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2 years ago

12

Which materials produce greater or fewer vibrations on a roller coaster?

1 answer:

mixer [17]2 years ago

3 0

Steel materials produce less vibrations whilst less dense and more malleable parts produce more vibrations, e.g. the wooden part of the coaster itself.

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A 5 kg book falls from a shelf. If it lands at a speed of 5.7 m/s, from what height did it fall?

Pachacha [2.7K]

Answer:

57m

Explanation:

speed = 5.7m/s

or 57/10 m/s

so definitely the time taken will be 10 secs

so speed = distance/time taken

5.7= distance/10

distance = 5.7× 10

Distance = 57m

7 0

3 years ago

Momentum is most similar to which other physics concept?

Dmitrij [34]

Answer:

Among force and impulse, There is a lesser calculation for impulse. Also, it has same dimensions as momentum. Hence that is most closely related.

3 0

2 years ago

In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he

lana [24]

Answer:8.28 km

Explanation:

Given

First it drifts $45^{\circ}$ 2.5 km

$r_1=2.5cos45 i+2.5sin45 j$

Secondly it drifts $60^{\circ}$ 4.70 km

$r_{12}=4.7cos60 i-4.7sin60 j$

After that it drifted along east direction 5.1 km

$r_{23}=5.1 i$

After that it drifts $55^{\circ}$ 7.2 km

$r_{34}=-7.2cos55 i-7.2sin55 j$

After that it drifts $5^{\circ}$ 2.8 km

$r_{54}=-2.8cos5 i+2.8sin5 j$

$r_{5O}$ = $\left [ 2.5cos45+4.7cos60+5.1-7.2cos55-2.8cos5\right ]\hat{i}$ + $\left [ 2.5sin45-4.7sin60-7.2sin55+2.8sin5\right ]\hat{j}$

$r_{5O}=2.299\hat{i}-7.95\hat{j}$

$|r_{5O}|=8.28 km$

for direction

$tan\theta =\frac{7.95}{2.299}=3.4580$

$\theta =73.87^{\circ}$ south of east

7 0

3 years ago

Read 2 more answers

During the contraction of the heart, 65 cm3 blood is ejected from the left ventricle into the aorta with a velocity of approxima

vova2212 [387]

Answer:

The value $F = 0.1396 \ N$

Explanation:

From the question we are told that

The volume blood ejected is $V = 65 \ cm^3 = 65*10^{-6} \ m^3$

The velocity of the blood ejected is $v = 103 \ cm/s = \frac{103}{100} = 1.03 \ m/s$

The density of blood is $\rho = 1060 \ kg/m^3$

The heart beat is $R = 59 \ bpm(beats \ per \ minute) = \frac{59}{60}= 0.9833\ bps$

The average force exerted by the blood on the wall of the aorta is mathematically represented as

$F = 2 * \rho * V * R * v$

=> $F = 2 * 1060 * 65*10^{-6} * 0.9833 * 1.03$

=> $F = 0.1396 \ N$

8 0

3 years ago

Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum th

Strike441 [17]

Answer:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$

Explanation:

The electric field equation of a electromagnetic wave is given by:

$E=E_{max}(kx-\omega t)$ (1)

E(max) is the maximun value of E, it means the amplitude of the wave.
k is the wave number
ω is the angular frequency

We know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).

By definition:

$k=\frac{2\pi}{\lambda}$

$k=8.98*10^{6} [rad/m]$

And the relation between λ and f is:

$c=\lambda f$

$f=\frac{c}{\lambda}$

$f=\frac{3*10^{8}}{700*10^{-9}}$

$f=4.28*10^{14}$

The angular frequency equation is:

$\omega=2\pi f$

$\omega=2\pi*4.28*10^{14}$

$\omega=2.69*10^{15} [rad/s]$

Therefore, the E equation, suing (1), will be:

$E=3.5(8.98*10^{6}x-2.69*10^{15}t)$ (2)

For the magnetic field we have the next equation:

$B=B_{max}(kx-\omega t)$ (3)

It is the same as E. Here we just need to find B(max).

We can use this equation:

$E_{max}=cB_{max}$

$B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}$

$B_{max}=1.17*10^{-8}T$

Putting this in (3), finally we will have:

$B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)$ (4)

I hope it helps you!

8 0

3 years ago