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3 years ago

Write 8.54 x 103 in standard notation a. 854,000 b. 85,400 c. 8,540 d. 854

1 answer:

5 0

B. 8540, (assuming you meant 8.54 x 10 to the power 3) you move the decimal point 3 places to the right, if the decimal point can't pass any more numbers add zeros. If it confuses you to think of it this way do 10 cubed first, i.e. 10 x 10 x10 = 1000 and then do 8.54 x 1000 if thats easier for you.

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When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees

nexus9112 [7]

(a) The man comes to a halt in 1.99 ms = 1.99 × 10⁻³ s, so his average acceleration slowing him from 6.33 m/s to a rest is

a = (6.33 m/s - 0) / (1.99 × 10⁻³ s) ≈ 3180 m/s²

so that the average net force on the man during the landing is

F = (70.8 kg) a ≈ 225,000 N

i.e. with magnitude 225,000 N.

(b) With knees bent, the man has an average acceleration of

a = (6.33 m/s - 0) / (0.147 s) ≈ 43.1 m/s²

and hence an average net force of

F = (70.8 kg) a ≈ 3050 N

(c) The net force on the man is

∑ F = n - w = m a

where

n = magnitude of the normal force, i.e. the force of the ground pushing up on the man

w = the man's weight, m g ≈ 694 N

m = the man's mass, 70.8 kg

g = mag. of the acceleration due to gravity, 9.80 m/s²

a = the man's acceleration

Using the acceleration in part (b), we have

n = m g + m a = m (g + a)

n = (70.8 kg) (9.80 m/s² + 43.1 m/s²) ≈ 3740 N

4 0

3 years ago

How does electric circuit wor

zaharov [31]

Electricity passes through it

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4 years ago

An electron emitted from a filament is travelling at 1.5 x 105 m/s when it enters an acceleration of an electron gun in a televi

Crank

Answer:

The acceleration of the electron is 1.457 x 10¹⁵ m/s².

Explanation:

Given;

initial velocity of the emitted electron, u = 1.5 x 10⁵ m/s

distance traveled by the electron, d = 0.01 m

final velocity of the electron, v = 5.4 x 10⁶ m/s

The acceleration of the electron is calculated as;

v² = u² + 2ad

(5.4 x 10⁶)² = (1.5 x 10⁵)² + (2 x 0.01)a

(2 x 0.01)a = (5.4 x 10⁶)² - (1.5 x 10⁵)²

(2 x 0.01)a = 2.91375 x 10¹³

$a = \frac{2.91375 \ \times \ 10^{13}}{2 \ \times \ 0.01} \\\\a = 1.457 \ \times \ 10^{15} \ m/s^2$

Therefore, the acceleration of the electron is 1.457 x 10¹⁵ m/s².

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3 years ago

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AnnZ [28]

After doing some research on Wiki, Google, and Easyscienceforkids, I have come to a conclusion that:

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3 years ago