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mariarad [96]
3 years ago
10

Determine the angle between the pipe segments BA and BC.

Physics
1 answer:
11111nata11111 [884]3 years ago
4 0
It is an obtuse angle
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PLEASE HELP WILL GIVE 25 POINTS!!!!!
zimovet [89]

The answer is B. 1,4, and 6

8 0
3 years ago
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Consider the potential energy diagram shown below. This graph shows the chemical potential energy in a reaction system over time
kolbaska11 [484]

Answer:

A. Endothermic reaction.

B. +150KJ.

C. 250KJ.

Explanation:

A. The graph represents endothermic reaction because the heat of the product is higher than the heat of the reactant.

B. Determination of the enthalpy change, ΔH for the reaction. This can be obtained as follow:

Heat of reactant (Hr) = 50KJ

Heat of product (Hp) = 200KJ

Enthalphy change (ΔH) =..?

Enthalphy change = Heat of product – Heat of reactant.

ΔH = Hp – Hr

ΔH = 200 – 50

ΔH = +150KJ

Therefore, the enthalphy change for the reaction is +150KJ

C. The activation energy for the reaction is the energy at the peak of the diagram.

From the diagram, the activation energy is 250KJ.

6 0
4 years ago
Sarah launches her purse straight up in air with a velocity of 30.2m/s<br> How high will it go?
yaroslaw [1]

Answer: 46.53

Explanation:

6 0
4 years ago
Jon's bathtub is rectangular and its base is 18 ft2. (a) How fast is the water level rising if Jon is filling the tub at a rate
Kitty [74]

Answer:

The water level in the bath tub is rising at a rate of 0.0111 ft/s

Explanation:

Volume of the bath tub = (Area of base) × (height)

Area of base = 18 ft² (constant)

Height = h (variable)

V = 18h

(dV/dt) = 18 (dh/dt)

If (dV/dt) = 0.2 ft³/s

0.2 = 18 (dh/dt)

(dh/dt) = (0.2/18)

(dh/dt) = 0.0111 ft/s

Hope this Helps!!!

6 0
3 years ago
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A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c
tensa zangetsu [6.8K]

Answer:

E = \frac{3kQ^2}{5R}

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

V = \frac{kq}{r}

now we can say that

q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)

q = \frac{Qr^3}{R^3}

now electric potential is given as

V = \frac{k\frac{Qr^3}{R^3}}{r}

V = \frac{kQr^2}{R^3}

now work done to bring a small charge from infinite to the surface of this sphere is given as

dW = V dq

dW = \frac{kQr^2}{R^3} dq

here we know that

dq = \frac{3Qr^2dr}{R^3}

now the total energy of the sphere is given as

E = \int dW

E = \int_0^R  \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})

E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)

E = \frac{3kQ^2}{5R}

7 0
4 years ago
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