1. a=(v-v0)/t
a=14 m/s / 2s
a=7 m/s^2
2. <span>a=(v-v0)/t
</span>a=(30 m/s - 0 m/s) / 12 s
a=2.5 m/s^2
3. <span>a=(v-v0)/t
a=(37 m/s - 22 m/s) / 2 s
a= 7.5 m/s^2
4. </span><span>a=(v-v0)/t
a=(12 km/s - 0 km/s)/8 s
a=1.5 km/s^2
etc</span>
Answer:
a. one-half as great
Explanation:
The power developed by the first lifter is one-half as great as that of the second person.
Power is defined as the rate at which work is done;
Power =
Since the two lifters do the same work at different time, let us estimate their power;
P₁ = P₂ =
We see that for P₁, power is half of the work done whereas in P₂ power is the same as the work done.
Therefore,
The power of the first weight lifter is one-half the second lifter.
Answer:
4
Explanation:
In order for the current to continue flowing through the circuit (and for the bulbs to continue shining), there must be a closed path containing the battery where current can flow. Let's see the effect of removing each bulb on the circuit:
- 1: when removing bulb 1 only, the current can still flow through the path battery-bulb 3- bulb 4
- 2: when removing bulb 2 only, the current can still flow through the path battery-bulb 3- bulb 4
- 3: when removing bulb 3 only, the current can still flow through the path battery-bulb 1-bulb 2- bulb 4
- 4: when removing bulb 4 only, the current can no longer flow. In fact, there is no closed path that contains the battery now, so the current will not flow and all the bulbs will stop shining.
Answer: Uranium phosphide (U3P4) P2U-6 - PubChem.
Explanation:
Answer:
p2 = 9.8×10^4 Pa
Explanation:
Total pressure is constant and PT = P = 1/2×ρ×v^2
So p1 + 1/2×ρ×(v1)^2 = p2 + 1/2×ρ×(v2)^2
from continuity we have ρ×A1×v1 = ρ×A2×v2
v2 = v1×A1/A2
and
r2 = 2×r1
then:
A2 = 4×A1
so,
v2 = (v1)/4
then:
p2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v2)^2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v1/4)^2
p2 = 3.0×10^4 Pa + 1/2×(1000 kg/m^3)×(12m/s)^2 - 1/2×(1000kg/m^3)×(12^2/16)
= 9.75×10^4 Pa
= 9.8×10^4 Pa
Therefore, the pressure in the wider section is 9.8×10^4 Pa