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k0ka [10]
1 year ago
12

A trained sea lion slides from rest down a long

Physics
1 answer:
andrey2020 [161]1 year ago
5 0

Answer:

1.5 m/s²

Explanation:

Draw a free body diagram.  There are three forces acting on the sea lion: gravity pulling down, normal force perpendicular to the ramp, and friction parallel to the ramp.

Sum of the forces perpendicular to the incline:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces parallel to the incline:

∑F = ma

mg sin θ − Nμ = ma

Substitute for N:

mg sin θ − (mg cos θ) μ = ma

g sin θ − g cos θ μ = a

a = g (sin θ − μ cos θ)

Given θ = 23° and μ = 0.26:

a = 9.8 (sin 23 − 0.26 cos 23)

a = 1.48

Rounded to two significant figures, the sea lion accelerates at 1.5 m/s².

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erastova [34]
•wind
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5 0
2 years ago
An ant travels 2.78 cm [W] and then turns and travels 6.25 cm [S 40 degrees E]. What is the ant's total displacement?
Sunny_sXe [5.5K]
Answer is 6.84 approx
reason:-
            (2.78^2+6.25^2)^1/2=6.84 approx
4 0
1 year ago
A block rests on a flat plate that executes vertical simple harmonic motion with a period of 0.74 s. What is the maximum amplitu
Mumz [18]

Answer:

maximum amplitude  = 0.13 m

Explanation:

Given that

Time period T= 0.74 s

acceleration of gravity g= 10 m/s²

We know that time period of simple harmonic motion given as

T=\dfrac{2\pi}{\omega}

0.74=\dfrac{2\pi}{\omega}

ω = 8.48 rad/s

ω=angular frequency

Lets take amplitude = A

The maximum acceleration given as

a= ω² A

The maximum acceleration should be equal to g ,then block does not separate

a= ω² A

10= 8.48² A

A=0.13 m

maximum amplitude  = 0.13 m

8 0
2 years ago
A seasoned mini golfer is trying to make par on a tricky number five hole. The golfer can complete the hole by hitting the ball
liberstina [14]

Answer:5.17 m/s

Explanation:

Given

let u be the speed at cliff initial point

range over cliff is 1.45 m

and range of projectile is given by

R=\frac{u^2\sin 2\theta }{g}

1.45=\frac{u^2\sin 90}{9.8}

u=3.77 m/s

Conserving Energy

E_{bottom}=E_{initial\ point\ at\ cliff}

Kinetic energy=Kinetic energy +Potential energy gained

Let v be the initial velocity

\frac{mv^2}{2}=mgh+\frac{mu^2}{2}

v^2=u^2+2gh

v=\sqrt{u^2+2gh}

v=\sqrt{3.77^2+2\time 9.8\times 0.64}

v=\sqrt{26.75}=5.17 m/s

5 0
1 year ago
The dogs of four-time Iditarod Trail Sled Dog Race champion Jeff King pull two 100-kg sleds that are connected by a rope. The sl
KonstantinChe [14]

Answer:

Acceleration, a=1.2\ m/s^2

Explanation:

Given that,

The dogs of four-time Iditarod Trail Sled Dog Race champion Jeff King pull two 100-kg sleds that are connected by a rope, m = 100 kg

Force exerted by the doges on the rope attached to the front sled, F = 240 N

To find,

The acceleration of the sleds.

Solution,

Let a is the acceleration of the sleds. The product of mass and acceleration is called force. Its expression is given by :

F = ma

a=\dfrac{F}{m}

a=\dfrac{240\ N}{2\times 100\ kg} (m = 2m)

a=1.2\ m/s^2

So, the acceleration of the sleds is 1.2\ m/s^2.

6 0
1 year ago
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