Ask question

Ask question

All categories

4 years ago

6

In which of the following situations would be the government had been most likely to become involved in union protests?

1 answer:

andreyandreev [35.5K]4 years ago

6 0

C. A violent or destructive protest

You might be interested in

A baseball is hit at ground level. The ball reaches its maximum height above ground level 3.0 s after being hit. Then 2.5 s afte

vaieri [72.5K]

Answer:

Part a)

$H = 44.1 m$

Part b)

$y = 13.48 m$

Part c)

$d = 8.86 m$

Explanation:

Part a)

As we know that ball will reach at maximum height at

t = 3 s

now we will have

$t = \frac{v sin\theta}{g}$

now we have

$3 = \frac{vsin\theta}{9.8}$

$v sin\theta = 29.4 m/s$

Now maximum height above ground is given as

$H = \frac{v^2sin^2\theta}{2g}$

$H = \frac{29.4^2}{2(9.8)}$

$H = 44.1 m$

Part b)

Height of the fence is given as

$y = (vsin\theta) t - \frac{1}{2}gt^2$

$y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)$

$y = 13.48 m$

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

$x = v_x t$

$97.5 = v_x (5.5)$

$v_x = 17.72 m/s$

now total time of flight is given as

$T = 3 + 3 = 6 s$

so range is given as

$R = v_x T$

$R = (17.72)(6)$

$R = 106.4 m$

so the distance from the fence is given as

$d = 106.4 - 97.5$

$d = 8.86 m$

7 0

3 years ago

A nasa spacecraft measures the rate r of at which atmospheric pressure on mars decreases with altitude. the result at a certain

Lesechka [4]

Answer: $4.21 \times 10^{-10} J/cm^4$

$1 kPa= 10^3 Pa$

$1 km=10^5 cm$

$1kPa/km=0.01 Pa/cm$

$1kPa/km=10^{-8} J/cm^4$

$\Rightarrow r= 0.0421 kPa/km= 0.0421 kPa/km \times \frac{10^{-8} J/cm^4}{1 kPa/km}= 0.0421 \times 10^{-8}J/cm^4=4.21 \times 10^{-10} J/cm^4$

3 0

3 years ago

Read 2 more answers

A 10.0 kg weather rocket generates a thrust of 230 NN . The rocket, pointing upward, is clamped to the top of a vertical spring.

blondinia [14]

Answer: 0.2m

Explanation: Firstly only the Rocket's Weight Compress the spring which can be found by

$F_r=M_r*g\\F_r=10*9.81\\F_r=98.1N$

According to Hooks Law

$F_r=k*x\\x=F_r/k\\x=98.1/480\\x=0.2m$

The part b and c of this question is done in the attachment

7 0

3 years ago

A gyroscope flywheel of radius 3.25 cm is accelerated from rest at 11.6 rad/s2 until its angular speed is 1820 rev/min. (a) What

kati45 [8]

Explanation:

The given data is as follows.

radius (r) = 3.25 cm, $\alpha = 11.6 rad/s^{2}$

Now, we will calculate the tangential acceleration as follows.

$a_{tangential} = \alpha \times r$

Putting the given values into the above formula as follows.

$a_{tangential} = \alpha \times r$

= $11.6 rad/s^{2} \times 3.25 cm$

= 37.7 $rad cm/s^{2}$

Thus, we can conclude that the tangential acceleration of a point on the rim of the flywheel during this spin-up process is 37.7 $rad cm/s^{2}$ .

6 0

3 years ago

An electric heating element is connected to a 110v circuit and a current of 3.2 a is flowing through the element. How much energ

Kisachek [45]

2.580 Wh.... .........

7 0

3 years ago

Read 2 more answers