50-cm3 chunk of iron is totally submerged in the water. Which one has the gre
ludmilkaskok [199]

Answer:

Iron

Explanation:

The buoyant force equals to the weight of water being displaced by the object.

Since, the volume of both iron and wood is equal and the wood is not completely submerged, but the iron block is completely submerged i.e more volume of the water is being displaced by the iron block.

Hence, the buoyant force is more on the iron.

4 0
3 years ago
What is the pressure exerted by 0.801 mol of co2 in a 12 l container?
Svetlanka [38]
We assume that the gas is an ideal gas so we can use the relation PV=nRT. Assuming that the temperature of the system is at ambient temperature, T = 298 K. We can calculate as follows:

PV = nRT
P = nRT / V
P = (0.801 mol ) (0.08205 L-atm / mol-K) (298.15 K) / 12 L
P = 1.633 atm
8 0
3 years ago
When you come out of a river you feel cold. why?​
GenaCL600 [577]

Answer:

the wind

Explanation:

5 0
3 years ago
A small object is placed at the top of an incline that is essentially frictionless. The object slides down the incline onto a ro
Georgia [21]

Answer

given,

time taken to stop by the object = 5 s

distance travel before stopping = 60 m

final velocity = 0

using equation of motion

v = u + at

0 = u - 5 a

a_x = \dfrac{u}{5}

s = u t + \dfrac{1}{2}at^2

s = 5u + \dfrac{1}{2}\times \dfrac{4}{5}\times u^2

60 = 2.5 u

u = 24 m/s

a_x = \dfrac{24}{5}

a_x = 4.8 m/s²

b) using energy conservation

\dfrac{1}{2}mv_i^2 + \dfrac{1}{2}mv_f^2 = mg(\Delta h)

\dfrac{1}{2}v_i^2= g(\Delta h)

(\Delta h) = \dfrac{0.5 \times 24^2}{9.8}

Δh = 29.38 m

4 0
3 years ago
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