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2 years ago

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

1 answer:

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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A motorcycle is following a car that is traveling at a constant speed on a straight highway. Initially, the car and the motorcyc

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Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at $v_c = 23m/s$ speed is

$s_c = \Delta t v_c = 23\Delta t$

The distance traveled by the motorcycle after Δt (seconds) at $m_m = 23 m/s$ speed and acceleration of a = 8 m/s2 is

$s_m = \Delta t v_m + a\Delta t^2/2$

$s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2$

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

$s_m = s_c + 58$

$23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58$

$4 \Delta t^2 = 58$

$\Delta t^2 = 14.5$

$\Delta t = \sqrt{14.5} = 3.807s$

(b)

$s_m = 23 \Delta t + 4 \Delta t^2$

$s_m = 23*3.807 + 58 = 145.581 m$

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2 years ago

A tiny sphere carrying a charge of 6. 5 µc sits in an electric field, at a point where the electric potential is 240 v. what is

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The sphere’s Electric potential energy is 1.6* $10^{-6}$ J

Given,

q=6. 5 µc, V=240 v,

We know that sphere’s Electric potential energy(E) = qV=6.5* $10^{-6} *240$ =1.6* $10^{-6}$ J

<h3>Electric potential energy</h3>

The configuration of a certain set of point charges within a given system is connected with the potential energy (measured in joules) known as electric potential energy, which is a product of conservative Coulomb forces. Two crucial factors—its inherent electric charge and its position in relation to other electrically charged objects—can determine whether an object has electric potential energy.

In systems with time-varying electric fields, the potential energy is referred to as "electric potential energy," but in systems with time-invariant electric fields, the potential energy is referred to as "electrostatic potential energy."

A tiny sphere carrying a charge of 6. 5 µc sits in an electric field, at a point where the electric potential is 240 v. what is the sphere’s potential energy?

Learn more about Electric potential energy here:

brainly.com/question/24284560

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1 year ago

While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find th

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Answer:

Explanation:

From the given information:

radius = 15 m

Time T = 23 s

a) Speed (v) = $\dfrac{2 \pi r}{T}$

$v = \dfrac{2\times \pi \times 15}{23}$

v = 4.10 m/s

b) The magnitude of the acceleration is:

$a = \dfrac{v^2}{r} \\ \\ a = \dfrac{(4.10)^2}{15}$

a = 1.12 m/s²

c) True weight = mg

Apparent weight = normal force

From the top;

the normal force = upward direction,

weight is downward as well as the acceleration.

true weight - normal force = ma

apparent weight =mg - ma

$\dfrac{apparent \ weight}{true \ weight} = \dfrac{(mg - ma)}{(mg)}$

$=1- \dfrac{1.12}{9.8}$

= 0.886 m/s²

From the bottom;

acceleration is upward, so:

apparent weight - true weight = ma

apparent weight = true weight + ma

$\dfrac{apparent \ weight }{true \ weight} =\dfrac{ mg + ma}{mg}$

$\dfrac{apparent \ weight }{true \ weight} =1+ \dfrac{a}{g} \\ \\ = 1 + \dfrac{1.12}{9.8}$

$= 1 + \dfrac{1.12}{9.8} \\ \\$

= 1.114 m/s²

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3 years ago

In a typical golf swing the club is in contact with the ball for about 0.0010 s. If the .045 kg ball experiences a force of 4000

Musya8 [376]

Answer:

v = 88.89 [m/s]

Explanation:

To solve this problem we must use the principle of conservation of momentum which tells us that the initial momentum of a body plus the momentum added to that body will be equal to the final momentum of the body.

We must make up the following equation:

$F*t = m*v$

where:

F = force applied = 4000 [N]

t = time = 0.001 [s]

m = mass = 0.045 [kg]

v = velocity [m/s]

$4000*0.001=0.045*v\\v=88.89[m/s]$

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3 years ago

Which of the following is an example of potential energy? View Available Hint(s) Which of the following is an example of potenti

Brums [2.3K]

Answer:

a bowling ball placed on the top shelf of a closet.

Explanation:

Potential energy is the energy that is stored in an object due to its position relative to some zero position. An object possesses gravitational potential energy if it is positioned at a height above (or below) the zero height.

Mathematically,

Potential energy, PE = M × g × h

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3 years ago