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3 years ago

CAN YALL HELP ME WITH DIS PLZ

Physics

2 answers:

Furkat [3]3 years ago

6 0

Answer:

it would dry your hair lol

Lera25 [3.4K]3 years ago

3 0

Um i think that that’s a thing that monitors the heat in your house, and when you set the heat to a specific level it will turn on the heaters until this device senses the temperature you set. If you pointed a hairdryer at it and it was on hot, it would measure the heat until it reaches the set temp. it would then stop the heaters

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Can someone help me with this Physics question please?

strojnjashka [21]

Answer:

Explanation:

The formula for this, the easy one, is

$N=N_0(\frac{1}{2})^{\frac{t}{H}$ where No is the initial amount of the element, t is the time in years, and H is the half life. Filling in:

$N=48.0(\frac{1}{2})^{\frac{49.2}{12.3}$ and simplifying a bit:

$N=48.0(.5)^4$ and

N = 48.0(.0625) so

N = 3 mg left after 12.3 years

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3 years ago

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saw5 [17]

A convex mirror makes a reflected light rays spread out.

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3 years ago

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~~~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

3 years ago

Light rays from an object after passing through a convex lens form an image at the focal point behind the lens (opposite side of

Delvig [45]

The actual position of the object is at a great distance, effectively infinite. The other options given in the question are not at all correct. The correct option among all the options that are given in the question is the last option or option "D". I hope that this answer has actually come to your great help.

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3 years ago

Can someone help me in this one :)

djyliett [7]

I believe it is “runs on parallel circuits”! my bad if incorrect

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3 years ago