What is the smallest element?

When the activation energy of an exothermic reaction decreases at a given temperature, the reaction rate increases because the _

fgiga [73]

When the activation energy of an exothermic reaction decreases at a given temperature, the reaction rate increases because the <span>number of successful effective collisions is higher. More of the reactants collide and are able to form products. Hope this answers the question. have a nice day.</span>

6 0

3 years ago

How is light reflected from a prism

Ivenika [448]

Answer:

White light entering a prism is bent, or refracted, and the light separates into its constituent wavelengths. Each wavelength of light has a different colour and bends at a different angle. The colours of white light always emerge through a prism in the same order—red, orange, yellow, green, blue, indigo, and violet.

7 0

2 years ago

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Explain why average velocity in one dimension can be positive or negative.

Softa [21]

Imagine an object is moving in one dimension on a number line, and for this we'll say that the numbers on the line are a metre apart. If the object moves from 2 m to 7 m, the change in position is 7-2=+5 metres. But if the object moves back from 7 m to 2 m, the change in position is 2-7=-5 metres. since $velocity = \frac{change in position}{time}$ , and time is always positive, velocity will be positive in one direction and negative in the other direction.

3 0

3 years ago

Read 2 more answers

Determine the components of reaction at the fixed support

denis-greek [22]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The components of reaction at the fixed support are

$A_{(x)} = 400 \ N$ , $A_{(y)} = -500 \ N$ , $A_{(z)} = 600 \ N$ , $M_x = 1225 \ N\cdot m$ , $M_y = 750 \ N\cdot m$ , $M_z = 0 \ N\cdot m$

Explanation:

Looking at the diagram uploaded we see that there are two forces acting along the x-axis on the fixed support

These force are 400 N and $A_{(x)}$ [ i.e the reactive force of 400 N ]

Hence the sum of forces along the x axis is mathematically represented as

$A_{(x)} - 400 = 0$

=> $A_{(x)} = 400 \ N$

Looking at the diagram uploaded we see that there are two forces acting along the y-axis on the fixed support

These force are 500 N and $A_{(y)}$ [ i.e the force acting along the same direction with 500 N ]

Hence the sum of forces along the x axis is mathematically represented as

$A_{(y)} + 500 = 0$

=> $A_{(y)} = -500 \ N$

Looking at the diagram uploaded we see that there are two forces acting along the z-axis on the fixed support

These force are 600 N and $A_{(z)}$ [ i.e the reactive force of 600 N ]

Hence the sum of forces along the x axis is mathematically represented as

$A_{(z)} - 600 = 0$

=> $A_{(z)} = 600 \ N$

Generally taking moment about A along the x-axis we have that

$\sum M_x = M_x - 500 (0.75 + 0.5) + 600 ( 1 ) = 0$

=> $M_x = 1225 \ N\cdot m$

Generally taking moment about A along the y-axis we have that

$\sum M_y = M_y - 400 (0.75 ) + 600 ( 0.75 ) = 0$

=> $M_y = 750 \ N\cdot m$

Generally taking moment about A along the z-axis we have that

$\sum M_z = M_z = 0$

=> $M_z = 0 \ N\cdot m$

8 0

2 years ago

A projectile is fired with an initial speed of 40 m/s at an angle of elevation of 30∘. Find the following: (Assume air resistanc

BabaBlast [244]

Answer:

a. 2.0secs

b. 20.4m

c. 4.0secs

d. 141.2m

e. 40m/s, ∅= -30°

Explanation:

The following Data are giving

Initial speed U=40m/s

angle of elevation,∅=30°

a. the expression for the time to attain the maximum height is expressed as

$t=\frac{usin\alpha }{g}$

where g is the acceleration due to gravity, and the value is 9.81m/s if we substitute values we arrive at

$t=40sin30/9.81\\t=2.0secs$

b. the expression for the maximum height is expressed as

$H=\frac{u^{2}sin^{2}\alpha }{2g} \\H=\frac{40^{2}0.25 }{2*9.81} \\H=20.4m$

c. The time to hit the ground is the total time of flight which is twice the time to reach the maximum height ,

Hence T=2t

T=2*2.0

T=4.0secs

d. The range of the projectile is expressed as

$R=\frac{U^{2}sin2\alpha}{g}\\R=\frac{40^{2}sin60}{9.81}\\R=141.2m$

e. The landing speed is the same as the initial projected speed but in opposite direction

Hence the landing speed is 40m/s at angle of -30°

3 0

2 years ago