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Scorpion4ik [409]

4 years ago

8

Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slid

e forward, crushing the cab, if the truck stops sud- denly in an accident or even in braking. Assume, for exam- ple, a 10 000-kg load sits on the flatbed of a 20 000-kg truck moving at 12.0 m/s. Assume the load is not tied down to the truck and has a coefficient of static friction of 0.500 with the truck bed. (a) Calculate the minimum stopping distance for which the load will not slide forward relative to the truck. (b) Is any piece of data unnecessary for the solution?

1 answer:

zhuklara [117]4 years ago

3 0

Answer:a) There is no relative motion between the truck and the load if they have the same acceleration.

b) mass of the load and the truck are unncessary

Explanation: a) There is no relative motion between the truck and the load if they have the same acceleration.

Let their accelerations be a. Then a is in the direction opposite to that of the velocity of the truck (only then can the truck stop).

Let the velocity of the truck be towards right and the acceleration be towards left.

Now make the free-body diagram for the load :

Weight (mg) acts vertically downwards

Normal reaction (N) acts vertically upwards

So N = mg

Force of friction (f) acts towards left (since the block has a tendency to slip towards right)

Maximum value of friction , f = uN = umg

So f = ma

or a = ug = 4 m/s^2

Now consider the truck :

Initial velocity, u = 11 m/s

Final velocity, v = 0 m/s

a = - 4m/s^2 (- sign shows that a is in the direction opposite to that of u)

So

s = (v^2 - u^2)/2a = 15.125 m

(b) Mass of the load and the truck are unnecessary.

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A skydiver free falls from rest. How long does it take for the skydiver to reach 7 m/s?

aev [14]

Answer:

<em>The skydiver needs 0.71 seconds to reach 7 m/s</em>

Explanation:

<u>Free Fall Motion </u>

When an object is dropped in free air (no friction) from a certain height h, it follows a free-fall motion, whose acceleration is due exclusively to gravity. The speed at a moment t when the object is dropped (from rest) is:

$V_f=gt$

We need to find How long does the skydiver needs to reach 7 m/s. We solve for t

$\displaystyle t=\frac{V_f}{g}$

$\displaystyle t=\frac{7}{9.8}$

$t=0.71\ sec$

The skydiver needs 0.71 seconds to reach 7 m/s

5 0

3 years ago

A projectile of mass 9.6 kg is launched from the ground with an initial velocity of 12.4 m/s at angle of 54° above the horizonta

Temka [501]

Answer:

The location is at (3.535, 1.162) m

Solution:

As per the question:

Mass of the projectile, m = 9.6 kg

Initial velocity, v = 12.4 m/s

Angle, $\theta = 54^{\circ}$

Mass of one fragment, m = 6.5 kg

Time taken by the fragment, t = 1.42 s

Height of the fragment, y = 5.9 m

Horizontal distance, x = 13.6 m

Now,

To determine the location of the second fragment:

Horizontal Range, $R = \frac{v^{2}sin2\theta}{g}$

$R = \frac{12.4^{2}sin2(54)}{9.8} = 14.92\ m$

Time of flight, t' = $\frac{2vsin\theta}{g} = \frac{2\times 12.4sin108}{9.8}= 2.406\ s$

Now, for the fragments:

Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg

Distance traveled horizontally:

$s_{x} = vcos\theta = 12.4cos54^{\circ}\times 1.42 = 10.35\ m$

Distance traveled vertically:

$s_{y} = vcos\theta - \frac{1}{2}gt^{2}$

$s_{y} = 12.4sin54^{\circ}\times 1.42 - \frac{1}{2}\times 9.8\times 1.42^{2} = 14.25 - 9.88 = 4.37\ m$

Now,

$s_{x} = \frac{mx + m'x'}{M}$

$10.35= \frac{6.5\times 13.6 + 3.1x'}{9.6}$

x' = 3.535 m

Similarly,

$s_{y} = \frac{my + m'y'}{M}$

$4.37= \frac{6.5\times 5.9 + 3.1y'}{9.6} = 1.162\ m$

The location of the other fragment is at (3.535, 1.162)

5 0

4 years ago

A 350 gram mass is tied to a string and spin in a horizontal circle with a radius of 11.0cm.The speed of the mass is held consta

SSSSS [86.1K]

Explanation:

a) The angular speed omega is given defined as

omega = #rev/period = 1 rev/(0.6 s) = 1.7 rev/s

= (1.7 rev/s)×(2pi/rev) = 10.5 rad/s

b) v = (omega)×r

= (10.5 rad/s)(0.110 m)

= 1.16 m/s

7 0

3 years ago

An object has a mass of 12kg. On Planet A, the object weighs 117.6 N. Is the force of gravity on Planet A equal to, less than or

kiruha [24]

Well I don't know !
Let's work it out:

   Weight = (mass) x (local acceleration of gravity)

   117.6 N = (12 kg) x (gravity)

   Gravity on Planet A = (117.6 N) / (12 kg)

   = 9.8 m/s² exactly

The gravity on Planet-A is so close to Earth gravity
that nobody could ever tell the difference without
making sensitive measurements.
They are essentially equal.

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3 years ago

If root-mean-square voltage of a supply is 240 V, calculate the maximum voltage of

Levart [38]

Answer:

Vp= Vr sq. root 2

= 240 × sq. root 2

=339.4 V

5 0

4 years ago