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2 years ago

3.4.3 Quiz: Buoyancy

Physics

1 answer:

lara31 [8.8K]2 years ago

8 0

The word that belongs after "it's" is "volume".

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2 (a) A plane from airport A flies 280 km to the east to airport B. The plane then travelled north to airport C, 190 km away. (i

Mekhanik [1.2K]

Answer:

See below

Explanation:

See attached diagram

280 km east then 190 km north

Use Pythagorean theorem to find the resultant displacement

d^2 = 280^2 + 190^2

d = 338.4 km

Angle will be arctan ( 190/280) = 34.16 °

4 0

1 year ago

Pleaseeee give answer ASAP. Major test

sergiy2304 [10]

Answer:

B, A, C

Explanation:

3 0

2 years ago

Trivia Time!! At what velocity does light move?

Maslowich

Explanation:

Light Light from a moving source travels at 300,000 km/sec (186,000 miles/sec).

Note:

The Speed of Light Is Constant.

7 0

2 years ago

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A 50 Kg box sits at rest on a 30 degree ramp where the coef of static friction is 0.5773. If your push was directed at an angle

emmasim [6.3K]

Given data:

m= 50 Kg,

W= m×g = 50 × 9.81 = 490.5 N

ramp angle (α) = 30 degrees,

coefficient of friction (μs) = 0.5773,

Push at an angle (Θ) = 40 degrees,

Determine: Push to get box move up (P)=?

From the figure,

Resolving the forces along the plane

W sinα + μs.R = P cos Θ --------------------- (i)

Resolving the forces perpendicular to the inclined plane

W cosα = R+Psin Θ => R= W cosα - Psin Θ -------------- (ii)

Solving (i) and (ii) and keeping μs = tan Φ, Φ = Θ 

Pmin = W sin( α +Θ )

= W[ sin α.Cos Θ + cos α.sin Θ]

= 490.5 [ (sin 30.cos40) + (cos30.sin 40)]

= 460.9 N

Minimum push required to move the box up the ramp is 460.9 N

7 0

3 years ago

A 4.5 g coin sliding to the right at 23.8 cm/s makes an elastic head-on collision with a 13.5 g coin that is initially at rest.

Airida [17]

Answer:

a) $v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )$ , b) $\Delta K = 9.559\times 10^{-5}\,J$

Explanation:

a) The final velocity of the 13.5 g coin is found by the Principle of Momentum Conservation:

$(4.5\times 10^{-3}\,kg)\cdot (23.8\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot (0\,\frac{m}{s} ) = (4.5\times 10^{-3}\,kg)\cdot (-11.9\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot v$

The final velocity is:

$v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )$

b) The change in the kinetic energy of the 13.5 g coin is:

$\Delta K = \frac{1}{2}\cdot (13.5\times 10^{-3}\,kg)\cdot \left[(11.9\times 10^{-2}\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}\right]$

$\Delta K = 9.559\times 10^{-5}\,J$

4 0

3 years ago

Read 2 more answers