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lesantik [10]
1 year ago
11

1. Which of the following regarding a collision is/are true? a. If you triple your speed your force of impact will be three time

s greater b. Speed has little effect on impact forces c. If you double your speed, the energy dissipated in a crash is four times greater d. If you double your speed, the energy dissipated in a crash is two times greater
Physics
2 answers:
34kurt1 year ago
7 0

Answer:

c

Explanation:

If you double your speed, the energy dissipated in a crash is four times greater

Because impact increases with square of increase in speed.

vlabodo [156]1 year ago
4 0

Answer:

c. If you double your speed, the energy dissipated in a crash is four times greater.

Explanation:

Given:

Impulse, P = force, f × time, t

= mass, m × velocity, v

Energy = kinetic energy, E = 1/2 × m × v^2

At v2 = 2v1

E1 = 1/2 × m × v1^2

= (m × v1^2)/2

E2 = 1/2 × m × (2v1^2)

= 1/2 × m × 4v1^2

= m × 2v1^2

(2 × E1)/v1^2 = E2/2v1^2

E1 = E2/4

E2 = 4 × E1

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A long coaxial cable (Fig. 2.26) carries a uniform volume charge density rho on the inner cylinder (radius a), and a uniform sur
Yuki888 [10]

Answer:

a) E = ρ / e0

b) E = ρ*a / (e0 * r)

c) E = 0

Explanation:

Because of the geometry, the electric field lines will all have a radial direction.

Using Gauss law

Q/e0 = \int \int E * dA

Using a Gaussian surface that is cylinder concentric to the cable, the side walls will have a flux of zero, because the electric field lines will be perpendicular. The round wall of the cylinder will have the electric field lines normal to it.

We can make this cylinder of different radii to evaluate the electric field at different points.

Then:

A = 2*π*r (area of cylinder per unit of length)

Q/e0 = 2*π*r*E

E = Q / (2*π*e0*r)

Where Q is the charge contained inside the cylinder.

Inside the cable core:

There is a uniform charge density ρ

Q(r) = ρ * 2*π*r

Then

E = ρ * 2*π*r / (2*π*e0*r)

E = ρ / e0 (electric field is constant inside the charged cylinder.

Between ther inner cilinder and the tube:

Q = ρ * 2*π*a

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E = ρ*a / (e0 * r)

Outside the tube, the charges of the core cancel each other.

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4 0
2 years ago
Suppose you look out the window of a skyscraper and see someone throw a tomato downward from above your window. your window is a
photoshop1234 [79]
Formula for height 
<span> r(t) = a/2 t² + v₀ t + r₀
</span><span> where 
</span><span> a = acceleration = -32 ft/sec² (gravity) 
</span><span> v₀ = initial velocity 
</span><span> r₀ = initial height 
</span><span> r(t) = -16t² + v₀ t + r₀
</span> <span>Tomato passes window (height = 450 ft) after 2 seconds: 
</span><span> r(2) = 450
</span><span> -16(4) + v₀ (2) + r₀ = 450 
</span><span> r₀ = 450 + 64 - 2v₀ 
</span><span> r₀ = 514 - 2v₀ 
</span><span> Tomato hits the ground (height = 0 ft) after 5 seconds: 
</span><span> r(5) = 0 
</span><span> -16(25) + v₀ (5) + r₀ = 0
</span> r<span>₀ = 16(25) - 5v₀ 
</span><span> r₀ = 400 - 5v₀ 
</span><span> 
 r₀ = 514 - 2v₀ and r₀ = 400 - 5v₀
</span> <span>514 - 2v₀ = 400 - 5v₀
</span><span> 5v₀ - 2v₀ = 400 - 514
</span> <span>3v₀ = −114 
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</span><span> Initial velocity = −38 ft/sec (so tomato was thrown down) 
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4 0
2 years ago
A pole-vaulter is nearly motionless as he clears the bar, set 4.2 m above the ground. he then falls onto a thick pad. the top of
scZoUnD [109]
Refer to the diagram shown below.

Neglect wind resistance, and use g = 9.8 m/s².

The pole vaulter falls with an initial vertical velocity of u = 0.
If the velocity upon hitting the pad is v, then
v² = 2*(9.8 m/s²)*(4.2 m) = 82.32 (m/s)²
v = 9.037 m/s

The pole vaulter comes to res after the pad compresses by  50 cm (or 0.5 m).
If the average acceleration (actually deceleration) is (a m/s²), then
0 = (9.037 m/s)² + 2*(a m/s²)*(0.5 m)
a = - 82.32/(2*0.5) = - 82 m/s²

Answer: - 82 m/s² (or a deceleration of 82 m/s²)

8 0
2 years ago
Read 2 more answers
Suppose a convex mirror has a focal length of 120 cm. A candle sits directly in front of the mirror. If the image of that candle
Andru [333]
We can solve the problem by using the mirror equation:
\frac{1}{f} = \frac{1}{d_o}+ \frac{1}{d_i}
where
f is the focal length
d_o is the distance of the object from the mirror
d_i is the distance of the image from the mirror

For the sign convention, the focal length is taken as negative for a convex mirror:
f=-120 cm
and the image is behind the mirror, so virtual, therefore its sign is negative as well:
d_i=-24 cm
putting the numbers in the mirror equation, we find the distance of the object from the mirror surface:
\frac{1}{d_o} = \frac{1}{f}- \frac{1}{d_i}= \frac{1}{-120 cm} - \frac{1}{-24 cm}= \frac{1}{30 cm}
So, the distance of the object from the mirror is d_o = 30 cm
8 0
2 years ago
Ten days after it was launched toward Mars in December 1998, the Mars Climate Orbiter spacecraft (mass 629 kg) was 2.87×106km fr
Andreas93 [3]

Answer:

3494444444.44444 J

-87077491.39453 J

Explanation:

M = Mass of Earth = 6.371\times 10^{6}\ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

R = Radius of Earth = 6.371\times 10^{6}\ m

h = Altitude = 2.87\times 10^9\ m

m = Mass of satellite = 629 kg

v = Velocity of spacecraft = 1.2\times 10^4\ km/h

The kinetic energy is given by

K=\frac{1}{2}629\times \left(1.2\times 10^4\times \dfrac{1000}{3600}\right)^2\\\Rightarrow K=3494444444.44444\ J

The spacecraft's kinetic energy relative to the earth is 3494444444.44444 J

Potential energy is given by

U=-\dfrac{GMm}{R_e+h}\\\Rightarrow U=-\dfrac{6.67\times 10^{-11}\times 5.97\times 10^{24}\times 629}{2.87\times 10^9+6.371\times 10^{6}}\\\Rightarrow U=-87077491.39453\ J

The potential energy of the earth-spacecraft system is -87077491.39453 J

4 0
1 year ago
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