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Trava [24]
4 years ago
12

Q- A body is acted upon by two forces 30N due east and 40N due North. Calculate

Physics
1 answer:
Dmitry_Shevchenko [17]4 years ago
4 0

Answer:

the following image will make you understand

Explanation:

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méthode if séparation muddy water

3 0
3 years ago
if the Horse and Rider have a combined mass of 572 kg what force would be required to accelerate them 5 kph per second
Vlad [161]

Answer:

   Force required to accelerate = 794.44 N

Explanation:

 Force required = Mass of horse x Acceleration of horse

 Mass of horse and rider, m=   572 kg

 Acceleration of horse and rider, a = 5 kph per second

                                      =\frac{5*1000}{60*60} =1.39 m/s^2

  Force required = ma

                             = 572 x 1.39 = 794.44 N

  Force required to accelerate = 794.44 N

8 0
3 years ago
Plz help me I'll mark you brainliest
My name is Ann [436]
Rain will be expected in Billings
5 0
3 years ago
Read 2 more answers
How do you calculate the net force when there are multiple forces in different directions?​
Artyom0805 [142]

To find F_{net} we need to use vector addition and use the x and y components. First we subtract vector 2 from vector 5 which results in a vector with a  length of 3 pointing directly east, then we use the distance formula to find the length of the net force F_{net} = \sqrt{(3)^2+(4)^2} \\  which gives F_{net} = 5. We now have a magnitude but we also need a direction, since vector 4 and vector 5 are perpendicular. Using \theta = \tan^{-1} (\frac{4}{3})   where tan^-1(y/x) we get an angle of 53 degrees. The resultant force vector is 5 distance with an angle of 53 degrees north east.

4 0
3 years ago
A car accelerates from rest at -3.00m/s^2. What is the velocity at the end of 5.0s? What is the displacement after5.0s?
Andrew [12]

Speed = (acceleration) x (time)
Velocity = (speed) in (direction of the speed)

Speed = (-3 m/s²) x (5 s) = 15 m/s
Velocity =
             (15 m/s) in the direction opposite to the direction you call positive
.

Displacement = (distance between start-point and end-point)
                           in the direction from start-point to end-point.

Distance = (1/2) (acceleration) (time)²
Distance = (1/2) (3 m/s²) (5 s)²
                 = (1/2) (3 m/s²) (25 s²)  =  37.5 meters

Displacement =
                     37.5 meters in the direction opposite to the direction you call positive.

5 0
3 years ago
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