Fill in the blanks with answe igits after the decimal point when rounding numer

Question

Fill in the blanks with answe igits after the decimal point when rounding numer correct answers. Only the correct answers count. Keep 2 erical values I. Two forces app the west. (a) The magnitude of the total force is the acceleration of the object is are applied to a 5 kg object, one is 6 N to the north and the other is &.N to (b) The magnitude of 2. A 32 N force, parallel to the incline, is required to push a certain crate at constant velocity up a frictionless incline that is 30 above the horizontal. (a) The mass of the crate (b) The acceleration of the crate is 3. A car is traveling at 20 m/s on a horizontal road. The brakes are applied and the skids to a stop in 4 s. (a) The coefficient of kinetic friction between the tires and roa (b) The magnitude of the acceleration of the car is 4. An object of mass 2 kg starts from rest and slides down an inclined plane 80 n 0.5 s. (a) The acceleration of the object down the incline is rce acting on the object along the incline is (i n object weights 40 N on the carth's surface. (a) The weight of the obie has one te nth the earth's mass and one half the earth's radius is he mass of the object at one earth's radius above earth's surface is

Explanation / Answer

Q-1 ) a) North and west are perpendicular forces
Fnet = sqrt (F1^2 + F2^2)
       = sqrt (6^2 + 8^2)
       = 10 N

b) a = Fnet/m
       = 10 N/5 Kg
       = 2 m/s^2

Q-2) a) 32 N ---> acts up the plane
m*g*sin thetha --> acts down the plane
for constant velocity,
m*g*sin thetha = 32
m*9.8*sin 30 = 32
m = 6.53 Kg

b) since it is moving with constant velocity, acceleration = 0 m/s^2

Q-3)

b) a = (vf-vi)/t
      = (0-20)/4
      = -5 m/s^2

a) a = *g
5 = *9.8
= 0.51

Q-4) a) S = ut + 1/2at^2
         0.8 = 0 + 1/2 a * 0.5^2
         a = 6.4 m/s^2

b) F_net = m*a

             = 2 * 6.4

             = 12.8 N

Q-5)

The weight of the object is,

    W = GM2/r2

if mass of the object 1/10 of mass of earth and its radius is equal to half of the radius of eartm,

hence, the above equation becomes as,

W '= G(1/10)MM / [r/2]2

      = (4/10) GM2/r2

      = (4/10)(40)

       = 16 N

The mass of the object does not change.

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