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3 years ago

5

All of these are likely to speed up the rate of a reaction except decreasing the surface area. increasing the temperature. incre

asing the pressure. adding a catalyst.

1 answer:

Vanyuwa [196]3 years ago

6 0

Answer:

decreasing the surface area.

Explanation:

The rate of reaction or the reaction rate may be defined as the speed or the rate at which the reaction occurs. It is the speed of formation of the products from the reactants. The following factors increases the rate of reaction :

1. by adding a catalyst

2. by increasing the temperature

3. by increasing the pressure

4. by increasing the surface area

5. by increasing concentration

Hence, decreasing the surface area of the reaction reduces the rate of reaction.

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Besides gravity, what factor keeps the moon and Earth in orbit?

earnstyle [38]

Answer:

interna

Explanation:

please mark as brainllest

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3 years ago

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A wave has a speed of 30 m/s, a frequency of 6 Hz, and a wavelength of 5 m. If the wavelength remains constant, and the frequenc

mixer [17]

Answer: 60m/s

Explanation:

The wavespeed is the distance covered by the wave in one second. It is measured in metre per second, and represented by the symbol V

Wavespeed (V) = Frequency F x wavelength λ

i.e V = F λ

In the first case:

Wavespeed = 30 m/s

Frequency of sound = 6Hz

Wavelength = 5m

In the second case:

Wavespeed = ?

Frequency of sound = (2x 6Hz = 12Hz)

Wavelength = 5m (remains constant)

Apply V = F λ

Wavespeed = 12 Hz x 5m

Wavespeed = 60m/s

Therefore, when frequency is doubled, the speed is also doubled. Thus, the new speed of the wave is 60m/s

8 0

3 years ago

A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1

Nesterboy [21]

Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain $d=2.46\times10^{3}\ m$

Height of island $h=1.80\times10^{3}\ m$

Distance of the enemy ship from the mountain $d'=6.10\times10^{2}\ m$

Initial velocity $v=2.55\times10^{2}\ m/s$

Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

$v_{x}=v\cos\theta$

Put the value into the formula

$v_{x}=2.55\times10^{2}\cos74.9$

$v_{x}=66.42\ m/s$

We need to calculate the vertical component of initial velocity

Using formula of vertical component

$v_{y}=v\sin\theta$

Put the value into the formula

$v_{y}=2.55\times10^{2}\sin74.9$

$v_{y}=246.19\ m/s$

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v_{x}}$

$t=\dfrac{2.46\times10^{3}}{66.42}$

$t=37.03\ sec$

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

$H= v_{y}t-\dfrac{1}{2}gt^2$

Put the value in the equation

$H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2$

$H=2397.4\ m$

We need to calculate the distance close to the peak

Using formula of distance

$H'=H-h$

Put the value into the formula

$H'=2397.4-1800$

$H'=597.4\ m$

Hence, The distance close to the peak is 597.4 m.

6 0

3 years ago

A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten

babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

$P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2$

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

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3 years ago

Analogue signals transmit information for such things as _____________.

ivann1987 [24]

Transmission of information in ANY form can be done digitally
or analoguely.

Beginning about 30 years ago, everything slowly started changing
to digital. Today, all commercial satellite communication, all optical
fiber communication, all internet communication, all computer
communication, all commercial cable communication, all commercial
television, and much of the telephone system, are all digital.

On your computer ... .pdf, .jpg, .mp3 etc. are all digital methods of
moving and storing information.

AM and FM radio are an interesting subject. They're all still analog.
They could easily be changed to all digital, and it would be a big
improvement, both for the broadcasters and for the listeners.
BUT ... every AM and FM radio that anybody has now would be
obsolete. Every single radio would either need to be replaced,
OR you'd need to add a digital decoder to every radio, like we
had to do with our TV sets a few years ago when television
suddenly became all digital. With AM and FM radios, the decoders
would be bigger, and would cost more, than most of the radios.

And that's why commercial radio broadcasting is still analog.

7 0

3 years ago

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