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3 years ago

As the temperature of a liquid increases, the solubility of a solid in the liquid

1 answer:

4 0

D, it depends. For water, the solubility increases as it get hotter, up to 100 degrees, but in some cases, liquids behave the opposite way.

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The diagram shows a screwdriver being used as a lever to open a tin

Musya8 [376]

Answer:

(a) Most reactive ${}$ Metal B

${}$ Metal D

${}$ Metal A

Least reactive ${}$ Metal C

(b) (i) Bubbles should form very slowly

(ii) No reaction takes place

Explanation:

(a) The given metals arranged in their order of reactivity are;

Most reactive ${}$ Metal B

${}$ Metal D

${}$ Metal A

Least reactive ${}$ Metal C

The other of reactivity is based on the nature of their reactivity of the metals in air

(b) (i) Based on the reactivity of the metals in air, whereby metal A reacts very slowly and an oxide is formed, we have that, based on the reactivity of the metal A, when mixed with dilute hydrochloric acid, bubbles should form very slowly

(ii) Similarly, given that metal C is unreactive, we have that when small pieces of metal C are added to dilute hydrochloric acid, no reaction takes place.

7 0

3 years ago

On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const

IceJOKER [234]

Answer:

Explanation:

Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

$x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} (1)$

Since the car starts from rest, v₀ =0.
We know the value of t = 5 sec., but we need to find the value of a.
Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

$a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2 (2)$

Replacing a and t in (1):

$x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2} = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m. (3)$

Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

$a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2 (4)$

Replacing v₀, at and t in (1), we have:

$x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m (5)$

Therefore, as the truck travels twice as far as the car, the right answer is a).

7 0

3 years ago

A car with mass 1500 kg moves with constant velocity of 36 m/s. The driver sees a group of cows in front and he immediately step

Crazy boy [7]

From laws of motion:

$S = ( \frac{v + u}{2} ) \times t$
Where S is the distance/displacement (as you would call it) which is unknown
v = final velocity which is 0m/s (this is because the car stops)
u = initial velocity which is 36m/s (from the data given)
t = time taken for the distance to be covered and it is 6s

Substitute the values, hence:
$S = ( \frac{0 + 36}{2} ) \times 6$
$S = (18) \times 6 \\ \\ S = 108m$

But this is merely the distance he travelled in the 6 seconds he was trying to stop the car.

Therefore, the distance between the car and the cows = 160-108
Distance = 52m

6 0

4 years ago

A golf ball has more mass than a tennis ball because it ____. (1 point) takes up more space. contains more matter. contains diff

Elis [28]

D.) Because it has a definite composition...

5 0

3 years ago

If a car is taveling with a speed 6 and comes to a curve in a flat road with radius ???? 13.5 m, what is the minumum value the c

Lemur [1.5K]

Answer:

$\mu_s \geq 0.27$

Explanation:

The centripetal force acting on the car must be equal to mv²/R, where m is the mass of the car, v its speed and R the radius of the curve. Since the only force acting on the car that is in the direction of the center of the circle is the frictional force, we have by the Newton's Second Law:

$f_s=\frac{mv^{2}}{R}$

But we know that:

$f_s\leq \mu_s N$

And the normal force is given by the sum of the forces in the vertical direction:

$N-mg=0 \implies N=mg$

Finally, we have:

$f_s=\frac{mv^{2}}{R} \leq \mu_s mg\\\\\implies \mu_s\geq \frac{v^{2}}{gR} \\\\\mu_s\geq \frac{(6\frac{m}{s}) ^{2}}{(9.8\frac{m}{s^{2}})(13.5m) }\\\\\mu_s\geq0.27$

So, the minimum value for the coefficient of friction is 0.27.

4 0

3 years ago