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Ostrovityanka [42]
3 years ago
3

The diagram shows a coiled wire that carries an electrical current.

Physics
2 answers:
KengaRu [80]3 years ago
4 0
<h3><u>Answer;</u></h3>

Magnetic field

<h3><u>Explanation;</u></h3>
  • Magnetic field is the area around a magnet in which magnetic force is felt. It describes the influence of magnets and electrical currents.
  • Magnetic fields are shown by continuous lines of force which emerge from north-seeking magnetic poles and enter to the south poles. Density of the lines indicates the strength of magnetic field.
Reptile [31]3 years ago
3 0
The answer is A i did the test.
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What type of relationship does this graph show?
aliya0001 [1]

Answer:

I'm pretty sure it shows a direct relationship

Explanation:

5 0
3 years ago
10 PTS! HELP!
wolverine [178]
Mechanical energy
I think
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3 years ago
Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo
djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for  the force of static friction:

f_s = \mu_s N --- (1)

where;

f_s = static friction force

\mu_s = coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

N = ma_{min}

From equation (1)

f_s = \mu_s ma_{min}

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

F = f_s

mg = \mu_s ma_{min}

a_{min} = \dfrac{mg }{ \mu_s m}

a_{min} = \dfrac{g }{ \mu_s }

where;

\mu_s = 0.383 and g = 9.8 m/s²

a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }

\mathbf{a_{min}= 25.59 \ m/s^2}

3 0
3 years ago
One strategy in a snowball fight is to throw
Oxana [17]

second question: How many seconds after the first snowball

should you throw the second so that they

arrive on target at the same time?

Answer in units of s.

Answer:

Part 1: 28°

Part 2: 1.367

Explanation:

Part 1:

Given: 62°  

Simple

θ = 90°- 62°

<u>θ = 28°</u>

Part 2:

Y-direction

Δy=v_{yo} t+\frac{1}{2} a_{y} t^{2}

0=[16.2sin(62)]t_{1}+1/2(-9.8)t_{1}^{2} \\

t_{1} =\frac{2[16.2sin(62)]}{9.8}

t_{1}=2.91913s

0=[16.2sin(28)]t_{2}+1/2(-9.8)t_{2}^{2}

t_{2} =\frac{2[16.2sin(28)]}{9.8}

t_{2}=1.55213s

Δt=t_{1}-t_{2}

Δt=2.91913-1.55213

<u>Δt= 1.367s</u>

Hope it helps :)

7 0
4 years ago
A 5kg cart moving to the right with a velocity of 16 m/s collides with a concrete wall and
Kryger [21]

Explanation:

mass, m = 5kg

initial velocity, u = 16m/s

final velocuty, v = -22m/s

change in momentum, ∆p = ?

∆p = m (v-u)

5(-22-16)

5(38)

∆p = 190kgm/s

check the calculations!

8 0
3 years ago
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