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3 years ago

Which one of the following is a decomposition reaction?

1 answer:

4 0

You would know a decomposition reaction occurred if the reactants separated. For example from AB → A+B.

Now if you look at your options only 1 works out for that equation. Letter A.
From the compound K2CO3 it split up to K2O +CO2

It cannot be letter B because synthesis/combination occurred. The same goes for letter C. Letter D, single displacement occurred.

Again, the answer is A.

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Are any minerals flammable

Irina18 [472]

Yes minerals are flammable. I almost caught my shirt on fire onetime

8 0

2 years ago

Read 2 more answers

And I need help with seven and eight only I will appreciate it

Mumz [18]

Answer:

7] Force = mass × acceleration

Force = 2 × 5

Force = 10 N

8] Velocity = acceleration due to gravity × time taken

Velocity = 9.8 × 12

Velocity = 117.6 m/s

8 0

2 years ago

A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio

Mariulka [41]

Answer:

Yes. Towards the center. 8210 N.

Explanation:

Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to $F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N$

Note that 95 km/h is equal to 26.3 m/s.

This is the centripetal force and equal to the x-component of the applied force.

$F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3$

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.

The amount of the friction force should be $8.21\times 10^3~N$

Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.

5 0

3 years ago

A juggler throws a ball straight up into the air. The ball remains in the air for a time (t) before it lands back in the juggler

natka813 [3]

Answer:

$9.8 m/s^2$ , downward

Explanation:

There is only one force acting on the ball during its motion: the force of gravity, which is given by

$F=mg$

where

m is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity (downward)

According to Newton's second law,

$F=ma$

where F is the net force on the object and a is its acceleration. Rearranging for a,

$a=\frac{F}{m}$

As we said, the only force acting on the ball is gravity, so F = mg and the acceleration of the ball is:

$a=\frac{mg}{m}=g$

Therefore, the ball has a constant acceleration of $9.8 m/s^2$ downward for the entire motion.

5 0

2 years ago

Calculate the force of gravity on the 0.60- kg mass if it were 1.3×107 m above Earth's surface (that is, if it were three Earth

nignag [31]

The gravitational force between two objects is given by:
$F=G \frac{m_1 m_2}{r^2}$
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation

In this problem, the first object has a mass of $m_1=0.60 kg$ , while the second "object" is the Earth, with mass $m_2=5.97 \cdot 10^{24}kg$ . The distance of the object from the Earth's center is $r=1.3 \cdot 10^7 m$ ; if we substitute these numbers into the equation, we find the force of gravity exerted by the Earth on the mass of 0.60 kg:
$F=G \frac{m_1m_2}{r^2}=(6.67\cdot 10^{-11}) \frac{(0.60 kg)(5.97 \cdot 10^{24} kg)}{(1.3 \cdot 10^7 m)^2}= 1.41 N$

5 0

3 years ago