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matrenka [14]
3 years ago
5

How much force do you push down on the Earth with? (Hint: 1kg is about 2 pounds).​

Physics
2 answers:
e-lub [12.9K]3 years ago
6 0

The force with which I push down on the Earth is PRECISELY equal to the force with which the Earth pushes UP on me.  

I weigh EXACTLY the same amount on the Earth as the Earth weighs on me.  Find THAT amount, and you'll have the answer to the question.

(Hint:  1 kg is a mass, and 2 pounds is a force, so they can never be equal.  1kg weighs about 2.2 pounds IF it happens to be on Earth.  But if the same 1kg is on the Moon, then it weighs about 5.8 ounces.  And if it happens to be on Mars, then it weighs about 13.3 ounces.  The same kilogram weighs different weights in different places.)

On a good day, I push down on the Earth with about 890 Newtons of force.  No matter how hard I try, I can never push more or less than that.

Gennadij [26K]3 years ago
3 0

Answer: 100lbs

Explanation: The Earth pushes you down at 100lbs, so you push down the earth by 100lbs which is enough to keep you firmly attached to the ground and not allow you to jump more than a couple of feet.

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Answer:

the current is 3.5 A

Explanation:

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As we know that

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= 3.5 A

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Would you expect to find a high or low level of resistance in a copper wire and why?
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4 years ago
A circuit is built based on the circuit diagram shown. What is the current in the 50 Ω resistor
In-s [12.5K]

Answer:

1.2 A

Explanation:

From the diagram attached, The three resistors are parallel because the each ends of the resistors are connected together. Since they are in parallel, the voltage across each resistor is the same. The voltage source connected in parallel to the resistors is 60 V. Therefore the voltage across the 50 Ω resistor is 60 V. Using ohm law:

Voltage (V) = Current (I) × Resistance (R)

V = IR

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5 0
3 years ago
A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p
777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s

Now, we can calculate the angular acceleration  (w0=0rad/s)

\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}

\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}

with this value we can compute the angular velocity

\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}

and the tangential velocity of point B, and then the acceleration of point B:

v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}

hope this helps!!

6 0
3 years ago
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