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kotykmax [81]
3 years ago
13

Consider a particle with initial velocity v⃗ that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis.

Part A What is the x component v⃗ x of v⃗ ?
Physics
1 answer:
kramer3 years ago
5 0

Answer:

  v_x = -6\ m/s

Explanation:

initial velocity

magnitude of velocity, v = 12 m/s

angle made of velocity with negative x-axis,θ = 60°

We need to calculate x- component of v

  v_x = v cos \theta

velocity is in negative x-direction, v = -12 m/s

now,

  v_x = -12\times cos 60^0

  v_x = -12\times 0.5

  v_x = -6\ m/s

Hence, the velocity x-component is equal to -6 m/s.

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While playing baseball with your friends your hands begin to sting after you ctach several fast balls.
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7 0
3 years ago
80N force ;
lutik1710 [3]

The power of the engine is 320 W.

<u>Explanation:</u>

Power may be defined as the rate of doing work (or) work done per unit time. One unit of energy is used to do the one unit of work.

                          Power = Work done / Time taken

Given, Force = 80 N,    height = 5 m , final velocity = 4 m/s

To calculate the power, we must know the time taken.

To find the time, use the distance and speed formula which is given by

                              Time = Distance / speed

Here distance = 5 m and speed = 4 m/s

                              Time = 5 / 4 = 1.25 s.

Now,          Power = work done / time

                              = (F * d) / t  = (80 * 5) / 1.25

                  Power = 320 W.

The standard unit of power is watt (W) which is joule per second.

                                                               

4 0
3 years ago
Given that the concentration of bovine carbonic anhydrase is 3.3 pmol ⋅ L − 1 and R max ( V max ) = 222 μmol ⋅ L − 1 ⋅ s − 1 , d
LuckyWell [14K]

Answer:

The turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1.

Explanation:

Given:

The concentration of bovine carbonic anhydrase = total enzyme concentration = Et = 3.3 pmol⋅L^–1 = 3.3 × 10^–12 mol.L^–1

The maximum rate of reaction = Rmax (Vmax) = 222 μmol⋅L^–1⋅s^–1 = 222 × 10^–6 mol.L^–1⋅s^–1

The formula for the turnover number of an enzyme (kcat, or catalytic rate constant) = Rmax ÷ Et = 222 × 10^–6 mol.L^–1⋅s^–1 ÷ 3.3 × 10^–12 mol.L^–1 = 67,272,727.27 s^–1

Therefore, the turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1

3 0
3 years ago
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