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2 years ago

You should practice all new exercises without weight resistance first? A. True B. False

Physics

2 answers:

trapecia [35]2 years ago

6 0

Answer:

Explanation:

Kaylis [27]2 years ago

4 0

Answer:

B. False

Sorry if wrong. :D

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Physics is explicitly involved in studying which of these activities?

Dmitrij [34]

Hey Friends
The answer to this question would be C
Hope i helped
~Katie

8 0

3 years ago

Read 2 more answers

The Cartesian coordinate of a point in the xy plane are (x,y)=(-3.50,-2.50)m. Find the poler coordinate of this point

Masja [62]

Answer:

The polar coordinate of $P(x,y) = (-3.50\,m,-2.50\,m)$ is $P (r,\theta) = (4.301\,m, 215.538^{\circ})$ .

Explanation:

Given a point in rectangular form, that is $P(x,y) = (x,y)$ , its polar form is defined by:

$P(x,y) = (r,\theta)$ (1)

Where:

$r$ - Norm, measured in meters.

$\theta$ - Direction, measured in sexagesimal degrees.

The norm of the point is determined by Pythagorean Theorem:

$r = \sqrt{x^{2}+y^{2}}$ (2)

And direction is calculated by following trigonometric relation:

$\theta = \tan^{-1} \frac{y}{x}$ (3)

If we know that $x = -3.50\,m$ and $y = -2.50\,m$ , then the components of coordinates in polar form is:

$r = \sqrt{(-3.50\,m)^{2}+(-2.50\,m)^{2}}$

$r \approx 4.301\,m$

Since $x < 0\,m$ and $y < 0\,m$ , direction is located at 3rd Quadrant. Given that tangent function has a period of 180º, we find direction by using this formula:

$\theta = 180^{\circ}+\tan^{-1} \left(\frac{-2.50\,m}{-3.50\,m} \right)$

$\theta \approx 215.538^{\circ}$

The polar coordinate of $P(x,y) = (-3.50\,m,-2.50\,m)$ is $P (r,\theta) = (4.301\,m, 215.538^{\circ})$ .

5 0

2 years ago

A boy is pedaling his bicycle at a velocity of 0.20km/ minute . How far will he travel in 2.5 hours

fredd [130]

30km. 24 the first two hours and 6 the half hour

4 0

3 years ago

Read 2 more answers

What is the process of change in a crust due to the motion of hot material underneath?(answer)?

frutty [35]

Im not exactly sure but I think the answer is techtonic plates collide

5 0

3 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du

ELEN [110]

With constant angular acceleration $\alpha$ , the disk achieves an angular velocity $\omega$ at time $t$ according to

$\omega=\alpha t$

and angular displacement $\theta$ according to

$\theta=\dfrac12\alpha t^2$

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

$21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}$

b. Under constant acceleration, the average angular velocity is equivalent to

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2$

where $\omega_f$ and $\omega_i$ are the final and initial angular velocities, respectively. Then

$\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}$

c. After 1.00 s, the disk has instantaneous angular velocity

$\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}$

d. During the next 1.00 s, the disk will start moving with the angular velocity $\omega_0$ equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle $\theta$ according to

$\theta=\omega_0t+\dfrac12\alpha t^2$

which would be equal to

$\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}$

5 0

3 years ago