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2 years ago

You accidentally drop an eraser out of the window of an apartment 15 m above the ground

Physics

1 answer:

docker41 [41]2 years ago

3 0

Answer:

hello, yes or nou sorry jaja

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Could I get help plz

Sloan [31]

Answer:

440hz

Explanation:

saxophone a plays at 430hz and a frequency of 5 beats per second can be heard so saxophone b is playing at a frequency 10hz louder than saxophone A making it 440hz

3 0

2 years ago

A photon has 3.4 × 10–18 joules of energy. Planck’s constant is 6.63 × 10–34 J•s.

hodyreva [135]

Ok i will answer for real this time. Please give me brainliest.
The Answerr is:
5.12*10^15. Since e=h*f, f=e/h. 3.4*10^(-18)/h.
i am so sorry i was doing a challenge and i needed answers to get 100 pts.
Hope I Helped
~TeenOlafLover <3

5 0

2 years ago

Which expression is equivalent to 5(2+7)? 0 2(5+7) O 2 + 7(5) O 5(2)+7 O 5(2)+5(7)

professor190 [17]

Answer:

5(2)+5(7)

Explanation:

5(2)+5(7)=10+35=45

6 0

3 years ago

The particles in an object move in (many/ two directions)

pentagon [3]

Answer:

They are negitive

Explanation:

Im pretty sure

8 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