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katovenus [111]
3 years ago
13

Water that soaks into Earth may become _____ under the surface.

Physics
1 answer:
larisa [96]3 years ago
5 0
If it soaks into the earths surface it becomes ground water

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Which statement is true of a mechanical wave?
svet-max [94.6K]
It must have a medium. It must travel in empty space. Mechanical waves are waves which needs medium of propogation.
7 0
1 year ago
The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr
MA_775_DIABLO [31]

Answer:

q=49Q/64

and

x =16R/15

Explanation:

See  attached figure.

E_{Q}= E due to sphere

E_{q}= E due to particule

E_{total}=E_{Q}-E_{q}=0  (1)

according to the law of gauss and superposition Law:

E_{Q}=E_{1}+E_{2}=E_{2} ; electric field due to the small sphere with r1=R/4

E_{Q}=kq_{2}/(r_{1}^{2})=

q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}

then: E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})  (2)

on the other hand, for the particule:

E_{q}=kq/(r_{p}^{2})

r_{p}=2R-R/4=7R/4   ⇒    E_{q}=16kq/(49R^{2})   (3)

We replace (2) y (3) in (1):

E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})

q=49Q/64

--------------------

if R<x<2R   AND E_{total}=E_{Q}-E_{q}=0

E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})

remember that  q=49Q/64

then:

Q(2R-x^{2})=49/64*x^{2}

solving:

x_{1} =16R/15

x_{2} =16R

but: R<x<2R  

so : x =16R/15

7 0
3 years ago
A negative charge of -2.0 C and a positive charge of 3.0 C are separated by 80 m. What is the electrostatic force between the tw
faltersainse [42]

Answer:

1. 8437500 N

2. The force between the two charges is attractive.

Explanation:

1. Determination of the force between the two charges.

Charge 1 (q₁) = –2.0 C

Charge 2 (q₂) = 3.0 C

Distance apart (r) = 80 m

Electrical constant (K) = 9×10⁹ Nm²/C²

Force (F) =?

F = Kq₁q₂ / r²

F = 9×10⁹ × 2 × 3 / 80²

F = 5.4×10¹⁰ / 6400

F = 8437500 N

Thus, the force of attraction between the two charges is 8437500 N

2. From the question given, the charges are:

Charge 1 (q₁) = –2.0 C

Charge 2 (q₂) = 3.0 C

We understood that like charges repels while unlike charges attract. Since the two charges (i.e –2 C and 3 C) has opposite signs, it means they will attract each other.

Thus the force between them is attractive.

6 0
3 years ago
What is the internal resistor of the cell in closed circuit?
Drupady [299]

Generally, the internal resistance of the new battery is small, about 0.2 euros, while the old battery is large, close to 1 euro ,

5 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

b

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
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