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RSB [31]
2 years ago
15

Course hero N4M.6 A board has one end wedged under a rock having a mass of 380 kg and is supported by another rock that touches

the bottom side of the board at a point 85 cm from the end under the rock. The board is 4.5 m long, has a mass of about 22 kg, and projects essentially horizontally out over a river. Is it safe for an adult with a mass of 62 kg to stand at the unsupported end of the board
Physics
1 answer:
aleksandrvk [35]2 years ago
8 0

Answer:

Therefore it is save to carry a 62kg adult

Explanation:

From the question we are told that:

Mass m=380kg

Height of supporting Rock X=85cm

Length of BoardL_r=4.5m

Mass of board M_b=22kg

Mass of adult M_a=62  

Generally the moment of balance about wedge part about  is mathematically given by

N -Q + R = Mg + mg

0.85*N - Mg*2.25 - mg*(2.25 + x) = 0

0.85*N  = + Mg*2.25 + mg*(2.25 + x)

where

N+R=4547

therefore

N = 570.70588 + 1608.3529 + 714.823 x

if N=0 at fallen person

x=3.04m

Therefore it is save to carry a 62kg adult

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Help please:Calculate the mechanical energy of a bird flying at a speed of 10 m / s at an altitude of 15 m. If its mass is 150 g
Sonja [21]

Mass of the bird(m) = 150 g = 0.15 kg

Speed (v) = 10 m/s

Kinetic Energy = \frac{1}{2}m v^{2} = \frac{1}{2} 0.15 (10)^{2} = 7.5 J

Altitude (h) = 15 m

Gravitational Potential Energy = (0.15)(9.81)(15) = 22.0725 J

Mechanical Energy = Kinetic Energy + Potential Energy = 7.5 + 22.0725

= 29.5725 J

4 0
3 years ago
A positively charged object is brought near but not in contact with the top of an uncharged gold leaf electroscope. The experime
Olin [163]

Answer:

The leaves of the electroscope move further apart.

Explanation:

This is what happens; when the positive object is brought near the top, negative charges migrating from the gold leaves to the top. This is because the negative charges in the gold are attracted by the positive charge. Thus, it leaves behind a net positive charge on the leaves, though the scope remains neutral overall. To that effect, the leaves repel each other and move apart. If a finger touches the top of the electroscope at the moment when the positive object remains near the top, it basically grounds the electroscope and thus the net positive charge in the leaves flows to the ground through the finger. However, the positive object continues to "hold" negative charges in place at the top. Ar this moment the gold leaves have lost their net positive charge, so they no longer repel, and they move closer together. If the positive object is moved away, the negative charges at the top are no longer attracted to the top, and they redistribute themselves throughout the electroscope, moving into the leaves and charging them negatively.

Thus, the leaves move apart from each other again and we now have a negatively charged electroscope. If a negatively charged object is now brought close to the top, but without touching, the negative charges already in the electroscope will be repelled down toward the leaves, thereby making them more negative, causing them to repel more, and hence move even further apart.

So, the leaves move further apart.

7 0
3 years ago
3. In the wave models, you imagined adding energy into the wave by
castortr0y [4]

Answer:

by moving your hands up and down your we're creating transverse wave which travel in a right angle shape towards your friend

7 0
3 years ago
a 30 kg child is sitting 2 meters from the center of a merry go round. The coefficients of static and kinetic friction between t
laiz [17]

Answer: A

Explanation:

From the question, the given parameters are given.

Mass M = 30kg

Radius r = 2 m

Coefficient of static friction μ = 0.8

Coefficient of kinetic friction μ = 0.6

Kinetic friction Fk = μ × mg

Fk = 0.6 × 30 × 9.8

Fk = 176.4 N

The force acting on the merry go round is a centripetal force F.

F = MV^2/r

This force must be greater than or equal to the kinetic friction Fk. That is,

F = Fk

F = 176.4

Substitute F , M and r into the centripetal force formula above

176.4 = (30×V^2)/2

Cross multiply

352.8 = 30V^2

V^2 = 352.8/30

V = sqrt (11.76) m/s

V = 5.24 m/s

Therefore, the maximum speed of the merry go round before the child begins to slip is sqrt (12) m/s approximately

4 0
3 years ago
during a typical afternoon thunderstorm in the summer, an area of 66.0 km2 receives 9.57 108 gal of rain in 18 min. how many inc
zheka24 [161]

Answer:

2.16 inch

Explanation:

area under water = 66 km²

= 66 x ( 3280.84 x 12 )² inch²

= 1.023 x 10¹¹ sq inch

volume of rain = 9.57 x 10⁸  gallon = 9.57 x 10⁸ x 231 inch³

= 2.21 x 10¹¹ inch³

If depth of rainfall be t

volume of rain = surface area x depth

= 1.023 x 10¹¹ x t

So ,

1.023 x 10¹¹ x t  = 2.21 x 10¹¹

t = 2.16 inch

5 0
2 years ago
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