Answer:
Explanation:
We shall apply conservation of momentum law in vector form to solve the problem .
Initial momentum = 0
momentum of 12 g piece
= .012 x 37 i since it moves along x axis .
= .444 i
momentum of 22 g
= .022 x 34 j
= .748 j
Let momentum of third piece = p
total momentum
= p + .444 i + .748 j
so
applying conservation law of momentum
p + .444 i + .748 j = 0
p = - .444 i - .748 j
magnitude of p
= √ ( .444² + .748² )
= .87 kg m /s
mass of third piece = 58 - ( 12 + 22 )
= 24 g = .024 kg
if v be its velocity
.024 v = .87
v = 36.25 m / s .
Yes for an object moving on a horizontal plane, R = mg (where mg = weight). therefore, for an object moving on a horizontal plane: F = μmg
Answer:
303 Ω
Explanation:
Given
Represent the resistors with R1, R2 and RT
R1 = 633
RT = 205
Required
Determine R2
Since it's a parallel connection, it can be solved using.
1/Rt = 1/R1 + 1/R2
Substitute values for R1 and RT
1/205 = 1/633 + 1/R2
Collect Like Terms
1/R2 = 1/205 - 1/633
Take LCM
1/R2 = (633 - 205)/(205 * 633)
1/R2 = 428/129765
Take reciprocal of both sides
R2 = 129765/428
R2 = 303 --- approximated
Answer:
71.85 m/s
Explanation:
Given the following :
Length of skid marks left by jaguar (s) = 290 m
Skidding Acceleration (a) = - 8.90m/s²
Final velocity of jaguar (v) = 0
Speed of Jaguar before it Began to skid =?
Hence, initial speed of jaguar could be obtained using the formula :
v² = u² + 2as
Where
v = final speed of jaguar ; u = initial speed of jaguar(before it Began to skid) ; a = acceleration of jaguar ; s = distance /length of skid marks left by jaguar
0² = u² + (2 × (-8.90) × 290)
0 = u² + (-5,162)
u² = 5162
Take the square root of both sides
u = √5162
u = 71.847 m/s
u = 71.85m/s
