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16 days ago

Killer whales are known to reach 32 ft in length and have a mass of over 8,000 kg. They are also very quick, able to accelerate

up to 30 mi/h in a matter of seconds. Disregarding the considerable drag force of the water, calculate the average power a killer whale named Shamu with mass 8.00 x 103 kg would need to generate to reach a speed of 12 m/s in 6 s

Physics

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A careful photographic survey of Jupiter’s moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to

Sav [3153]

Answer:

529.15 m/s

Explanation:

h = Highest point = 70000 m

g = Gravitational acceleration = 2 m/s²

m = Sulfur's mass

Since both potential and kinetic energies are conserved

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 2\times 70000}\\\Rightarrow v=529.15\ m/s$

The velocity at which the liquid sulfur exited the volcano is 529.15 m/s

7 0

3 months ago

A car traveling at 70 mph70 mph down the interstate collides with a bug trying to cross the highway. Which of the following stat

Sav [3153]

The force exerted by the car on the bug is identical to the force the bug applies back on the car.

5 0

1 month ago

13–82. the 8-kg sack slides down the smooth ramp. if it has a speed of 1.5 m> s when y = 0.2 m, determine the normal reaction

Softa [3030]

The second question necessitates a figure to provide an answer. For the initial question
The acceleration of the sack is
1.5² - 0² = 2a(0.2)
a = 5.63 m/s²
The ramp's reaction force is
F = 8 kg (5.63 m/s²)
F = 45 N
Differentiate the kinematic equation with respect to time to find the velocity's rate of increase.

4 0

2 months ago

Two vectors A⃗ and B⃗ are at right angles to each other. The magnitude of A⃗ is 4.00. What should be the length of B⃗ so that th

serg [3582]

Answer:

B= $\sqrt{65}$ ≅8.06

Explanation:

Applying the Pythagorean theorem:

$C^{2}$ = $A^{2}$ + $B^{2}$

Here, C denotes the hypotenuse length, while A and B signify the lengths of the other two sides of the triangle. We can calculate B's length knowing the hypotenuse is 9 and A is 4.

$9^{2}$ = $4^{2}$ + $B^{2}$

81= 16+ $B^{2}$

81-16= $B^{2}$

B= $\sqrt{65}$ ≅8.06

8 0

3 months ago

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 V and 16.2 V/m

inna [3103]

Answer:

A. 30.7 cm

B. $1.7*10^{-10}C$

C. The electric field points outward from the charge

Explanation:

$because, E=\frac{v}{d} \\\\d= \frac{4.98}{16.2}\\\\ d = 0.307m\\\\d = 30.7 cm$

Using Gauss's law

$EA = \frac{Q}{e}\\\\ where, e = pertittivity. space= 8.85* 10^{-12} Fm^{-1} \\\\A = surface. area. with.radius 0.307m\\Q= eEA = (8.85*10^{-12})(16.2)(4\pi)(0.307)^{2}\\\\= 1.7*10^{-10}C$

C. The electric field emanates away from the point of charge if the charge is positive.

3 0

3 months ago