This is the upthrust on an object which is placed inside a fluid
This force act upwards and always push upwards
so the correct answer is given as
D. A force within a fluid that pushes upward
this force is always due to pressure difference at two levels of
at lower level since pressure is more that is why the force is upwards and this upthrust is known as Buoyancy
Answer:
If conditions are just right, you can see Polaris from just south of the equator. Although Polaris is also known as the North Star, it doesn't lie precisely above Earth's North Pole. If it did, Polaris would have a declination of exactly 90 degree.
Explanation:
Answer:
Net force: 20 N to the right
mass of the bag: 20.489 kg
acceleration: 0.976 m/s^2
Explanation:
Since the normal force and the weight are equal in magnitude but opposite in direction, they add up to zero in the vertical direction. In the horizontal direction, the 195 N tension to the right minus the 175 force of friction to the left render a net force towards the right of magnitude:
195 N - 175 N = 20 N
So net force on the bag is 20 N to the right.
The mass of the bag can be found using the value of the weight force: 201 N:
mass = Weight/g = 201 / 9.81 = 20.489 kg
and the acceleration of the bag can be found as the net force divided by the mass we just found:
acceleration = 20 N / 20.489 kg = 0.976 m/s^2
Answer:
TRUE
Explanation:
We currently live in the digital age, where almost everything is digitized, including strategic information. Thus, each individual and especially corporations that have sensitive data must use protection mechanisms against cyber attacks. One of the measures most recommended by experts is encryption, which consists of a set of rules that aims to encode data information so that only the sender and the receiver can decipher it.
Answer:
Explanation:
Given:
U1 = 1.6 m/s
U2 = -1.1 m/s
M1 = 1850 kg
M2 = 1400 kg
V1 = 0.27 m/s
Using momentum- collision equation,
M1U1 + M2U2 = M1V1 + M2V2
1850 × 1.6 - 1400 × 1.1 = 1850 × 0.27 + 1400 × V2
1420 = 499.5 + 1400V2
V2 = 0.6575 m/s
B.
KE = 1/2 × MV^2
KEa1 + KEa2 = KEb1 + KEb2
Delta KE = KE2 - KE1
KEa1 = 2368 J
KEb1 = 847 J
KEa2 = 67.433 J
KEb2 = 302.6 J
KE1 = KEa1 + KEb1
= 3215 J
KE2 = 370.033 J
Delta KE = -2845 J.
