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3 years ago

Which BEST describes the speed and velocity of the graph?

Physics

2 answers:

Neko [114]3 years ago

6 0

I think the answer is A (sorry if it isn't).

Kazeer [188]3 years ago

6 0

This is tricky, but the correct answer is A.

Choice-D is also true. We can't COMPLETELY determine velocity
from the graph, because an essential part of velocity is the DIRECTION
of motion, and the graph doesn't tell us anything about the direction.

But even though we can't COMPLETELY determine velocity from
the graph, we DO know that velocity is changing, because the speed
is definitely changing, and speed is the other big part of velocity.

The graph shows clearly that speed is changing. That right there
tells us that velocity is also changing, and we can pick choice-A.

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PLEASE ANSWER AS QUICK AS POSSIBLE ANSWER WORTH 20 POINTS

Slav-nsk [51]

Answer:

The density of balsa wood is about 170 kg/m^3.

Explanation:

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5 0

4 years ago

The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for 6.00 s, at which time it

8_murik_8 [283]

Answer:

16.035 revolutions

Explanation:

Part 1:

t = 6 s, f0 = 0 , f = 5 rps,

Let the number of revolutions be n1.

Use first equation of motion for rotational motion

w = w0 + α t

2 x 3.14 x 5 = 0 + α x 6

α = 5.233 rad/s^2

Let θ1 be the angle turned.

Use second equation of motion for rotational motion

θ1 = w0 t + 12 x α x t^2

θ1 = 0 + 0.5 x 5.233 x 6 x 6 = 94.194 rad

n1 = θ1 / 2π = 94.194 / 2 x 3.14 = 15 revolutions

Part 2:

f0 = 5 rps, f = 0, t = 13 s

Let the number of revolutions be n2.

Use first equation of motion for rotational motion

w = w0 + α t

0 = 2 x 3.14 x 5 + α x 13

α = - 2.415 rad/s^2

Let θ2 be the angle turned.

Use third equation of motion for rotational motion

w^2 = w0^2 + 2 x α x θ2

0 = 2 x 3.14 x 5 - 2 x 2.415 x θ2

θ2 = 6.5 rad

n2 = θ2 / 2π = 6.5 / 2 x 3.14 = 1.035 revolutions

total revolutions n = n1 + n2 = 15 + 1.035 = 16.035 revolutions

3 0

3 years ago

(1 point) Find a linearly independent set of vectors that spans the same subspace of R3R3 as that spanned by the vectors ⎡⎣⎢3−1−

Natalka [10]

Answer:

Explanation:

$A=\left[\begin{array}{ccc}3&-9&-3\\-1&2&0\\-2&3&-1\end{array}\right] \\\\R_2\rightarrow 3R_2+R_1,R_3\rightarrow 3R_3+2R_1\\\\=\left[\begin{array}{ccc}3&-9&-3\\0&-3&-3\\0&-9&-9\end{array}\right] \\\\R_3\rightarrow 3R_3-9R_2\\\\=\left[\begin{array}{ccc}3&-9&-3\\0&-3&-3\\0&0&0\end{array}\right]$

This is the row echelon form of A. This means that only two of the vectors in our set are linearly independent. In other words, the first two vectors alone will span the same subspace of $R^4$ as all three vectors.

Therefore, the linearly independent spanning set for the subspace is

$\left[\begin{array}{ccc}3\\-1\\-2\end{array}\right] \left[\begin{array}{ccc}-9\\2\\3\end{array}\right] \left[\begin{array}{ccc}3\\0\\-1\end{array}\right]$