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3 years ago

The graph below shows the salmon population, in thousands of pounds, in the Columbia River, from 1866 to 1983.

Physics

2 answers:

cupoosta [38]3 years ago

6 0

A drastic drop in the salmon population occurred between 1910 and 1912.
This suggests that a large amount of contaminants were present in the river.

Answer:
d. A large amount of contaminants were present in the river between 1910 and 1912.

Softa [21]3 years ago

3 0

Answer:

d. A large amount of contaminants were present in the river between 1910 and 1912.

Explanation:

If you observe the graph and the options you can see that the salmons definetely breed between 1866 and 1879 since the population in pound of salmons increased to almost its historical maximum, and the other options are the activation of the hydropower units, and since both options, 1931 and 1970 after the given events in one it decreased the population and in the other it increased, so they cancel eachother, so the only option that matches the graph would be that a larg amount of contaminants were present in the river between 1910 and 1912, since the population decresead largely.

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A scientist heated a tank containing 50 g of water. The specific heat of water is 4.18 J/gºC. The temperature of the water incre

Amanda [17]

I got 5,225 by 50x4.18= 209(25)=5,225

6 0

3 years ago

A ball is thrown straight up at time t=0 with an initial speed of 19m/s. Take the point of release to be y0=0 and upwards to be

Wittaler [7]

First we write the corresponding kinematics equations:
a = -g
v = -g * t + vo
y = -g * ((t ^ 2) / 2) + vo * t + yo
Substituting the values:
y = - (9.81) * (((0.50) ^ 2) / 2) + (19) * (0.50) + (0) = 8.27m
answer:
the displacement at the time of 0.50s is 8.27m

4 0

4 years ago

A uniform, solid sphere of radius 2.50 cm and mass 4.75 kg starts with a purely translational speed of 3.00 m/s at the top of an

allsm [11]

Answer:

The final translational seed at the bottom of the ramp is approximately 4.84 m/s

Explanation:

The given parameters are;

The radius of the sphere, R = 2.50 cm

The mass of the sphere, m = 4.75 kg

The translational speed at the top of the inclined plane, v = 3.00 m/s

The length of the inclined plane, l = 2.75 m

The angle at which the plane is tilted, θ = 22.0°

We have;

$K_i$ + $U_i$ = $K_f$ + $U_f$

K = (1/2)×m×v²×(1 + I/(m·r²))

I = (2/5)·m·r²

K = (1/2)×m×v²×(1 + 2/5) = 7/10 × m×v²

U = m·g·h

h = l×sin(θ)

h = 2.75×sin(22.0°)

∴ 7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) = 7/10 × 4.75× $v_f$ ² + 0

7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) ≈ 77.93

∴ 77.93 ≈ 7/10 × 4.75× $v_f$ ²

$v_f$ ² = 77.93/(7/10 × 4.75)

$v_f$ ≈ √(77.93/(7/10 × 4.75)) ≈ 4.84

The final translational seed at the bottom of the ramp, $v_f$ ≈ 4.84 m/s.

3 0

3 years ago

In reaching her destination, a backpacker walks with an average velocity of 1.41 m/s, due west. This average velocity results, b

V125BC [204]

The total average velocity $v=+1.41 m/s$ (I assume west as positive direction) is given by the total displacement, S, divided by the total time taken, t:
$v= \frac{S}{t}= \frac{S_1+S_2}{t_1+t_2}$
where:
-The total displacement S is the algebraic sum of the displacement in the first part of the motion ( $S_1=6.30 km=6300 m$ , due west) and of the displacement in the second part of the motion ( $S_2$ , due east).
-The total time taken t is the time taken for the first part of the motion, $t_1$ , and the time taken for the second part of the motion, $t_2$ . $t_1$ can be found by using the average velocity and the displacement of the first part:
$t_1= \frac{S_1}{v_1}= \frac{6300 m}{2.49 m/s}=2530 s$

$t_2$ , instead, can be written as $\frac{S_2}{v_2}$ , where $v_2=-0.630 m/s$ is the average velocity of the second part of the motion (with a negative sign, since it is due east).

Therefore, we can rewrite the initial equation as:
$v=1.41 = \frac{6300+S_2}{2530- \frac{S_2}{0.630} }$
And by solving it, we find the displacement in the second part of the motion (i.e. how far did the backpacker move east):
$S_2=-844 m=-0.844 km$

4 0

3 years ago

A bomb of mass 4 kg, initially at rest, explodes breaking into two fragments of 1 kg and 3 kg. Which one of the following statem

Kazeer [188]

Answer: The 1 kg fragment will have three times the speed of the 3kg fragment.

Explanation:Here for the bomb, its chemical energy gets converted into the mechanical energy.

According to the law of conservation of momentum, the two bodies will have equal momentum and to satisfy this condition the lighter mass will have the higher velocity.

∵ momentum, p = mass × velocity

∴The 1 kg fragment will have three times the speed of the 3kg fragment.

6 0

4 years ago