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makvit [3.9K]
3 years ago
10

What is the first step dealing with an injured relationship

Physics
2 answers:
RideAnS [48]3 years ago
7 0
Well there is two if u are badly injured go check it out if u are a bit hurt rest it while having ice on it
Alex_Xolod [135]3 years ago
6 0
Communicate your issues and agree on a solution
You might be interested in
If a boat and its riders have a mass of 900 kg and the boat drifts in at 1.4 m/s how much work does sam do to stop it?
Rama09 [41]
The initial kinetic energy of the boat and its rider is
K_i =  \frac{1}{2} mv_i^2 =  \frac{1}{2}(900 kg)(1.4 m/s)^2=882 J

After Sam stops it, the final kinetic energy of the boat+rider is
K_f = 0 J
because its final velocity is zero.

For the law of conservation of energy, the work done by Sam is the variation of kinetic energy of the system:
W=K_f-K_i =0-882 J=-882 J
where the negative sign is due to the fact that the force Sam is applying goes against the direction of motion of the boat.
6 0
3 years ago
Please help me
Grace [21]

Answer:

a. 60 N*s

b. 60 (kg*m)/s

c. 3 m/s

Explanation:

Givens:

m = 20 kg

v_i = 0 m/s

t = 10 s

F = 6 N

a) Impulse:

I = F*t

I = 6 N*10 s

I = 60 N*s

b) Momentum:

p = v*m

F = m(a)

a = F/m

a = 6 N/20 kg

a = 0.3m/s^2

a = (v_f -v_i)/t

v_f = (0.3 m/s^2)*10 s

v_f = 3.0 m/s

p = 3 m/s*20 kg

p = 60 (kg*m)/s

c. Final velocity

a = (v_f -v_i)/t

v_f = (0.3 m/s^2)*10 s

v_f = 3.0 m/s

6 0
3 years ago
Who sponsored Felix Baumgartner in the second space jump that took placed in<br> 2008?
KengaRu [80]

Alban Geissler, who developed the SKYRAY carbon fiber wing with Christoph Aarns, suggested after Baumgartner's jump that the wing he used was a copy of two prototype SKYRAY wings sold to Red Bull (Baumgartner's sponsor) two years earlier. - wiki

8 0
3 years ago
Was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s? Why?/Why Not?
frosja888 [35]

Answer:

The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

Explanation:

From the question given above, the following data were obtained:

Height to which the target is located = 50 m

Initial velocity (u) = 20 m/s

To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:

Initial velocity (u) = 20 m/s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = 10 m/s²

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 20² – (2 × 10 × h)

0 = 400 – 20h

Collect like terms

0 – 400 = – 20h

– 400 = – 20h

Divide both side by – 20

h = – 400 / – 20

h = 20 m

Thus, the the maximum height to which the cannon ball attained is 20 m.

From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

3 0
3 years ago
Astronomers observe the motion of four planets that orbit a star similar to the Sun. Each planet follows an elliptical orbit aro
Sergeu [11.5K]

Answer:

planet that is farthest away is planet X

kepler's third law

Explanation:

For this exercise we can use Kepler's third law which is an application of Newton's second law to the case of the orbits of the planets

          T² = (\frac{4\pi ^2}{ G M_s} a³ = K_s a³

           

Let's apply this equation to our case

          a = \sqrt[3]{ \frac{T^2}{K_s} }

for this particular exercise it is not necessary to reduce the period to seconds

Plant W

             10² = K_s  a_{w}^3

             a_w = \sqrt[3]{ \frac{100}{ K_s} }

             a_w = \frac{1}{ \sqrt[3]{K_s} }  4.64

Planet X

             a_x = \sqrt[3]{ \frac{640^3}{K_s} }

             a_x = \frac{1}{ \sqrt[3]{K_s} } 74.3

Planet Y

              a_y = \sqrt[3]{ \frac{80^2}{K_s}  }

              a_y = \frac{1}{ \sqrt[3]{K_s} } 18.6

Planet z

              a_z = \sqrt[3]{ \frac{270^2}{K_s} }

              a_z = \frac{1}{ \sqrt[3]{K_s} } 41.8

From the previous results we see that planet that is farthest away is planet X

where we have used kepler's third law

3 0
3 years ago
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