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4 years ago

Eq-36 when paddling a canoe at night, which piece of equipment should be carried to help avoid a collision?

1 answer:

7 0

Flashlight should be carried to help avoid a collision, when paddling a canoe at night.
There are some rules and skills for paddling a canoe. So, before paddling you should know everything about it. Canoeing is not easy as it looks and not everyone can paddle a canoe. A strong stroke is a fundamental rule of Canoeing.

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A luggage handler pulls a suitcase of mass 19.6 kg up a ramp inclined at an angle 24.0 ∘ above the horizontal by a force F⃗ of m

Dvinal [7]

(a) 638.4 J

The work done by a force is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement

Here we want to calculate the work done by the force F, of magnitude

F = 152 N

The displacement of the suitcase is

d = 4.20 m along the ramp

And the force is parallel to the displacement, so $\theta=0^{\circ}$ . Therefore, the work done by this force is

$W_F=(152)(4.2)(cos 0)=638.4 J$

b) -328.2 J

The magnitude of the gravitational force is

W = mg

where

m = 19.6 kg is the mass of the suitcase

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$W=(19.6)(9.8)=192.1 N$

Again, the displacement is

d = 4.20 m

The gravitational force acts vertically downward, so the angle between the displacement and the force is

$\theta= 90^{\circ} - \alpha = 90+24=114^{\circ}$

Where $\alpha = 24^{\circ}$ is the angle between the incline and the horizontal.

Therefore, the work done by gravity is

$W_g=(192.1)(4.20)(cos 114^{\circ})=-328.2 J$

c) 0

The magnitude of the normal force is equal to the component of the weight perpendicular to the ramp, therefore:

$R=mg cos \alpha$

And substituting

m = 19.6 kg

g = 9.8 m/s^2

$\alpha=24^{\circ}$

We find

$R=(19.6)(9.8)(cos 24)=175.5 N$

Now: the angle between the direction of the normal force and the displacement of the suitcase is 90 degrees:

$\theta=90^{\circ}$

Therefore, the work done by the normal force is

$W_R=R d cos \theta =(175.4)(4.20)(cos 90)=0$

d) -194.5 J

The magnitude of the force of friction is

$F_f = \mu R$

where

$\mu = 0.264$ is the coefficient of kinetic friction

R = 175.5 N is the normal force

Substituting,

$F_f = (0.264)(175.5)=46.3 N$

The displacement is still

d = 4.20 m

And the friction force points down along the slope, so the angle between the friction and the displacement is

$\theta=180^{\circ}$

Therefore, the work done by friction is

$W_f = F_f d cos \theta =(46.3)(4.20)(cos 180)=-194.5 J$

e) 115.7 J

The total work done on the suitcase is simply equal to the sum of the work done by each force,therefore:

$W=W_F + W_g + W_R +W_f = 638.4 +(-328.2)+0+(-194.5)=115.7 J$

f) 3.3 m/s

First of all, we have to find the work done by each force on the suitcase while it has travelled a distance of

d = 3.80 m

Using the same procedure as in part a-d, we find:

$W_F=(152)(3.80)(cos 0)=577.6 J$

$W_g=(192.1)(3.80)(cos 114^{\circ})=-296.9 J$

$W_R=(175.4)(3.80)(cos 90)=0$

$W_f =(46.3)(3.80)(cos 180)=-175.9 J$

So the total work done is

$W=577.6+(-296.9)+0+(-175.9)=104.8 J$

Now we can use the work-energy theorem to find the final speed of the suitcase: in fact, the total work done is equal to the gain in kinetic energy of the suitcase, therefore

$W=\Delta K = K_f - K_i\\W=\frac{1}{2}mv^2\\v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(104.8)}{19.6}}=3.3 m/s$

6 0

3 years ago

How do we find euphoria...

Sonja [21]

Answer:

listen)) is the experience (or affect) of pleasure or excitement and intense feelings of well-being and happiness. Certain natural rewards and social activities, such as aerobic exercise, laughter, listening to or making music, and dancing, can induce a state of euphoria.

Explanation:

4 0

3 years ago

A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit

makvit [3.9K]

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

2.87 = -1.442 + 1.8 u

u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.

7 0

4 years ago

A spaceship maneuvering near planet zeta is located at r⃗ =(600i^−400j^+200k^)×103km, relative to the planet, and traveling at v

slava [35]

Answer:

The spaceship's position when the engine shuts off = $(708.15 i - 444.1 j + 200 k)*10^3km$

Explanation:

Initial location of spaceship = (600 i - 400 j + 200 k)* $10^3km$ = (600 i - 400 j + 200 k)* $10^6m$

Initial velocity = 9500 i m/s

Acceleration = (40 i - 20 k) $10^3m/s^2$

Time = 35 minute = 35 * 60 = 2100 seconds

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Substituting

$s= (9500 i)*2100+\frac{1}{2}*(40 i - 20 k)*2100^2\\ \\ s=9500*2100 i+20*2100^2i-10*2100^2j\\ \\ s=19.95*10^6i+88.2*10^6i-44.1*10^6j\\ \\ \\s=(108.15i-44.1j)*10^6m$

So final position = $((600 i - 400 j + 200 k)+(108.15i-44.1j))*10^6=(708.15 i - 444.1 j + 200 k)*10^6m$

= $(708.15 i - 444.1 j + 200 k)*10^3km$

The spaceship's position when the engine shuts off = $(708.15 i - 444.1 j + 200 k)*10^3km$

3 0

3 years ago

Read 2 more answers

a disk of a radius 50 cm rotates at a constant rate of 100 rpm. what distance in meters will a point on the outside rim travel d

trasher [3.6K]

50π m ≈ 157 m
Explanation:
100 rev/min (2π rad/rev) / (60 sec/min) = 3⅓π rad/s
d = ωrt = 3⅓π(0.50)(30) = 50π m ≈ 157 m

8 0

3 years ago