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3 years ago

12

If a train travels at a speed of 40 km/h for 3 hours, how far will it travel

2 answers:

VikaD [51]3 years ago

8 0

120 km/3 hours. 40/1=?/3 1x3=3 hours so 40x3=120 km

Alik [6]3 years ago

4 0

Answer:

120 meters

Explanation:

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please help. Sound travels at 330 m/s. If a lightning bolt strikes the ground 5 m away from you, how long will it take for the s

Olenka [21]

Answer:

0.015 seconds

Explanation:

5/330=0.015

3 0

2 years ago

Why does the density of a substance remain the same for different amount of the substance

tresset_1 [31]

Think of it this way:
-- Any time you have something that means (some number) PER UNIT,
it doesn't matter how many units there are on the table or in the bucket,
because that amount doesn't change the (number) PER UNIT.

-- If oranges cost $1 PER POUND, it doesn't matter how many pounds
you buy, the whole bagful is still $1 PER POUND.

-- If a certain salad dressing has 40 calories PER Tablespoon, it doesn't
matter whether you eat a drop of it or drink the whole jar. You still get
40 calories PER Tablespoon.

-- Density means '(mass) PER unit of volume'. Whether you have a tiny
chip of the substance or a whole truckload of it, there's still the same
amount of mass IN EACH unit of volume.

6 0

3 years ago

A 2430 pound roller coaster starts from rest and is launched such that it crests a 105 ft high hill with a speed of 59 mph. The

Rudik [331]

Answer:

Explanation:

Given

Weight of roller coaster is $W=2430\ pound$

mass of roller coaster $m=\frac{W}{g}=\frac{2430}{32.2}=75.45$

Distance traveled by roller coaster $d=396\ ft$

drag force $f_d=85\ pounds$

velocity at top $v=59 mph\approx 86.53\ ft/s$

Suppose E is the initial energy

Conserving Energy at bottom and top

$E=\frac{1}{2}mv^2+mgh+f_d\cdot d$

$E=0.5\times 75.45\times 86.53^2+2430\times 105+85\times 396$

$E=2.9\times 10^5\ foot-pound$

5 0

3 years ago

A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the

irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

$x=V_{o} cos \theta t$ (1)

$y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2}$ (2)

Where:

$x$ is the horizontal distance between the cannon and the ball

$V_{o}=23 m/s$ is the cannonball initial velocity

$\theta=0\°$ since the cannonball was shoot horizontally

$t$ is the time

$y=0$ is the final height of the cannonball

$y_{o}=1.96 m$ is the initial height of the cannonball

$g=9.8 m/s^{2}$ is the acceleration due gravity

Isolating $t$ from (2):

$t=\sqrt{-\frac{2(y-y_{o})}{g}}$ (3)

$t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}}$ (4)

$t=0.63 s$ (5)

Substituting (5) in (1):

$x=(23 m/s) cos(0\°) 0.63 s$ (6)

Finally:

$x=14.49 m$

5 0

2 years ago

Suppose Earth orbited a star whose mass was double the mass of the sun. If the radius of Earth’s orbit remained the same as it i

Vaselesa [24]

Answer:

Two times as much

Explanation:

The equation for gravitational force is: Fg = GMm/r^2 with G being the universal gravitational constant.

So to make things easier we'll set r equal to 1 since it's a constant as well as G.

Then we're left with Fg=Mm with M being the mass of the sun and m being the mass of the earth.

So if m is constant and supposedly equals 1 then Fg=M so Fg is proportional to M therefore if M doubles then Fg doubles.

7 0

2 years ago