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2 years ago

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A dynamite blast blows a heavy rock straight up from the ground with a launch velocity of 160ft/sec. a) Write the height functio

n.

1 answer:

olga2289 [7]2 years ago

8 0

Answer:

y = 160t - 16t²

Explanation:

Using y = ut -1/2gt² where y = height of heavy rock above the ground, u = initial launch velocity of rock = 160 ft/s, g = acceleration due to gravity = 32 ft/s² and t = time.

So, substituting the values of the variables into y, we have

y = (160 ft/s)t -1/2 × 32 ft/s² × t²

y = (160 ft/s)t - (16 ft/s²)t²

y = 160t - 16t²

So, the height function is y = 160t - 16t²

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Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2

Bess [88]

Answer:

The acceleration of $M_2$ is $a = 0.7156 m/s^2$

Explanation:

From the question we are told that

The mass of first block is $M_1 = 2.25 \ kg$

The angle of inclination of first block is $\theta _1 = 43.5^o$

The coefficient of kinetic friction of the first block is $\mu_1 = 0.205$

The mass of the second block is $M_2 = 5.45 \ kg$

The angle of inclination of the second block is $\theta _2 = 32.5^o$

The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

The acceleration of $M_1 \ and\ M_2$ are same

The force acting on the mass $M_1$ is mathematically represented as

$F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as

$F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

$M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

$F_1 = F_2$

So

$T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

$a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

substituting values $a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}$

=> $a = 0.7156 m/s^2$

3 0

3 years ago

If the original pressure is 5 atm and original volume is 100 L, and the new volume is 20 L, what is the new pressure?

Nana76 [90]

P1V1=P2V2
5*100=P2*20
500=20P
P=25

7 0

3 years ago

The 5.5 million vinyl long-playing (LP) records sold in the United States per year pales in comparison with the 1.26 billion dig

miv72 [106K]

Answer:

decline

Explanation:

Based on the scenario being described within the question it can be said that these types of firms are in the decline stage of the product life cycle. This stage refers to when a product has already passed it's peak potential and sales begin to decline until production is ultimately halted and the product dies off. Which is exactly what is happening to the LP's since everyone has moved on to digital downloads.

4 0

3 years ago

A balloon is filled to a volume of 7.00*10^2 mL at a temperature of 20.0°C. The balloon is then cooled at constant pressure to a

wolverine [178]

The final volume of the gas is 238.9 mL

Explanation:

We can solve this problem by using Charle's law, which states that for a gas kept at constant pressure, the volume of the gas (V) is proportional to its absolute temperature (T):

$\frac{V}{T}=const.$

Which can be also re-written as

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where

$V_1, V_2$ are the initial and final volumes of the gas

$T_1, T_2$ are the initial and final temperature of the gas

For the gas in the balloon in this problem, we have:

$V_1 = 7.00\cdot 10^2 mL = 700 mL$ is the initial volume

$T_1=20.0^{\circ}C+273=293 K$ is the initial absolute temperature

$V_2$ is the final volume

$T_2 = 1.00\cdot 10^2 K = 100 K$ is the final temperature

Solving for $V_2$ ,

$V_2 = \frac{V_1 T_2}{T_1}=\frac{(700)(100)}{293}=238.9 mL$

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6 0

3 years ago

A quarter is flipped from a height of 1.45 m above the ground. How much time will it take to reach the ground if the person flip

Artemon [7]

It will take the quarter 0.151 seconds to reach the ground.

<u>Given the following data:</u>

Height = 1.45 meters

Initial velocity = 10.32 m/s

We know that acceleration due to gravity (a) for an object is equal to 9.8 meter per seconds square.

To find how much time it will take the quarter to reach the ground, we would use the second equation of motion.

Mathematically, the second equation of motion is given by the formula;

$S = ut + \frac{1}{2} at^2$

Where:

S is the height or distance covered.

u is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting the values into the formula, we have;

$1.45 = 10.32(t) + \frac{1}{2} (9.8)t^2\\\\1.45 = 10.32t + 4.9t^2\\\\4.9t^2 + 10.32t - 1.45 = 0$

The standard form of a quadratic equation is:

$ax^2 + bx + c = 0$

a = 4.9, b = 10.32 and c = 1.45

We would solve the above quadratic equation by using the quadratic equation formula;

$x = \frac{-b\; \pm \;\sqrt{b^2 - 4ac}}{2a}$

Substituting the values, we have;

$t = \frac{-10.32\; \pm \;\sqrt{10.32^2\; - \;4(4.9)(1.45)}}{2(4.9)}\\\\t = \frac{-10.32\; \pm \;\sqrt{106.5024\; - \;28.42}}{9.8}\\\\t = \frac{-10.32\; \pm \;\sqrt{78.0824}}{9.8}\\\\t = \frac{-10.32\; \pm \;8.84}{9.8}\\\\t = \frac{-10.32\; + \;8.84}{9.8}\\\\t = \frac{1.48}{9.8}$

<em>Time, t = 0.151 seconds.</em>

Therefore, it will take the quarter 0.151 seconds to reach the ground.

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3 0

2 years ago