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Vinvika [58]
3 years ago
7

A spaceship is traveling at 24,000 m/sec. At T=5 sec, the rocket trusts are turned on. At T=55 sec, the spaceship reaches a spee

d of 29,500 m/sec. Whats the spaceships acceleration?
Physics
1 answer:
slava [35]3 years ago
8 0

Answer:

<em>480m/s²</em>

Explanation:

Acceleration is the change in velocity of a body with respect to time;

Acceleration = change in velocity/change in time

change in velocity = 29,500 - 24,000

change in velocity= 5,500

Change in time = 55 - 5

change in time = 50secs

Substitute into the formula;

spaceships acceleration = 24000/50

spaceships acceleration = 480 m/s²

<em>Hence the spaceships acceleration is 480m/s²</em>

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Answer:

magnitude = 304.14 km/h

direction: 9.46^o West of North

Explanation:

The final plane's vector velocity will be the result of the vector addition of one pointing North of length 300 km/h, another one pointing West of length 50 km/h.

To find the magnitude of the final velocity vector (speed) we need to apply the Pythagorean theorem in a right angle triangle with sides: 300 and 50, and find its hypotenuse:

|v|=\sqrt{300^2+50^2}=\sqrt{92500}  = 304.14 km/h

The actual direction of the plane is calculated using trigonometry, in particular with the arctan function, since the tangent of the angle can be written as:

tan(\theta)=\frac{50}{300} = \frac{1}{6} \\\theta = arctan(\frac{1}{6} ) = 9.46^o

So the resultant velocity vector of the plane has magnitude = 304.14 km/h,

and it points 9.46^o West of the North direction.

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4 years ago
In the mixtures lab, you waited to see if the liquid would form lumpy or fluffy masses, which would indicate it was a _____.
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B) colloid because i took the test and that the answer

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3 years ago
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An electric motor does 900j of work for 8 hours. Calculate the power used​
nataly862011 [7]

Answer:

0.031 W

Explanation:

The power used is equal to the rate of work done:

P=\frac{W}{t}

where

P is the power

W is the work done

t is the time taken to do the work W

In this problem, we have:

W = 900 J is the work done by the motor

t = 8 h is the time taken

We have to convert the time into SI units; keeping in mind that

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We have

t=8\cdot 3600 =28,800 s

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6 0
3 years ago
A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out
anzhelika [568]

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

KE_{initial} + PE_{initial} = KE_{final} + PE_{final}

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

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2gH = v^{2}

v = \sqrt{2\times 9.8\times 5} = 9.89\ m/s

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v = velocity at the hill's bottom

Now,

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mv = (m + m')v'

65.0\times 9.89 = (65.0 + 20.0)v'

v' = 7.56\ m/s

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h = \frac{1}{2}gt^{2}

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2\times 2}{9.8} = 0.638\ s

Now, the horizontal distance is given by:

x = v't = 7.56\times 0.638 = 4.823\ m

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