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2 years ago

Question 1 of 20

Physics

2 answers:

jolli1 [7]2 years ago

5 0

Answer:

Explanation:

notka56 [123]2 years ago

4 0

Answer:

B. The magnet has magnetized the right side of the block.

Explanation:

a pe x

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After a laser bean passes through two thin parallel slits, thefirst completely dark fringes occur at ± 15.00with the original di

PSYCHO15rus [73]

Answer:

143 °

Explanation:

a ) If d be the distance between slits , λ be wavelength of light used and at angle θ nth dark fringe is formed then

d sinθ = ( 2n+1) λ/2

for first dark fringe

d sinθ = λ/2

d /λ = 1/ 2 sinθ

1 / 2 sin15

= 1.93

b )

For intensity of fringe at angle θ, the relation is

I = I₀ cos²θ

I / I₀ = cos²θ/2

Given I / I₀ =0. 1

0.1 = cos²θ/2

θ/2 = 71.5

θ = 143 °

4 0

2 years ago

A very long wire generates a magnetic field of 0.0020x 10^-4 T at a distance of 10 mm. What is the magnitude of the current? A)

zimovet [89]

Answer:

2*10^-5

Explanation:

B=IL

I=B/L

I=0.0020*10^-4/10

I=2*10^5

6 0

3 years ago

~~~NEED HELP ASAP~~~ Please solve each section and show all work for each section.

anastassius [24]

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for N as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$

8 0

2 years ago

A 4.50-kg wheel that is 34.5 cm in diametet rotates through an angle of 13.8 rad as it slows down uniformly from 22.0 rad/s to 1

lisabon 2012 [21]

Answer:

$\alpha =10.93radian/sec^2$

Explanation:

We have given given the final angular velocity $\omega _{final}=13.5rad/sec$

And $\omega _{initial}=22rad/sec$

Displacement $\Theta =13.8radian$

We have to find the angular acceleration $\alpha$

According to law of motion $\omega _{final}^2=\omega _{initial}^2+2\alpha \Theta$

So $13.5^2=22^2+2\times \alpha \times 13.8$

$\alpha =-10.93radian/sec^2$

In question we have tell about magnitude only so $\alpha =10.93radian/sec^2$

4 0

3 years ago

A patient arrives at an emergency room complaining of pain in her ankle. The nurse examines the patient’s ankle, looking for ski

Bond [772]

Answer:

Superficial anatomy.

Explanation:

Superficial anatomy can be defined as the physical examination of the external parts of a living organism such as ankle, nose, skin, knee, toes, fingers, cornea etc.

Hence, superficial anatomy is also popularly referred to as surface anatomy.

In this scenario, a patient arrives at an emergency room complaining of pain in her ankle. The nurse examines the patient’s ankle, looking for skin discoloration or swelling. The nurse is relying mostly on his knowledge of superficial anatomy by studying or examining the patient's ankle for any sign of decoration or swelling on the skin.

5 0

3 years ago