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3 years ago

A train travels at a speed of 70km/h and travels a distance of 630 km. How long did it take the train to complete its journey?

Physics

2 answers:

sashaice [31]3 years ago

7 0

Answer:

9 hr

Explanation:

Speed = distance/time

Let the time taken by the train be t.

=> 70 = 630/t

=> t = 630/70

=> t = 9

it take 9 hr to complete its journey

lakkis [162]3 years ago

7 0

Answer:

9 hours

Explanation:

Using the formula to solve motion problems: d=r*t (Distance = Rate * Time)

Given:

Distance = 630 km

Rate (Speed) = 70 km/h

Time = Distance / Time

Time = 630/70

= 9

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3 years ago

A hill is 132 m long and makes an angle of 12.0 degrees with the horizontal. As a 54 kg jogger runs up the hill, how much work d

TEA [102]

Answer:

14523.55J

Explanation:

The work done by the jogger against gravity is given by the following equation;

$W=mgh.................(1)$

where m is the mass, g is acceleration due to gravity taken as $9.8m/s^2$ and h is the height of the hill.

Since the length of the hill is 132m and it is inclined at 12 degrees to the horizontal, the height is thus given as follows;

$h=132sin12^o\\h=27.44m$

Substituting this into equation (1) with all other necessary parameters, we obtain the following;

$W=54*9.8*27.44\\W=14523.55J$

4 0

3 years ago

A marble is dropped off the top of a building. Find it's velocity and how far below to its dropping point for the first 10 secon

Yuri [45]

consider the motion in y-direction

v₀ = initial velocity = 0 m/s

a = acceleration = g = - 9.8 m/s²

t = time

v = final velocity at any time "t"

velocity at any time is given as

v = v₀ + at

v = 0 + (- 9.8) t

v = (- 9.8) t

so at t = 1 , v = (- 9.8) (1) = - 9.8 m/s

at t = 2 , v = (- 9.8) (2) = - 19.6 m/s

at t = 3 , v = (- 9.8) (3) = - 29.4 m/s and so on

Y₀ = initial position where the marble was dropped from = 0 m

Y = final position at any time "t"

final position at any time "t" is given as

Y = Y₀ + v₀ t + (0.5) a t²

Y = 0 + (0) t + (0.5) (-9.8) t²

Y = - 4.9 t²

at t = 1 , Y = - 4.9 (1)² = - 4.9 m

at t = 2 , Y = - 4.9 (2)² = - 19.6 m

at t = 3 , Y = - 4.9 (3)² = - 44.1 m

and So on

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3 years ago

A horizontal clothesline is tied between 2 poles, 12 meters apart. When a mass of 1 kilograms is tied to the middle of the cloth

nadya68 [22]

Answer:

The tension on the clotheslines is $T = 8.83 \ N$

Explanation:

The diagram illustrating this question is shown on the first uploaded image

From the question we are told that

The distance between the two poles is $d = 12 \ m$

The mass tie to the middle of the clotheslines $m = 1 \ kg$

The length at which the clotheslines sags is $l = 4 \ m$

Generally the weight due to gravity at the middle of the clotheslines is mathematically represented as

$W = mg$

let the angle which the tension on the clotheslines makes with the horizontal be $\theta$ which mathematically evaluated using the SOHCAHTOA as follows

$Tan \theta = \frac{ 4}{6}$

=> $\theta = tan^{-1}[\frac{4}{6} ]$

=> $\theta = 33.70^o$

So the vertical component of this tension is mathematically represented a

$T_y = 2* Tsin \theta$

Now at equilibrium the net horizontal force is zero which implies that

$T_y - mg = 0$

=> $T sin \theta - mg = 0$

substituting values

$T = \frac{m*g}{sin (\theta )}$

substituting values

$T = \frac{1 *9.8}{2 * sin (33.70 )}$

$T = 8.83 \ N$

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3 years ago