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Vladimir79 [104]

2 years ago

10

Lila is a track and field athlete. She must complete four laps around a circular track. The track itself is a 400 meter track an

d it took her 6 minutes total to complete. Which best describes her speed and velocity? (There are 60 seconds in 1 minute.) A. Her speed is 4.4 m/s, and her velocity is 0 m/s. B.Her speed is 1.1 m/s, and her velocity is 0 m/s. C.Her speed is 0 m/s, and her velocity is 2400 m/s. D. Her speed is 4.4 m/s, and her velocity is 4.4 m/s.

1 answer:

trapecia [35]2 years ago

6 0

Answer: A. Her speed is 4.4 m/s, and her velocity is 0 m/s.

Explanation: i took the test on edgenuity

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While the negatively charged rod is near the disk without touching it, a hand briefly touches the end of the post. Then the nega

Paraphin [41]

Answer:

that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.

Explanation:

Let us carefully analyze the situation, when the bar is facing the index post a load of equal magnitude, but opposite sign on its surface, these two charges are in balance; When the hand touches the pole, it creates a path to the ground where the charges that were induced on the pole can be balanced with the charge coming from the ground, leaving a zero charge on the pole.

Now if the hand is removed, there can be no exchange of charges with the earth. When the bar is removed, the induced loads are redistributed in the post, but the excess loads that came from the earth that have the same value and are of a sign opposite to the induced ones remain, you want to sign that they are of the same sign as the charges of the bar.

In summary, after the process, the post has a load of equal magnitude and sign (negative) that of the bar.

If we assume that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.

4 0

2 years ago

The space shuttle releases a satellite into a circular orbit 630 km above the Earth.

solniwko [45]

Answer:

7,539 m/s

Explanation:

Let's use this equation to find the gravitational acceleration of this space shuttle:

$\displaystyle g=\frac{GM}{r^2}$

We know that G is the gravitational constant: 6.67 * 10^(-11) Nm²/kg².

M is the mass of the planet, which is Earth in this case: 5.972 * 10^24 kg.

r is the distance from the center of Earth to the space shuttle: radius of Earth (6.3781 * 10^6 m) + distance above the Earth (630 km → 630,000 m).

Plug these values into the equation:

$\displaystyle g=\frac{(6.67\cdot 10^-^1^1 \ Nm^2kg^-^2)(5.972\cdot 10^2^4 \ kg)}{[(6.3781\cdot 10^6 \ m)+(630000 \ m)]^2}$

Remove units to make the equation easier to read.

$\displaystyle g=\frac{(6.67\cdot 10^-^1^1 )(5.972\cdot 10^2^4 )}{[(6.3781\cdot 10^6)+(630000 )]^2}$

Multiply the numerator out.

$\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(6.3781\cdot 10^6)+(630000 )]^2}$

Add the terms in the denominator.

$\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(7008100)]^2}$

Simplify this equation.

$\displaystyle g=8.11045189 \ \frac{m}{s^2}$

The acceleration due to gravity g = 8.11045189 m/s². Now we use the equation for acceleration for an object in circular motion which contains v and r.

$\displaystyle a = \frac{v^2}{r}$

a = g, v is the velocity that the space shuttle should be moving (what we are trying to solve for), and r is the radius we had in the previous equation when solving for g.

Plug these values into the equation and solve for v.

$\displaystyle 8.11045189 \ \frac{m}{s^2} = \frac{v^2}{7008100 \ m}$

Remove units to make the equation easier to read.

$\displaystyle 8.11045189 = \frac{v^2}{7008100}$

Multiply both sides by 7,008,100.

$56838857.89=v^2$

Take the square root of both sides.

$v=7539.154985$

The shuttle should be moving at a velocity of about 7,539 m/s when it is released into the circular orbit above Earth.

5 0

2 years ago

A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left

Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

Length of the massless beam; $L = 4.00m$

Distance of support from the left end; $x = 3.00m$

First mass; $m1 = 31.3 kg$

Distance of beam from the left end( m₁ is attached to ); $x_1 = ?$

Second mass; $m_2 = 61.7 kg$

Distance of beam from the right of the support( m₂ is attached to ); $x_1 = 0.273m$

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, $m_1g( x-x_1) = m_2gx_2$

we divide both sides by $g$

$m_1( x-x_1) = m_2x_2$

Next, we make $x_1$ , the subject of the formula

$x_1 = x - [ \frac{m_2x_2}{m_1} ]$

We substitute in our given values

$x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]$

$x_1 = 3.00m - 0.538m$

$x_1 = 2.46m$

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

6 0

2 years ago

An 888.0 kg elevator is moving downward with a velocity of 0.800 m/s. It decelerates uniformly and comes to a stop in a distance

bagirrra123 [75]

Answer:

The value of tension on the cable T = 1065.6 N

Explanation:

Mass = 888 kg

Initial velocity ( u )= 0.8 $\frac{m}{sec}$

Final velocity ( V ) = 0

Distance traveled before come to rest = 0.2667 m

Now use third law of motion $V^{2}$ = $u^{2}$ - 2 a s

Put all the values in above formula we get,

⇒ 0 = $0.8^{2}$ - 2 × a ×0.2667

⇒ a = 1.2 $\frac{m}{sec^{2} }$

This is the deceleration of the box.

Tension in the cable is given by T = F = m × a

Put all the values in above formula we get,

T = 888 × 1.2

T = 1065.6 N

This is the value of tension on the cable.

5 0

2 years ago

A weightlifter raises a 50kg weight to a height of 2m in 2 minutes. What was the power spent by the weightlifter?

belka [17]

Answer:

117.72kW

Explanation:

Given data

Mass m= 50kg

height x = 2m

time taken = 2 minutes= 129 seconds

let us find the work done

WD= force * distance

WD= mgx

WD= 50*9.81*2

WD= 981 Joules

Let us find the power

Power= work * time

Power= 981*120

Power= 117720

Power= 117.72 kW

Hence the power spent is 117.72kW

8 0

2 years ago