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worty [1.4K]
2 years ago
10

If on the average every man lives for 70 years, for how many microseconds is this life span​

Physics
1 answer:
MakcuM [25]2 years ago
7 0

Answer:

3.1556926 × 10^13 microseconds

Explanation:

Hope this helped!

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5) What type of substance can seep<br> groundwater and make it unusable?
sergeinik [125]

Answer:

I believe <u>chemicals</u>, though I might be wrong.

Explanation:

7 0
2 years ago
Suppose that the resistance between the walls of a biological cell is 6.8 × 109 ω. (a) what is the current when the potential di
GenaCL600 [577]
(a) We can find the current flowing between the walls by using Ohm's law:
I= \frac{\Delta V}{R}
where \Delta V=69 mV=0.069 V is the potential difference and R=6.8\cdot 10^9 \Omega is the resistance. Substituting these values, we get
I=1.01 \cdot 10^{-11} A

(b) The total charge flowing between the walls is the product between the current and the time interval:
Q=I \Delta t
The problem says \Delta t=0.86 s, so the total charge is
Q=(1.01\cdot 10^{-11} A)(0.86 s)=8.73 \cdot 10^{-12} C

The current consists of Na+ ions, each of them having a charge of e=1.6 \cdot 10^{-19} C. To find the number of ions flowing, we can simply divide the total charge by the charge of a single ion:
N= \frac{Q}{e} = \frac{8.73 \cdot 10^{-12}C}{1.6 \cdot 10^{-19}C} = 5.45 \cdot 10^7 ions
4 0
3 years ago
Force due to gravity. While JWST is in orbit, the Earth will be at a distance of 1.494 x 109 m from the telescope, and the Sun w
Alona [7]

Answer:

the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

Explanation:

Given that;

distance of earth from the telescope r1 = 1.494 × 10⁹ m

and the Sun will be 1.49598 × 10¹¹ m further away

so r2 = r1 + 1.49598 × 10¹¹

r2 = 1.494 × 10⁹ m + 1.49598 × 10¹¹ m = 1.51092 × 10¹¹ m

mass of sun Ms = 1.9884 × 10³⁰ kg

mass of earth Me = 5.945× 10²⁴ kg

mass of JWST Mj = 6500 kg

What is the gravitational force JWST will feel from the Sun (strength and direction)?

the gravitational force of the sun will be attractive based on Newton law of gravitational force; so

Fjs = GMjMs / r2²

constant G = 6.674 × 10⁻¹¹ Nm²/kg²

Force on the JWST by the sun will be;

Fjs = GMjMs / r2² { leftward}

we substitute

Fjs = [(6.674 × 10⁻¹¹ Nm²/kg²)(6500 kg )(1.9884 × 10³⁰ kg)] / (1.51092 × 10¹¹ m)²

=  (8.62587804 × 10²³) / ( 2.28287925 × 10²² )

= 37.785 Newton { leftward }

Therefore, the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

7 0
3 years ago
Scott travels north 5 miles, then goes west 3 miles, and then goes south for 2 miles.
Yuki888 [10]
Scott traveled 10 miles
5 0
2 years ago
A 1090 kg car has four 12.7 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to ro
max2010maxim [7]

Answer:

\frac{KE_{Rotational}}{KE_{Total}} = 0.018

Explanation:

To develop this exercise we proceed to use the kinetic energy equations,

In the end we replace

KE_{Total}=KE_{Translational}+KE_{Rotational}

KE_{Total}=\frac{1}{2}m_{car}+4*\frac{1}{2}*I*(\frac{v}{r})^2

Here

I=\frac{1}{2}m_{wheels}*r^2 meaning the 4 wheels,

So replacing

KE_{Rotational}=4\frac{1}{2}*(\frac{1}{2}m_{wheels}*r^2)*(\frac{v}{r})^2=m*v^2

So,

\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}*v^2}{\frac{1}{2}m_{car}*v^2+m_{wheels}*v^2}

\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}}{\frac{1}{2}m_{car}+m_{wheels}}

\frac{KE_{Rotational}}{KE_{Total}} =  \frac{10}{545+10}

\frac{KE_{Rotational}}{KE_{Total}} = 0.018

3 0
3 years ago
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