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poizon [28]
3 years ago
14

Which of the following statements describes a compound machine?

Physics
2 answers:
ASHA 777 [7]3 years ago
6 0
From the word compound, the compound machine is already a combination of two or more types of simple machine. Thus, the answer is letter C. Because of its complexity, it is able to perform several other functions than a simple one. 
larisa86 [58]3 years ago
6 0

It is C. It is made up of more than one simple machine

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30 km/h is _________m/s?<br> A.)8.3<br> B.)5.6<br> C.)13.9<br> D.)11.1
Juli2301 [7.4K]

Answer:

Correct  Option :-  A

Explanation:

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2 years ago
What important characteristics of planets and moon are studied by astronomers?
kondor19780726 [428]

mass density orbit time temperature surface conditions

distyance from sun

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3 years 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:

a)

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

b)

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
The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the t
Nina [5.8K]

Answer:

The speed of the boxes are 1 m/s.

Explanation:

Given that,

Mass of box = 1 kg

Mass of another box = 2 kg

Suppose 1 kg box moves with 3 m/s speed.

We need to calculate the speed of the boxes

Using formula of conservation of momentum

m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v

Where, u = initial velocity

v = final velocity

Put the value into the formula

1\times3+2\times0=(1+2)v

v=\dfrac{3}{3}

v=1\ m/s

Hence, The speed of the boxes are 1 m/s.

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Arada [10]
I think it's A. Pressure and heat built up in the magma chamber.
4 0
3 years ago
Read 2 more answers
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