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3 years ago

15

What does it mean to say water dissociates

1 answer:

Lerok [7]3 years ago

4 0

It means to brake up like
HA-H+ + A-
THE H+ IS the acid hydrogen while the A- is rest of the acid

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are the properties of a specific element the same as the properties of a molecule made up of that element ? why or why not

konstantin123 [22]

Because it’s science

8 0

4 years ago

A racing car travels with a constant tangential speed of 80.8 m/s around a circular track of radius 602 m. (a) Find the magnitud

KatRina [158]

Answer:

10.84 m/s2 radially inward

Explanation:

As the car is traveling an a constant tangential speed of 80.8 m/s, the total acceleration only consists of the centripetal acceleration and no linear acceleration. The formula for centripetal acceleration with respect to tangential speed v = 80.8 m/s and radius r =602 m is

$a = \frac{v^2}{r} = \frac{80.8^2}{602} = 10.84 m/s^2$

b) The direction of this centripetal acceleration is radially inward

6 0

4 years ago

You throw a stone straight down from the top of a tall tower. It leaves your hand moving at 8.00 m/s, Air resistance can be negl

ser-zykov [4K]

Answer:

The velocity after 1.5 s is 22.7 m/s downwards.

Explanation:

Initial velocity = - 8 m/s

acceleration, a = - 9.8 m/s2

time, t = 1.5 s

Use first equation of motion

v = u + at

v = - 8 - 9.8 x 1.5

v = - 8 - 14.7

v = - 22.7 m/s

Thus, the velocity after 1.5 s is 22.7 m/s downwards.

5 0

3 years ago

A 11kg block slides up a 30° inclined plane at a constant velocity. The coefficient of friction between the block and the plane

astra-53 [7]

Answer:

The magnitude of the applied force is 94.74 N

Explanation:

Mass of the block, m = 11 kg

Angle of inclination of the plane, $\theta = 30^{\circ}$

Friction coefficient, $\mu_{k} = 0.2$

Now,

Normal force that acts on the block is given by:

$F_{N} = mgcos\theta + Fsin\theta$ (1)

Now, to maintain the equilibrium parallel to ramp the forces must be balanced.

Thus

$Fcos\theta = \mu_{k}F_{N}$ (2)

From eqn (1) and (2)

$Fcos\theta = \mu_{k}(mgcos\theta + Fsin\theta)$

$F(cos\theta - \mu_{k}sin\theta) = \mu_{k}mgcos\theta$

$F = \frac{\mu_{k}mgcos\theta}{cos\theta - \mu_{k}sin\theta}$

$F = \frac{0.2\times 11\times 9.8cos30^{\circ}}{cos30^{\circ} - 0.2\times sin30^{\circ}}$

F = 94.74 N

3 0

4 years ago

A car traveling at 25m/sec, stops in 0.103 km. what is the deceleration provided by the brakes?

lozanna [386]

Answer:

-3.03 m/s²

Explanation:

v² = v₀² + 2a(x − x₀)

where v is the final velocity,

v₀ is the initial velocity,

a is the acceleration,

x is the final position,

and x₀ is the initial position.

Given:

v = 0 m/s

v₀ = 25 m/s

x = 103 m

x₀ = 0 m

Find: a

v² = v₀² + 2a(x − x₀)

(0 m/s)² = (25 m/s)² + 2a (103 m − 0 m)

a = -3.03 m/s²

Round as needed.

7 0

3 years ago