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3 years ago

What is the unique geological feature found on Mercury surface?

Physics

1 answer:

Cloud [144]3 years ago

3 0

Answer:

The surface of Mercury has landforms that indicate its crust may have contracted. They are long, sinuous cliffs called lobate scarps. These scarps appear to be the surface expression of thrust faults, where the crust is broken along an inclined plane and pushed upward.

Explanation:

I hope this helps a little bit.

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A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d

Dmitry [639]

Answer:

the distance in meters traveled by a point outside the rim is 157.1 m

Explanation:

Given;

radius of the disk, r = 50 cm = 0.5 m

angular speed of the disk, ω = 100 rpm

time of motion, t = 30 s

The distance in meters traveled by a point outside the rim is calculated as follows;

$\theta = \omega t\\\\\theta = (100 \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m$

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m

6 0

3 years ago

A capacitor is charged until its stored energy is 7.54 J. A second capacitor is then connected to it in parallel. If the charge

Ivan

Answer:

2 J

Explanation:

A charged capacitor of capacitance $C_1$ with energy of 7.54 J, is connected in parallel with another capacitor $C_2$ , so the charge is equally distributed between them.

(a) The energy stored in the capacitor before it being connected to the other capacitor is:

$U_O=q_0^2/2C_1=7.54 J\\$

The energy stored in the electric field is the sum of the energies of the two capacitors:

$U=U_1+U_2\\U=q_1^2/2C_1+q_2^2/2C_2$

since the charge equally distributed, $q_1$ = $q_2$ = $q_o/2$ . and since they are connected in parallel the potential difference on both of them is the same :

$V_1=V_2\\q_1/C_1=q_2/C_2\\q_0/C_1=q_0/C_2\\C_1=C_2=C_3\\$

hence,

$U=q_0^2/8C+q_0^2/8C\\U=q_0^2/4C\\U=2J$

8 0

3 years ago

A block has a volume of 0.09 m3 and a density of 4,000 kg/m3. What's the force of gravity acting on the

SCORPION-xisa [38]

Density = (mass) / (volume)
4,000 kg/m³ = (mass) / (0.09 m³)
(4,000 kg/m³) x (0.09 m³) = mass
mass = 360 kg
force of gravity = (mass) x (acceleration of gravity) = (360 kg) x (9.8 m/s²) = (360 x 9.8) kg-m/s² = 3,528 newtons .

5 0

3 years ago

A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.77

Vikki [24]

The formula for the rotational kinetic energy is

$KE_{rot} = \frac{1}{2}(number \ of\ propellers)( I)( omega)^{2}$

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

$I=m L^{2}=(2.67 \ kg) (0.777 \ m)^{2} =2.07459 \ kgm^{2}$

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

$KE_{rot} =( \frac{1}{2} )(5)(2.07459 \ kgm^{2}) (60\ rad/s)^{2}$

$KE_{rot} =18,671.31 \ J$

6 0

3 years ago

A 65 kg bungee jumper leaps from a bridge. She is tied to a bungee cord that is 12 m long when unstretched, and falls a total of

Nimfa-mama [501]

Answer:

(a) k = 30.33 N/m

(b) a = 9.8 m/s²

Explanation:

First, we need to find the force acting on the bungee jumper. Since, this is a free fall motion. Therefore, the force must be equal to the weight of jumper:

F = W = mg

F = (65 kg)(9.8 m/s²)

F = 637 N

(a)

Now applying Hooke's Law:

F = k Δx

where,

k = spring constant = ?

Δx = change in length of bungee cord = 33 m - 12 m = 21 m

Therefore,

637 N = k(21 m)

k = 637 N/21 m

k = 30.33 N/m

(b)

Since, this is free fall motion. Thus, the maximum acceleration will be the acceleration due to gravity.

a = g

a = 9.8 m/s²

7 0

3 years ago