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kykrilka [37]
3 years ago
10

Artistic works in the animal style often contain ________ and images of abstracted animals.

Physics
1 answer:
nikdorinn [45]3 years ago
5 0
The answer to the above question is "Interlace".

Artistic works in the animal style often contain INTERLACE  and images of abstracted animals. In the medieval art, this artistic work contains decorative elements. Some elements are knotted while others are braided.

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A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a
KatRina [158]

Answers:

a) 0.80 kg

b) 2.12 s

c) 1.093 m/s^{2}

Explanation:

We have the following data:

k=7 N/m is the spring constant

A=12.5 cm \frac{1 m}{100 cm}=0.125 m is the amplitude of oscillation

V=32 cm/s=0.32 m/s is the velocity of the block when x=\frac{A}{2}=0.0625 m

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

U_{o}+K_{o}=U_{f}+K_{f} (1)

Where:

U_{o}=k\frac{A^{2}}{2} is the initial potential energy

K_{o}=0  is the initial kinetic energy

U_{f}=k\frac{x^{2}}{2} is the final potential energy

K_{f}=\frac{1}{2} m V^{2} is the final kinetic energy

Then:

k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2} (2)

Isolating m:

m=\frac{k(A^{2}-x^{2})}{V^{2}} (3)

m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}} (4)

m=0.80 kg (5)

<h3>b) Period</h3>

The period T is given by:

T=2 \pi \sqrt{\frac{m}{k}} (6)

Substituting (5) in (6):

T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}} (7)

T=2.12 s (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration a_{max} is when the force is maximum F_{max}, as well :

F_{max}=m.a_{max}=k.x_{max} (9)

Being x_{max}=A

Hence:

m.a_{max}=kA (10)

Finding a_{max}:

a_{max}=\frac{kA}{m} (11)

a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg} (12)

Finally:

a_{max}=1.093 m/s^{2}

5 0
3 years ago
Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges
Nostrana [21]

Answer:

r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

E_1 = E_2

Let the position where net field is zero will lie at distance "r" from q1

\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}

now we will have

\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}

now square root both sides

\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}

now we have

\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1

so we have

r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})

8 0
3 years ago
Work and Power Practice Calculations
Elza [17]

Answer:

1. W = F d = 20 N * 6 m = 120 J

2. F = W / d = 60 J / 2 m = 30 N

3. d = W / F = 350 J / 85 N = 4.12 m

4. P = W / t = F d / t = 45 N * 9 m / 10 s = 40.5 Watts

5. W = P t = 500 W * 120 sec = 60,000 J

6. t = W / P = 550 J / 310 W = 1.77 sec

5 0
2 years ago
If you need 40.0 Nm of torque in order to loosen a nut on a wn
KonstantinChe [14]

Answer:

0.301 m

Explanation:

Torque = Force × Radius

τ = Fr

40.0 Nm = 133 N × r

r = 0.301 m

The mechanic must apply the force 0.301 m from the nut.

6 0
3 years ago
A swimmer, capable of swimming at a speed of 1.0 m/s in still water (i.e., the swimmer can swim with a speed of 1.0 m/s relative
Vesnalui [34]
If the current takes him downstream we must find the resultant vector of the velocities: V res= \sqrt{1^{2}+0.91^{2}  } = \sqrt{1.8281}= 1.3520747 Then if the river is 3000 m-wide the swimmer will have to pass:
 1.3520747 · 300 = 4056.14 m                t = 4056.14 m : 1 m/s
a ) It takes 4056.15 seconds ( 1 hour 7 minutes and 36 seconds ) to cross the river.  
b ) 0.91 · 3000 = 2730 m
He will be 2730 m downstream.
5 0
3 years ago
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