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2 years ago

8

What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 1

0 24 kg and radius of the earth is 6.4 × 10 6 m.)

1 answer:

Free_Kalibri [48]2 years ago

7 0

Answer:

Explanation:

just use the gravational force equation which is G x m of earth x m of object divided by r squared (which is radius of earth)

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3 years ago

Read 2 more answers

A driven RLC circuit is being driven by an AC emf source with a maximum current of 2.75 A and maximum voltage of 150 V. The curr

weqwewe [10]

Answer:

(a). Z = 54.54 ohm

(b). R = 36 ohm

(c). The circuit will be Capacitive.

Explanation:

Given data

I = 2.75 A

Voltage = 150 V

$\phi = 0.85$ rad = 48.72°

(a). Impedance of the circuit is given by

$Z = \frac{V}{I}$

$Z = \frac{150}{2.75}$

Z = 54.54 ohm

(b). We know that resistance of the circuit is given by

$R = \frac{Z}{\sqrt{1 + \tan^{2}\phi } }$

Put the values of Z & $\phi$ in above formula we get

$R = \frac{54.54}{\sqrt{1 + \tan^{2} ( \ 48.72) } }$

R = 36 ohm

(c). Since the phase angle is negative so the circuit will be Capacitive.

3 0

3 years ago

If you wish to observe features that are around the size of atoms, say 5.5 × 10^-10 m, with electromagnetic radiation, the radia

Tanya [424]

Answer:

a) 5.5×10^17 Hz

b) visible light

Explanation:

Since the wavelength of the electromagnetic radiation must be about the size of the about itself, this implies that;

λ= 5.5 × 10^-10 m

Since;

c= λ f and c= 3×10^8 ms-1

f= c/λ

f= 3×10^8/5.5 × 10^-10

f= 5.5×10^17 Hz

The electromagnetic wave is visible light

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3 years ago