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Levart [38]
3 years ago
7

Evaluate How can knowing about Earth's history help you to choose your dig site?

Physics
1 answer:
alexandr1967 [171]3 years ago
5 0

Answer:if you know the history, you know the best dig site in a way where you know your roots and where the best place is.

Explanation:let’s say that if someone was going on a trip to a foreign country, they would do a little research to know what they were expecting. Similarly, if you were picking a Destiination for your dig site, you would have to do a little research for example, like the earths history to know where the greatest treasures you could find and make the most discoveries while in your dig site.

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HELP PLSSS I HAVE AN EXAM MONDAY AND I THINK THIS IS GONNA BE ON ITTTT
il63 [147K]
<h2>Answer:</h2>

(a) 3.2 x 10²s

(b) 0.9 m/s (S 13 E)

(c) 2.9 x 10²m

<h2>Explanation:</h2>

The sketch illustrating the scenario has been attached to this response.

As shown;

The fish swims due east with a velocity V_{x} = 0.2m/s

The river current has a velocity V_{y} due South = 0.9m/s

The resultant of the velocity is V

The width of the river is x = 64m

(a) To calculate how long it took the fish to get across the river, we know that velocity is the rate of change in distance, therefore we can use the relation;

V = \frac{d}{t}      -------------(i)

Where;

V = velocity of the fish = V_{x} = 0.2m/s

d = distance from the start to the end = width of the river = x = 64m

t = time taken to move for that distance

Make t subject of the formula in equation (i);

t = \frac{d}{V}

Substitute the values of d and V into the equation;

t = \frac{64m}{0.2m/s}

t = 320 s

t = 3.20 x 10²s

Therefore, the time taken for the fish to get across the river is 3.20 x 10²s

(b) The resulting vector of the fish is V whose magnitude is the algebraic sum of vectors  V_{x} and  V_{y}, and direction is given by θ. i.e

<em>The magnitude of the resulting vector is;</em>

|V| = \sqrt{(V_x)^2 + (V_y)^2}

|V| = \sqrt{(0.2)^2 + (0.9)^2}

|V| = \sqrt{(0.04) + (0.81)}

|V| = \sqrt{(0.85)}

|V| = 0.92m/s

|V| ≅ 0.9m/s

<em>The direction of the resulting vector θ and is given by;</em>

tan θ = \frac{V_y}{V_x}

tan θ = \frac{0.9}{0.2}

tan θ = 4.5

θ = tan⁻¹ ( 4.5)

θ = 77.47° South of East.

θ  ≈ 77.5° South of East.

Subtracting θ = 77.5° from 90° gives its value East of South

i.e

90 - 77.5 = 12.5° East of South

<em>This can also be written as S12.5°E</em>

<em>Approximating to the nearest whole number gives </em>S 13 E

Therefore, the resulting velocity of the fish is 0.9m/s in the direction S13°E

(c) When the fish arrives on the opposite bank, its distance from being at the point directly across from where it started is the product of the velocity of the river current and the time taken by the fish to get across the river. This point is equivalent to k as shown in the diagram.

Therefore;

distance = velocity of river current x time taken

distance = 0.9m/s x 3.20 x 10²s

distance = 2.88 x 10²m

distance ≅ 2.9 x 10²m

<em>Notice that the velocity of the river current is used since that's the velocity of the fish on the y-axis.</em>

<em />

<em />

7 0
3 years ago
Explain the why a constant speed has a slope of 0 on a graph of speed v.time.
Studentka2010 [4]
Because if I do 0*time you get the answer as 0
6 0
3 years ago
Which of the following statements describes an interaction between the geosphere and biosphere?
atroni [7]
A soil acidity affects plant growth
7 0
2 years ago
1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal.
harkovskaia [24]

Answer:

(a) The net force is 80.394 N

    The acceleration of the crate is 0.804 m/s²

(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, F_y = Fsin \theta = 250sin45 = 176.78 \ N

Applied force in x-direction, F_x = Fcos \theta = 250cos45 = 176.78 \ N

The normal force is calculated as;

N + Fy -W = 0

N = W - Fy

N = (100 x 9.8) - 176.78

N = 980 - 176.78 = 803.22 N

The frictional force is given by;

Fk = μN

Fk = 0.12 x 803.22

Fk = 96.386 N

(a) The net force is given by;

F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N

Apply Newton's second law of  motion;

F = ma

a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2

(b) the velocity of the crate after 5.0 s

F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s

7 0
3 years ago
A 1500 kg car is parked at the top of a hill 5.2 m high. At the bottom of the hill, what is the kinetic energy, in Joules, of th
irga5000 [103]

Answer:

10.1

Explanation:

I think it's right?

4 0
3 years ago
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