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Anastaziya [24]
3 years ago
6

Determine the automobile’s braking distance from 90 km/h when it is going up a 5° incline. (Round the final value to one decimal

place.) The automobile’s braking distance from 90 km/h when it is going up a 5° incline is:
Physics
1 answer:
NeX [460]3 years ago
8 0

Answer:

366 m

Explanation:

u = 90 km/h = 25 m/s,

theta = 5 degree

acceleration, a = g Sin theta = 9.8 x Sin 5 = 0.854 m/s^2

The final velocity os zero and let the braking distance be s.

Use third equation of motion

v^2 = u^2 - 2 a s

0 = 25 x 25 - 2 x 0.854 x s

s = 366 m

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Explanation:

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Maximum weight in this case will be 50 N

(b)acceleration of magnitude 2.98 m/s^2

T-mg=ma

50=m\left ( g+a\right )

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3 years ago
A pin-supported, vertically-oriented 1-m long thin rod is struck by a pellet at m down from the pin at the top. The mass of the
Natasha_Volkova [10]

Answer:

the angular velocity of the rod immediately after being struck by the pellet, provided that the pellet gets lodged in the rod is = 0.5036 k` rad/s

Explanation:

Using the conservation of momentum of approach.

From the question; the pellet is hitting at a distance of 0.4 m down from the point of rotation of the rod.

So, the angular momentum of the system just before the collision occurs  with respect to the axis of the rotation is expressed by the formula:

L_i ^ { ^  \to } = mp ( r_y } ^ { ^  \to }  * v_{pi}  ^ { ^  \to } )    ----- equation (1)

The position vector can now be :

x ^ { ^  \to } = - 0.4 \ j \ m

Also, given that :

v_{p,i}  ^ { ^  \to } = (280 \ i - 350 \  j)  \ m/s

Replacing the value into above equation (1); we have:

L_i ^ { ^  \to } =0.012 ((- \ 0.4 \  j) *(280 \ i - 350 \ j ))

L_i ^ { ^  \to } =0.012 * 112 \  k    (by using cross product )

L_i ^ { ^  \to } = 1.344 k` \  \  kg  m^2 s^{-1}

However; the moment of inertia of the rod about the axis of rotation is :

I_{rod} = \frac{1}{3}m_rl^2  \\ \\ I_{rod} = \frac{1}{3}*8*1^2 \\ \\ I_{rod} = \frac{8}{3} \ \  kg  \ m^2

Also, the moment of inertia of the pellet about the axis of rotation is:

I_{pellet} = m_pr_y^2 \\ \\ I_{pellet} = 0.012 *0.4^2  \\ \\ I_{pellet} = 1.92*10^{-3} kg . m^2

So, the moment of inertia of the rod +pellet system is:

I = I_{rod}+I_{pellet}

I =( \frac{8}{3}+1.92 *10^{-3} )kg. m^2

I = 2.6686 \  kg. m^2

The final angular momentum is :

L_f ^ {^ \to}  = I \omega { ^ {^ \to} } = 2.6686 \  \omega  ^ {^ \to}

The angular velocity of the rod \omega is determined by equating the angular momentum just before the collision with the final angular momentum (i.e after the collision).

So;

L_f ^ {^ \to} = L_i ^ { ^ \to}

2.6686 \omega ^ {^ \to} = 1.344 \ k ^ {^ \to}

\omega  ^ {^ \to} = \frac{1.344 \ k`}{2.6686}

= 0.5036 k` rad/s

Hence; the angular velocity of the rod immediately after being struck by the pellet, provided that the pellet gets lodged in the rod is = 0.5036 k` rad/s

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