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Alex
3 years ago
6

You can speed up a chemical reaction by using a recyclable substance called areactant

Physics
2 answers:
lyudmila [28]3 years ago
8 0
The recyclable substance that speeds up chemical reactions is normally called catalyst or when it comes to reactions in living systems- an enzyme.
Vilka [71]3 years ago
5 0
The answer is True. You can speed up chemical reactions by using a recyclable substance called a reactant.










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5) A 20.0 kg cart with no friction wheels sits on a table. A light string is attached to it and runs over a low friction pulley
ella [17]

Answer:

1) Please find attached, created with Microsoft Visio

2) The acceleration of the masses connected by the light string is 0.00735 m/s²

3) The tension in the cord is 0.147 N

4) The time it would take the block to go 1.2 m to the edge of the table is approximately 18.07 s

5) The velocity of the cart as soon as it gets to the edge of the table is 0.042 m/s

Explanation:

1) Please find attached, the required free body diagram, showing the tension, weight and frictional (zero friction) forces acting on the cart and the mass created with Microsoft Visio

2) The acceleration of the masses connected by the light string is given as follows;

F = Mass, m × Acceleration, a

The mass of the truck, M = 20.0 kg

The mass attached to the string, hanging rom the pulley, m = 0.0150 kg

The force, F acting on the system = The pulling force on the cart = The tension on the cable = The weight of the hanging mass = 0.0150 × 9.8 = 0.147 N

The pulling force acting on the cart, F = M × a

∴ F = 0.147 N = 20.0 kg × a

a = 0.147 N/(20.0 kg) = 0.00735 m/s²

The acceleration of the truck = a = 0.00735 m/s²

3) The tension in the cord = F = 0.147 N

4) The time, t, it would take the block to go 1.2 m to the edge of the table is given by the kinematic equation, s = u·t + 1/2·a·t²

Where;

s = The distance to the edge of the table = 1.2 m

u = The initial velocity = 0 m/s (The cart is assumed to be initially at rest)

a = The acceleration of the cart = 0.00735 m/s²

t = The time taken

Substituting the known values, gives;

s = u·t + 1/2·a·t²

1.2 = 0 × t + 1/2 ×0.00735 × t²

1.2 = 1/2 ×0.00735 × t²

t² = 1.2/(1/2 ×0.00735) ≈ 326.5306

t = √(1.2/(1/2 ×0.00735)) ≈ 18.07

The time it would take the block to go 1.2 m to the edge of the table = t ≈ 18.07 s

5) The velocity, v, of the cart as soon as it gets to the edge of the table is given by the kinematic equation, v² = u² + 2·a·s as follows;

v² = u² + 2·a·s

u = 0 m/s

v² = 0² + 2 × 0.00735 × 1.2 = 0.001764

v = √(0.001764) = 0.042

The velocity of the cart as soon as it gets to the edge of the table = v = 0.042 m/s.

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Two Velocities in a Traveling Wave? Wave motion is characterized by two velocities: the velocity with which the wave moves in th
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For the answer to the question above,
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Answer:

speed =  \frac{distance}{time}  \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   =  \frac{1000}{400}  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: = 2.5m {s}^{ - 1}

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Answer:

1. Convection

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Hope this helps!

Explanation:

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