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3 years ago

The small increase in global oil production coupled with an increased demand for those resources leads to _______.

2 answers:

5 0

The correct answer for this question is this one: "b. increased prices." The small increase in global oil production coupled with an increased demand for those resources leads to increase in prices.

Here are the following choices.
a. lack of modern energy sources
b. increased prices
c. decreased competition
d. decreased prices

Snowcat [4.5K]3 years ago

5 0

Answer:

b. increased prices

for edgenunity

Explanation:

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Questions 8 out of 20

MrMuchimi

Answer:

potential energy

Explanation:

8 0

3 years ago

The suspension system of a 2100 kg automobile "sags" 8.5 cm when the chassis is placed on it. Also, the oscillation amplitude de

Gennadij [26K]

Answer:

Part a)

$k = 6.06 \times 10^4 N/m$

Part b)

$b = 1795.4 kg/s$

Explanation:

Part a)

as the mass of the suspension system is given as

$m = 2100 kg$

also we have

$x = 8.5 cm$

so now for force balance we have

$mg = kx$

$(525)(9.81) = k(0.085)$

$k = 6.06 \times 10^4 N/m$

Part b)

Now we know that amplitude decreases by 63% in each cycle

so after one cycle the amplitude will become 37% of initial amplitude

so it is given as

$A = 0.37 A_o$

also we know

$A = A_o e^{-bt/2m}$

$0.37 A_o = A_o e^{-bt/2m}$

$\frac{bt}{2m} = 1$

$b = \frac{2m}{t}$

here t = time period of one oscillation

so it is

$t = 2\pi\sqrt{\frac{m}{k}}$

$t = 2\pi\sqrt{\frac{525}{6.06 \times 10^4}}$

$t = 0.58 s$

now damping constant is

$b = \frac{2(525)}{0.58}$

$b = 1795.4 kg/s$

7 0

3 years ago

easy bio A person is standing on a level floor. His head, upper torso, arms, and hands together weigh 438 N and have a center of

goblinko [34]

Answer:

1.034 m above the floor

Explanation:

The location of center of body for a compound body, when the weights are given is calculated as:

$\bar x = \frac{W_1x_1+W_2x_2+W_3x_3+W_4x_4+W_5x_5+.......+Wnx_n}{W_1+W_2W_3W_4W_5+.....+W_n}$

where,

$\bar x$ is the center of gravity of the entire body

W = weight of the individual body

x = center of gravity of the individual body

Thus on substituting the values we get,

$\bar x = \frac{438\times 1.28+144\times 0.760+87\times 0.250}{438+144+87}$

$\bar x = \frac{691.83}{669}$

$\bar x =1.034m$

Hence, the center of gravity of the entire body lies 1.034 m above the floor

6 0

3 years ago

What school did Ronald McNair go to and what kind of science did he work in

alina1380 [7]

Answer:

McNair graduated as valedictorian of Carver High School in 1967. In 1971, he received a Bachelor of Science degree in engineering physics, magna cu.m laude, from the North Carolina Agricultural and Technical State University in Greensboro, North Carolina.

5 0

3 years ago

Please help !! ASAP

umka2103 [35]

Answer:

22) 5.6 Kg.m/s

23) 0.067 Kg

24) 1.16 m/s

Explanation:

Momentum is found by multiplying mass and velocity

P=mv where P is momentum, m is mass of object and v is the velocity

Given

Mass, m=2 Kg

Rate, v=2.8 m/s

Relationship

P=mv

Solution

P=mv=2*2.8=5.6 Kg.m/s

Given

Momentum, p=0.1 Kg.m/s

Rate, v=1.5 m/s

Relationship

P=mv and making m the subject then

$m=\frac {P}{v}$

Solution

$m=\frac {0.1 Kgm.s}{1.5 m/s}\approx 0.067 Kg$

Given

Momentum, $p=4.9\times 10^{8} Kg.m/s$

Mass, $m=4.23\times10^{8} Kg$

Relationship

P=mv and making v the subject then

$v=\frac {P}{m}$

Solution

$m=\frac {4.9\times 10^{8} Kgm.s}{4.23\times10^{8} }\approx 1.16 m/s$

5 0

3 years ago