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Luba_88 [7]
3 years ago
10

If the magnitude of the resultant force is to be 500n, directed along the positive y axis, determine the magnitude of force f an

d its direction
Physics
1 answer:
scoundrel [369]3 years ago
5 0
The answer is 0=45.2 degrees
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Ivan walks 10 meters west to the water fountain, then runs 2 meters east to his class to avoid a lockout. His displacement would
otez555 [7]

Answer:

<u>-8</u>

Explanation:

if he starts at ten and takes 10 steps left he'll be at -10... then if he takes 2 steps to the right , he's at -8 on the number-line

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3 years ago
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Two technicians are discussing service information. Technician A says that online service information is only available from the
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The correct one is A. Technician Only.
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According to the rules,  all vehicle manufacturers are required to make their service information available for everyone online for 'reasonable prices'.
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4 0
3 years ago
Of the three sciences, physics, chemistry, and biology, the most complex is
Agata [3.3K]

Answer:

I would probably say C to be completely honest

Explanation:

If you agree make sure to give me a like

6 0
2 years ago
Find deacceleration An engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. When
earnstyle [38]

Answer:

The deceleration is  a = -0.7273 \ m/s^2  

Explanation:

From the question we are told that

  The distance of the car from the crossing is d = 360 \  m

   The speed is u  = 16 \  m/s

    The reaction time of the engineer is  t = 0.53 \ s

Generally the distance covered during the reaction time is  

      d_r = u * t

=>   d_r = 16 * 0.53

=>   d_r = 8.48 \ m

Generally distance of the car from the crossing after the  engineer reacts is

      D = d- d_r        

=> D = 360 - 8.48      

=> D = 352 \ m

Generally from kinematic equation

      v^2 = u^2 + 2as

Here v is the final velocity of the car which is  0 m/s

So

        0^2 = 16^2 + 2 * a * 352

=>    a = -0.7273 \ m/s^2  

3 0
2 years ago
A child is riding a merry-go-round that has an instantaneous angular speed of 12 rpm. If a constant friction torque of 12.5 Nm i
sammy [17]

Answer:

-0.25 rad/s^2

Explanation:

The equivalent of Newton's second law for rotational motions is:

\tau = I \alpha

where

\tau is the net torque applied to the object

I is the moment of inertia

\alpha is the angular acceleration

In this problem we have:

\tau = -12.5 Nm (net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)

I=50.0 kg m^2 is the moment of inertia

Solving for \alpha, we find the angular acceleration:

\alpha = \frac{\tau}{I}=\frac{-12.5 Nm}{50.0 kg m^2}=-0.25 rad/s^2

3 0
3 years ago
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