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pshichka [43]
3 years ago
10

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 44.1m/s^2 . The acce

leration period lasts for time 7.00s until the fuel is exhausted. After that, the rocket is in free fall. Find the maximum height y_max reached by the rocket. Ignore air resistance and assume a constant acceleration due to gravity equal to 9.80 rm{m/s^2} .
Physics
1 answer:
ollegr [7]3 years ago
7 0
The acceleration and distance is related to the following expression:
y=v0*t + a*t^2/2 ; v0=0 
y=44.1*100/2 = 2205m 
hence, the speed will be 
v=0 + a*t = 441m/s 
from that height it will just be subjected to the gravitational acceleration 
0=v_acc^2 -2g*y_free 
y_free = v_acc^2/2g = 9922.5m 
<span>y_max = y_acc+y_free = 441+9922.5 =10363.5m</span>
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if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu
RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

F = \frac{k|q_1||q_2|}{r^2}

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N

Therefore, the mutual force between the two point charges is 319.64 N

4 0
3 years ago
An amusement park ride consists of a rotating circular platform 11.1 m in diameter from which 10 kg seats are suspended at the e
frozen [14]

To solve this problem we will use the relationship given between the centripetal Force and the Force caused by the weight, with respect to the horizontal and vertical components of the total tension given.

The tension in the vertical plane will be equivalent to the centripetal force therefore

Tsin\theta= \frac{mv^2}{r}

Here,

m = mass

v = Velocity

r = Radius

The tension in the horizontal plane will be subject to the action of the weight, therefore

Tcos\theta = mg

Matching both expressions with respect to the tension we will have to

T = \frac{\frac{mv^2}{r}}{sin\theta}

T = \frac{mg}{cos\theta}

Then we have that,

\frac{\frac{mv^2}{r}}{sin\theta} =  \frac{mg}{cos\theta}

\frac{mv^2}{r} = mg tan\theta

Rearranging to find the velocity we have that

v = \sqrt{grTan\theta}

The value of the angle is 14.5°, the acceleration  (g) is 9.8m/s^2 and the radius is

r = \frac{\text{diameter of rotational circular platform}}{2} + \text{length of chains}

r = \frac{11.1}{2}+2.41

r = 7.96m

Replacing we have that

v = \sqrt{(9.8)(7.96)tan(14.5\°)}

v = 4.492m/s

Therefore the speed of each seat is 4.492m/s

6 0
3 years ago
I need help quickly pls
sukhopar [10]

Answer:

Ok I'm not 100% on this one but, try 3 lifes sorry if u get it wrong D:

Explanation:

4 0
3 years ago
Use pascal and pressure in the same sentence
kozerog [31]
The unit of pressure is the Pascal (Pa), equivalent to \frac{\text{kg}}{\text{ms}^{2}}.
8 0
3 years ago
A beam of visible light is passed through a plexiglass container filled with water. If a beam of light comes in at a slight angl
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B) The beam of light will bend three times: small angle of refraction through container wall; larger angle of refraction through water; small angle passing out of a container.
Eliminate
8 0
3 years ago
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