Keep your lane position, and sound your horn while braking in a controlled manner. Sudden panic stops are not a good idea, as they could spook the animal, causing it to suddenly dart into the path of another vehicle.

4 0

3 years ago

Read 2 more answers

Some chlorine atoms have an atomic mass of 37, while others have an

lara [203]

Answer: D

Explanation: :) Just took the quizz

8 0

2 years ago

Read 2 more answers

(a) In what direction would the ship in Exercise 3.57 have to travel in order to have a velocity straight north relative to the

shtirl [24]

Answer:

Direction of ship: 9.45° West of North

Ship's relative speed: 7.87m/s

Explanation:

A. Direction of ship: since horizontal of the velocity of boat relative to the ground is 0

Vx=0

Therefore, -VsSin∅+VcCos∅40°

Sin∅ = Vc/Vs × Cos 40°

Sin∅ = 1.5/7 ×Cos40°

Sin∅= 0.164

∅= Sin-¹ (0.164)

∅= 9.45° W of N

B. Ship's relative speed:

Vy= VsCos∅ + Vcsin40°

= 7Cos9.45° + 1.5sin40°

= 7×0.986 + 1.5×0.642

= 7.865

= 7.87m/s

4 0

3 years ago

A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i

tresset_1 [31]

Answer:

a) Acceleration of the car is given as

$a_{car} = -21 m/s^2$

b) Acceleration of the truck is given as

$a_{truck} = 10.15 m/s^2$

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

$m_1v_1 + m_2v_2 = (m_1 + m_2)v$

now we have

$1400 (6.32) + 2900(0) = (1400 + 2900) v$

$v = 2.06 m/s$

a) For acceleration of car we know that it is rate of change in velocity of car

so we have

$a_{car} = \frac{v_f - v_i}{t}$

$a_{car} = \frac{2.06 - 6.32}{0.203}$

$a_{car} = -21 m/s^2$

b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

$a_{truck} = \frac{v_f - v_i}{t}$

$a_{truck} = \frac{2.06 - 0}{0.203}$

$a_{truck} = 10.15 m/s^2$

5 0

3 years ago