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sladkih [1.3K]
3 years ago
15

While John is traveling along an interstate highway, he notices a 150 mi marker as he passes through town. Later John passes a 1

12 mi marker. What is the distance between town and John’s current location? Answer in units of mi. 003 (part 2 of 2) 10.0 points What is John’s current position?
Physics
2 answers:
Lerok [7]3 years ago
7 0

Answer:

Part a)

distance = 112 miles

Part b)

current position = 112 miles from the position of town

Explanation:

Part a)

Since the distance marker is showing the distance between the town and the position of john at all time

so here we have

d = 112 miles

Part b)

Current position of John is given as

r = 112 miles

from the position of the town

lisov135 [29]3 years ago
7 0

Answer:

a) When John is in the town (and suppose that he is travelling in a straigth line), he sees a 150 mi marker (means that he is 150 mi away from somewhere, let's suppose is a city), and latter, he sees a 112 mi marker (so he is now 112mi from the city) this means that he is closer to the city, and further to the town, then, the distance between john and the town will be 150 mi - 112 mi = 38 mi

b) currently, john is 38 mi ahead of the town and 112 mi away from the city.

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Juli2301 [7.4K]

The car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.

<h3>Average velocity of the car</h3>

The average velocity of the car is calculated as follows;

x(t) = a + bt + ct2

v = dx/dt

v(t) = b + 2ct

v(0) = -10.1 m/s + 2(1.1)(0) = -10.1 m/s

v(10) = -10.1 + 2(1.1)(10) = 11.9 m/s

<h3>Average velocity</h3>

V = ¹/₂[v(0) + v(10)]

V = ¹/₂ (-10.1  + 11.9 )

V = 0.9 m/s

Thus, the car’s velocity as a function of time is b + 2ct and the car’s average velocity during this interval is 0.9 m/s.

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3 0
1 year ago
B)A man walks 95 km, East, then 55 km, north. Calculate his RESULTANT
Varvara68 [4.7K]

The resultant displacement of the man is 109.77 km in the direction N60°E.

<h3>Displacement</h3>

Displacement is the distance travelled in a specified direction.

To calculate displacement, the straight line from starting point to end point of travel is taken and calculated.

<h3>Resultant displacement of the man </h3>

In the example above, a man walks 95 km, East, then 55 km, north.

The two distances form a right-angled triangle with two sides 95 and 55 units. The hypotenuse gives the resultant displacement, D.

Using Pythagoras rule:

D^2 = 95^2 + 55^2

D^2 = 12050

D = 109.77

Thus, the resultant displacement is 109.77 km

To calculate the direction:

Let the direction be y

y + x = 90°

tan x = 55/95

tanx x = 0.578

x = 30°

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Therefore, the resultant displacement of the man is 109.77 km in the direction N60°E.

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A 3-kg rock is thrown upward with a force of 200 N at a location where the local gravitational acceleration is 9.79 m/s2 . Deter
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Explanation:

If we make an analysis of the net force F_{net} of the rock that was thrown upwards, we will have the following:

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F_{up}=200N is the force with which the rock was thrown

W is the weight of the rock

Being the weight the relation between the mass m=3kg of the rock and the acceleration due gravity g=9.79m/s^{2} :

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W=29.37 N (3)

Substituting (3) in (1):

F_{net}=200N-29.37 N  (4)

F_{net}=170.63 N  (5) This is the net Force on the rock

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Finding the acceleration a:

a=\frac{F_{net}}{m}  (7)

a=\frac{170.63 N}{3kg} (8)

Finally:

a=56.87m/s^{2}

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