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3 years ago

5

The number of employees for a certain company has been decreasing each year by 4%. If the company currently has 650 employees an

d this rate continues, find
the number of employees in 20 years.
The number of employees in 20 years will be approximately
(Round to the nearest whole number as needed)
HELP PLEASE!

1 answer:

nika2105 [10]3 years ago

8 0

The answer would probably be 649.96

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A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci

ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity ( $v$ ), measured in meters per second, by using this kinematic equation:

$v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s}$ (1)

Where:

$a$ - Acceleration, measured in meters per square second.

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$ - Initial velocity, measured in meters per second.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $a = 2.35\,\frac{m}{s^{2}}$ and $\Delta s = 595\,m$ , the final velocity of the rocket is:

$v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}$

$v\approx 52.882\,\frac{m}{s}$

The time associated with this launch ( $t$ ), measured in seconds, is:

$t = \frac{v-v_{o}}{a}$

$t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }$

$t = 22.503\,s$

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

$v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o})$ (2)

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

If we know that $v_{o} = 52.882\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s^{2}}$ and $s_{o} = 595\,m$ , then the maximum height reached by the rocket is:

$v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})$

$s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}$

$s = 737.577\,m$

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

$s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$ (2)

Where:

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $s_{o} = 595\,m$ , $v_{o} = 52.882\,\frac{m}{s}$ , $s = 0\,m$ and $a = -9.807\,\frac{m}{s^{2}}$ , then the time needed by the rocket is:

$0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}$

$-4.904\cdot t^{2}+52.882\cdot t +595 = 0$

Then, we solve this polynomial by Quadratic Formula:

$t_{1}\approx 17.655\,s$ , $t_{2} \approx -6.872\,s$

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0

3 years ago

Two power lines run parallel for a distance of 222 m and are separated by a distance of 40.0 cm. if the current in each of the t

earnstyle [38]

1) Magnitude of the force:

The magnetic field generated by a current-carrying wire is
$B= \frac{\mu_0I}{2 \pi r}$
where
$\mu_0$ is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated

Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance
$r=40.0 cm=0.40 m$ ,
which is the distance at which the other wire is located:
$B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T$

Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
$F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N$

2) direction of the force:
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)

Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive

A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.

6 0

3 years ago

Two charged objects separated by some distance attract each other. If the charges on both objects are doubled with no change in

Serggg [28]

Answer:

(a) The force between them quadruples

Explanation:

According to coulomb's law, initial force between the two charged objects is given as;

$F_1=\frac{Kq_1q_2}{r^2}$

where;

k is coulomb's constant

q₁ is the charge on the first object

q₂ is the charge on the second object

r is the distance between the two objects

When the charges on both objects are doubled, then;

q₁ = 2q₁

q₂ = 2q₂

Force between the two charged objects will become

$F_2 = \frac{K2q_12q_2}{r^2} = \frac{4Kq_1q_2}{r^2} = 4(\frac{Kq_1q_2}{r^2}) = 4F_1$

Therefore, the force between them quadruples

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3 years ago

What is the lowest gauge pressure that the average person can achieve inside his lungs? express your answer with the appropriate

eimsori [14]

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spin [16.1K]

The correct answer is A

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