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3 years ago

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A gull is flying horizontally 10.80 m above the ground at 6.00 m/s. The bird is carrying a clam in its beak and plans to crack t

he clamshell by dropping it on some rocks below. (Ignore air resistance.) With what speed relative to the gull does the clam smash into the rocks?

1 answer:

UkoKoshka [18]3 years ago

8 0

Answer:

v_y = 14.55 m/s

Explanation:

given,

height at which gull is flying = 10.80 m

speed of the gull = 6 m/s

acceleration due to gravity = 9.8 m/s²

Relative to the seagull, the x-speed is 0,

because the seagull has the same x-speed.

Only the y-speed counts:

$v_y^2 = u^2 + 2 g h$

$v_y^2 = 0^2 + 2 g h$

$v_y = \sqrt{2gh}$

$v_y = \sqrt(2\times 9.8 \times 10.8)$

$v_y = \sqrt(211.68)$

v_y = 14.55 m/s

hence, the speed at which the clam smash the rock is v_y = 14.55 m/s

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Answer:

The 10-meter long rod of an SR-71 airplane expands 0.02 meters (2 centimeters) when plane flies at 3 times the speed of sound.

Explanation:

From Physics we get that expansion of the rod portion is found by this formula:

$\Delta l = \alpha\cdot l_{o}\cdot (T_{f}-T_{o})$ (Eq. 1)

Where:

$\Delta l$ - Expansion of the rod portion, measured in meters.

$\alpha$ - Linear coefficient of expansion for titanium, measured in $\frac{1}{^{\circ}C}$ .

$l_{o}$ - Initial length of the rod portion, measured in meters.

$T_{o}$ - Initial temperature of the rod portion, measured in Celsius.

$T_{f}$ - Final temperature of the rod portion, measured in Celsius.

If we know that $\alpha = 5\times 10^{-6}\,\frac{1}{^{\circ}C}$ , $l_{o} = 10\,m$ , $T_{o} = 0\,^{\circ}C$ and $T_{f} = 400\,^{\circ}C$ , the expansion experimented by the rod portion is:

$\Delta l = \left(5\times 10^{-6}\,\frac{1}{^{\circ}C} \right)\cdot (10\,m)\cdot (400\,^{\circ}C-0\,^{\circ}C)$

$\Delta l = 0.02\,m$

The 10-meter long rod of an SR-71 airplane expands 0.02 meters (2 centimeters) when plane flies at 3 times the speed of sound.

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3 years ago

On a trip to the Colorado Rockies, you notice that when the freeway goes steeply down a hill, there are emergency exits every fe

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Answer:

The maximum speed that the truck can have and still be stopped by the 100m road is the speed that it can go and be stopped at exactly 100m. Since there is no friction, this problem is similar to a projectile problem. You can think of the problem as being a ball tossed into the air except here you know the highest point and you are looking for the initial velocity needed to reach that point. Also, in this problem, because there is an incline, the value of the acceleration due to gravity is not simply g; it is the component of gravity acting parallel to the incline. Since we are working parallel to the plane, also keep in mind that the highest point is given in the problem as 100m. Solving for the initial velocity needed to have the truck stop after 100m, you should find that the maximum velocity the truck can have and be stopped by the road is 18.5 m/s.

Explanation:

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Which other element is most likely to be a non-reactive gas

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Answer:

C

Explanation:

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A concave mirror produces four times magnified image of ab object placed at 10 cm in front of the mirror. find the position of i

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Answer:

40 cm

Explanation:

We are given that

Magnification of concave mirror, m=4

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We have to find the position of image.

We know that

Magnification of concave mirror, m= $-\frac{v}{u}$

Using the formula

$4=-\frac{v}{-10}$

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$v=4\times 10$

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Hence, the image is formed at distance 40 cm behind the mirror.

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3 years ago

A heavier mass m1 and a lighter mass m2 are 17.0 cm apart and experience a gravitational force of attraction that is 9.70 10-9 N

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Answer:

The value of m1 is 4.2 kg and value of m2 is 1 kg.

Explanation:

Note:- The value of force is assumed to be 9.70*10^-9 N.

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where F is the force, G is the universal gravitational constant, m1 and m2 are masses and r is the distance between the masses.

Then the product of the masses is,

m1m2= Fr^2/G

Given F=9.70*10^-9 N, r is 17.0 cm which is equal to 0.17 m.

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m1m2= 4.20 kg^2

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The sum of masses is m1+m2=5.20 kg

4.20/m2+m2=5.20

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After solving the above quadratic equation, mass m2 has values 4.2 kg and 1 kg.

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