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gtnhenbr [62]
3 years ago
5

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?
Physics
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 gauge pressure in Pascals inside a honey droplet is 416 Pa

Explanation:

Given;

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radius of the honey droplet, R = 0.05 cm = 0.0005 m

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Apply Laplace's law for a spherical membrane with two surfaces

Gauge pressure =  P₁ - P₀ = 2 (2γ / r)

Where;

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Gauge pressure = 4γ / r

Gauge pressure = 4 (0.052) / (0.0005)

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The magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is th
Darya [45]

Answer:

1.92 x 10⁻¹²J

Explanation:

The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e

F = ma

Where;

a = \frac{v^2}{r}             [v = linear velocity, r = radius of circular path]

=> F = m\frac{v^2}{r}           ------------(i)

We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;

F = q v B sin θ       ----------(ii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = the angle between the velocity and the magnetic field.

Combining equations (i) and (ii) gives

m\frac{v^2}{r} = q v B sin θ           [θ = 90° since the proton is orbiting at the maximum orbital radius]

=> m\frac{v^2}{r} = q v B sin 90°

=> m\frac{v^2}{r} = q v B

Divide both side by v;

=> m\frac{v}{r} = qB

Make v subject of the formula

v = \frac{qBr}{m}

From the question;

B = 1.25T

m = mass of proton = 1.67 x 10⁻²⁷kg

r = 0.40m

q = charge of a proton = 1.6 x 10⁻¹⁹C

Substitute these values into equation(iii) as follows;

v = \frac{(1.6*10^{-19})(1.25)(0.4)}{(1.67*10^{-27})}

v = 4.79 x 10⁷m/s

Now, the kinetic energy, K, is given by;

K = \frac{1}{2}mv²

m = mass of proton

v = velocity of the proton as calculated above

K = \frac{1}{2}(1.67*10^{-27} * (4.79 * 10^7)^2 )

K = 1.92 x 10⁻¹²J

The kinetic energy is 1.92 x 10⁻¹²J

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sammy [17]

Answer:

2.06 m/s

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From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.

Momentum before collision

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Momentum after collision

The momentum after collision will be given by (9+27)*0.9=32.4

Relating the two then 9v+13.5=32.4

9v=18.5

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Rounded off, v is approximately 2.06 m/s

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