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Bad White [126]
3 years ago
10

The Asian Silver Carp is an invasive species of fish which has infested a number of American rivers. The disturbance of passing

boats has been known to make these fish leap large distances out of the water. One such fish leaps with an initial velocity at the water's surface of 11.0 m/s at an angle of 34.0° above the horizontal. How high above the water's surface, in meters, does the fish reach?
Physics
1 answer:
Zarrin [17]3 years ago
4 0

Answer:

1.93 m

Explanation:

Initial velocity of fish u = 11 m /s

Angle of jump = 34 degree.

Vertical component of its velocity = u sin34 = 11 x.5592 = 6.15 m /s

Considering its motion in vertical direction ,

u = 6.15 m/s

g = - 9.8 m /s

maximum height attained = h

From the formula

v² = u² - 2gh

0 = 6.15x 6.15 - 2 x 9.8 h

h = (6.15 x 6.15 )/ 2 x 9.8

= 1.93 m .

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The resistance and the magnitude of the current depend on the path that the current takes. The drawing shows three situations in
maria [59]

Answer: (a). Resistance = 0.4286ohms and Current (I) = 7A

(b). Resistance (R) = 0.027 ohms and Current (I) = 111.1A

(c). Resistance (R) = 0.1071 ohms and Current (I) = 28A

Explanation:

From the question, given that;

ρ = 1.5*10-2ῼ.m

Lo = 7cm = 0.07m

V = 3V

From the formula R = ρL/A, where A is the area of cross section, L is the length of material and ρ is the resistivity.

(A)

L = 4Lo and A = 2Lo*Lo

R = ρL/A

R = ρ4Lo/(2Lo*Lo)

R = 2ρ /Lo = 2*1.5*10-2/0.07

R = 0.4286 ῼ

From this the current becomes;

I = V/R = 3/ 0.4286 = 6.99 = 7A

(B)

L = Lo and A = 4Lo * 2Lo

R = ρL/A

R = ρLo/ (4Lo*2Lo) after eliminating Lo from both sides we get,

R = ρ/8Lo = 1.5*10-2 / 8*0.07

R = 0.027

Current (I) = V/R = 3/0.027 = 111.1A

(C)

L = 2Lo and A = Lo * 4Lo

R = ρL/A

R = ρ2Lo/ (Lo*4Lo) eliminating Lo from both sides we get,

R = ρ/2Lo = 1.5*10-2 / 2*0.07 = 0.1071

The current becomes;

I = V/R = 3/0.1071 = 28A

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3 years ago
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On a cold winter day, a penny (mass 2.50 g) and a nickel (mass 5.00 g) are lying on the smooth (frictionless) surface of a froze
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3 years ago
the gravitational force between two objects is 1600 and what will be the gravitational force between the objects if the distance
Xelga [282]

I believe this is what you have to do:

The force between a mass M and a point mass m is represented by

F = G\frac{Mm}{r^{2} }

So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁

So F₁ = G(Mm/r^2)

Now the distance has doubled so lets account for this in F₂:

F₂ = G(Mm/(2r)^2)

Now square the 2 that gives you four and we can pull that out in front to give

F₂ = \frac{1}{4} G(Mm/r^2)

Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations

now we see that:

F₂ = \frac{1}{4} F₁

So the second force will be 0.25 (1/4) x 1600 or 400 N.



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3 years ago
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<span>If Jack is filing married-filing-separate he would report $76,000 gross income as head of household.</span>
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