The correct answer for the question that is being presented above is this one: "(d) the magnitude of the normal force on the box from the ramp." A box is on a ramp that is at angle θ to the horizontal. As θ is increased from zero, and before the box slips, do the following increase, decrease, or remain the same and <span> the magnitude of the normal force on the box from the ramp.</span>
Answer: they could get a piece of the coral and run tests on it
Explanation:
Answer:
Explanation:
Hydrogen and hydroxide ions reacts to form water in a neutralisation reaction
Answer:
Explanation:
Force = q ( v x B)
- 5.6 x 10⁻⁹ (v x - 1.25 k )
- 3.4x 10⁻⁷i + 7.4 x 10⁻⁷j
Let v = ai+bj +ck
Force = - 5.6 x 10⁻⁹ [(ai+bj +ck) x - 1.25 k )]
= - 5.6 x 10⁻⁹ ( 1.25aj - 1.25bi )
= - 7 a j + 7 b i
( 7bi - 7aj ) x 10⁻⁹
Comparing with given force
7b x 10⁻⁹ b = - 3.4 x 10⁻⁷
b = - 48.57
- 7 a x 10⁻⁹ = 7.4 x 10⁻⁷
a = - 105.7
velocity
= -105.7 i - 48.57 j + ck
b ) Component along k can not be obtained .
c ) v . F = ( -105.7 i - 48.57 j + ck ) . −(3.40×10−7N) ˆı +(7.40×10−7N) ˆȷ
= 105.7 x 3.4 x 10⁻⁷ - 48.57 x 7.4 x 10⁻⁷
= 359.38 x 10⁻⁷ - 359.38 x 10⁻⁷
=0
angle between v and F = 90 degree
Answer:
a) 33.6 min
b) 13.9 min
c) Intuitively, it takes longer to complete the trip when there is current because, the swimmer spends much more time swimming at the net low speed (0.7 m/s) than the time he spends swimming at higher net speed (1.7 m/s).
Explanation:
The problem deals with relative velocities.
- Call Vr the speed of the river, which is equal to 0.500 m/s
- Call Vs the speed of the student in still water, which is equal to 1.20 m/s
- You know that when the student swims upstream, Vr and Vs are opposed and the net speed will be Vs - Vr
- And when the student swims downstream, Vr adds to Vs and the net speed will be Vs + Vr.
Now, you can state the equations for each section:
- distance = speed × time
- upstream: distance = (Vs - Vr) × t₁ = 1,000 m
- downstream: distance = (Vs + Vr) × t₂ = 1,000 m
Part a). To state the time, you substitute the known values of Vr and Vs and clear for the time in each equation:
- (Vs - Vr) × t₁ = 1,000 m
- (1.20 m/s - 0.500 m/s) t₁ = 1,000 m⇒ t₁ = 1,000 m / 0.70 m/s ≈ 1429 s
- (1.20 m/s + 0.500 m/s) t₂ = 1,000 m ⇒ t₂ = 1,000 m / 1.7 m/s ≈ 588 s
- total time = t₁ + t₂ = 1429s + 588s = 2,017s
- Convert to minutes: 2,0147 s ₓ 1 min / 60s ≈ 33.6 min
Part b) In this part you assume that the complete trip is made at the velocity Vs = 1.20 m/s
- time = distance / speed = 1,000 m / 1.20 m/s ≈ 833 s ≈ 13.9 min
Part c) Intuitively, it takes longer to complete the trip when there is current because the swimmer spends more time swimming at the net speed of 0.7 m/s than the time than he spends swimming at the net speed of 1.7 m/s.
