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3 years ago

15 PTS!!!!

Physics

2 answers:

faltersainse [42]3 years ago

7 0

B. group I think.
:))))))))

Anon25 [30]3 years ago

6 0

I thinks its period because they are on the last column to the right

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In the equation for centripetal force, which expression represents the centripetal acceleration of the object? mv2 StartFraction

Sphinxa [80]

Answer: $\frac{V^{2}}{r}$

Explanation:

According to Newton's 2nd Law of motion the force $F$ is proportional to the mass $Fm$ and acceleration $a$ :

$F=m.a$ (1)

On the other hand, the equation for the Centripetal force is:

$F=\frac{mV^{2}}{r}$ (2)

Where:

$V$ is the velocity

$r$ is the radius of the circular motion

Making (1) and (2) equal:

$m.a=\frac{mV^{2}}{r}$ (3)

Hence:

$a=\frac{V^{2}}{r}$ This is the expression for the centripetal acceleration

It should be noted, this acceleration is directed toward the center of the circumference of the circular motion (that's why it's called centripetal acceleration).

3 0

3 years ago

Read 2 more answers

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics

spayn [35]

Answer:

$v_0 = 3.53~{\rm m/s}$

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

$x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}$

For the y-direction gives

$v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t$

Combining both equation yields the y_component of the final velocity

$v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}$

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

$\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}$

6 0

3 years ago

For safety reasons, a worker’s eye travel time in a certain operation must be separated from the manual elements that follow. Th

iogann1982 [59]

Answer:

The answer is 12.67 TMU

Explanation:

Recall that,

worker’s eyes travel distance must be = 20 in.

The perpendicular distance from her eyes to the line of travel is =24 in

What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element = ?

Now,

We solve for the given problem.

Eye travel is = 15.2 * T/D

=15.2 * 20 in/24 in

so,

= 12.67 TMU

Therefore, the MTM -1 of normal time that should be allowed for the eye travel element is = 12.67 TMU

7 0

3 years ago

Use the SI prefixes in Table 3 of this chapter to convert these hypothetical units of measure into appropriate quantities: a. 10

8090 [49]

A. 10 rations = 1 deca-ration.

b. 2000 mockingbirds = 2 x 10³ = 2 kilo-mockingbirds.

c. 10⁻⁵ phones = 1 micro-phones.

d. 10⁻⁹ goats = 1 nano-goats.

e. 1018 miners = 1.018 x 10³ = 1.018 kilo-miners.

4 0

3 years ago

You are a visitor aboard the New International Space Station, which is in a circular orbit around the Earth with an orbital spee

Alchen [17]

Answer:

The minimum total speed is 11.2km/s

Explanation:

We are been asked to find the escape velocity.

Escape velocity is defined as the minimum initial velocity that will take a body(projectile)away above the surface of a planet(earth) when it's projected vertically upwards.

The formula to calculate the escape velocity is Ve = √2gR

For the earth g = 9.8m/s2 , R = 6.4*10^6

Substituting into the equation Ve = √2*9.8*6.4*10^6 = 11.2*10^3m/s

=11.2km/s

5 0

3 years ago

Read 2 more answers