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3 years ago

Select the correct answer.

Physics

2 answers:

wolverine [178]3 years ago

5 0

The answer is C in this question.

KiRa [710]3 years ago

3 0

Answer:

C. 6.3 × 102 kilogram meters/second

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Subduction occurs because the sea floor is: more dense than the continental floor less dense than the continental floor more buo

svetoff [14.1K]

A)more dense than the continental floor

7 0

3 years ago

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Which of the following examples illustrates static friction in action?

lilavasa [31]

I think its d. but im not sure

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2 years ago

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Radar uses radio waves of a wavelength of 2.9 m . The time interval for one radiation pulse is 100 times larger than the time of

Mandarinka [93]

Answer:

145 m

Explanation:

Given:

Wavelength (λ) = 2.9 m

we know,

c = f × λ

where,

c = speed of light ; 3.0 x 10⁸ m/s

f = frequency

thus,

$f=\frac{c}{\lambda}$

substituting the values in the equation we get,

$f=\frac{3.0\times 10^8 m/s}{2.9m}$

f = 1.03 x 10⁸Hz

Now,

The time period (T) = $\frac{1}{f}$

T = $\frac{1}{1.03\times 10^8}$ = 9.6 x 10⁻⁹ seconds

thus,

the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s

Time between pulses = (100T×10) = 9.6 x 10⁻⁶ s

Now,

For radar to detect the object the pulse must hit the object and come back to the detector.

Hence, the shortest distance will be half the distance travelled by the pulse back and forth.

Distance = speed × time = 3 x 10^8 m/s × 9.6 x 10⁻⁷ s) = 290 m {Back and forth}

Thus, the minimum distance to target = $\frac{290}{2}$ = 145 m

6 0

2 years ago

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of

wariber [46]

Answer:

For left = 0 N/C

For right = 0 N/C

At middle = $-7.6836 * 10^{-11} \vec{i}$ N/C

Explanation:

Given data :-

б = $6.8 * 10^{-22}$ C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

$E = \frac{\sigma }{2\varepsilon }$

1) To the left of the plate

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

2) To the right of them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

3) Between them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(-\vec{i})$ = $(\frac{\sigma }{\varepsilon })(-\vec{i})$ = $-\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} } \vec{i}$ = $-7.6836 * 10^{-11} \vec{i}$ N/C

5 0

2 years ago

Ship A is located 4.1 km north and 2.3 km east of ship B. Ship A has a velocity of 22 km/h toward the south and Ship B has a vel

castortr0y [4]

Answer:

a) x component = -31.25 km/hr

b) y component = 46.64 km/hr

Explanation:

Given data:

A position is 4km north and 2.5 km east to B

Ship A velocity = 22 km/hr

ship B velocity = 40 km/hr

A velocity wrt to velocity of B

$\vec{V_{AB}} =\vec{V_A} - \vec{V_B}$

$\vec{V_A} = 22 km/hr$

$\vec{V_B} = 40 cos38\hat{i} + 40sin 38 \hat{j}$

$= 31.52\hat{i} + 24.62 \hat{j}$

putting respective value to get velocity of A with respect to B

$\vec{V_{AB}} = -22 \hat {j} - (31.52\hat{i} + 24.62 \hat{j})$

$\vec{V_{AB}} = -31.52\hat{i} - 46.62\hat{j}$

a) x component = -31.25 km/hr

b) y component = 46.64 km/hr

6 0

3 years ago