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scoray [572]
3 years ago
6

Choose the type of literary device being used in the example below. I ran as fast as a cheetah but I still missed the bus, which

is chaos on wheels.
Physics
1 answer:
Feliz [49]3 years ago
5 0
Simile, as the example compares the person with a cheetah using “as” (a simile uses “like” or “as” for comparisons between two things).
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I need help ASAP! It's urgent ​
djverab [1.8K]

Answer:

hope this helps you

3 0
3 years ago
In humans, which part of the brain contains 80 percent of its weight? cerebrum cerebellum spiral cord brain stem
scZoUnD [109]
I believe it's the Cerebrum
6 0
2 years ago
the electric current in a wire is 1.5 A. How many electrons flow past a given point in a time of 2 s?​
34kurt

Answer:

The quantity of electrons that flows past a given point is 3.0 C.

Explanation:

An electric current (I) is the ratio of the quantity of charges (Q) that flows through a point to the time taken (t).

i.e            I = \frac{Q}{t}

It is measured in Ampere's by the use of an ammeter in the laboratory. The quantity of charge that flow through a given point is measured in Coulombs, while time is measured in seconds.

Given that;     I = 1.5A and t = 2s, find Q.

                          Q = It

                             = 1.5 × 2

                             = 3.0 C

The quantity of electrons that flows past a given point is 3.0 C.

6 0
3 years ago
In a star nuclear fusion occurs in the
Anika [276]

Answer:

C. Core

Explanation:

8 0
2 years ago
A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m
nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

F_T \cdot r = I \cdot \alpha  = m \cdot r^2  \cdot \alpha

Where;

F_T = The tangential force

I =  The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

I_m \cdot \alpha_m  = I_m \cdot \dfrac{v_m}{r \cdot t}

I_m = The rotational inertia of the merry-go-round

\alpha _m = The angular acceleration of the merry-go-round

v _m = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

I_b \cdot \alpha_b  = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Where;

I_b = The rotational inertia of the boy

\alpha _b = The angular acceleration of the boy

v _b = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t}

Which gives;

v_m = \dfrac{m_b \cdot r^2  \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2  \cdot v_b}{I_m}

From which we have;

v_m =  \dfrac{20 \times 3^2  \times 5}{600} =  1.5

The velocity of the merry-go-round, v_m, after the boy hops on the merry-go-round = 1.5 m/s.

5 0
2 years ago
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