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3 years ago

WHERE IS THE HYPOTHESIS IN THIS PASSAGE?

Physics

1 answer:

yuradex [85]3 years ago

4 0

Answer:

pretty sure for this one, "if the density affected the liquid's ability to retain heat."

Explanation:

a hypothesis should be in "if, then, because.." format, and while this bit of the passage doesnt include all three, it does include one! that bit of the passage would be an incomplete hypothesis.

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A Capacitor is a circuit component that stores energy and can be charged when current flows through it. A current of 3A flows th

ddd [48]

Answer:

$8\mu C$

Explanation:

t = Time taken = $2\mu s$

i = Current = 3 A

q(0) = Initial charge = $2\mu C$

Charge is given by

$q=\int_0^t idt+q(0)\\\Rightarrow q=\int_0^{2\mu s} 3dt+2\mu C\\\Rightarrow q=3(2\mu s-0)+2\mu C\\\Rightarrow q=6\mu C+2\mu C\\\Rightarrow q=8\mu C$

The magnitude of the net electric charge of the capacitor is $8\mu C$

4 0

3 years ago

Sam makes six star-shaped cookies out of dough. He puts all of the cookies on a balance and measures the total weight as 36 gram

son4ous [18]

Answer:

The answer is "36 grams".

Explanation:

In this question, the weight of the ball is not mentioned but is the weight of the cookies is declared, which is equal to 36 grams, and all the cookies are squeezes into the ball and after that, it calculates the overall weight so, let assume that ball weight is =0 and then the overall weight is:

$=\text{weight of ball + cookies weight}\\\\=0+36 \ grams \\\\=36 \ grams$

5 0

3 years ago

An object is floating in equilibrium on the surface of a liquid. The object is then removed and placed in another container, fil

Anestetic [448]

Answer:

The fraction of its volume inside liquid is increased .

Explanation:

According to principle pf floatation , an object floats on the surface of water

when the weight of liquid displaced by it becomes equal to weight of the object . weight of the liquid depends upon the density of the liquid .

In the second case , when the body is dipped into liquid of lesser density , in order to balance the weight of body , more volume of liquid will be displaced so that weight of displaced liquid becomes equal to object's weight . So the body floats with greater depth inside liquid . The fraction of its volume inside liquid is increased .

4 0

3 years ago

A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The

nordsb [41]

Answer:

$t=6.96s$

Explanation:

From this exercise, our knowable variables are hight and initial velocity

$v_{oy}=96ft/s$

$y_{o}=112ft$

To find how much time does the ball strike the ground, we need to know that the final position of the ball is y=0ft

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

$0=112ft+(96ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}$

Solving for t using quadratic formula

$t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}$

$a=-\frac{1}{2} (32.2)\\b=96\\c=112$

$t=-0.999s$ or $t=6.96s$

Since time can't be negative the answer is t=6.96s

7 0

3 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$