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4vir4ik [10]
3 years ago
10

A 1000 kg weather rocket is launched straight up.The rocket motor provides a constant acceleration for 16 s, then the motor stop

s. The rocket altitude 20 s after launch is 5100 m. You can ignore any effects of air resistance. What is the rocket's speed as it passes through a cloud 5100 m above the ground?
Not sure what i'm doing.
Physics
1 answer:
ohaa [14]3 years ago
3 0
I’d say 10200m as the answer not sure though
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Under what conditions can an object have forces acting on it, but its motion does not change ?
Lady_Fox [76]

Answer: An object at rest has zero velocity - and (in the absence of an unbalanced force) will remain with a zero velocity. Such an object will not change its state of motion.

Explanation: I hoped that helped!!

7 0
3 years ago
Light of wavelength 500nm is passed through a diffraction grating which has 400 lines per mm. What is the angular spectrum betwe
Citrus2011 [14]

Answer:

because of the gravity of the earth

8 0
2 years ago
A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When
natita [175]

Answer:

a) \mu_{k} = 0.704, b) R = 0.312\,m

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s

\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}

\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}

\mu_{k} = 0.704

b) The speed of the block is determined by using the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}

v = x\cdot \sqrt{\frac{k}{m} }

v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }

v \approx 9.829\,\frac{m}{s}

The radius of the circular loop is:

\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}

\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}

R = 0.312\,m

5 0
3 years ago
Object A and Object B are 100 meters apart. If Object A gains some
satela [25.4K]

The gravitational force between the two objects A) It increases.

Explanation:

The gravitational force between two objects is given by:

F=G\frac{m_1 m_2}{r^2} (1)

where

G is the gravitational constant

m_1, m_2 are the masses of the two objects

r is the separation between the objects

In this problem, object A and object B are initially at a distance of

r = 100 m

And at that distance, the force between them is

F

Later, object A gains some mass. We notice from eq.(1) that the gravitational force is directly proportional to the mass: therefore, if the mass of either of the two objects increases, then the gravitational force between them also increases. Therefore, the new force will be larger than the original force:

F' > F

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

7 0
3 years ago
PLEASE HELP URGETNT
Angelina_Jolie [31]

Answer:

I think (d) is right answer

6 0
3 years ago
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