A block of weight 9.8N is placed on a table. The table surface exerts an upward force of 10N on the block. Assume `g=9.8m//s^(2)`.

Monjur Alahi

A block of weight 9.8N is placed on a table. The table surface exerts an upward force of 10N on the block. Assume `g=9.8m//s^(2)`.
A. The block exerts a force of 10N on the table
B. The block exerts a force of 19.8N on the table
C. The block exerts a force of 9.8N on the table
D. The block has an upward acceleration.


0 like 0 dislike
3 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

Correct Answer - B::D
Since `vec(a)_(1)nevec(a)_(2)` there is slipping between the two
`f = mu N`

0 like 0 dislike
3 views
Talk Doctor Online in Bissoy App
If `N_(2)` gas is bubbled through water at `293 K`, how many millimoles of `N_(2)` gas would dissolve in`1 L` of water. Assume that` N_(2)` exerts a p

`"Mole fraction of" N_(2)(x_(N_2))=("Partial pressure of" N_(2))/(K_(H)"for"N_(2))` ` =((0.987 bar))/((76.48 k bar))=((0.987 bar))/((76480 bar))=1.29xx10^(-5)` `x_(N_2)=n_(N_2)/(n_(N_2)+n_(H_2)o)=n_(N_2)/(n_(H_2)o)` `(N_(2)) "in the denominator has been negleeted as the gas is very little soubel in water").`...

2 Answers 1 views
If `O_(2)` gas in bubbled through water at 293 K, how many millimoles of `O_(2)` gas would dissove in 1 littre of water ? Assume that `O_(2)` exerts a

Correct Answer - 1.57 milli that `O_(2)` Step I. Calculation of `x_(O_(2))` `x_(O_(2))=rho_(O_(2))/k_(H)=((0.98"bar"))/((43860"bar"))=2.83xx10^(-5)` Step II. Calculation os milli-moles of `O_(2)` If n moles os `O_(2)` are present in 1 L of...

2 Answers 1 views
A plano-convex lens has a maximum thickness of 6 cm. When placed on a horizontal table with the curved surface in contact with the table surface, then

Correct Answer - D (d) As,n=`("Real depth")/("Apparent depth")` `rArrn=6/4=3/2` Also, `n_(1)/u-n_(2)/v=(n_(1)~n_(2))/R=1.5/6-4/17=(1.5-1)/R` Radius of curvature, R=34 cm

2 Answers 1 views
A force 250N is required to lift a 1200N weight through a pulley system in order to lift the weight through 1m. The rope has to be pulled through 5. F

Effeciency `cap%("out put")/("in put")xx100=("work done by load")/("work down by effort")xx100=(LX)/(EY)xx100` Where given `E=250N, Y=5m, L=1200N, X=1m` `cap%=(1200xx1)/(250xx5)xx100=96%`

2 Answers 1 views
A block of mass 0.1 is held against a wall applying a horizontal force of 5N on block. If the coefficient of friction between the block and the wall i

Correct Answer - C `f_("max")=0.5xx5=2.5N` `F_("req")=mg=0.98=0.98N`

2 Answers 1 views
Assertion: Frictional force is the component of contact force parallel to the surface. Reason: Friction force always opposes the motion of a body.

Correct Answer - C Friction force opposes relative motion.

2 Answers 1 views
Assertion: A body of weight 10N (W) is at rest on an inclined plane `(mu=sqrt(3)/(2))` making an angle of `30^(@)` with the horizontal. The force of f

Correct Answer - B `f=mg sin theta=10xx(1)/(2)=5N` `f_("max")=mu mg cos theta=7.5N` NO explanation for friction `lt f` (limmiting)

2 Answers 1 views
Assertion: When a force `vecF` attempts to slide a body along a surface, a frictional force parallel to the surface and directed so as to oppose the s

Correct Answer - A Bond tension opposes relative motion.

2 Answers 2 views
Equal weight of X (atomic weight = 36) and Y(atomic weight = 24) are reacted to form compound `X_(3)Y_(5)`, which of the following is/are correct?

Correct Answer - B::D `{:(,3X,+,5YrarrX_(3)Y_(5),,),("Mass","ag",,"ag",,),("Mole",(a)/(36),,(a)/(24),,),("Mole",(a)/(36)xx(1)/(3),,(a)/(24)xx(1)/(5),"Y is the limiting reagent",):}` `therefore " Mass of "X_(3)Y_(5)=(1)/(5)xx((a)/(24))(36xx3+5xx24)=1.9a`

2 Answers 1 views
The rough surface of a horizontal table offers a definite maximum opposing force to initiate the motion of a block along with the table,

(A) Component of F2 adds up to weight to increase the normal reaction.

2 Answers 2 views
X