Removing the dimensional prefix and express in scientific notation ( include unit)7.4nm
6.78 x 10^8 MHz
5.36 x 10^6 kV
1500mA

Calculate the following numbers into scientific notation (include units)

7800nm + 95pm =
2500nm - 7pm =
(65x10^4m) x (4.5x10^-6s-1)=
(24x10^5m) / (2x10^-8s) =

convert the following to the prefix commonly used (use units)
55 000 000 Hz
4.8x10^4V
0.0068s
0.00000000234m

Answers

Answer 1

Answer: 7.4nm = 7.4 x 10⁻⁹m
6.78 x 10^8 MHz = 6.78 x 10¹⁴ Hz
5.36 x 10^6 kV = 5.26 x 10⁹V
1500mA=1.5 A

7800nm + 95pm = 7800.095 x 10⁻⁹ m
2500nm - 7pm = 2499.993 x 10⁻⁹ m
(65x10^4m) x (4.5x10^-6s-1) = 2.925 m/s
(24x10^5m) / (2x10^-8s) = 1.2 x 10¹⁴ m/s (that's 400,000 'c' !)

55 000 000 Hz = 55 MHz
4.8x10^4V = 48 KV
0.0068s = 6.8 mS
0.00000000234m = 2.34 nm

Answer 2

Answer: 7.4 x 10^9 m
6.78 x 10^11 Hz
5.36 x 10^9 V
1.5 x 10^3 mA

1nm = 10^9m
and 1pm=10^11m

55MHz
48kV
234pm

PLEASE MARK BEST

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Answers

Answer:

$(Ie)/(lm) = 1.10*10^(-3)$

Explanation:

GIVEN DATA:

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Explanation:

Given

Initial angular speed of engine( $\omega _E$ )=8325 rpm

Final angular speed of engine( $\omega _E_f$ )=12125 rpm

Initial angular speed of Motorcycle( $\omega _M$ )=0 rpm

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$I_E\omega _E+I_M\omega _M=I_E\omega _E_f+I_M\omega _M_f$

$I_E\omega _E+=I_E\omega _E_f+I_M\omega _M_f$

$I_E\left ( \omega _E-\omega _E_f\right )=I_M\omega _M_f$

$(I_E)/(I_M)=(\omega _M_f)/(\omega _E-\omega _E_f)$

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