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Leviafan [203]
2 years ago
7

2. A car is sitting at the top of a hill that is 14 m high. The car has a mass of 53 kg. The car has

Physics
1 answer:
Mamont248 [21]2 years ago
5 0
The car has gravitational potential energy (Eg)

Eg = mgh
Eg = (53kg) (9.81N/kg) (14m)
Eg = 7 279.02 J
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Why are some fasteners double or triple threaded?
OLga [1]
<span>Fasteners are double or triple threaded with the idea of durability. People look for durability in a fastener and if they receive one with triple threaded or double threaded they will feel more safe and at ease knowing it has extra strength added.</span>
3 0
3 years ago
A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread non uniformly throu
Aloiza [94]
In other words a infinitesimal segment dV caries the charge 
<span>dQ = ρ dV </span>

<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>

<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>

<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
<span>= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) </span>
<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>
6 0
3 years ago
4. A car, initially traveling east with a speed of
PilotLPTM [1.2K]

Answer:

150 m

Explanation:

Given,

u=5m/s

a=2m/s2

t=10s

v=?

s=?

Now,

v=u+at

=5+2×10

=5+20

=25m/s

So,

s=u+v/2×t

=5+25/2×10

=30/2×10

=15×10

=150m

5 0
2 years ago
Mechanical waves propagate or move through a medium because
lys-0071 [83]
The answer to that would be that 

they require so its mandatory for mechanical waves to travel through a medium
8 0
3 years ago
An 89 kg man drops from rest on a diving board −3.1 m above the surface of the water and comes to rest 0.5 s after reaching the
OLga [1]

To solve this problem we will use the linear motion kinematic equations, for which the change of speed squared with the acceleration and the change of position. The acceleration in this case will be the same given by gravity, so our values would be given as,

m= 89 kg\\x = 3.1 m\\t = 0.5s\\a = g = 9.8m/s^2

Through the aforementioned formula we will have to

v_f^2-v_i^2 = 2ax

The particulate part of the rest, so the final speed would be

v_f^2 = 2gx

v_f=\sqrt{2(9.8)(3.1)}

v_f = 7.79m/s

Now from Newton's second law we know that

F = ma

Here,

m = mass

a = acceleration, which can also be written as a function of velocity and time, then

F = m\frac{dv}{dt}

Replacing we have that,

F = (89)\frac{7.79}{0.5}

F = 1386.62N

Therefore the force that the water exert on the man is 1386.62

3 0
3 years ago
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