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3 years ago

Terrane accretion generally occurs along a ________ boundary between a continental plate and an oceanic plate.

1 answer:

8 0

Answer;
Terrane accretion generally occurs along a convergent boundary between a continental plate and an oceanic plate.

Explanation;
Terrane accretion occur at convergent plate boundaries, however, it may be possible for a terrane to be brought from an exotic location along a transform plate boundary.
Additionally, it is also likely for a new divergent plate boundary to develop that rifts a continent apart.

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A person of mass 75 kg stands at the center of a rotating merry-go-round platform of radius 3.3 m and moment of inertia

vivado [14]

Answer:

(a) 0.426 rad/s

(b) Before 296.225 J and After 148.5 J

Explanation:

At the center of merry-go-round, the person's weight is 75 Kg with moment of inertia of 820 kgm2 and the initial angular velocity is 0.85 rad/s

At the edge, the new angular velocity gained depends on the new moment of inertia

Here, the final moment of inertia is given by Initial moment of inertia + $MR^{2}$ where M is the mass and R is the radius of merry-go-round

Final moment of inertia= $820+(75*3.3^{2})=1636.75$

From the law of conservation of angular momentum as torque is zero

$I_i\times\omega_1=I_f\times\omega_2$ and making $\omega_2$ the subject then

$\omega_2=\frac {I_i\times\omega}{I_f}$

where $I_i$ and $I_f$ are initial and final moment of inertia, $\omega_1$ and $\omega_2$ are initial and final angular velocity.

Substituting the provided values

$\omega_2=\frac {0.85\times 820}{1636.75}=0.425843898\approx 0.426 rad/s$

(b)

Initial rotational kinetic energy is given by

$0.5I_i\times \omega_1^{2}$

$KE_i=0.5\times 820\times 0.85^{2}=296.225 J$

The final rotational kinetic energy is given by

$0.5I_f\times \omega_2^{2}$

$KE_i=0.5\times 1636.75\times 0.426^{2}\approx 148.5 J$

7 0

3 years ago

4. A tankful of liquid has a volume of 0.2m3. What is the volume in (a) lities (b) cm3 (c)ml

yan [13]

Explanation:

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5 0

3 years ago

Albert observes that Emmy's watch is ticking eanwhile, Emmy observes Albert's watch and compares it with her watch. Emmy observe

alukav5142 [94]

Answer:

1) b. more quickly than

Explanation:

It is stated in the question that Albert's watch is ticking more slowly than more quickly than at the same rate as her own watch.

Answer:

2) a. more slowly than

Explanation:

Albert's watch is ticking more slowly than more quickly than at the same rate as her own watch.

5 0

3 years ago

3. In a physics lab, 0.500-kg cart (Cart A) moving rightward with a speed of 100 m/s collides with a 1.50-kg cart (Cart B) movin

Alex787 [66]

Answer:

The speed of the two carts after the collision is 10 m/s.

Explanation:

Hi there!

The momentum of the system Cart A - Cart B is conserved because there is no external force acting on the system at the instant of the collision. Then, the momentum of the system before the collision will be equal to the momentum of the system after the collision. The momentum of the system is calculated as the sum of momenta of cart A and cart B:

initial momentum = mA · vA1 + mB · vB1

final momentum = (mA + mB) · vAB2

Where:

mA = mass of cart A = 0.500 kg

vA1 = velocity of cart A before the collision = 100 m/s

mB = mass of cart B = 1.50 kg.

vB1 = velocity of cart B before the collision = - 20 m/s

vAB2 = velocity of the carts that move as a single object = unknown.

(notice that we have considered leftward as negative direction)

Since the momentum of system remains constant:

initial momentum = final momentum

mA · vA1 + mB · vB1 = (mA + mB) · vAB2

Solving for vAB2:

(mA · vA1 + mB · vB1) / (mA + mB) = vAB2

(0.500 kg · 100 m/s - 1.50 kg · 20 m/s) / (0.500 kg + 1.50 kg) = vAB2

vAB2 = 10 m/s

The speed of the two carts after the collision is 10 m/s.

6 0

3 years ago

A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed

Vaselesa [24]

Answer:

E = 1580594.95 N/C

Explanation:

To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:

$\int EdS=\frac{Q_{in}}{\epsilon_o}$ (1)

dS: differential of the Gaussian surface

Qin: charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85 × 10-12 C2/N ∙ m2

The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:

$\int EdS=ES=E(4\pi r^2)$ (2)

Qin is calculate by using the charge density:

$Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho$ (3)

Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.

The charge density is given by:

$\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}$

Next, you use the results of (3), (2) and (1):

$E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})$

Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:

$E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}$

hence, the electric field is 1580594.95 N/C

7 0

3 years ago