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viktelen [127]
3 years ago
9

Which of the following are properties of acids?

Physics
2 answers:
KATRIN_1 [288]3 years ago
6 0

Reacts with certain metals

Corrosive

How would one know what acid tastes like?

vladimir2022 [97]3 years ago
4 0
The properties of acids are: (A), (C) and (D).

Explanation:
Acids taste sour, and most acids are poisonous. Furthermore, acids react with certain metals to produce hydrogen gas. Few acids are corrosive as well. The option (B) is not valid because acids react with BASES (not with other acids) to produce salt and water. Hence A, C and D are the correct options.
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A fireworks rocket is fired vertically upward. At its maximum height of 90.0 m , it explodes and breaks into two pieces, one wit
Alex73 [517]

Answer:

Ai. Speed of the fragment with mass mA= 1.35 kg is 34.64 m/s

Aii. Speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. 475.3 m

Explanation:

A. Determination of the speed of each fragment.

I. Determination of the speed of the fragment with mass mA = 1.35 kg

Mass of fragment (m₁) = 1.35 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₁) =?

KE = ½m₁u₁²

810 = ½ × 1.35 × u₁²

810 = 0.675 × u₁²

Divide both side by 0.675

u₁² = 810 / 0.675

u₁² = 1200

Take the square root of both side.

u₁ = √1200

u₁ = 34.64 m/s

Therefore, the speed of the fragment with mass mA = 1.35 kg is 34.64 m/s

II. I. Determination of the speed of the fragment with mass mB = 0.270 kg

Mass of fragment (m₂) = 0.270 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₂) =?

KE = ½m₂u₂²

810 = ½ × 0.270 × u₂²

810 = 0.135 × u₂²

Divide both side by 0.135

u₂² = 810 / 0.135

u₂² = 6000

Take the square root of both side.

u₂ = √6000

u₂ = 77.46 m/s

Therefore, the speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. Determination of the distance between the points on the ground where they land.

We'll begin by calculating the time taken for the fragments to get to the ground. This can be obtained as follow:

Maximum height (h) = 90.0 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

90 = ½ × 10 × t²

90 = 5 × t²

Divide both side by 5

t² = 90/5

t² = 18

Take the square root of both side

t = √18

t = 4.24 s

Thus, it will take 4.24 s for each fragments to get to the ground.

Next, we shall determine the horizontal distance travelled by the fragment with mass mA = 1.35 kg. This is illustrated below:

Velocity of fragment (u₁) = 34.64 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₁) =?

s₁ = u₁t

s₁ = 34.64 × 4.24

s₁ = 146.87 m

Next, we shall determine the horizontal distance travelled by the fragment with mass mB = 0.270 kg. This is illustrated below:

Velocity of fragment (u₂) = 77.46 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₂) =?

s₂ = u₂t

s₂ = 77.46 × 4.24

s₂ = 328.43 m

Finally, we shall determine the distance between the points on the ground where they land.

Horizontal distance travelled by the 1st fragment (s₁) = 146.87 m

Horizontal distance travelled by the 2nd fragment (s₂) = 328.43 m

Distance apart (S) =?

S = s₁ + s₂

S = 146.87 + 328.43

S = 475.3 m

Therefore, the distance between the points on the ground where they land is 475.3 m

3 0
3 years ago
If I keep F constant in F=ma, what is the relationship between m and a?
Jobisdone [24]

Answer:

If F is a constant, we can take f = 1

f = m×a

ma = 1

therefore we can say that force is hence proportinal to the product of mass and acceleration.

6 0
2 years ago
A copper block rests 30.0 cm from the center of a steel turntable. The coefficient of static friction between the block and the
PIT_PIT [208]

Answer:

refer to the above attachment

3 0
1 year ago
The mass of a radioactive substance follows a continuous exponential decay model, with a decay rate parameter of 1% per day. A s
Bumek [7]

Answer:

2,38kg

Explanation:

Mass in function of time can be found by the formula: m_{(t)} =m_{0} e^{-kt}, where m_{0} is the initial mass, t is the time and k is a constant.

Given that a sample decay 1% per day, that means that after first day you have 99% of mass.

m_{(1)} =m_{0} e^{-k(1)}, but m_{(1)}=\frac{99m_{0} }{100}, so we have \frac{99m_{0} }{100}=m_{0}e^{-k}, then k=-ln(\frac{99}{100})=0.01

Now using k found we must to find m_{(5)}.

m_{(5)}=m_{0}e^{-(0.01)5}=2.5kge^{-0.05} =2.5x0.951=2.38kg

6 0
3 years ago
The group of test subject that are NOT given the experimental treatment is called what?
-BARSIC- [3]

Answer:

The group that remains unaltered is called the control group.

5 0
3 years ago
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