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MA_775_DIABLO [31]

3 years ago

12

A resistor of 5 Ω is placed in a circuit. The voltage drop across the resistor is 12 V. What is the current through the resistor

?

1 answer:

Kaylis [27]3 years ago

5 0

<span>Data:

</span>R = 5 Ω
U = 12 V
i = ?
<span>
Formula:

</span> $U = R*i$ → $i = \frac{U}{R}$ <span>

Solving:

</span> $i = \frac{U}{R}$
$i = \frac{12}{5}$
$\boxed{i = 2.4A}$

Answer:

<span>The current through the resistor is 2.4 Amperes</span><span>

</span>

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An airplane is flying 340 km/hr at 12o east of north. the wind is blowing 40 km/hr at 34o south of east. what is the plane's act

jok3333 [9.3K]

Define an x-y coordinate system such that
The positive x-axis = the eastern direction, with unit vector $\hat{i}$ .
The positive y-axis = the northern direction, with unit vector $\hat{j}$ .

The airplane flies at 340 km/h at 12° east of north. Its velocity vector is
$\vec{v}_{1} = 340(sin(15^{o})\hat{i} + cos(15^{o})\hat{j} ) = 88\hat{i} + 328.4\hat{j}$

The wind blows at 40 km/h in the direction 34° south of east. Its velocity vector is
$\vec{v}_{2} =40(cos(34^{o})\hat{i} - sin(24^{o})]\hat{j}) = 33.1615\hat{i} -22.3677\hat{j})$

The plane's actual velocity is the vector sum of the two velocities. It is
$\vec{v}=\vec{v}_{1}+\vec{v}_{2} = 121.1615\hat{i}+306.0473\hat{j}$

The magnitude of the actual velocity is
v = √(121.1615² + 306.0473²) = 329.158 km/h

The angle that the velocity makes north of east is
tan⁻¹ (306.04733/121.1615) = 21.6°

Answer:
The actual velocity is 329.2 km/h at 21.6° north of east.

5 0

3 years ago

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Gala2k [10]

Answer:

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Explanation:

6 0

2 years ago

A 37 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar

Ne4ueva [31]

Answer:

The frictional force acting on the block is 14.8 N.

Explanation:

Given that,

Weight of block = 37 N

Coefficients of static = 0.8

Kinetic friction = 0.4

Tension = 24 N

We need to calculate the maximum friction force

Using formula of friction force

$f=\mu mg$

Put the value into the formula

$f=0.8\times37$

$f=29.6\ N$

So, the tension must exceeds 29.6 N for the block to move

We need to calculate the frictional force acting on the block

Using formula of frictional force

$f = \mu N$

Put the value in to the formula

$f=0.4\times37$

$f=14.8\ N$

Hence, The frictional force acting on the block is 14.8 N.

6 0

3 years ago

Question 2 (ID=81813)

zaharov [31]

Answer:

The answer is B

Explanation:

5/2=2.5

2.5x2=5

Hope this helps ik its kinda confusing lol

7 0

2 years ago

Read 2 more answers

The kicker now kicks the ball with the same speed as in the number of 4,but at 60.0°from the horizontal or 30.0° from the vertic

Shkiper50 [21]

Answer:

i) 0.7

ii) 1.39

iii) 0.6

Next time, when compiling a Physics question, ensure you put the unit of each measurement.

Explanation:

i) T = time of flight = $\frac{2uSin(A)}{g}$

where u = speed = 4, A = 60 and g = acceleration due to gravity = 10 (It is a constant);

Subsituting the values, we have: T = $\frac{2(4)Sin(60)}{10}$ = 0.7

ii) distance travel = Range = R = $\frac{u^{2}Sin(2A) }{g}$

where u = speed = 4, A = 60 and g = acceleration due to gravity = 10 (It is a constant);

Subsituting values, we have: R = $\frac{4^{2}Sin(2*60) }{10}$ = 1.39

iii) Maximum Height = H = $\frac{u^{2}(Sin(A))^{2} }{2g}$

where u = speed = 4, A = 60 and g = acceleration due to gravity = 10 (It is a constant);

Subsituting values, we have: $\frac{4^{2}(Sin60)^{2} }{2*10}$ = 0.6

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3 years ago