If a road does not have a bicycle lane, where must a bicyclist ride their bicycle?

1 answer:

5 0

A bicyclist can ride their bicycle still on the road. Bicycle riders be able to take the public ways which has the similar rights and accountability as motorists and are subject to the same guidelines and protocols. The law says that individuals who ride bikes should ride as nearby to the right side of the road as likely excluding under the following conditions: when passing, preparing for a left go, evading risks, if the lane is too constricted to share, or if oncoming a place where a right turn is approved. In a road which has a bike lane the bicyclists roving slower than road traffic must custom the bike way excluding when creating a left turn, passing, evading hazardous settings, or impending a place where a right turn is approved.

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A bullet with a mass of 4.5 g is moving with a speed of 300 m/s (with respect to the ground) when it collides with a rod with a

Law Incorporation [45]

Answer:

ω = 5.41 rad/s

Explanation:

Since the rod is rotating around its axis, angular momentum will play part in this question.

The conservation of angular momentum implies that

$L_1 = L_2$

So, the initial angular momentum is

$L_1 = m_b v_b \frac{L}{4} = (4.5\times 10^{-3}~{\rm kg})(300~{\rm m/s})(\frac{0.25}{4}~{\rm m}) = 0.0844~{\rm kg.m^2/s}$

The final angular momentum includes the rod and the bullet together. So,

$L_2 = I\omega\\I = I_{rod} + I_{bullet} = \frac{1}{12}m_r L^2 + m_b(\frac{L}{4})^2 = \frac{1}{12}3(0.25)^2 + (4.5\times 10^{-3})(\frac{0.25}{4})^2 = 0.0156~{\rm kg.m^2}\\L_2 = L_1 = I\omega\\0.0844 = 0.0156\omega\\\omega = 5.41~{\rm rad/s}$

5 0

3 years ago

The volume of a gas is 200.0 mL at 275 K and 92.1 kPa. Find its volume at STP.

ycow [4]

To solve this question we will use ideal gas equation:
$p*V=n*R*T$
Where:
p = pressure
V = volume
n = number of moles
R = gas constant
T = temperature

We can rearrange formula to get:
$\frac{p*V}{T} =n*R$
We are working woth same gas so we can write following formula. Index 1 stands for conditions before change and index 2 stands for conditions after change.
$\frac{ p_{1}*V_{1} }{T_{1}} = \frac{ p_{2}*V_{2} }{T_{2}}$

We are given:
p1=92.1kPa = 92100Pa
V1=200mL = 0.2L
T1=275K
p2= 101325Pa
T2=273K
V2=?

We start by rearranging formula for V2. After that we can insert numbers:
${ p_{1}*V_{1} *T_{2}} = { p_{2}*V_{2}*T_{1} } \\ \\ V_{2}= \frac{p_{1}*V_{1} *T_{2}}{ p_{2}*T_{1}} \\ \\ V_{2}= \frac{92100*0.2*273}{101325*275} \\ \\ V_{2}= 0.18L=180mL$

7 0

4 years ago

A worker at the top of a 588-m-tall television transmitting tower accidentally drops a heavy tool. If air resistance is negligib

liq [111]

The final velocity of the tool is 107.4 m/s

Explanation:

We can solve this problem by using the principle of conservation of energy.

In fact, if air resistance is negligible, the total mechanical energy of the tool is conserved during the fall, so we can write:

$K_i + U_i = K_f + U_f$