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3 years ago

13

How do contact forces such as friction differ from gravitational and magnetic forces?

1 answer:

Sloan [31]3 years ago

7 0

"Contact" means "touching". A contact force can't act on an object unless
they're touching. Friction is a contact force.

Gravitational forces pull masses together even if they're not touching.
Magnetic forces pull a piece of iron toward a magnet even if they're not touching.

Gravitational and magnetic forces are "non-contact" forces.

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A man weighing 180 lbf pushes a block weighing 100 lbf along a horizontal plane. the dynamic coefficient of friction between the

Talja [164]

The first thing you should know is that the work is defined as:
W = F * d
Where
F = force
d = displacement
We have then
(a) the block
F = (0.2) * (100) = 20
d = 100
W = (20) * (100) = 2000 ft.lbf
(b) the man as the system.
F = (0.2) * (100 + 180) = 56
d = 100
W = (56) * (100) = 5600 ft.lbf
answer:
(a) 2000 ft.lbf
(b) 5600 ft.lbf

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3 years ago

The Achilles tendon, which connects the calf muscles to the heel, is the thickest and strongest tendon in the body. In extreme a

Tanzania [10]

Answer:

1) tensile stress = 76.648 Mpa

2) extension = 0.0215 m

Explanation:

Detailed explanation and calculation is shown in the image below

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4 years ago

A batter swings and hits a pitched baseball far over the left field wall. at the moment the baseball contacts the bat, which obj

Lyrx [107]

The baseball bat not the baseball

7 0

3 years ago

The question states: two large, parallel conducting plates are 12cm

AnnZ [28]

Answer:

1. 24375 N/C

2. 2925 V

Explanation:

d = 12 cm = 0.12 m

F = 3.9 x 10^-15 N

q = 1.6 x 10^-19 C

1. The relation between the electric field and the charge is given by

F = q E

So, $E=\frac{F}{q}$

$E=\frac{3.9 \times 10^{-15}}{1.6 \times 10^{-19}}$

E = 24375 N/C

2. The potential difference and the electric field is related by the given relation.

V = E x d

where, V be the potential difference, E be the electric field strength and d be the distance between the electrodes.

By substituting the values, we get

V = 24375 x 0.12 = 2925 Volt

6 0

3 years ago

A 13.0 kg wheel, essentially a thin hoop with radius 1.80 m, is rotating at 469 rev/min. It must be brought to a stop in 16.0 s.

Stella [2.4K]

Answer:

Explanation:

Given

mass of wheel m=13 kg

radius of wheel=1.8 m

N=469 rev/min

$\omega =\frac{2\pi \times 469}{60}=49.11 rad/s$

t=16 s

Angular deceleration in 16 s

$\omega =\omega _0+\alpha \cdot t$

$\alpha =\frac{\omega }{t}=\frac{49.11}{16}=3.069 rad/s^2$

Moment of Inertia $I=mr^2=13\times 1.8^2=42.12 kg-m^2$

Change in kinetic energy =Work done

Change in kinetic Energy $=\frac{I\omega ^2}{2}-\frac{I\omega _0^2}{2}$

$\Delta KE=\frac{42.12\times 49.11^2}{2}=50,792.34 J$

(a)Work done =50.79 kJ

(b)Average Power

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7 0

3 years ago