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k0ka [10]
3 years ago
5

In a crash test, a truck with mass 2500 kg traveling at 24 m/s smashes head-on into a concrete wall without rebounding. The fron

t end crumples so much that the truck is 0.72 m shorter than before.
a. What is the average speed of the truck during the collision (that is, during the interval between first contact with the wall and coming to a stop?
b. About how long does the collision last?
c. What is the magnitude of the average force exerted by the wall on the truck during the collision?
Physics
1 answer:
Molodets [167]3 years ago
7 0

Answer:

Explanation:

Given

mass of truck m=2500\ kg

velocity of truck u=24\ m/s

Crumbled distance d=0.72\ m

using equation of motion

v^2-u^2=2as

where u=initial speed

v=final speed

a=acceleration

s=displacement

0-(24)^2=2\times (a)\times (0.72)

a=-400\ m/s^2

(b)using v=u+at

here v=0

-u=-400\times t

t=\frac{24}{400}=0.06\ s

Impulse imparted is    J=F_{avg}\times \Delta t=\Delta P

F_{avg}=\frac{2500\times 24}{0.06}

F_{avg}=1000\ kN

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4 years ago
A bowling ball with a circumference of 27 in. weighs 14.8 lb and has a radius of gyration of 3.43 in. If the ball is released wi
Marrrta [24]

Answer:

Distance covered is 37.63 ft

Solution:

As per the question:

Circumference of the bowling ball, C = 27 in

Weight of the bowling ball, w = 14.8 lb

Radius of gyration, k = 3.43 in

Velocity of release of the ball, v' = 23 ft/s = 276 in/s

Coefficient of friction, \mu = 0.19

Now,

We know that the circumference of the circle is given by:

C = 2\pi R

27 = 2\pi R

R = \frac{27}{2\pi } = 4.29\ in

Also, in case of rolling, we know that:

Angular velocity, \omega = \frac{v}{R}

v = \omega R

Now, applying the conservation of angular momentum along the floor:

Initial angular momentum = Final angular momentum

m\omega' = m\omega + I\omega

Moment of Inertia, I = \frac{2}{5}mR^{2} = mk^{2}

mv'R = mvR + mk^{2}\times \frac{v}{R}

v'R = vR + k^{2}\times \frac{v}{R}

276\times 4.29 = 4.29v + 3.43^{2}\times \frac{276}{4.29}

1184.04 - 756.903 = 4.29v

v = 99.56 in/s

Now,

Friction force, f = \mu mg

Also, acceleration of the ball can be computed as:

\mu mg = - ma

a = - \mu g = - 0.19\times 386.09 = - 73.357\ in/s^{2}

Now, the distance, d covered by the ball before rolling without slipping:

v^{2} = v'^{2} + 2ad

99.56^{2} = 276^{2} + 2\times (- 73.357)d

99.56^{2} - 276^{2} = 2\times (- 73.357)d

d = 451.65 in = 37.53 ft

7 0
3 years ago
A +3.4 x 10-6 C test charge experiences forces from two other nearby charges: a 3 N force due east and a 15 N force due west. Wh
Alecsey [184]

Answer:

3.53×10⁶ N/c due west

Explanation:

From the question

E = F'/q........................ Equation 1

Where E =  Electric Field, F = Net Force, q = Charge.

But,

F' = F₂-F₁...................... Equation 2

Substitute equation 2 into equation 1

E = (F₂-F₁)/q................ Equation 3

Given: F₁ = 3 N due east, F₂ = 15 N due west,  q = 3.4×10⁻⁶ C

Substitute these values into equation 1

E = (15-3)/(3.4×10⁻⁶)

E = 12/(3.4×10⁻⁶)

E = 3.53×10⁶ N/c due west

4 0
3 years ago
energy transfered electrically is measured in _____. what one word completes the sentence?give the fullname not symbol.
Dmitry [639]

Answer:

joules

Explanation:

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4 0
3 years ago
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