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Mademuasel [1]
2 years ago
13

From your ____ chest to your___ shoulders you must have been wearing

Physics
1 answer:
natita [175]2 years ago
7 0
Sorry I don’t know the exact answer
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Which of the following is a typical property of an ionic compound
Yanka [14]

The answer is C.

Ionic compounds are those that bring together anions and cations bonded together by ionic bonds. The electrostatic forces of the different charges are significant in the bonds that make them strong hence require high energy to break them (high melting point).  Due to the regular structure of ionic compound that tend to form lattices in solid form, when struck, they shatter along the lines of weakness of the lattice.  

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3 years ago
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____ are solid while lines stretching across one or more lanes in the same direction, indicating the proper place to come to a s
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Stop lines are solid white lines painted across the traffic lanes at intersections and pedestrian crosswalks, indicating the exact place to stop.
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3 years ago
The aluminum atom______electrons to form an ion.<br> The ion that is formed is______.
Alchen [17]

The aluminum atom_loses_____electrons to form an ion.

The ion that is formed is_Al³⁺_____.

aluminium has the electronic configuration as 1s² 2s² 2p⁶ 3s² 3p¹

from the electronic configuration , we see that aluminium can attain stability by losing 3 electrons from outer shell.

after losing 3 electrons , the ion formed is given as Al³⁺

hence the correct options to fill in the blanks are lose  and Al³⁺


3 0
3 years ago
Read 2 more answers
A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t
svetlana [45]

Answer:

(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

Explanation:

Given that,

Initial spinning = 50.0 rad/s

Time = 20.0

Distance = 2.5 m

Mass of pole = 4 kg

Angle = 60°

We need to calculate the angular acceleration

Using formula of angular velocity

\omega=-\alpha t

\alpha=-\dfrac{\omega}{t}

\alpha=-\dfrac{50.0}{20.0}

\alpha=-2.5\ rad/s^2

The angular acceleration is -2.5 rad/s²

We need to calculate the number of revolution

Using angular equation of motion

\theta=\omega_{0}t+\dfrac{1}{2}\alpha t

Put the value into the formula

\theta=50\times20-\dfrac{1}{2}\times2.5\times20^2

\theta=500\ rad

The number of revolution is 500 rad.

(II). We need to calculate the torque

Using formula of torque

\vec{\tau}=\vec{r}\times\vec{f}

\tau=r\times f\sin\theta

Put the value into the formula

\tau=2.5\times4\times 9.8\sin60

\tau=84.87\ N-m

Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

8 0
3 years ago
An 85.0-N crate of apples sits at rest on a ramp that runs from the ground to the bed of a truck. The ramp is inclined at 28.0°
irina [24]

Answer:

A  75.1 N and a direction of 152° to the vertical.

B 85.0 N at 0° to the vertical.

Explanation:

A) The interaction partner of this normal force has what magnitude and direction?

The interaction partner of this normal force is the component of the weight of the crate perpendicular to the ramp. <u>It has a magnitude of 85cos28° = 75.1 N and a direction of 180° - 28° = 152° to the vertical(since it is directed downwards perpendicular to the ramp).</u>

B) The normal and frictional forces are perpendicular components of the contact force exerted on the crate by the ramp. What is the magnitude and direction of the contact force?

Since this force has to balance the weight of the crate, its magnitude is 85.0 N. Its direction has to be vertically opposite to that of the weight.

Since the weight is 180° to the vertical (since it is directed downwards), this force is 0° to the vertical.

<u>So, this force has a magnitude of 85.0 N and a direction of 0° to the vertical.</u>

8 0
2 years ago
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