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valentinak56 [21]

3 years ago

5

Label the following descriptions with (M) for Mass or (W) for Weight. measured with a spring scale measured in Newtons _

___measured with a triple beam balance ____changes if gravity changes ____measured in grams or kilograms ____does not change if gravity changes

1 answer:

Valentin [98]3 years ago

8 0

1. mass
2. weight
3. weight
4. weight
5. mass
6. mass

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The respiratory system

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what do scientific models predict will happen to Earths oceans if current trends continue unmitigated?

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3 years ago

In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression,

eduard

(a) 4.06 cm

In a simple harmonic motion, the displacement is written as

$x(t) = A cos (\omega t + \phi)$ (1)

where

A is the amplitude

$\omega$ is the angular frequency

$\phi$ is the phase

t is the time

The displacement of the piston in the problem is given by

$x(t) = (5.00 cm) cos (5t+\frac{\pi}{5})$ (2)

By putting t=0 in the formula, we find the position of the piston at t=0:

$x(0) = (5.00 cm) cos (0+\frac{\pi}{5})=4.06 cm$

(b) -14.69 cm/s

In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:

$v(t) = x'(t) = -\omega A sin (\omega t + \phi)$ (3)

Differentiating eq.(2), we find

$v(t) = x'(t) = -(5 rad/s)(5.00 cm) sin (5t+\frac{\pi}{5})=-(25.0 cm/s) sin (5t+\frac{\pi}{5})$

And substituting t=0, we find the velocity at time t=0:

$v(0)=-(25.00 cm/s) sin (0+\frac{\pi}{5})=-14.69 cm/s$

(c) -101.13 cm/s^2

In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:

$a(t) = v'(t) = -\omega^2 A cos (\omega t + \phi)$

Differentiating eq.(3), we find

$a(t) = v'(t) = -(5 rad/s)(25.00 cm/s) cos (5t+\frac{\pi}{5})=-(125.0 cm/s^2) cos (5t+\frac{\pi}{5})$

And substituting t=0, we find the acceleration at time t=0:

$a(0)=-(125.00 cm/s) cos (0+\frac{\pi}{5})=-101.13 cm/s^2$

(d) 5.00 cm, 1.26 s

By comparing eq.(1) and (2), we notice immediately that the amplitude is

A = 5.00 cm

For the period, we have to start from the relationship between angular frequency and period T:

$\omega=\frac{2\pi}{T}$

Using $\omega = 5.0 rad/s$ and solving for T, we find

$T=\frac{2\pi}{5 rad/s}=1.26 s$

4 0

3 years ago

An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi

alexdok [17]

Actual displacement that he required to move

$d_1 = 5.4 km$ towards North

Displacement that he moved due to snow is

$d_2 = 8.1 km$ at 47 degree North of East

now in vector component form we can say

$d_1 = 5.4 \hat j$

$d_2 = 8.1 cos47 \hat i + 8.1 sin47 \hat j$

$d_2 = 5.52 \hat i + 5.92 \hat j$

now the displacement that is more required to reach the destination is given as

$d = d_1 - d_2$

$d = 5.4\hat j - (5.52 \hat i + 5.92\hat j)$

$d = -5.52 \hat i - 0.52 \hat j$

so the magnitude of the displacement is given as

$d = \sqrt{5.52^2 + 0.52^2}$

$d = 5.54 km$

its direction is given as

$\theta = tan^{-1}\frac{0.52}{5.52} = 5.38 degree$

so it is 5.54 km towards 5.38 degree North of West.

4 0

3 years ago