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Natali5045456 [20]
4 years ago
12

Why is there convection in the outer core and what is the result of this?

Physics
1 answer:
mars1129 [50]4 years ago
6 0

Answer:

re believed to influence the Earth's magnetic field. ... As heat is transferred outward toward the mantle, the net trend is for the inner boundary of the liquid region to freeze, causing the solid inner core to grow

Explanation:

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A 20 kg shopping cart moving at a velocity of 0.5 m/s collides with a store wall and
MAXImum [283]

Answer:

<h2>10 kg.m/s</h2>

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 20 × 0.5

We have the final answer as

<h3>10 kg.m/s</h3>

Hope this helps you

7 0
3 years ago
What is the highest temperature ever recorded on earth
Llana [10]
Therefore the world's record high temperature of 134.0°F (56.7°C) is held by Furnace Creek Ranch in Death Valley, California. That global high temperature was attained on July 10, 1913.
4 0
3 years ago
A constant horizontal force F is applied to a 5 kg box on a frictionless horizontal surface. The box starts from rest and moves
sergejj [24]
65(58)+85)6463(9786+)9_43=
7 0
3 years ago
Suppose 8.50 ✕ 10^5 J of energy are transferred to 1.63 kg of ice at 0°C. The latent heat of fusion and specific heat of water a
PolarNik [594]

Answer:

(a) 5.43 x 10⁵ J

(b) 3.07 x 10⁵ J

(c) 45 °C

Explanation:

(a)

L_{f} = Latent heat of fusion of ice to water = 3.33 x 10⁵ J/kg

m = mass of ice = 1.63 kg

Q_{f} = Energy required to melt the ice

Energy required to melt the ice is given as

Q_{f} = m L_{f}

Q_{f} = (1.63) (3.33 x 10⁵)

Q_{f} = 5.43 x 10⁵ J

(b)

E = Total energy transferred = 8.50 x 10⁵ J

Q  = Amount of energy remaining to raise the temperature

Using conservation of energy

E = Q_{f} + Q

8.50 x 10⁵ = 5.43 x 10⁵ + Q

Q = 3.07 x 10⁵ J

(c)

T₀ = initial temperature = 0°C

T = Final temperature

m = mass of water = 1.63 kg

c = specific heat of water = 4186 J/(kg °C)

Q = Amount of energy to raise the temperature of water = 3.07 x 10⁵ J

Using the equation

Q = m c (T - T₀)

3.07 x 10⁵ = (1.63) (4186) (T - 0)

T = 45 °C

5 0
3 years ago
If a steel cylindrical specimen is stressed nominally to 53 MPa, what stress level exists at the tip of an elliptical surface fl
ElenaW [278]

Answer:

227.9MPa

Explanation:

Length of the flaws is given by

2b = 5.8microns

b = 2.9 × 10⁻⁶m

The relation between the radius of curvature and length and width of the elliptical flaw

r = \frac{a^2}{b}

a = \sqrt{rb}

Equation for stress at the tip of an elliptical surface flaw

\sigma _t = \sigma(1 + 2\frac{b}{a}  )\\\\\sigma _t = \sigma(1 + 2\frac{b}{\sqrt{rb} })\\\\\sigma _t = \sigma(1 + 2\frac{\sqrt{b} }{\sqrt{r} })\\\\\sigma _t = \sigma(1 + 2\sqrt{\frac{b}{r} })

\sigma _t = 53 \times 10^6 (1 + 2\sqrt{\frac{2.9\times10^-^6 }{1065 \times 10^-^9} } \\\\\sigma _t = 227.9 \times 10^6\\\\\sigma _t  = 227.9MPa

4 0
3 years ago
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