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babymother [125]
3 years ago
5

A bear, searching for food wanders 35.0 meters east then 20.0 meters north. Calculate the angle of the bear's displacement (w/re

spect to his starting position).
Physics
1 answer:
mr_godi [17]3 years ago
6 0

Answer:

29.7^{\circ} north of east

Explanation:

Using a reference system in which east corresponds to the x-coordinate and north corresponds to the y-coordinate, the angle of the bear's displacement can be calculated by using the formula:

tan \theta = \frac{d_y}{d_x}

where

d_y = 20.0 m is the displacement along the y-direction

d_x=35.0 m is the displacement along the x-direction

Substituting, we find

tan \theta=\frac{20.0 m}{35.0 m}=0.571\\\theta=tan^{-1}(0.571)=29.7^{\circ}

where the angle is measured with respect to the x-direction (east).

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Relate RNA to translation and transcription
vlabodo [156]

Answer:

RNA plays a role in both Translation and transcription.

Explanation:

RNA translation uses mRNA (or messenger RNA) in order to bring the coding to the ribosome.

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4 0
3 years ago
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.

7 0
3 years ago
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Answer:

<u><em>on  flow properties and free-flowing and cohesive. </em></u>

Explanation:

the power Free flowing powders do not cling together, as cohesive powders stick to each other and form that do not disperse well during mixing

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How do streams flow?
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When rain falls on the land, it either seeps into the ground or becomes runoff, which flows downhill into rivers and lakes.
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