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2 years ago

Which is true about Earth's energy budget?

Physics

1 answer:

REY [17]2 years ago

5 0

Answer:

I think is B but I'm not sure

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The specific weight of sea water is 10.1 kN/m^3. Convert to lbs/in^3.

Viktor [21]

Answer:

0.03719 lbs/in³

Explanation:

Specific weight is given by multiplying the density of an object to the acceleration due to gravity.

$\gamma =\rho g\\\Rightarrow \rho=\frac{gamma}{g}\\\Rightarrow \rho=\frac{10.1\times 10^3}{9.81}\\\Rightarrow \rho=1029.562\ kg/m^3$

$1\ kg=2.20462\ lb$

$1\ m=39.3701\ in$

$\\\Rightarrow 1029.562\ kg/m^3=\frac{1029.562\times 2.20462}{39.3701^3}=0.03719\ lbs/in^3$

So,

$10.1\ kN/m^3=0.03719\ lbs/in^3$

8 0

3 years ago

A circus act involves a trapeze artist and a baby elephant. They are going to balance on a teeter-totter that is 10 meters long

elena-s [515]

Answer:

Explanation:

Balance point will be achieved as soon as the weight of the baby elephant creates torque equal to torque created by weight of woman about the pivot. torque by weight of woman

weight x distance from pivot

= 500x 5

= 2500 Nm

torque by weight of baby woman , d be distance of baby elephant from pivot at the time of balance

= 2500x d

for equilibrium

2500 d = 2500

d = 1 m

So elephant will have to walk up to 1 m close to pivot or middle point.

6 0

3 years ago

Question 25 of 30

VARVARA [1.3K]

Answer:

it's B. circuit a and b are series circuit while c is parallel

7 0

2 years ago

it takes 90 j of work to stretch a spring 0.2 m from its equilibrium position. How muc work is needed to stretch it an additiona

Vinvika [58]

Work needed: 720 J

Explanation:

The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by

$E=\frac{1}{2}kx^2$

where

k is the spring constant

x is the stretching of the spring from the equilibrium position

In this problem, we have

E = 90 J (work done to stretch the spring)

x = 0.2 m (stretching)

Therefore, the spring constant is

$k=\frac{2E}{x^2}=\frac{2(90)}{(0.2)^2}=4500 N/m$

Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of

x = 0.2 + 0.4 = 0.6 m

Substituting,

$E'=\frac{1}{2}kx^2=\frac{1}{2}(4500)(0.6)^2=810 J$

Therefore, the additional work needed is

$\Delta E=E'-E=810-90=720 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

7 0

3 years ago

A circuit has a current of 1.2 A. If the voltage decreases to one third of its original amount while the resistance remains cons

Katen [24]

When resistance is constant, current is proportional to voltage. When 1/3 the voltage is applied, 1/3 the current will result.

(1/3)*(1.2 A) = 0.4 A

The resulting current will be 0.4 A.

8 0

3 years ago