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wlad13 [49]
3 years ago
12

What evidence (data) would support the claim that an increase inhuman population contributes to the emission of greenhouse gases

?
A. greenhouse gas production has leveled off once the global population reached a certain number

B. amount of greenhouse gases produced varies with the level of development a country has reached

C. greenhouse gas emissions per individual humas has decreased over the past 30 years

D. 10 countries with the largest populations are the countries that produce the most greenhouse gases
Physics
1 answer:
Natali [406]3 years ago
5 0
Hey there!

The answer would be D. 10 countries with the largest populations are the countries that produce the most greenhouse gases.

This is the only statements that supports the idea of large populations affecting the emissions of greenhouse gases.

Hope this helps!
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A rod of length 35.50 cm has linear density (mass per length) given by λ = 50.0 + 23.0x where x is the distance from one end, an
hram777 [196]

Answer:

(a)20.65g

(b)0.19m

Explanation:

(a) The total mass would be it's mass per length multiplied by the total lenght

0.355(50 + 23*0.355) = 20.65 g

(b) The center of mass would be at point c where the mass on the left and on the right of c is the same

Hence the mass on the left side would be half of its total mass which is 20.65/2 = 10.32 g

c(50 + 23c) = 10.32

23c^2 + 50c - 10.32 = 0

c \approx 0.19m

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3 years ago
Find the magnitude of the buoyant force on the block. The acceleration of gravity is 9.8 m/s 2 . Answer in units of N.
harkovskaia [24]
The answer is 2248.02 N.
6 0
2 years ago
N experiment is performed in deep space with two uniform spheres, one with mass 27.0 and the other with mass 107.0 . They have e
Reptile [31]

Answer:

Explanation:

Apply the law of conservation of energy

KE_i+PE_i=KE_f+PE_f

Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)

from the law of conservation of the linear momentum

m_1v_1=m_2v_2

Therefore,

Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)

=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]

v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]

Substitute the values in the above result

v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]

=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s

B)  the speed of the sphere with mass 107.0 kg is

v_2=\frac{m_1v_1}{m_2}

=[\frac{27}{107} ](1.2664 \times 10^-^5)\\\\=3.195\times 10^-^6m/s

C)  the magnitude of the relative velocity with which one sphere is

v_r=v_1+v_2\\\\=1.2664\times 10^-^5+3.195\times10^-^6\\\\=15.859\times10^-^6m/s

D) the distance of the centre is proportional to the acceleration

\frac{x_1}{x_2} =\frac{a_1}{a_2} \\\\=\frac{m_2}{m_1} \\\\=3.962

Thus,

x_1=3.962x_2

and

x_2=0.252x_1

When the sphere make contact with eachother

Therefore,

x_1+x_2+2r=41\\x_1+0,252x_1+2r=41\\1.252x_1+2r=41\\x_1=32.747-1.597r

And

x_1+x_2+2r=41\\3.962x_2+x_2+2r+41\\4.962x_2+2r=41\\x_2=8.262-0.403r

The point of contact of the sphere is

32.747-1.597r=8.262-0.403r\\\\r=\frac{24.485}{1.194} \\\\=20.506m

3 0
3 years ago
A mass m0 is attached to a spring and hung vertically. The mass is raised a short distance in the vertical direction and release
iragen [17]

Answer:

The frequency of the oscillations in terms of fo will be f2=fo/3

E xplanation:

T= 2pie\frac sqrt {m}{k}

 \frac {{f2}/times {fo}}=1:3

⇒f2=fo\3

Here frequency f is inversely poportional to square root of mass m.

so the value of remainder of frequency f2 and fo is equal to 1:3.

⇒\frac{f2} {f1} = \frac sqrt{m1}[m2}

⇒\frac{f2}{fo} = 1:3

⇒f2=\frac{fo} {3}

6 0
3 years ago
A truck is moving north word at a constant speed the momentum of the car is
nataly862011 [7]

Answer:

its constant i think

Explanation:

or its stable dunno which term will they be using

6 0
2 years ago
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