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Kay [80]
3 years ago
9

The layer of earth that has the lightest elements is the

Physics
1 answer:
Sonja [21]3 years ago
8 0
If I remember correctly (from my studies long time ago) the layers are from the outer to the center:
SiAl : Silicon-Aluminum
SiMa : Silicon-Magnesium (although should be Mg)
NiFe : Nickel-Iron

The SiMa layer should have the lightest elements (Magnesium is lighter than Aluminum)
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A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 joules (with the poten
lions [1.4K]

Answer:

 h = 20 m

Explanation:

given.

height, h = 10 m

Potential energy at 10 m = 50 J

Kinetic energy at 10 m = 50 J

maximum height the ball will reach, H = ?

Total energy of the system

T E = 50 J + 50 J

T E = 100 J

now,

A h = 10 m

P E = m g h

50 = m g x 10

mg = 5 ..............(1)

at the top most Point the only Potential energy will be acting on the body.

now, TE = Potential energy

 100 = m g h

5 h = 100

 h = 20 m

hence, the maximum height reached by the ball is equal to 20 m.

5 0
3 years ago
How this induced force can be increased? Write three ways.
Ludmilka [50]
Strength of induced current increased when strength of magnetic field increases. It will also increase when the number of turns are increased or if the speed of conductor increases
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3 years ago
What is the mass of a basketball that has Connecticut energy of 102 J and it's travelling at 5 m/s
Stella [2.4K]

Answer:

8.16Kg

Explanation:

K.E= 1/2 mv²

102=1/2 m 5²

m=8.16 Kg

8 0
3 years ago
At what position or positions on the x-axis is the electric field zero?
ElenaW [278]

Answer:

The electric field will be zero at x = ± ∞.

Explanation:

Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.

We know that,

The electric field is

E=\dfrac{kq}{r^2}

The electric field vector due to charge one

\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})

The electric field vector due to charge second

\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})

We need to calculate the electric field

Using formula of net electric field

\vec{E}=\vec{E_{1}}+\vec{E_{2}}

\vec{E_{1}}+\vec{E_{2}}=0

Put the value into the formula

\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0

\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})

(\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}

\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}

Put the value into the formula

\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}

2.0+x=x

If x = ∞, then the equation is be satisfied.

Hence, The electric field will be zero at x = ± ∞.

4 0
3 years ago
In an electrolytic cell, to which electrode will a positive ion migrate and undergo reduction? the anode, which is negatively ch
ahrayia [7]
The reaction of reduction always undergoes with the cathode, the positive ion will migrate towards the cathode with the negative charge whilst the anode always has oxidation reaction. These two types of reaction does not change.
4 0
3 years ago
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