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Naddika [18.5K]
3 years ago
9

Which of the following would be an example of mechanical energy?

Physics
2 answers:
Mkey [24]3 years ago
6 0

Answer:

b

Explanation:

Wittaler [7]3 years ago
4 0

Answer:

B

Explanation:

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To protect her new two-wheeler, Iroda Bike
goldfiish [28.3K]

Answer:

The length of chain she is allowed is 1.169 ft

Explanation:

The given parameters are;

The linear density of the chain = 0.83 lb/ft

The weight limit of the chain she wants = 1.4 lb

The formula for linear density = Weight/length

Therefore, in order to keep the chain below 1.4 lb, we have;

Linear density = Weight/length

Therefore;

The maximum length she wants = Weight/(Linear density)

Which gives;

The maximum length she wants = 1.4 lb/(0.83 lb/ft) =1.169 ft

Therefore;

The length of chain she is allowed = 1.169 ft.

6 0
3 years ago
Two point charges 3q and −8q (with q > 0) are at x = 0 and x = L, respectively, and free to move. A third charge is placed so
riadik2000 [5.3K]

Answer:

Explanation:

The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.

So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.

force on it due to rest of the charges will be equal and opposite so

k3q Q / x² =k 8q Q / (L+x)²

8x² = 3 (L+x)²

2√2 x = √3 (L+x)

2√2 x - √3 x = √3 L

x(2√2 - √3 ) = √3 L

x = √3 L / (2√2 - √3 )

Let us consider the balancing force on 3q

force on it due to -Q and -8q will be equal

kQ . 3q / x² = k3q  8q / L²

Q = 8q  (x² / L²)

so charge required = - 8q  (x² / L²)

and its distance from x on negative x side = √3 L / (2√2 - √3 )

3 0
3 years ago
Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as
MAXImum [283]

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m

or in kilometers,

h_2 = 47 km

the first altitude in kilometers is

h_1 = 155 km

so the difference between the two altitudes is

\Delta h = 155 km - 47 km = 108 km

8 0
3 years ago
What conclusions can you draw about an object that either has an OVERALL negative charger OR an OVERALL positive charge?
sveticcg [70]

Answer:

The object is also positively charged because same or alike charges repel

Explanation:

6 0
3 years ago
A 5.75 × 107 kg battle ship originally at rest fires a 1100-kg artillery shell horizontally with a velocity of 550 m/s. show ans
prohojiy [21]

Answer:

-0.01052 m/s

Explanation:

M = mass of ship = 5.75\times 10^7\ kg

m = mass of shell = 1100 kg

v = velocity of shell = 550 m/s

u = recoil velocity of ship

As linear momentum is conserved

(M - m)u=-mv\\\Rightarrow u=-\frac{mv}{M - m}\\\Rightarrow u=-\frac{1100\times 550}{5.75\times 10^7+1100}\\\Rightarrow u=-0.01052\ m/s

The recoil velocity of the ship taking the firing direction to be the positive direction is -0.01052 m/s

6 0
3 years ago
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