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inn [45]
3 years ago
13

On a safari, a team of naturalists sets out toward a research station located 4.63 km away in a direction 38.7 ° north of east.

After traveling in a straight line for 2.13 km, they stop and discover that they have been traveling 25.9 ° north of east, because their guide misread his compass. What are (a) the magnitude and (b) the direction (as a positive angle relative to due east) of the displacement vector now required to bring the team to the research station?
Physics
2 answers:
Talja [164]3 years ago
5 0

This is a vector subraction problem.

maxonik [38]3 years ago
3 0

Vector  subraction problem


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A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff
strojnjashka [21]

(a) 6.43\cdot 10^5 J

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

E=K+U

The initial kinetic energy is:

K=\frac{1}{2}mv^2

where m = 58.0 kg is the mass of the projectile and v=140 m/s is the initial speed. Substituting,

K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J

The initial potential energy is given by

U=mgh

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J

So, the initial mechanical energy is

E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J

(b) -1.67 \cdot 10^5 J

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J

while the potential energy is

U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J

So, the mechanical energy is

E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J

And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:

W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J

And the work is negative because air friction is opposite to the direction of motion of the projectile.

(c) 88.1 m/s

The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:

W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J

When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:

E = K

The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:

E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J

And this is only kinetic energy:

E=K=\frac{1}{2}mv^2

So, we can solve to find the final speed:

v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s

4 0
3 years ago
PLEASE HELP THIS IS TIMED!
Monica [59]
3.5 because yea periodt
4 0
3 years ago
Light travels at a speed of 3.00 x 10^11 cm/s. What is the speed of light in kilometers/hour?
rewona [7]

As 1 km = 1000 m = 1000,00 cm,

So, 1 cm = (1/1000,00) km

1 hour = 60 × 60 s = 3600 s

So,  1 s = (1 / 3600) hour

The light travels at a speed of 3.00 \times 10^{11}  \ cm/s.

In kilometer/hour,

3.00 \times 10^{11}  \ cm/s = 3.00 \times 10^{11} \frac{(1/1000,00) km}{ (1 / 3600) hour} = 108 \times 10^8 \ km/hour


4 0
4 years ago
Please help asap asap<br> wdtwygcbhyeegdlu tcgbihjlnkdm;fnbkhjg hv
kicyunya [14]

Answer:

B

Explanation:

Bbbb

6 0
3 years ago
A railroad cart with a mass of m1 = 11.6 t is at rest at the top of an h = 10.9 m high hump yard hill.
leonid [27]

The final common speed of the two carts will be 69.3 m/sec.The momentum conservation principle is applied.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

Unit conversion:

1 metric tons = 1000 kg

Given data;

m₁= 11.6 metric ton =11600 kg

m₂ = 23.2 metric ton =  23200 kg

Let v represent the combined velocity of the two carts once they are connected, and let u represent the starting velocity of cart 1 when it reaches the bottom.

Considering energy conservation;

\rm  m_1 g h = \frac{1}{2} m_1 \times  u^2 \\\\ u^2 = 2gh\\\\ u^2 = 2  \times 9.8 \times 10.6 \\\\  u = 207.972 \ m/s

From the conservation of momentum principle;

\rm m_1 \times  u = (m_1 + m_2) v\\\\ 11600 \times  207.972 = (11600 + 23200) \times v \\\\ v = 69.3  \ m/s

Hence, the final common speed of the two carts will be 69.3 m/sec.

To learn more about the law of conservation of momentum refer;

brainly.com/question/1113396

#SPJ1

7 0
2 years ago
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