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Assoli18 [71]
2 years ago
13

A solid wooden cube, 30cm on each side can be totally submerged in water if it is pushed downward with a force of 54N. What is t

he density of the wood?
Physics
1 answer:
Stels [109]2 years ago
4 0

Answer:

the density of the wooden cube is 204.1 kg/m³

Explanation:

Given;

applied force, F = 54 N

length of each side of the solid wooden cube, L = 30 cm = 0.3 m

mass of the wooden cube is calculated as;

F = mg

m = F/g

m = 54/9.8

m = 5.51 kg

The volume of the wooden cube is calculated as;

V = L³

V = (0.3)³

V = 0.027 m³

The density of the wooden cube is calculated as;

ρ = m/V

ρ = (5.51 kg) / (0.027 m³)

ρ = 204.1 kg/m³

Therefore, the density of the wooden cube is 204.1 kg/m³

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PLEASE SOLVE QUICKLY!!!<br> Solve for A, B, and C from graph
hram777 [196]

A = 59.35cm

B = 196.56g

C = 74.65g

<u>Explanation:</u>

We know,

x = \frac{L}{\frac{W}{F} +1}

and L = x+y

1.

Total length, L = 100cm

Weight of Beam, W = 71.8g

Center of mass, x = 49.2cm

Added weight, F = 240g

Position weight placed from fulcrum, y = ?

L-y = \frac{L}{\frac{W}{F}+1 } \\100 - y = \frac{100}{\frac{71.8}{49.2}+1 } \\100 - y = \frac{100}{1.46+1}\\\\100 - y = \frac{100}{2.46} \\100-y = 40.65\\\\y = 59.35cm

Therefore, position weight placed from fulcrum is 59.35cm

2.

Total length, L = 100cm

Center of mass, x = 47.8 cm

Added weight, F = 180g

Position weight placed from fulcrum, y = 12.4cm

Weight of Beam, W = ?

47.8 = \frac{100}{\frac{W}{180}+1 }\\\47.8  = \frac{100}{\frac{W+180}{180} } \\\\47.8 = \frac{100 X 180}{W+180}\\ \\47.8W + 47.8 X 180 = 18000\\47.8W  = 18000 - 8604\\W = \frac{9396}{47.8}\\ W = 196.56g

Therefore, weight of the beam is 196.56g

3.

Total length, L = 100cm

Center of mass, x = 50.8 cm

Position weight placed from fulcrum, y = 9.8cm

Weight of Beam, W = 72.3g

Added weight, F = ?

50.8 = \frac{100}{\frac{72.3}{F}+1 }\\\ 50.8  = \frac{100}{\frac{72.3+F}{F} } \\\\50.8 = \frac{100 X F}{72.3+F}\\ \\50.8 X 72.3 + 50.8 X F = 100F\\\\3672.84 = 100F-50.8F\\3672.84 = 49.2F\\F = 74.65g

Therefore, Added weight F is 74.65g

A = 59.35cm

B = 196.56g

C = 74.65g

4 0
3 years ago
How long would light take to travel from one hill to another
zhuklara [117]

186282 miles. Hope it helps

7 0
3 years ago
Long sand ridges oriented at right angles to the wind are called _____ dunes
S_A_V [24]

The answer is:

Transverse dune

The explanation:

Transverse dune : is abundant barchan dunes It  may merge into barchanoid ridges, which then grade into linear .

The transverse dunes is called that because they lie transverse, or across, the wind direction, with the wind blowing perpendicular to the ridge crest.

It is large, very asymmetrical, elongated dune lying at right angles 90° to the prevailing wind direction.

Transverse dunes have a gently sloping windward side and a steeply sloping leeward side.

They general form in areas of sparse vegetation and abundant sand are transverse dunes.

3 0
3 years ago
Read 2 more answers
what happens to the current in a circuit if the resitance of the components in the circuit is increased​
Anit [1.1K]

Answer:

The current decreases.

Explanation:

Current and resistance are inversely proportional. The equation connecting current, resistance and voltage is V = IR, where V is voltage, I is current and R is resistance.

Rearranging this equation, you get:

I = \frac{V}{R}

and

R = \frac{V}{I}

If the value of voltage in both equations remains constant, and the value of R decreases, the value of I will increase. Conversely, if in the second equation R = \frac{V}{I} , the value of V remains constant the value of I decreases, then the value of R, resistance will increase.

Thus, it can be seen that the current will decrease as resistance increases and vice versa.

7 0
3 years ago
An 8.00 kg mass moving east at 15.4 m/s on a frictionless horizontal surface collides with a 10.0 kg object that is initially at
andrew-mc [135]

Answer:

9.3m/s

Explanation:

Based on the law of conservation of momentum

Sum of momentum before collision = sum of momentum after collision

m1u1 +m2u2 = m1v1+m2v2

m1 = 8kg

u1 = 15.4m/s

m2 = 10kg

u2 = 0m/s(at rest)

v1 = 3.9m/s

Required

v2.

Substitute

8(15.4)+10(0) = 8(3.9)+10v2

123.2=31.2+10v2

123.2-31.2 = 10v2

92 = 10v2

v2 = 92/10

v2 = 9.2m/s

Hence the velocity of the 10.0 kg object after the collision is 9.2m/s

6 0
3 years ago
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