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lesya692 [45]
3 years ago
15

Balance the following equations: CUCO3+H2SO4- CUSO4+H2O+CO2​

Physics
1 answer:
Setler79 [48]3 years ago
6 0

Answer:

It is already balanced

Explanation:

It is already balanced

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Calculate the kinetic energy of a 5.2kg object moving at 2.4m/s.
dedylja [7]

Answer:2.5

Explanation:

5 0
3 years ago
A forces of 5 N accelerates an object. The object’s mass is 50 g. What is the acceleration of the object? (Formula: F=ma)
Harrizon [31]

Answer:

100 m / s^

F =ma from this divided F and ma by m to get acceleration

a = F / m

a = ?

F = 5 N

m = 50 g = 0. 0 5 kg ; if 1 kg = 1000g how much kg of 50 g by criss cross method we get 0.05 kg

so; 5N / 0. 05 kg = 5kg m / s^ / 0.05 kg

kg cancel by kg

the result is 100 m/ s^

7 0
2 years ago
A spring loaded toy shoots straight upward with a velocity of 4.5 m/s. Determine the maximum height it reaches. Determine the ti
Angelina_Jolie [31]

Recall that

{v_y}^2-{v_{0y}}^2=2a_y(y-y_0)

At its maximum height y_{\mathrm{max}}, the toy will have 0 vertical velocity, so that

-\left(4.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)y_{\mathrm{max}}

\implies y_{\mathrm{max}}=1.0\,\mathrm m

For the toy to reach this maximum height, it takes time t such that

\dfrac{0+v_0}2t=y_{\mathrm{max}}\implies t=0.46\,\mathrm s

which means it takes twice this time, i.e. t=0.92\,\mathrm s, for the toy to reach its original position.

The velocity of the toy when it falls 1.0 m below its starting point is

{v_y}^2-\left(4.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)(0-1.0\,\mathrm m)

\implies{v_y}^2=39.85\,\dfrac{\mathrm m^2}{\mathrm s^2}

\implies v_y=-6.4\,\dfrac{\mathrm m}{\mathrm s}

where we took the negative square root because we expect the toy to be moving in the downward direction.

6 0
3 years ago
What is responsible for the positive charge in an atom?
Ivenika [448]

Answer:proton as they bear positive charge and located at the nucleus

5 0
3 years ago
The length of second hand of clock is 14cm, an ant sits on the top of second hand. find the following
Anastaziya [24]

Answer:

The answer is below

Explanation:

i) Since the length of the second clock (radius) is 14 cm = 0.14 m, the distance covered by the second hand in one revelution is:

Distance covered = 2πr = 2π(0.14) = 0.88 m

The time taking to complete one revolution = 60 seconds, hence;

Speed = distance covered in one revolution / time take o complete a revolution

Speed = 0.88 m / 60 s = 0.0147 m/s

ii) Distance covered in 150 s = speed * 150 s = 0.0147 * 150 = 2.2 m

iii) Displacement in 150 seconds = distance from initial position to final position

At 150 s, the hand has covered 2 revolutions and moved 30 s. Hence:

Displacement in 150 seconds = speed * 30 s = 0.0147 * 30 = 0.44 m

4 0
3 years ago
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