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drek231 [11]
3 years ago
5

What is the ideal mechanical advantage of a machine that changes only the direction of the input force?

Physics
1 answer:
Galina-37 [17]3 years ago
6 0
Ideal M.A. is 1 I.e, load =effort
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Look at the first person’s answer. Cause I know I’m wrong
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How are electromagnetic waves used in the military?
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12. A rocket, initially at rest on the ground, accelerates vertically. It accelerates uniformly until it
vodomira [7]

Answer:

We kindly invite you to read carefully the explanation and check the image attached below.

Explanation:

According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:

v_{f} = v_{o}+a\cdot (t-t_{o}) (1)

Where:

v_{o} - Initial velocity, measured in meters per second.

v_{f} - Final velocity, measured in meters per second.

a - Acceleration, measured in meters per square second.

t_{o} - Initial time, measured in seconds.

t - Final time, measured in seconds.

Now we obtain the kinematic equations for thrust and free fall stages:

Thrust (v_{o} = 0\,\frac{m}{s}, a = 30\,\frac{m}{s^{2}}, t_{o} = 0\,s, 0\,s\le t< 30\,s)

v = 30\cdot t (2)

Free fall (v_{o} = 900\,\frac{m}{s}, a = -9.807\,\frac{m}{s}, t_{o} = 30\,s, 30\,s \le t \le 120\,s)

v = 900-9.81\cdot (t-30) (3)

Now we created the graph speed-time, which can be seen below.

5 0
2 years ago
Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.
Artist 52 [7]

Answer:

a) C.M =(\bar x, \bar y)=(0.767,0.7)m

b) (x_4,y_4)=(-1.917,-1.75)m

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}

\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}

Where M represent the sum of all the masses on the system.

And the center of mass C.M =(\bar x, \bar y)

Part a

m_1= 3 kg, m_2=5kg,m_3=7kg represent the masses.

(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5) represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m

C.M =(\bar x, \bar y)=(0.767,0.7)m

Part b

For this case we have an additional mass m_4=6kg and we know that the resulting new center of mass it at the origin C.M =(\bar x, \bar y)=(0,0)m and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m

If we solve for a we got:

(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0

a=-\frac{(5kg*2.3m)}{6kg}=-1.917m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m

(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0

And solving for b we got:

b=-\frac{(7kg*1.5m)}{6kg}=-1.75m

So the coordinates for this new particle are:

(x_4,y_4)=(-1.917,-1.75)m

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Which factor below does NOT affect how fast a solute dissolves in a solvent?
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There is no factor on your list of choices that has any effect.
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