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ankoles [38]
3 years ago
12

A person walks 30m due north, then 50m at 35 degrees. Find their total displacement.

Physics
1 answer:
Harman [31]3 years ago
5 0
Using the cosine rule (a^2 = b^2 + c^2 - 2bc cos A), we can work out the displacement:
Displacement = a
b = 30
c = 50
A = 180 - 35 = 145 degrees.

a^2= 900 + 2500 -1500*-0.81915...
      = 3400 + 1228.728...
      = 4628.72...
a = 68.034...
   = 68.0m (to 3s.f.).

To work out the angle from starting place, use another configuration of the cosine rule: 
cos(C)= \frac{a^{2} +b^{2}-  c^{2}  }{2ab}:
cos (C)= \frac{4628.72...+900-2500}{2*68*30}
           = 3028.7.../4080
           = 0.7423...
C = 42.069... degrees
   = 042 bearing
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2 years ago
IF EARTH HAD NO ATMOSPHERE, WOULD A FALLING OBJECT EVER REACH TERMIONAL VELOCITY?
EleoNora [17]
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3 0
4 years ago
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3 0
3 years ago
slader How much energy is required to move a 1040 kg object from the Earth's surface to an altitude four times the Earth's radiu
andrew-mc [135]

Answer:

ΔU = 5.21 × 10^(10) J

Explanation:

We are given;

Mass of object; m = 1040 kg

To solve this, we will use the formula for potential energy which is;

U = -GMm/r

But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.

Thus;

ΔU = -GMm((1/r_f) - (1/r_i))

Where;

M is mass of earth = 5.98 × 10^(24) kg

r_f is final radius

r_i is initial radius

G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Since, it's moving to altitude four times the Earth's radius, it means that;

r_i = R_e

r_f = R_e + 4R_e = 5R_e

Where R_e is radius of earth = 6371 × 10³ m

Thus;

ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)

× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))

ΔU = 5.21 × 10^(10) J

7 0
3 years ago
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agasfer [191]

Explanation:

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Therefore, the light gets refracted more in the glass than air to water.

4 0
3 years ago
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