I’m still not sure on this one please help

How can the rate of a reaction be decreased?

andre [41]

Answer:

Option A

Lowering the amount of reactants

Explanation:

To reduce the rate of chemical reaction, one can reduce temperature or surface area. The addition of catalysts increases rate of reaction but decreasing the amount of reactants decreases rate of reaction. Therefore, from the choices provided, choice A is correct.

6 0

3 years ago

What are the top three osha cited ladder violations

bonufazy [111]

Answer:

Explanation:

1. FALL PROTECTION-GENERAL REQUIREMENTS (29 CFR 1926.501) 6,010 VIOLATIONS

2. HAZARD COMMUNICATION (29 CFR 1910.1200). 3,671

3. SCAFFOLDING (29 CFR 1926.451). 2,813

6 0

3 years ago

A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image

Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

$=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}$

$=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}$

Re-arrange to get i

$\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m$

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

$m=\frac{h'}{h} =-\frac{i}{o}$

substitute the values to get the height of image h'

$\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm$

5 0

3 years ago

Cart 1 of mass m is traveling with speed 2vo in the +x-direction when it has an elastic collision with cart 2 of

iogann1982 [59]

Answer:

Explanation:

Momentum conservation

$m2v_0+2mv_0=mv_1+2mv_2 \quad (1/m) \quad 4v_0=v_1+2v_2\\$

Kinetic energy conservation

$\displaystyle \frac{1}{2}m(2v_0)^2+\frac{1}{2}2mv_0^2=\frac{1}{2}mv_1^2+\frac{1}{2}2mv_2^2 \quad (1/m) \quad 6v_0^2=v_1^2+2v_2^2$

Solve the system

6 0

3 years ago

Consider a concave spherical mirr or that has focal length f = +19.5 cm.

lidiya [134]

The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

<h3>What is concave mirror?</h3>

A concave mirror has a reflective surface that is curved inward and away from the light source.

Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.

<h3>Object distance of the concave mirror</h3>

Apply mirrors formula as shown below;

1/f = 1/v + 1/u

where;

f is the focal length of the mirror
v is the object distance
u is the image distance

when image height = object height, magnification = 1

u/v = 1

v = u

Substitute the given parameters and solve for the distance of the object from the mirror's vertex

1/f = 1/v + 1/v

1/f = 2/v

v = 2f

v = 2(19.5 cm)

v = 39 cm

Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

Learn more about concave mirror here: brainly.com/question/27841226

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7 0

1 year ago