Explanation: reading the first words kind of give it away as soon as staying positive you compliment, not criticize, face angrily, or refuse.
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A 58-kg young swings a baseball bat, which causes a 0.140-kg baseball to relocate toward third base v a velocity that 38.0 m/s.
As us know, resistance is the proportion of voltage used and current flowing through the circuit. So,
R = V/I
By error calculation
∆R/R = <(∆V/V)100> + <(∆I/I)100>
V = 100 ± 6% V
I = 10 ± 0.2% A
∆R/R= (5/100)×100 + (0.2/10)×100
So, percent error in resistance (R) = ± 7%.
A hot-air balloon of diameter 12 mm rises vertically at a continuous speed the 14 m/s. A passenger accidentally autumn his camera fr
The balloon is 66.62 m high
The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is climbing at a continuous speed the 14 m/s. When the camera is dropped, it initially has the same speed together the balloon (vo=14 m/s). The camera has an upward motion for some time till it runs out of speed. Then, it falls to the ground. The elevation of an item that was introduced from one initial elevation yo and also speed vo is
The worths are
We must uncover the values of t such the the height of the camera is 0 (when it access time the ground)
Multiplying through 2
Clearing the coefficient that
Plugging in the offered values, we reach come a second-degree equation
The equation has two roots, however we only keep the positive root
Once we know the time of flight of the camera, we usage it to recognize the elevation of the balloon. The balloon has a constant speed vr and it currently was 15 m high, hence the new height is
4 months ago
A 20 g sphere of clay traveling east at 3.0 m/s collides with a 30 g sphere of clay traveling north at 2.0 m/s. What space the speed a
The speed and the direction the the resulting 50 g sphere of clay is 1.70 m/s in ~ a direction 45° north of east.
below momentum is conserved.
Initial inert = final momentum
Let eastern represent hopeful x axis and also north represent hopeful y axis.
A 20 g round of clay traveling east at 3.0 m/s collides with a 30 g ball of clay traveling north at 2.0 m/s
Mass that clay 1 = 20g = 0.02 kg
Velocity the clay 1 = 3 i m/s
Mass of clay 2 = 30g = 0.03 kg
Velocity of clay 3 = 2 j m/s
Initial momentum = 0.02 x 3 ns + 0.03 x 2 j
Initial momentum = 0.06 i + 0.06 j
Combined fixed of clay = 50 g = 0.05 kg
We require to discover the velocity of merged clay.
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Final momentum = 0.06 i + 0.06 j
0.05 x Velocity = 0.06 ns + 0.06 j
Velocity = 1.2 i + 1.2 j
The speed and the direction of the resulting 50 g sphere of clay is 1.70 m/s in ~ a direction 45° north of east.
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