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BARSIC [14]
4 years ago
5

(PLEASE HELP) What area of the Earth's mantle is the densest?

Physics
2 answers:
Nady [450]4 years ago
7 0
The area nearest to the earth's core is the densest.. 
So your answer would be letter choice ( D ) . . . 

Hope it Helped :) 
Alex_Xolod [135]4 years ago
7 0

Answer: yes the answer is D

Explanation:

The asthenosphere is the part of the mantle that flows and moves the plates of the Earth. The mantle is the layer located directly under the sima. It is the largest layer of the Earth, 1800 miles thick. The mantle is composed of very hot, dense rock.

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A toaster draws 8 A of current with a voltage of 120 V. Which is the power used by the toaster?
dexar [7]
U=120 \text{ V}\\
I=8\text{ A}\\
P=U\cdot I\\\\
P=120\cdot8=960 \text{ W}

7 0
3 years ago
A single-phase electrical load draws 600KVA at 0.6 power factor lagging. a) Find the real and reactive power absorbed by the loa
Whitepunk [10]

Answer:

(a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

Explanation:

Given that,

Power factor = 0.6

Power = 600 kVA

(a). We need to calculate the reactive power

Using formula of reactive power

Q=P\tan\phi...(I)

We need to calculate the \phi

Using formula of \phi

\phi=\cos^{-1}(Power\ factor)

Put the value into the formula

\phi=\cos^{-1}(0.6)

\phi=53.13^{\circ}

Put the value of Φ in equation (I)

Q=600\tan(53.13)

Q=799.99\ kVAR

(b). We draw the power triangle

(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95

Using formula of reactive power

Q'=600\tan(0.95)

Q'=9.94\ KVAR

We need to calculate the difference between Q and Q'

Q''=Q-Q'

Put the value into the formula

Q''=799.99-9.94

Q''=790.05\ KVAR

Hence, (a). The reactive power is 799.99 KVAR.

(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.

8 0
3 years ago
A painter stands a horizontal platform which has a mass of 20kg and is 5m long, the platform is suspended by two vertical ropes
Lerok [7]

Answer:

R = 715.4 N

L =  166.6 N

Explanation:

ASSUME the painter is standing right of center

Let L be the left rope tension

Let R be the right rope tension

Sum moments about the left end to zero. Assume CCW moment is positive

R[5] - 20(9.8)[5/2] - 70(9.8)[5/2 + 2] = 0

R = 715.4 N

Sum moments about the right end to zero

20(9.8)[5/2] + 70(9.8)[5/2 - 2] - L[5] = 0

L =  166.6 N

We can verify by summing vertical forces

116.6 + 715.4 - (70 + 20)(9.8) ?=? 0

                                             0 = 0  checks

If the assumption about which side of center the paint stood is incorrect, the only difference would be the values of L and R would be swapped.

5 0
3 years ago
3 hours into minutes
otez555 [7]

Answer:

180

60 minutes in a hour

60x3 is 180minutes

6 0
2 years ago
1. An object (m = 500 g) with an initial speed of 0.2 m/s collides with another object (m = 1.5 kg) which was at rest before the
Anettt [7]

Answer:

Explanation:

1 )

We shall apply conservation of momentum law to solve the problem.

mv = ( M +m) V , m and M are masses of small and large object , v is the velocity of small object before collision and V is the velocity of both the objects together after collision .

.5 x .2 = (1.5 + .5)V

V = .05 m /s

2 ) We shall use formula for velocity of object after elastic collision as follows

v₁ = \frac{(m_1-m_2)}{(m_1+m_2)} u_1+\frac{2m_2u_2}{(m_1+m_2)}

m₁ and m₂ are masses of first and second object u₁ and u₂ are their initial velocity and v₁ and v₂ are their final velocity.

Putting the values

= \frac{(200-1000)}{(1000+200)} \times 1 +\frac{2 \times1000\times0}{(1000+200)}

= - .66 m /s

Since the sign is negative so it will be in opposite direction .

4 0
3 years ago
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