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3 years ago

In middle ages people believed that abnormal behavior was cause by ?

Physics

2 answers:

Svetach [21]3 years ago

8 0

C, they didn't know any better

liraira [26]3 years ago

5 0

Your answer is c) Demonic Possession. I just took the test, I hope this helps you pass!

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What is the average velocity of a car if it travels from position 25m to a position of -7m in 34 seconds?

zubka84 [21]

Answer:

vp = 0.94 m/s

Explanation

Formula

Vp = position/ time

position: Initial position - Final position

Position = 25 m - (-7 m) = 25 m + 7 m = 32 m

Then

Vp = 32 m / 34 seconds

Vp = 0.94 m/s

6 0

3 years ago

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What is the origin of magma

sveta [45]

Magma forms by partial melting of upper mantle and crust. Partial melt means that only a fraction of the available material forms a melt, and that the remainder stays solid. The partial melt rises because of its lower density and ascends through he crust.

6 0

3 years ago

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a cylinder has a radius of 2.1cm and a length of 8.8cm .total charge 6.1 x 10^-7C is distributed uniformly throughout. the volum

7nadin3 [17]

The answer for the following problem is explained below.

Therefore the volume charge density of a substance (ρ) is 0.04 × $10^{-1}$ C.

Explanation:

Given:

radius (r) =2.1 cm = 2.1 × $10^{-2}$ m

height (h) =8.8 cm = 8.8 × $10^{-2}$ m

total charge (q) =6.1× $10^{-7}$ C

To solve:

volume charge density (ρ)

We know;

ρ =q ÷ v

volume of cylinder = π ×r × r × h

volume of cylinder =3.14 × 2.1 × 2.1 × $10^{-4}$ × 8.8 × $10^{-2}$

volume of cylinder (v) = 122.23 × $10^{-6}$

ρ =q ÷ v

ρ = 6.1× $10^{-7}$ ÷ 122.23 × $10^{-6}$

ρ = 0.04 × $10^{-1}$ C

Therefore the volume charge density of a substance (ρ) is 0.04 × $10^{-1}$ C.

3 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;

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

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

$0.7344 \omega_3^2 = 2.128$

$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

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

7 0

3 years ago

Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.

Burka [1]

Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.

At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .

At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .

At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .

7 0

2 years ago

Read 2 more answers