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Agata [3.3K]
3 years ago
12

6

Physics
1 answer:
asambeis [7]3 years ago
6 0

Answer:

TAJUK

Explanation:

Sebab saya suka makan ayam goreng, esok saya nak pesan daripada kedai pak abu, terima kasih bosku

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ELEN [110]
C is correct.  The work-force relation is given by W=F·d, where F is force vector, and d is the displacement vector.  The dot is the dot product, which is a measure of how parallel the two vectors are.  It can be restated as the product of two vector magnitudes times the cosine of the angle between them.  Therefore work is a scalar, not a vector, since the dot product returns a scalar.  
W=Fdcos(\theta)
3 0
4 years ago
A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten
babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

8 0
4 years ago
A light spring of constant 176 N/m rests vertically on the bottom of a large beaker of water.
irakobra [83]

Answer:

14.0 cm

Explanation:

Draw a free body diagram of the block.  There are three forces: weight force mg pulling down, elastic force k∆L pulling down, and buoyancy ρVg pushing up.

Sum of forces in the y direction:

∑F = ma

ρVg − mg − k∆L = 0

(1000 kg/m³) (4.63 kg / 648 kg/m³) (9.8 m/s²) − (4.63 kg) (9.8 m/s²) − (176 N/m) ∆L = 0

∆L = 0.140 m

∆L = 14.0 cm

8 0
3 years ago
Which is the force that slows moving things down?
BlackZzzverrR [31]

i think centripetal Force

7 0
4 years ago
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Which of the following statements is true of space exploration?
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You don't have a following space exploration
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