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3 years ago

Fill in the blanks to complete each statement about mountains.

Physics

2 answers:

valina [46]3 years ago

8 0

Range, Belt.

Just got it right on e,,d,,g,,e

ella [17]3 years ago

4 0

Answer:

Range, Belt.

Explanation:

Mountain ranges are several mountains formed together. Mountain ranges combined forms a mountain system. Systems combined form mountain belts.

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There is strong evidence that Europa, a satellite of Jupiter, has a liquid ocean beneath its icy surface. Many scientists think

dangina [55]

Answer:

4.44 rpm

Explanation:

$\omega$ = Angular speed

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Europa = $\frac{3138000}{2}\ m$

R = Radius of arm = 6 m

The acceleration due to gravity is given by

$g=\frac{GM}{r^2}\\\Rightarrow g=\frac{6.67\times 10^{-11}\times 4.8\times 10^{22}}{\left(\frac{3138000}{2}\right)^2}\\\Rightarrow g=1.3\ m/s^2$

Here the centripetal acceleration of the arm and acceleration due to gravity are equal

$a_c=\omega^2R$

$a_c=g\\\Rightarrow \omega^2R=1.3\\\Rightarrow \omega^2\times 6=1.3\\\Rightarrow \omega=\sqrt{\frac{1.3}{6}}\\\Rightarrow \omega=0.46547\ rad/s$

Converting to rpm

$1\ rad/s=\frac{60}{2\pi}\ rpm$

$0.46547\ rad/s=0.46547\times \frac{60}{2\pi}\ rpm=4.44\ rpm$

The angular speed of the arm is 4.44 rpm

8 0

3 years ago

1. Determine the magnitude of two equal but opposite charges if they attract one another with a force of 0.7N when at distance o

adell [148]

Answer:

q = 2.65 10⁻⁶ C

Explanation:

For this exercise we use Coulomb's law

F = $k \frac{q_1q_2}{r^2}$

In this case they indicate that the load is of equal magnitude

q₁ = q₂ = q

the force is attractive because the signs of the charges are opposite

F = $k \ \frac{q^2}{r^2}$

q = $\sqrt{\frac{F \ r^2}{k} }$

we calculate

q = $\sqrt{\frac{0.7 \ 0.3^2 }{9 \ 10^9} }$

q = $\sqrt{7 \ 10^{-12} }$ Ra 7 10-12

q = 2.65 10⁻⁶ C

7 0

3 years ago

A student pushes a box with a total mass of 50 kg. What is the net force on the box if it accelerates 1.5 m/s

Harrizon [31]

We now that follow newton rules f=ma so net force equal to mass*acceleration=>f=50*1.5=75 N

3 0

3 years ago

Read 2 more answers

The magnitude of the angular momentum of the two-satellite system is best represented by

zloy xaker [14]

The magnitude of the angular momentum of the two-satellite system is best represented as, L=m₁v₁r₁-m₂v₂r₂.

<h3>What is angular momentum.?</h3>

The rotational analog of linear momentum is angular momentum also known as moment of momentum or rotational momentum.

It is significant in physics because it is a conserved quantity. the total angular momentum of a closed system remains constant. Both the direction and magnitude of angular momentum are conserved.

The magnitude of the angular momentum of the two-satellite system is best represented as;

L=∑mvr

L=m₁v₁r₁-m₂v₂r₂

Hence, the magnitude of the angular momentum of the two-satellite system is best represented as, L=m₁v₁r₁-m₂v₂r₂.

To learn more about the angular momentum, refer to the link;

brainly.com/question/15104254

#SPJ4

3 0

1 year ago

Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

Fantom [35]

a) Angular acceleration: $17.0 rad/s^2$

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

$W=mg$

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

$W_p =mg sin \theta$

where $\theta$ is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

$\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta$

where L is the length of the pencil.

The relationship between torque and angular acceleration $\alpha$ is

$\tau = I \alpha$ (1)

where

$I=\frac{1}{3}mL^2$

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for $\alpha$ , we find:

$\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}$

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when $\theta=10^{\circ}$ is

$\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2$

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude $mg$

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

$\tau = Fd$

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

$\tau =mg\frac{L}{2} cos \theta$

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0

3 years ago