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1 year ago

A 50 kg rocket generates 990 N of thrust. What will be its acceleration if it is launched straight up?

Physics

1 answer:

Ilia_Sergeevich [38]1 year ago

4 0

Answer:

The acceleration of the rocket is 10 m/s².

Explanation:

Let the acceleration of the rocket be $a$ m/s².

Given:

Mass of the rocket is, $m=50\ kg$

Thrust force acting upward is, $F_{th}=990\ N$

Acceleration due to gravity is, $g=9.8\ m/s^2$

Now, force acting in the downward direction is due to the weight of the rocket and is given as:

$W=mg=50\times 9.8=490\ N$

Now, net force acting on the rocket in upward direction is given as:

$F_{net}=F_{th}-W\\F_{net}=990-490=500\ N$

Therefore, from Newton's second law, net force acting on the rocket is equal to the product of mass and acceleration.

$F_{net}=ma\\500=50a\\a=\frac{500}{50}=10\ m/s^2$

Therefore, the acceleration of the rocket is 10 m/s².

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the torque requirement of 97 Newton meter.

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1 year ago

A 1.0-m-long copper wire of diameter 0.10 cm carries a current of 50.0 A to the east. Suppose we apply to this wire a magnetic f

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Answer:

The classification of that same issue in question is characterized below.

Explanation:

The given values are:

Current, I = 50.0 A

Diameter, d = 0.10 cm

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As we know,

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So,

⇒ $B\times I\times L=M\times g$

On putting the estimated values, we get

⇒ $B\times 50\times 1=7.037\times 10^{-3}\times 9.81$

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(b)...

As we know,

⇒ $m=\delta\times L\times \frac{\pi \ d^2}{4}$

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1 year ago

Which lanyard provides an impact force of less than 1,800 pounds, as recommended by good practices?

enyata [817]

Answer:

Energy absorbing lanyard as per OSHA

Explanation:

Energy absorbing lanyard if working over 6 feet in height so you don't break your back when you fall.

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1 year ago

An object at rest is suddenly broken apart into two fragments by an explosion one fragment acquires twice the kinetic energy of

satela [25.4K]

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1 year ago

During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s

Ray Of Light [21]

Answer:

part a : The dry unit weight is 0.0616  $lb/in^3$ 

part b : The void ratio is 0.77

part c : Degree of Saturation is 0.43

part d : Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}$

Here

$\gamma_d$ is the dry unit weight which is to be calculated
γ is the bulk unit weight given as

$\gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3$

w is the moisture content in percentage, given as 12%

Substituting values

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3$

The dry unit weight is 0.0616  $lb/in^3$ 

Part b

Void Ratio

The void ratio is given as

$e=\frac{G_s \gamma_w}{\gamma_d} -1$

Here

e is the void ratio which is to be calculated

$\gamma_d$ is the dry unit weight which is calculated in part a
$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

Substituting values

$e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766$

The void ratio is 0.77

Part c

Degree of Saturation

Degree of Saturation is given as

$S=\frac{G w}{e}$

Here

e is the void ratio which is calculated in part b

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

$S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261$

Degree of Saturation is 0.43

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}$

Here

$\gamma_{zav}$ is the zero air unit weight which is to be calculated

$\gamma_w$ is the water unit weight which is 62.4 $lb/ft^3$ or 0.04 $lb/in^3$

G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

$\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\$

Now as the volume is known, the the overall weight is given as

$weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb$

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

4 0

1 year ago