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mr Goodwill [35]
3 years ago
13

An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 16.0 m if her initial speed is 3.60 m/

s. what is the free-fall acceleration on the planet? (ignore air resistance.)
Physics
1 answer:
Airida [17]3 years ago
7 0
The maximum height that is reached by an object thrown or jumped at a certain initial velocity can be calculated through the equation,

   2ad = Vf² - Vi²

Vf is zero (0) in this equation because this is the point at the velocity at the maximum height of the object.

Substituting the known values,
  2(a)(-16) = 0 - (3.6)²

The value of a from the equation is 0.405 m/s².

<em>Answer: 0.405 m/s²</em>
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A box is pulled 6 meters across the ground at a constant velocity by a horizontally applied force of 50 newtons. At the same tim
Vikentia [17]

Answer:

(a) Friction force = 50 N

(b) Work done by friction = 300 j

(c) Net work done = 0 j

Explanation:

We have given that the box is pulled by 6 meter so d = 6 m

Force applied on the box F = 60 N

We have have given that velocity is constant so acceleration will be zero

So to applied force will be utilized in balancing the friction force

So friction force F_{friction}=50N

Work done by friction force W_{friction}=F_{friction}\times d=50\times 6=300j

Work done by applied force W=F\times d=50\times 6=300j

So net work done = 300-300 = 0 j

7 0
3 years ago
Two 2.0kg bodies A and B collide The velocities before the collision are U1=15i+30j and U2=-10j+5.0j After the collision V1=-5.0
Svetllana [295]

Use the law of conservation of momentum. Since the momentum is a linear measure, we can treat each of the dimension separately:

i-direction:

m_1u_{1i}+m_2u_{2i}=m_1v_{1i}+m_2v_{2i}\\v_{2i} = \frac{m_1u_{1i}+m_2u_{2i}-m_1v_{1i}}{m_2}=\frac{(2\cdot 15-2\cdot10+2\cdot5)kg\frac{m}{s}}{2kg}=10\frac{m}{s}

j-direction:

m_1u_{1j}+m_2u_{2j}=m_1v_{1j}+m_2v_{2j}\\v_{2j} = \frac{m_1u_{1j}+m_2u_{2j}-m_1v_{1j}}{m_2}=\frac{(2\cdot 30+2\cdot5-2\cdot20)kg\frac{m}{s}}{2kg}=15\frac{m}{s}

Answer: Final velocity is: (10i + 15j) m/s

Change in the kinetic energy:

\Delta E_k = E_{ku}-E_{kv} = \frac{1}{2}m(u_1^2+u_2^2-v_1^2-v_2^2)=\\=\frac{1}{2}2kg(1125+125-425-325)\frac{m^2}{s^2}=500J

Answer: The system lost 500J worth of kinetic energy in the collision

4 0
3 years ago
The eiffel tower is 300 meters tall. Disregarding air friction, at what velocity would an object be traveling when it reaches th
xeze [42]

Answer:

Hope it helps you :)

Explanation in the pic above.

8 0
2 years ago
Which ecosystem contains 32% of the world's producers
tangare [24]
The Answer Is D 32 % Of The worlds production is In the marine ecosystem
4 0
3 years ago
hooke's law is described mathematically using the formula fsp=-kx. which statement is correct about spring force, fsp? A. It is
bogdanovich [222]

Answer:

A. It is always a positive force

Explanation:

Hooke's law describes the relation between an applied force and extension ability of an elastic material. The law states that provided the elastic limit, e, of a material is not exceeded, the force, F, applied is proportional to the extension, x, provided temperature is constant.

i.e F = - kx

where k is the constant of proportionality, and the minus sign implies that the force is a restoring force.

The applied force can either be compressing or stretching force.

6 0
3 years ago
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