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raketka [301]
3 years ago
12

The tortoise and the hare are having a race over a course of length 0.52 km. The tortoise crawls along at a constant 0.075 m/s w

hile the hare sprints along at a constant 16.7 m/s (when he is not sleeping). The race starts and the hare runs halfway to the finish line and decides to take a nap while the tortoise plods along. Waking from his nap, the hare sees that the tortoise has made so much progress that he might lose the race. He takes off and just catches up to the toroise at the finish line so that the race ends in a tie. Be careful with units.
How many hours did the hare sleep before waking up?
Physics
1 answer:
EastWind [94]3 years ago
3 0

Answer:the hare slept for 1 hr 55minutes.

Explanation: distance is changed to meters = 520m

Speed=distance/time

Speed of tortoise is 0.075m/s=520m/t

Crops multiplying t= 520/0.075 = 6933.3 seconds.

Speed for hare 16.7m/s =260m/t

ie 260m is half of the distance.

t =15.56seconds

Therefore the time slept by the hare = 6933.3- 15.56= 6917.74 seconds.

Converting to hours = 1hr. 55minutes is the time the hare slept.

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The total mass of the train and its passengers is 750000kg. The train is traveling at a speed of 84m/s. The driver applies the b
fiasKO [112]

Answer:

|F| = 393750  N

Explanation:

Given that,

Total mass of the train, m = 750000 kg

Initial speed, u = 84 m/s

Final speed, v = 42 m/s

Time, t = 80 s

We need to find the net force acting on the train. The formula for force is given by :

F = ma

F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{750000\times (42-84)}{80}\\\\F=-393750\ N

So, the magnitude of net force is 393750  N.

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3 years ago
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3 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
3 years ago
A proton moves with a speed of 1.00 x 106 m/s perpendicular to a magnetic field, B. As a result, the proton moves in circle of r
Softa [21]

Answer:

0.0109 m ≈ 10.9 mm

Explanation:

proton speed = 1 * 10^6 m/s

radius in which the proton moves = 20 m

<u>determine the radius of the circle in which an electron would move </u>

we will apply the formula for calculating the centripetal force for both proton and electron ( Lorentz force formula)

For proton :

Mp*V^2 / rp  = qp *VB   ∴  rp = Mp*V / qP*B    ---------- ( 1 )

For electron:

re = Me*V/ qE * B -------- ( 2 )

Next: take the ratio of equations 1 and 2

re / rp = Me / Mp                                 ( note: qE = qP = 1.6 * 10^-19 C )

∴ re ( radius of the electron orbit )

= ( Me / Mp ) rp

= ( 9.1 * 10^-31 / 1.67 * 10^-27 ) 20

= ( 5.45 * 10^-4 ) * 20

= 0.0109 m ≈ 10.9 mm

5 0
3 years ago
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