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salantis [7]
3 years ago
13

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.

The bottom of the spring, whose spring constant is 480 N/mN/m , is anchored to the ground. Part A Initially, before the engine is ignited, the rocket sits at rest on top of the spring. How much is the spring compressed

Physics
1 answer:
blondinia [14]3 years ago
7 0

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

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If the distance between two masses is tripled, the gravitational force between changes by a factor of
maw [93]

A. 1/9

Explanation:

The gravitational force between two objects is given by

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1 and m2 are the two masses

r is the distance between the two masses

From the formula, we see that the magnitude of the force is inversely proportional to the square of the distance: therefore, if the distance is tripled (increased by a factor 3), the magnitude of the force changes by a factor

\frac{1}{r^2}=\frac{1}{3^2}=\frac{1}{9}

6 0
3 years ago
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A bus driver drove her bus for 5 miles. She drove that distance in a total of 30 minutes. What was the truck driver's average sp
nexus9112 [7]

Answer:

\mathrm{b.\:}0.17\:\mathrm{miles/minute}

Explanation:

Average speed is given by S=\frac{d}{t} where d is total distance and t is time.

Plugging in given values, we get:

S=\frac{5\:\mathrm{miles}}{30\:\mathrm{minutes}}=\fbox{$\mathrm{b.\:}0.17\:\mathrm{miles/minute}$}.

4 0
3 years ago
URGENT!!! sorry this is my first time using brainly! but here is my question: Two ropes are attached to a wagon, one horizontal
Elan Coil [88]

The components of the net force on the cart is determined as 67.66 N.

<h3>Component of net force on the cart</h3>

The component of net force on the cart is determined by resolving the forces into x and y -components.

T1 = 30 N

T2 = 40 N

T1x = -30cos(0) = 30 N

T1y = 30sin(0) = 0

T2x = 40 x cos(30) = 34.64 N

T2y = 40 x sin(3) = 20 N

∑X = 30 N + 34.64 N = 64.64 N

∑Y = 0 + 20 N = 20 N

<h3>Resultant force</h3>

R = √(64.64² + 20²)

R = 67.66 N

Learn more about net force here: brainly.com/question/25239010

#SPJ1

3 0
2 years ago
Give an example of an unbalanced forces acting on an object
777dan777 [17]
Forces<span> that are equal in size but opposite in direction are called </span>balanced forces<span>. </span>Balanced forces<span> do not cause a change in motion. When </span>balanced forces act on an object<span> at rest, the </span>object<span> will not move. If you push against a wall, the wall pushes back with an equal but opposite </span><span>force</span>
4 0
3 years ago
Read 2 more answers
Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.400mm wide. The diffraction pattern is observed
kogti [31]

Answer:

a)y_{first}=5.3mm

b)y_{second}=10.6-5.3 =5.3 mm  

Explanation:

a)

The width of the central bright in this diffraction pattern is given by:

y=\frac{m\lambda D}{a} when m is a natural number.

here:

  • m is 1 (to find the central bright fringe)                
  • D is the distance from the slit to the screen
  • a is the slit wide
  • λ is the wavelength

So we have:

y_{first}=\frac{633*10^{9}*3.35}{0.0004}

y_{first}=5.3mm

b)

Now, if we do m=2 we can find the distance to the second minima.

y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}

y_{2}=10.6 mm

Now we need to subtract these distance, to get the width of the first bright fringe :

y_{second}=10.6-5.3 =5.3 mm    

I hope it heps you!

     

4 0
3 years ago
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