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2 years ago

13

To calculate acceleration you must know both the objects velocity and_____

1 answer:

amm18122 years ago

8 0

You need to know the time as well.

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A tank is 10 m long, 6 m wide, 3 m high, and contains kerosene with density 820 kg/m3 to a depth of 2.5 m. (Use 9.8 m/s2 for the

Valentin [98]

Answer:

Explanation:

a )

Pressure on the bottom

= hdg

= 2.5 x 820 x 9.8

= 20090 N / m²

b )

force = area x pressure

= 20090 x 10 x 6 N

= 1205400 N

c )

Force on 10m x 2.5 m face

center of gravity of the tank will lie at height 2 .5 / 2 = 1.25 m

Pressure = hdg

= 1.25 x 820 x 9.8

= 10045 N/m²

force = pressure x area

= 10045 x 10 x 2.5

= 251125 N .

7 0

2 years ago

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some fuel cells are powered by hydrogen. scientists are looking into the decomposition of water (h2o) to make hydrogen fuel with

sesenic [268]

Answer : First of all, to harness the solar energy and utilize it for the decomposition of water to make hydrogen fuel cell seems to be easy but practically it has many hurdles in its way. It might not work well, when the days would not be sunny. Also, the production of hydrogen fuel would require decomposing water which may need a high energy of activation, which would be more than the energy that will be produced from the fuel cell.

4 0

3 years ago

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A ball of mass is released from rest at a height of 30 how fast is it going when it hits the ground

elena55 [62]

Answer:

If the height is in metres, the speed is 24.25m/s

7 0

2 years ago

In a mail-sorting facility, a 2.50-kg package slides down an inclined plane that makes an angle of 20.0° with the horizontal. Th

lawyer [7]

Answer:

The coefficient of kinetic friction is 0.382.

Explanation:

Given:

Angle of inclination is, $\theta=20.0$ °

Mass of package is, $m=2.50\ kg$

Initial speed of package is, $u=2.00\ m/s$

Final speed of the package at the bottom is, $v=0\ m/s$

Distance of travel along the incline is, $d=12.0\ m$

Acceleration due to gravity is, $g=9.8\ m/s^2$

Let the coefficient of kinetic friction be $\mu$ .

Now, the frictional force will be acting along the incline but in the direction opposite to the direction of motion.

So, the net acceleration acting on the package will be up the incline and is equal to:

$a=\mu g\cos\theta-g\sin\theta$ ----------------- 1

Now, using equation of motion, we have:

$v^2-u^2=2ad\\\\0-(2.00)^2=2a(12.0)$

Solving for 'a', we get:

$-4.00=24.0a\\\\a=-\frac{4}{24}=-\frac{1}{6}\ m/s^2$

Now, plug in the value of 'a' in equation (1). This gives,

$\mu g\cos\theta-g\sin\theta=\frac{1}{6}$ ( Neglecting negative sign)

Plug in all the given values and solve for $\mu$ . This gives,

$9.8(-sin(20)+\mu cos(20))=\frac{1}{6}\\\\-0.342+\mu\times 0.94=0.017\\\\0.94\mu=0.342+0.017\\\\0.94\mu=0.359\\\\\mu=\frac{0.359}{0.94}=0.382$

Therefore, the coefficient of kinetic friction is 0.382.

5 0

3 years ago

A basketball is shot from 2 meters up at an angle of 60° above the x axis at an initial velocity of 9 m/s. What is the maximum h

Nikitich [7]

Hope this helps a little

initial distance up = 2

initial velocity component up = 9 sin 60 = 7.79

v = 9 sin 60 - 9.8 t

when v = 0, we are there

9.8 t = 7.79

t = .795 seconds to top

h = 2 + 7.79(.795) - 4.9(.795^2)

7 0

3 years ago

Read 2 more answers