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Elena-2011 [213]

3 years ago

8

This diagram shows a baseball’s motion. Which statement is best supported by the diagram? Kinetic energy becomes potential energ

y between points R and S. Maximum potential energy is at point P. Maximum potential energy is at point R. Potential energy becomes kinetic energy between points P and Q.

2 answers:

LenKa [72]3 years ago

5 0

<h2>C. Maximum potential energy is at point R</h2>

The baseball is in maximum height in the air, meaning it has maximum potential energy, becuase it is about become kinetic energy, or it is about to move, which is what potential means.

Ivahew [28]3 years ago

4 0

The correct answer is:

C. Maximum potential energy is at point R

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A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an

maksim [4K]

Answer:

The value is $\alpha = 0.002 Np/m$

Explanation:

From the question we are told that

The first amplitude of the wave is $E_{max}1 = 98.02 \ V/m$

The first depth is $D_1 = 10 \ m$

The second amplitude is $E_{max}2 = 81.87 \ (V/m)$

The second depth is $D_2 = 100 \ m$

Generally from the spatial wave equation we have

$v(x) = Ae^{-\alpha d}cos(\beta x + \phi_o)$

=> $\frac{v(x)}{v(x)} =\frac{ Ae^{-\alpha d}cos(\beta x + \phi_o)}{ Ae^{-\alpha d}cos(\beta x + \phi_o)}$

So considering the ratio of the equation for the two depth

$\frac{A}{A_S} = \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}$

=> $\frac{98.02}{81.87} = \frac{e^{-10 \alpha }}{e^{-100 \alpha }}$

=> $\alpha = \frac{0.18}{90}$

=> $\alpha = 0.002 Np/m$

4 0

4 years ago

In what way has technology used aboard the International Space Station benefitted humans back on Earth?

FrozenT [24]

Answer:

Robotic arms used aboard the ISS are now used in delicate surgeries on Earth.

Explanation:

The ISS allows users to address hardware product development gaps, advanced manufacturing, and emerging technology proliferation. Microgravity-enabled material production capabilities and advanced manufacturing facilities are demonstrating scientific and commercial merit for Earth benefit

8 0

3 years ago

Read 2 more answers

Assume that the stopping distance of a van varies directly with the square of the speed. A van traveling 40 miles per hour can s

Daniel [21]

Answer:

d = 100.8 ft

Explanation:

As we know that initial speed of the van is 40 miles then the stopping distance is given as 70 feet

here we know that

$v_f^2 - v_i^2 = 2 ad$

so here we have

$0^2 - 40^2 = 2 a (70 feet)$

now again if the speed is increased to 48 mph then let say the stopping distance is "d"

so we will have

$0^2 - 48^2 = 2 a (d)$

now divide the above two equations

$\frac{40^2}{48^2} = \frac{70 feet}{d}$

$d = 100.8 ft$

4 0

4 years ago

A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike

Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g= $9.50\times 10^{3} kg$

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

$mu+ MU=mv+ MV$

Substitute the values

$9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V$

$3.61i=2.375j+0.150V$

$3.61 i-2.375j=0.150V$

$V=\frac{1}{0.150}(3.61 i-2.375j)$

$V=24.07i-15.83j$

Magnitude of velocity of stone

= $\sqrt{(24.07)^2+(-15.83)^2}$

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

$\theta=tan^{-1}(\frac{y}{x})$

= $tan^{-1}(\frac{-15.83}{24.07})$

$\theta=tan^{-1}(-0.657)$

=33.3 degree below the horizontal

(b)

Initial kinetic energy

$K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2$

$K_i=685.9 J$

Final kinetic energy

$K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2$

= $\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2$

$K_f=359.12 J$

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

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3 years ago

When the biker is at the top of the ramp, he has a speed of 10 m/s at

oksian1 [2.3K]

Answer:

25000

Explanation:

100 x 25 x 10 = 25000

8 0

3 years ago