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Troyanec [42]
4 years ago
9

a 68 kg skydiver with a parachute falls at constant velocity for 100 m how much work does the earth do on the skydiver

Physics
2 answers:
Ulleksa [173]4 years ago
8 0
<span>The earth does no work on her if she's falling in the direction of frictional force. But suppose it does work on the woman i.e in the opposite direction, N. Then we have F = mg where g is acceleration due to grav. So F = 68 * 9.8 = 666.4 Newton. Then work done = force * distance = 666.4 * 100 = 666400</span>
Vanyuwa [196]4 years ago
5 0

Answer:

666.4×100=666400

This is the ans.

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inn [45]

Answer:

Explanation:

Thermal Energy and Heat

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another? To answer this question, you need to think about the

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5 0
4 years ago
What is the molecular formula for the hydrocarbon in the mass spectrum above?
PtichkaEL [24]
What is the structure of a hydrocarbon that has $\mathrm{M}^{+}=120$ in its mass spectrum and has the following $1 \mathrm{H}$ NMR spectrum? 7.25 $\delta(5 \mathrm{H}, \text { broad singlet); } 2.90 \delta(1 \mathrm{H}, \text { septet, } J=7 \mathrm{Hz}) ; 1.22 \delta(6 \mathrm{H},\text { doublet, }$ $J=7 \mathrm{Hz})$
6 0
3 years ago
0.160-kg hockey puck is moving on an icy, frictionless, horizontal surface. At t 0, the puck is moving to the right at 3.00 m/s.
balandron [24]

Answer: (a)10.812\ m/s\ (b)\ 0.75\ m/s\ \text{left}

Explanation:

Given

Mass of hockey puck m=0.160\ kg

Initial velocity of hockey puck is u=3\ m/s

First a horizontal force of 25\ N is applied to the right for 0.05\ s

acceleration associated with it is

\Rightarrow a=\dfrac{25}{0.160}\\\\\Rightarrow a=156.5\ m/s^2

Using equation of motion i.e.

\Rightarrow v=u+at\\\Rightarrow v=3+156.25\times 0.05\\\Rightarrow v=3+7.812\\\Rightarrow v=10.812\ m/s

(b) When a force of 12\ N is applied for 0.05s

Using equation of motion i.e.

\Rightarrow v=3-\dfrac{12}{0.160}\times 0.05\\\\\Rightarrow v=3-75\times 0.05\\\Rightarrow v=3-3.75\\\Rightarrow v=-0.75\ m/s\\\Rightarrow v=0.75\ \text{towards left}

6 0
3 years ago
In your experiment, you measure a total deflection of 4.12 cm when an electric field of 1.10×103V/m is established between the p
Bond [772]

Answer:

B_0 = 1.69 \times 10^{-4}\ T

Explanation:

given,

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using formula

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v_0 = 6496355.63 m/s

v_0 = \dfrac{E}{B_0}

B_0 = \dfrac{E}{v_0}

B_0 = \dfrac{1.1\times 10^{3}}{6496355.63}

B_0 = 1.69 \times 10^{-4}\ T

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3 years ago
Push or pull of one object to another
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The term is called "force."
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