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timurjin [86]
3 years ago
7

The player who is serving continues to serve until he/she or their team commits a fault.

Physics
1 answer:
Roman55 [17]3 years ago
7 0

Answer:

true

Explanation:

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How much force would be required to accelerate a 2kg object at a rate of 15m/ sec?
Anna007 [38]
Im saying a or b but i pick B
5 0
3 years ago
Isaac Newton believed that God created the heavens and the Earth.<br><br> True<br> False
Lostsunrise [7]

Answer:

false dhdhehdudhdh

Explanation:

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7 0
4 years ago
A small rubber wheel is used to drive a large pottery wheel. The two wheels are mounted so that their circular edges touch. The
Goshia [24]

Answer:

0.629\ \text{rad/s}^2 counterclockwise

9.98\ \text{s}

Explanation:

r_1 = Small drive wheel radius = 2.2 cm

\alpha_1 = Angular acceleration of the small drive wheel = 8\ \text{rad/s}^2

r_2 = Radius of pottery wheel = 28 cm

\alpha_2 = Angular acceleration of pottery wheel

As the linear acceleration of the system is conserved we have

r_1\alpha_1=r_2\alpha_2\\\Rightarrow \alpha_2=\dfrac{r_1\alpha_1}{r_2}\\\Rightarrow \alpha_2=\dfrac{2.2\times 8}{28}\\\Rightarrow \alpha_2=0.629\ \text{rad/s}^2

The angular acceleration of the pottery wheel is 0.629\ \text{rad/s}^2.

The rubber drive wheel is rotating in clockwise direction so the pottery wheel will rotate counterclockwise.

\omega_i = Initial angular velocity = 0

\omega_f = Final angular velocity = 60\ \text{rpm}\times \dfrac{2\pi}{60}=6.28\ \text{rad/s}

t = Time taken

From the kinematic equations of linear motion we have

\omega_f=\omega_i+\alpha_2t\\\Rightarrow t=\dfrac{\omega_f-\omega_i}{\alpha_2}\\\Rightarrow t=\dfrac{6.28-0}{0.629}\\\Rightarrow t=9.98\ \text{s}

The time it takes the pottery wheel to reach the required speed is 9.98\ \text{s}

4 0
3 years ago
Two cars are facing each other. Car A is at rest while car B is moving toward car A with a constant velocity of 20 m/s. When car
lapo4ka [179]

Answer:

Let's define t = 0s (the initial time) as the moment when Car A starts moving.

Let's find the movement equations of each car.

A:

We know that Car A accelerations with a constant acceleration of 5m/s^2

Then the acceleration equation is:

A_a(t)  = 5m/s^2

To get the velocity, we integrate over time:

V_a(t) = (5m/s^2)*t + V_0

Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:

V_a(t) = (5m/s^2)*t

To get the position equation we integrate again over time:

P_a(t) = 0.5*(5m/s^2)*t^2 + P_0

Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:

P_a(t) = 0.5*(5m/s^2)*t^2

Now let's find the equations for car B.

We know that Car B does not accelerate, then it has a constant velocity given by:

V_b(t) =20m/s

To get the position equation, we can integrate:

P_b(t) = (20m/s)*t + P_0

This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:

P_b(t) = (20m/s)*t + 100m

Now we can answer this:

1) The two cars will meet when their position equations are equal, so we must have:

P_a(t) = P_b(t)

We can solve this for t.

0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0

This is a quadratic equation, the solutions are given by the Bhaskara's formula:

t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)}  }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}

We only care for the positive solution, which is:

t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s

Car A reaches Car B after 11.48 seconds.

2) How far does car A travel before the two cars meet?

Here we only need to evaluate the position equation for Car A in t = 11.48s:

P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m

3) What is the velocity of car B when the two cars meet?

Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s

4)  What is the velocity of car A when the two cars meet?

Here we need to evaluate the velocity equation for Car A at t = 11.48s

V_a(t) = (5m/s^2)*11.48s = 57.4 m/s

7 0
3 years ago
What is your weighted grade, if you have a 100% on homework and a 88% on tests. {homework is worth 40% of grade and tests are wo
Sphinxa [80]
You would have a 92.8, or a 93.
8 0
3 years ago
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