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dmitriy555 [2]
3 years ago
9

You are lost at night in a large, open field. Your GPS tells you that you are 122.0 m from your truck, in a direction 58.0o east

of south. You walk 72.0 m due west along a ditch. How much farther, and in what direction, must you walk to reach your truck?

Physics
1 answer:
Drupady [299]3 years ago
3 0

Answer:

The person is 187[m] farther and 70° south to east.

Explanation:

We can solve this problem by drawing a sketch of the location of the person and the truck, then we will draw the displacement vectors and finally the length of the vector and the direction of the vector will be measured in order to give the correct indication of where the person will have to move.

First we establish an origin of a coordinate system.

We can see in the attached schema that the red vector is the displacement vector from the last point to where the truck is located.

The length of the vector is 187 [m], and the direction is 70 degrees south to East.

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The height (in meters) of a projectile shot vertically upward from a point 2 m above ground level with an initial velocity of 22
alexgriva [62]
1) The law of motion of the projectile is
h(t) = 2+22.5 t-4.9 t^2
To find the velocity, we should compute the derivative of h(t):
v(t)=h'(t)=22.5-2\cdot 4.9t=22.5-9.8t
So now we can calculate the speed at t=2 s and t=4 s:
v(2.0s)=22.5-9.8\cdot2.0 =2.9 m/s
v(4.0s)=22.5-9.8\cdot 4.0s=-16.7 m/s
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.

2) The projectile reaches its maximum height when the speed is equal to zero:
v(t)=0
So we have
22.5-9.8 t=0
And solving this we find
t=2.30 s

3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
h(2.30 s)=2+22.5\cdot 2.30 -4.9 \cdot (2.30s)^2 = 27.83 m

4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
2+22.5t -4.9 t^2 = 0
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
t=4.68 s
And this is the time at which the projectile hits the ground.

5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
v(4.68 s)=22.5-9.8\cdot 4.68 =-23.36  m/s
with negative sign, because it is directed downward.
8 0
3 years ago
A vocalist with a bass voice can sing as low as 92 Hz.
Inessa05 [86]

Answer:

  • 3.26 x 10 to the power of 6

Explanation:

c = lambda × frequency

5 0
3 years ago
A 91.5 kg football player running east at 2.73 m/s tackles a 63.5 kg player running east at 3.09 m/s. what is their velocity aft
vivado [14]

Their velocity afterwards is 2.88 m/s east

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, for an isolated system (= no external force), the total momentum must be conserved before and after the collision. So we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1 + m_2)v

where: in this case:

m_1 = 91.5 kg is the mass of the first player

u_1 = 2.73 m/s is the initial velocity of the first player (choosing east as positive direction)

m_2 = 63.5 kg is the mass of the second player

u_2 = 3.09 m/s is the initial velocity of the second player

v is their combined velocity afterwards

Solving for v, we find:

v = \frac{m_1 u_1+m_2 u_2}{m_1+m_2}=\frac{(91.5)(2.73)+(63.5)(3.09)}{91.5+63.5}=2.88 m/s

And the sign is positive, so the direction is east.

Learn more about momentum here:

brainly.com/question/7973509  

brainly.com/question/6573742  

brainly.com/question/2370982  

brainly.com/question/9484203  

#LearnwithBrainly

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Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

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angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

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