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Elena-2011 [213]

2 years ago

12

a speed swimmer love to race around the parks pine which is 25m around if she can swim 20 laps in 7200s what is her average spee

d?

1 answer:

aleksley [76]2 years ago

8 0

Answer:

the average speed of the swimmer is 0.069 m/s.

Explanation:

Given;

complete distance around the park pine, d = 25 m

total lap completed, = 20 laps

time of laps completion, t = 7200 s

The total distance completed by the swimmer = 20 x 25 = 500 m

The average speed of the swimmer = distance / time

= (500 m) / (7200 s)

= 0.069 m/s.

Therefore, the average speed of the swimmer is 0.069 m/s.

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A uniform solid disk with a mass of 24.3 kg and a radius of 0.364 m is free to rotate about a frictionless axle. Forces of 90.0

a_sh-v [17]

Answer:

a. $-12.7 Nm$

b. $-7.9 rad/s^2$

Explanation:

I have attached an illustration of a solid disk with the respective forces applied, as stated in this question.

Forces applied to the solid disk include:

$F_1 = 90.0N\\F_2 = 125N$

Other parameters given include:

Mass of solid disk, $M = 24.3kg$

and radius of solid disk, $r = 0.364m$

a.) The formula for determining torque ( $T$ ), is $T = r * F$

Hence the net torque produced by the two forces is given as a summation of both forces:

$T = T_{125} + T_{90}\\= -r(125)sin90 + r(90)sin90\\= 0.364(-125 + 90)\\= -12.7 Nm$

b.) The angular acceleration of the disk can be found thus:

using the formula for the Moment of Inertia of a solid disk;

$I_{disk} = {\frac{1}{2}}Mr^2$

where $M$ = Mass of solid disk

and $r$ = radius of solid disk

We then relate the torque and angular acceleration ( $\alpha$ ) with the formula:

$T = I\alpha \\-12.7 = ({\frac{1}{2}}Mr^2)\alpha \\\alpha = -{\frac{12.7}{1.61}} = -7.9 rad/s^2$

4 0

3 years ago

A physics student skis down a hill, accelerating at a constant

ikadub [295]

<h3>Answer:</h3>

225 meters

<h3>Explanation:</h3>

Acceleration is the rate of change in velocity of an object in motion.

In our case we are given;

Acceleration, a = 2.0 m/s²

Time, t = 15 s

We are required to find the length of the slope;

Assuming the student started at rest, then the initial velocity, V₀ is Zero.

<h3>Step 1: Calculate the final velocity, Vf</h3>

Using the equation of linear motion;

Vf = V₀ + at

Therefore;

Vf = 0 + (2 × 15)

= 30 m/s

Thus, the final velocity of the student is 30 m/s

<h3>Step 2: Calculate the length (displacement) of the slope </h3>

Using the other equation of linear motion;

S = 0.5 at + V₀t

We can calculate the length, S of the slope

That is;

S = (0.5 × 2 × 15² ) - (0 × 15)

= 225 m

Therefore, the length of the slope is 225 m

6 0

2 years ago

An archer fires an arrow at a castle 230 m away. The arrow is in flight for 6 seconds, then hits the wall of the castle and stic

Vika [28.1K]

Answer:

(a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Explanation:

Given that,

Horizontal distance = 230 m

Time t = 6 sec

Vertical distance = 16 m

We need to calculate the horizontal component

Using formula of horizontal component

$R =u\cos\theta t$

Put the value into the formula

$\dfrac{230}{6} = u\cos\theta$

$u\cos\theta=38.33$ .....(I)

We need to calculate the height

Using vertical component

$H=u\sin\theta t-\dfrac{1}{2}gt^2$

Put the value in the equation

$16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2$

$u\sin\theta=\dfrac{16+9.8\times18}{6}$

$u\sin\theta=32.06$ .....(II)

Dividing equation (II) and (I)

$\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}$

$\tan\theta=0.8364$

$\theta=\tan^{-1}0.8364$

$\theta=39.90^{\circ}$

(a). We need to calculate the initial speed

Using equation (I)

$u\cos\theta\times t=38.33$

Put the value into the formula

$u =\dfrac{230}{6\times\cos39.90}$

$u=49.96\ m/s$

(b). We have already calculate the angle.

Hence, (a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

5 0

3 years ago

A wheel of mass 4kg is pulled up a plane inclined at 30° to the horizontal by a force of 45N applied to the axle and parallel to

Len [333]

Answer:

v = 10 m/s

Explanation:

Let's assume the wheel does not slip as it accelerates.

Energy theory is more straightforward than kinematics in my opinion.

Work done on the wheel

W = Fd = 45(12) = 540 J

Some is converted to potential energy

PE = mgh = 4(9.8)12sin30 = 235.2 J

As there is no friction mentioned, the remainder is kinetic energy

KE = 540 - 235.2 = 304.8 J

KE = ½mv² + ½Iω²

ω = v/R

KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²

v = √(2KE / (m + I/R²))

v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6

v = 10.07968...

5 0

2 years ago

If the time of impact in a collision is extended by 3 times, by hour much is

vodomira [7]

Answer:

The force of the impact would be smaller

Explanation: Examples:

If the force is big then the time would be small (2500N of Force = 10 seconds)

If the force is small then the time would be big (250N of Force = 50 seconds)

Impulse/Collision -> [Ft] = [M (vf-vo)] <- Change in momentum

4 0

2 years ago