Answer:
40 kg.m/s
Explanation:
Momentum, p is defined as the product of mass and velocity of an object. Numerically, it is represented as, p=mv where m is mass of the object and v is the velocity in which the object moves, with keen observation on the direction before and after collision. Substituting 10 kg for m and 4 m/s for v then momentum, P=10*4=40 kg.m/s
Gs*rs^2 = gm*rm^2
<span>rm = rs*√gs/gm </span>
<span>rm = 6370*√9.83/(9.83-0.009) = 6372.92 </span>
<span>mountain observatory is placed at an altitude worth 2920 m asl
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Explanation:
It is known that relation between pressure and density is as follows.
P =
where, P = pressure
= density
g = acceleration due to gravity
h = height
Putting the given values into the above formula as follows.
P =
=
= 110495000 Pa
Now, relation between pressure and force is as follows.
P =
or, F = PA
F =
=
Thus, we can conclude that a force of can be experienced at such depth.
Answer:
Option 10. 169.118 J/KgºC
Explanation:
From the question given above, the following data were obtained:
Change in temperature (ΔT) = 20 °C
Heat (Q) absorbed = 1.61 KJ
Mass of metal bar = 476 g
Specific heat capacity (C) of metal bar =?
Next, we shall convert 1.61 KJ to joule (J). This can be obtained as follow:
1 kJ = 1000 J
Therefore,
1.61 KJ = 1.61 KJ × 1000 J / 1 kJ
1.61 KJ = 1610 J
Next, we shall convert 476 g to Kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
476 g = 476 g × 1 Kg / 1000 g
476 g = 0.476 Kg
Finally, we shall determine the specific heat capacity of the metal bar. This can be obtained as follow:
Change in temperature (ΔT) = 20 °C
Heat (Q) absorbed = 1610 J
Mass of metal bar = 0.476 Kg
Specific heat capacity (C) of metal bar =?
Q = MCΔT
1610 = 0.476 × C × 20
1610 = 9.52 × C
Divide both side by 9.52
C = 1610 / 9.52
C = 169.118 J/KgºC
Thus, the specific heat capacity of the metal bar is 169.118 J/KgºC
Given below the arrangement of loading on the larger boat by two tug boats.
F₁ = 5 N
F₂ = 4 N
Angle between them θ = 90⁰
Resultant between two vectors,
Substituting
So magnitude of the net force on the block = 6.403 N