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suter
1 month ago
13

An object is at rest on the ground. The object experiences a downward gravitational force from Earth. Which of the following pre

dictions is correct about why the object does not accelerate downward? Select two answers. Justify your selections.A) The bonded molecules of the object are repelled upward by the bonded molecules of the ground with the same magnitude as the gravitational force downward on the object.B)The normal force is exerted upward on the object from the ground with the same magnitude as the gravitational force downward on the objectC) The bonded molecules of the object are attracted downward by the bonded molecules of the ground with the same magnitude as the gravitational force downward on the object.D) The force of friction is exerted upward on the object from the ground with the same magnitude as the gravitational force downward on the object.
Physics
2 answers:
Yuliya22 [3.3K]1 month ago
6 0

Answer:

A) and B) are valid.

Explanation:

When an object remains at rest, it is indicative that no net force acts upon it.

The downward gravitational force from Earth must be counterbalanced by an upward force of equal magnitude in order to maintain rest.

This upward force is provided by the normal force, which adjusts to satisfy Newton’s 2nd Law and is always perpendicular to the surface supporting the object (in this instance, the ground).

At the molecular level, this normal force comes from the ground's bonded molecules acting like tiny springs, compressed by the object’s molecules, providing an upward restorative force.

Thus, statements A) and B) are true.

serg [3.5K]1 month ago
4 0

Answer:

The bonded molecules in the object are pushed upwards by the bonded molecules in the ground, matching the gravitational force pulling down on the object.

The ground exerts a normal force upwards on the object, equal to the gravitational force acting downwards on it.

Explanation:

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- the index finger shows the direction of E
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7 0
1 month ago
Ben walks 500 meters from his house to the corner store. He then walks back toward his house, but continues 200 meters past his
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The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente
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J(r) = Br. We know that the area of a small segment, dA, is represented as 2 π dr. Thus, I = J A and dI = J dA. Plugging in the values gives us dI = B r. 2 π dr which simplifies to dI= 2π Br² dr. Now, integrating the above equation: Given that B= 2.35 x 10⁵ A/m³, with r₁ = 2 mm and r₂ equal to 2 + 0.0115 mm, or 2.0115 mm.
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A camera operator is filming a nature explorer in the Rocky Mountains. The explorer needs to swim across a river to his campsite
inna [3103]

Answer:

a. Angle= 28.82°

b. Approved. Although he might feel cold, he should be able to cross.

Explanation:

Velocity Vector

Velocity is a measure of how quickly something is moving in a specific direction. It is represented as a vector that has both magnitude and direction. If an object can only move in one direction, then speed can serve as the scalar equivalent of that velocity (only focusing on magnitude).

a.

The explorer aims to swim across a river to reach his campsite, as depicted in the image below. The river's velocity is vr and the explorer's swimming speed in still water is ve. If he were to swim straight towards the campsite, he would end up downstream due to the river's current. Therefore, he must swim at an angle that allows him to overcome the current while still moving towards his goal. This angle relative to the shore is what we need to determine. The explorer's speed can be broken down into its horizontal (vx) and vertical (vy) components. In order to counteract the river's flow:

v_{ey}=v_r

We can calculate the vertical component of the explorer's swimming speed as

v_{ey}=|v_e|cos\alpha

Thus

v_r=|v_e|cos\alpha

Finding the value of \alpha

\displaystyle cos\alpha=\frac{v_r}{|v_e|}

\displaystyle cos\alpha=\frac{0.665}{0.759}=0.876

Then the angle is given by

\alpha=28.82^o

b.

The component of the explorer's velocity that goes horizontally is

v_{ex}=0.759sin28.82^o

v_{ex}=0.366\ m/s

This represents the actual velocity directed towards the campsite

Considering that

\displaystyle v=\frac{x}{t}

To find t

\displaystyle t=\frac{x}{v}

Calculating the duration for the explorer to cross the river

\displaystyle t=\frac{29.3}{0.366}

t=80\ sec

As this time is under the hypothermia threshold (300 seconds), the conclusion is

Approved. Although he will feel cold, he should manage to cross successfully.

3 0
1 month ago
For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.300 kg of Italia
serg [3582]

Answer:

Part a)

A = 0.0581 m

Part b)

T = 0.37 s

Explanation:

A slice is dropped onto the plate from a height of 0.250 m,

therefore the speed of the slice upon impact is calculated as

v = \sqrt{2gh}

We know that

v = \sqrt{2(9.81)(0.250)}

v = 2.21 m/s

Now applying the conservation of momentum:

mv = (m + M)v_f

m = 0.300 kg

M = 0.400 kg

From this equation, we find:

0.300 (2.21) = (0.300 + 0.400) v_f

v_f = 0.95 m/s

0.400 (9.81) = 200 x_1

When the slice rests on the plate, the new mean position can be expressed as

x_1 = 0.01962 m

(0.300 + 0.400)9.81 = 200 x_2

We also determine that the speed of SHM is represented as

x_2 = 0.0343 m

Here, we derive values from

v = \omega\sqrt{A^2 - x^2}

\omega = \sqrt{\frac{k}{m + M}}

\omega = \sqrt{\frac{200}{0.300 + 0.400}}

\omega = 16.9 rad/s

a = x_2 - x_1 = 0.0343 - 0.01962 = 0.0147 m

Using the previous formula gives:

0.95 = 16.9\sqrt{A^2 - 0.0147^2}

A = 0.0581 m

Part b)

The time period for the scale is computed as

T = \frac{2\pi}{\omega}

T = \frac{2\pi}{16.9}

T = 0.37 s

8 0
1 month ago
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