Answer:
Objects are not pulled towards you due to friction
Explanation:
For example, if you consider a notebook, there is a friction between the notebook and the surface on which the notebook is placed. This frictional force cannot be overcome by the gravitational force as gravitational forces are extremely weak forces so the notebook stays in place.
<span><span> High School </span> <span> Physics </span> <span> 5+3 pts </span> </span><span><span>Previous question </span> </span> <span><span>Next question </span> </span> <span>Naomi wants to know the effect of different colors of light on the growth of plants. She believes that plants can survive best in white light. She buys 25
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Answer:
4.02 km/hr
Explanation:
5 km/hr = 1.39 m/s
The swimmer's speed relative to the ground must have the same direction as line AC.
The vertical component of the velocity is:
uᵧ = us cos 45
uᵧ = √2/2 us
The horizontal component of the velocity is:
uₓ = 1.39 − us sin 45
uₓ = 1.39 − √2/2 us
Writing a proportion:
uₓ / uᵧ = 121 / 159
(1.39 − √2/2 us) / (√2/2 us) = 121 / 159
Cross multiply and solve:
159 (1.39 − √2/2 us) = 121 (√2/2 us)
220.8 − 79.5√2 us = 60.5√2 us
220.8 = 140√2 us
us = 1.115
The swimmer's speed is 1.115 m/s, or 4.02 km/hr.
Answer:
i) E = 269 [MJ] ii)v = 116 [m/s]
Explanation:
This is a problem that encompasses the work and principle of energy conservation.
In this way, we establish the equation for the principle of conservation and energy.
i)

![W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]](https://tex.z-dn.net/?f=W_%7B1-2%7D%3D%20%28F%2Ad%29%20-%20%28m%2Ag%2Ah%29%5C%5CW_%7B1-2%7D%3D%28500000%2A2.5%2A10%5E3%29-%2840000%2A9.81%2A2.5%2A10%5E3%29%5C%5CW_%7B1-2%7D%3D%20269%2A10%5E6%5BJ%5D%20or%20269%20%5BMJ%5D)
At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.
Er = 269*10^6[J]
ii ) With the energy calculated at the previous point, we can calculate the speed developed.
![E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]](https://tex.z-dn.net/?f=E_%7Bk2%7D%3D0.5%2Am%2Av%5E2%5C%5C269%2A10%5E6%3D0.5%2A40000%2Av%5E2%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B269%2A10%5E6%7D%7B0.5%2A40000%7D%20%7D%5C%5C%20v%3D116%5Bm%2Fs%5D)
<u>Given </u><u>:</u><u>-</u>
- An elevator is moving vertically up with an acceleration a.
<u>To </u><u>Find</u><u> </u><u>:</u><u>-</u>
- The force exerted on the floor by a passenger of mass m .
<u>Solution</u><u> </u><u>:</u><u>-</u>
As the man is in a accelerated frame that is <u>non </u><u>inertial</u><u> frame</u><u> </u>, we would have to think of a pseudo force .
- The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
For the FBD refer to the attachment . From that ,
<u>Hence</u><u> </u><u>option</u><u> </u><u>d </u><u>is </u><u>correct</u><u> </u><u>choice </u><u>.</u>
<em>I </em><em>hope</em><em> this</em><em> helps</em><em> </em><em>.</em>