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r-ruslan [8.4K]
1 year ago
5

An orienteer runs 400m directly east and then 500m to the northeast (at a 45 degree andle from due east and from due north). Pro

vide a graphic solution to show final displacement with respect to the starting position

Physics
1 answer:
Aleks04 [339]1 year ago
6 0

Answer:

Final displacement with respect to the starting position is 832.37 meter

Explanation:

Lets consider that the orienteer start to run on the point of (0,0) point of a coordinate system. When he runs towards to east side about 400m, he will be at the point A(400,0). After he runs to the northeast (at a 45 degree angle from due east and from due north), approximately he will make 353.55 meter to east and 353.55 meter to the north side and he will be at the point of B(753.55, 353.55)

From the starting point total displacement will be 832.37 meter. Please check the attached graphical solution.

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A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha
Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3  s

the formula for the time period of the pendulum is given by

T = 2\pi \sqrt{\frac{I}{mgd}}    .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

I = \frac{mL^{2}}{12}+ md^{2}

Substituting the values in equation (1)

3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}

9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )

12d² -26.84 d + 2.25 =  0

d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}

d=\frac{26.84\pm 24.75}{24}

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0
1 year ago
the millersburg ferry (m=13000.0 kg loaded) puts its engines in full reverse and stops in 65 seconds. if the speed before brakin
kenny6666 [7]

The braking force is -400 N

Explanation:

We can solve this problem by using the impulse theorem, which states that the impulse applied on the ferry (the product of force and time) is equal to its change in momentum:

F \Delta t = m(v-u)

where in this problem, we have:

F is the force applied by the brakes

\Delta t = 65 s is the time interval

m = 13,000 kg is the mass of the ferry

u = 2.0 m/s is the initial velocity

v = 0 is the final velocity

And solving for F, we find the force applied by the brakes:

F=\frac{m(v-u)}{\Delta t}=\frac{(13000)(0-2.0)}{65}=-400 N

where the negative sign indicates that the direction is backward.

Learn more about impulse:

brainly.com/question/9484203

#LearnwithBrainly

4 0
1 year ago
A helicopter flies 250 km on a straight path in a direction 60° south of east. The east component of the helicopter’s displaceme
GaryK [48]

Given that,

Distance in south-west direction = 250 km

Projected angle to east = 60°

East component = ?

since,

cos ∅ = base/hypotenuse

base= hyp * cos ∅

East component = 250 * cos 60°

East component = 125 km

8 0
2 years ago
Read 2 more answers
In ocean waves, water particles move ________ and energy moves ________.
morpeh [17]
In ocean waves, water particles move with mechanical energy and energy moves with gravity

Not sure but hope it helps!
6 0
1 year ago
Read 2 more answers
A rock is dropped from the top of a tall building. The rock's displacement in the last second before it hits the ground is 46 %
olasank [31]

Answer:

height is 69.68 m

Explanation:

given data

before it hits the ground =  46 % of entire distance

to find out

the height

solution

we know here acceleration and displacement that is

d = (0.5)gt²     ..............1

here d is distance and g is  acceleration and t is time

so when object falling it will be

h = 4.9 t²   ....................2

and in 1st part of question

we have (100% - 46% ) = 54 %

so falling objects will be there

0.54 h = 4.9 (t-1)²       ...................3

so

now we have 2 equation with unknown

we equate both equation

1st equation already solve for h

substitute h in the second equation and find t

0.54 × 4.9 t² = 4.9 (t-1)²  

t = 0.576 s and  3.771 s

we use here 3.771 s because  0.576 s is useless displacement in the last second before it hits the ground is 46 % of the entire distance it falls

so take t = 3.771 s

then h from equation 2

h = 4.9 t²

h = 4.9 (3.771)²

h =  69.68 m

so height is 69.68 m

6 0
2 years ago
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