Answer:
Explanation:
Given that,
Spring constant k=200N/m
Compression x = 15cm = 0.15m
Attached mass m =2kg
Coefficient of kinetic friction uk= 0.2
The energy in the spring is given as
U =½kx²
U = ½ × 200 × 0.15²
U = 2.25J
Force in the spring is given by Hooke's law
F = ke
F = 200×0.15
F = 30N
The weight of body which is equal to the normal is give as
W = mg
W = 2 × 9.81
W = 19.62N
W = N = 19.62 Newton's 2nd Law
From law of friction,
Fr = uk•N
Fr = 0.2 × 19.62
Fr = 3.924
Using newton second law again
Fnet = F - Fr
Fnet = 30 - 3.924
Fnet = 26.076
Work done by net force is given as
W = Fnet × d
W = 26.076d
Then, the work done by this net force is equal to the energy in the spring
W = U
26.076d = 2.25
d = 2.25/26.076
d = 0.0863m
Which is 8.63cm
So the box will slide 8.63cm before stopping
Answer:
Explanation:
Small grains are negatively charged by the wind while big grains is positively charged and remains at the ground . This process creates an electric field due to the presence of oppositely charged particles.
When ever electric field exists it is directed from a positive charge to a negative charge so the here electric field is towards an upwards direction.
Answer:
Victor will always be able to select 4 of those cards with the following property
Explanation:
Number of trading cards = 100
victor selects 21 cards
let the 4 cards be labelled : A,B,C and D
The average power level of : A,B,C,D = ( A + B + C + D )/ 4 = P
let the two pairs be : ( A + B ) and ( C + D )
note: average power of each pair = P and this shows that
( A + B ) = ( C + D ) for Victor to select 4 cards out of the 21 cards that exhibit the same property
we have to check out the possible choices of two cards out of 21 cards yield distinct sums.
= C(21,2)=(21x20)/2 = 210.
from the question the number of distinct sums that can be created using 101 through 200 is < 210 .
hence it is impossible to get 210 distinct sums therefore Victor will always be able to select 4 of those cards
Answer:
The 50-W bulb glows more brightly than the 100-W bulb
Explanation:
The bulb has a rating of 100 W under 110 V . So it will glow with full brightness when it is fed 110 V . When bulbs are in parallel combination , each bulb receives 110 V . So they glow with full brightness .
When they are in series combination , 110 supply voltage gets distributed between the , thus , reducing the voltage appearing on each of them less than 110 V . So their brightness is reduced.
resistance = V
Bulb having high wattage rating has low resistance resulting in higher current . In the second case , both have same current as they are in series combination . So more heat will be generated in bulb having more resistance . Since 50 W bulb has higher resistance , it will glow brighter than 100 W bulb.
It bounces in random directions
It bounces back toward a single spot.
It passes through the mirror and moves in random directions.
It passes through the mirror and moves in a straight line.
