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Molodets [167]
4 years ago
6

When you weigh yourself, how does the support force of the scale acting on you compare to the gravitational force between you an

d Earth?

Physics
1 answer:
Alex Ar [27]4 years ago
8 0

Answer with explanation:

We know that when we weigh ourselves on a scale the scale measures the Reaction force that acts between the scale and the person who wishes to determine his weight.

The free body diagram of the person on a weighing scale is represented in the attached figure.

As we can see for equilibrium

\sum F_{v}=0\\\\\Rightarrow R-W=0\\\\\therefore R=W

where,

R is the reaction force between the support and the person

W is the weight of the person which by definition is the force that the earth exerts on the person.

Thus according to the above relation we conclude that that support force of the scale is equal to the gravitational force in magnitude but is opposite in direction.

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Two blocks in contact with each other are pushed to the right across a rough horizontal surface by the two forces shown. If the
ale4655 [162]

I assume the blocks are pushed together at constant speed, and it's not so important but I'll also assume it's the smaller block being pushed up against the larger one. (The opposite arrangement works out much the same way.)

Consider the forces acting on either block. Let the direction in which the blocks are being pushed by the positive direction.

The 2.0-kg block feels

• the downward pull of its own weight, (2.0 kg) <em>g</em>

• the upward normal force of the surface, magnitude <em>n₁</em>

• kinetic friction, mag. <em>f₁</em> = 0.30<em>n₁</em>, pointing in the negative horizontal direction

• the contact force of the larger block, mag. <em>c₁</em>, also pointing in the negative horizontal direction

• the applied force, mag. <em>F</em>, pointing in the positive horizontal direction

Meanwhile the 3.0-kg block feels

• its own weight, (3.0 kg) <em>g</em>, pointing downward

• normal force, mag. <em>n₂</em>, pointing upward

• kinetic friction, mag. <em>f₂</em> = 0.30<em>n₂</em>, pointing in the negative horizontal direction

• contact force from the smaller block, mag. <em>c₂</em>, pointing in the <u>positive</u> horizontal direction (this is the force that is causing the larger block to move)

Notice the contact forces form an action-reaction pair, so that <em>c₁</em> = <em>c₂</em>, so we only need to find one of these, and we can get it right away from the net forces acting on the 3.0-kg block in the vertical and horizontal directions:

• net vertical force:

<em>n₂</em> - (3.0 kg) <em>g</em> = 0   ==>   <em>n₂</em> = (3.0 kg) <em>g</em>   ==>   <em>f₂</em> = 0.30 (3.0 kg) <em>g</em>

• net horizontal force:

<em>c₂</em> - <em>f₂</em> = 0   ==>   <em>c₂</em> = 0.30 (3.0 kg) <em>g</em> ≈ 8.8 N

4 0
3 years ago
Four equal masses m are located at the corners of a square of side L, connected by essentially massless rods. Find the rotationa
skelet666 [1.2K]

Answer:

Explanation:

a )

Moment of inertial of four masses about axis that coincides with one side :

Out of four masses . location of two masses will lie on the axis so their moment of inertia will be zero .

Moment of inertia of the two remaining masses

= m L² + m L²

= 2 mL²

b )

Axis that bisects two opposite sides

Each of the four masses will lie at a distance of L / 2 from this axis so moment of inertia of the four masses

= 4 x m x ( L/2 )²

= 4 x  mL² / 4

= m L² .

3 0
3 years ago
Band of which of these colours are not seen in a spectrum
Ede4ka [16]
The only colours that are in the spectrum are red, orange, yellow, green, blue, indigo and violet. hope this helps!
7 0
3 years ago
Answer and explanation please, much appreciated!! ​
NNADVOKAT [17]

Answer:

Done

Explanation:

I already answered this problem when you had previously asked for it. Kindly refer that solution.

3 0
2 years ago
A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of
Lynna [10]

We have the relation

\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}

where v_{A \mid B} denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}

v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}

Let's assume the river flows South-to-North, so that

\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath

and let -90^\circ < \theta < 90^\circ be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)

Then the velocity of the boat relative to the Earth is

\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath

The crossing is 153.0 m wide, so that for some time t we have

153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s

which is minimized when \theta=0^\circ so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}

The boat's position \vec x at time t is

\vec x = \vec v_{B\mid E} t

so that after 30.0 s, the boat's final position on the other side of the river is

\vec x(30.0\,\mathrm s) = (153\,\mathrm m) \,\vec\imath + (111\,\mathrm m)\,\vec\jmath

and the boat would have traveled a total distance of

\|\vec x(30.0\,\mathrm s)\| = \sqrt{(153\,\mathrm m)^2 + (111\,\mathrm m)^2} \approx \boxed{189\,\mathrm m}

3 0
2 years ago
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