The chances of being involved in a fatal crash is higher at night then during the day

When two sticks are laid end-to-end they cover a length of 8.32 feet. One stick is 2.93 ft longer than the other. What is the le

77julia77 [94]

To solve this problem we will start by defining the length of the shortest stick as 'x'. And the magnitude of the longest stick, according to the statement as

$x+2.93$

Both cover a magnitude of 8.32 ft, therefore

$x +(x+2.97) = 8.32$

Now solving for x we have,

$x + (x + 2.93) = 8.32$

$2x + 2.93 = 8.32$

$2x = 8.32 - 2.93$

$x = \frac{ 8.32 - 2.93}{2}$

$x = 2.695 ft$

Therefore the shorter stick is 2.695ft long.

7 0

3 years ago

a 300kg motorboat is turned off as it approaches a dock and coasts towards it at .5 m/s. Isaac, whose mass is 62 kg jumps off th

Zolol [24]

-- Before he jumps, the mass of (Isaac + boat) = (300 + 62) = 362 kg,
their speed toward the dock is 0.5 m/s, and their linear momentum is

Momentum = (mass) x (speed) = (362kg x 0.5m/s) = 181 kg-m/s

relative to the dock. So this is the frame in which we'll need to conserve
momentum after his dramatic leap.

After the jump:

-- Just as Isaac is coiling his muscles and psyching himself up for the jump,
he's still moving at 0.5 m/s toward the dock. A split second later, he has left
the boat, and is flying through the air at a speed of 3 m/s relative to the boat.
That's 3.5 m/s relative to the dock.

His momentum relative to the dock is (62 x 3.5) = 217 kg-m/s toward it.

But there was only 181 kg-m/s total momentum before the jump, and Isaac
took away 217 of it in the direction of the dock. The boat must now provide
(217 - 181) = 36 kg-m/s of momentum in the opposite direction, in order to
keep the total momentum constant.

Without Isaac, the boat's mass is 300 kg, so

(300 x speed) = 36 kg-m/s .

Divide each side by 300: speed = 36/300 = 0.12 m/s , away from the dock.
=======================================

Another way to do it . . . maybe easier . . . in the frame of the boat.

In the frame of the boat, before the jump, Isaac is not moving, so
nobody and nothing has any momentum. The total momentum of
the boat-centered frame is zero, which needs to be conserved.

Isaac jumps out at 3 m/s, giving himself (62 x 3) = 186 kg-m/s of
momentum in the direction toward the dock.

Since 186 kg-m/s in that direction suddenly appeared out of nowhere,
there must be 186 kg-m/s in the other direction too, in order to keep
the total momentum zero.

In the frame of measurements from the boat, the boat itself must start
moving in the direction opposite Isaac's jump, at just the right speed
so that its momentum in that direction is 186 kg-m/s.
The mass of the boat is 300 kg so
(300 x speed) = 186

Divide each side by 300: speed = 186/300 = 0.62 m/s away from the jump.

Is this the same answer as I got when I was in the frame of the dock ?
I'm glad you asked. It sure doesn't look like it.

The boat is moving 0.62 m/s away from the jump-off point, and away from
the dock.
To somebody standing on the dock, the whole boat, with its intrepid passenger
and its frame of reference, were initially moving toward the dock at 0.5 m/s.
Start moving backwards away from that at 0.62 m/s, and the person standing
on the dock sees you start to move away from him at 0.12 m/s, and that's the
same answer that I got earlier, in the frame of reference tied to the dock.

yay !

By the way ... thanks for the 6 points. The warm cloudy water
and crusty green bread are delicious.

4 0

3 years ago

You are walking along a small country road one foggy morning and come to an intersection. While you are crossing, you hear an am

emmainna [20.7K]

Answer:

64.945 miles per hour

Explanation:

Since the frequency of sound heard is higher than actual frequency, the ambulance is moving towards you!

The frequency of sound waves as heard from a distance for a sound wave coming towards one at v₀ m/s and whose real frequency is f₀ is given by

+f = f₀/[1 - (v₀/v)]

+f = frequency of sound as heard from the distance away = 8.61 KHz

f₀ = real frequency of sound = 7.87 KHz