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2 years ago

Please help me I have to send for my teacher

Physics

1 answer:

expeople1 [14]2 years ago

8 0

Answer:

for the fill in the blanks

1- static

2-kinetic

3-coeffiecient

4-opposite to

5-sin theta

6-cos theta

im not sure however what to do with the top part or if its even part of what you need help with

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What is the best unit to use when measuring the mass of a mineral sample?

Ksivusya [100]

Measuring density: Measure the mass (in grams) of each mineral sample available to you. The mass of each sample is measured using a balance or electronic scale. Record mass on a chart.

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2 years ago

If a rock is thrown upward on the planet Mars with a velocity of 15 m/s, its height above the ground (in meters) after t seconds

s2008m [1.1K]

(a) The velocity (in m/s) of the rock after 1 second is 11.28 m/s.

(b) The velocity of the rock after 2 seconds is 7.56 m/s.

(d) The velocity of the block at the maximum height is 0.

<h3>Velocity of the rock</h3>

The velocity of the rock is determined as shown below;

Height of the rock after 1 second; H(t) = 15(1) - 1.86(1)² = 13.14 m

v² = u² - 2gh

where;

g is acceleration due to gravity in mars = 3.72 m/s²

v² = (15)² - 2(3.72)(13.14)

v² = 127.23

v = √127.23

v = 11.28 m/s

<h3>Velocity of the rock when t = 2 second</h3>

v = dh/dt

v = 15 - 3.72t

v(2) = 15 - 3.72(2)

v(2) = 7.56 m/s

<h3>Time for the rock to reach maximum height</h3>

dh/dt = 0

15 - 3.72t = 0

t = 4.03 s

<h3>Velocity of the rock when it hits the surface</h3>

v = u - gt

v = 15 - 3.72(4.03)

v = 0

Learn more about velocity at maximum height here: brainly.com/question/14638187

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2 years ago

Does the theory of relativity show that Newtonian mechanics is wrong?

valina [46]

Answer:

Einstein extended the rules of Newton for high speeds. For applications of mechanics at low speeds, Newtonian ideas are almost equal to reality. That is the reason we use Newtonian mechanics in practice at low speeds.

Explanation:

<em>But on a conceptual level, Einstein did prove Newtonian ideas quite wrong in some cases, e.g. the relativity of simultaneity. But again, in calculations, Newtonian ideas give pretty close to correct answer in low-speed regimes. So, the numerical validity of Newtonian laws in those regimes is something that no one can ever prove completely wrong - because they have been proven correct experimentally to a good approximation.</em>

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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:

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

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3 years ago

A horticulturist us planning a garden that is 4 m x 20 m. Which dimensions could she use to create a scale model to plan it

Radda [10]

Answer: 40 cm x 200 cm

Explanation:

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3 years ago

Please help me I have to send for my teacher​

Please help me I have to send for my teacher