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3 years ago

Determine which factors will increase the reaction rate and which factors will decrease te reaction rate.

Physics

1 answer:

Elodia [21]3 years ago

8 0

Answer: increase reaction rate side : increase surface area

Increase temperature

Add a catalyst

Decrease reaction rate side :

Decrease surface area

Decrease temperature

Decrease concentration

Explanation:

Got it right

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1. A 6-volt battery produces a current of 0.5 amps. What is the power in the circuit?

allochka39001 [22]

Answer:

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Explanation:

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2 years ago

What are the seven types of electromagnetic waves?

Sunny_sXe [5.5K]

Radio waves. Giant satellite-dish antennas pick up long-wavelength, high-frequency radio waves. ...
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5 0

3 years ago

A baseball is thrown straight up from a building that is 25 meters tall with an initial velocity v = 10 m/s. How fast is it goin

Yanka [14]

Answer:-24,5m/s

Explanation: what we have here is a UALM with these gravity as acceleration (-9.8 m/s^2). The initial position is 25 m and initial speed is 10m/s.

Speed and gravity are increasing in the opposite direction, speed upwards and gravity downwards, while the position is also upwards, depending on your reference system.

The first thing I need to know is the maximum high it will reach.

Hmax=- S(0)^2/2g=

S= speed.

0= initial

G= gravity

Hm= 100/19,6= 5.1 m

So, the ball will go 5,1 m higher than the initial position, and from there it will fall free.

Then, I need to know how long it takes to fall. For that we use UALM equation:

X(t)= X(0) + S(0)*t + (A*t^2)/2.

X: position

S: speed

A: acceleration

T:time

0: initial

0 = 25m +10*t -(9.8 * t^2)/2

Solving the quadratic equation we get

T= 3,5 sec. ( Negative value for time is impossible)

So now we know that the ball to go up and then fall needs 3,5 sec.

Let's see how long it takes to go up:

30,1=25+10*t-4,9*t^2

0=-5,1+10*t-4,9*t^2

T= 1 sec. So it will take 1 sec to the ball to reach the maximum high and 0=speed and then it'll fall during the resting 2,5 sec

Finally, to know the speed just before it touches the ground, we use the following formula:

A= (St-S0)/t

-9.8m/s^2 = (St- 0m/s)/ 2,5s

-24,5 m/s= St

-24,5 m/s is the speed at 3,5 sec, which is the time just before falling

3 0

3 years ago

Sound waves that enter the external acoustic meatus eventually encounter the __________, which then vibrates at the same frequen

7nadin3 [17]

Answer:

tympanic membrane (eardrum)

Explanation:

The sound waves spread through the air and reach the outer ear, into which they penetrate through the ear canal. In doing so, they stimulate the eardrum, which closes the inner end of the duct. By vibrating this membrane, the vibration of a chain of ossicles located in the middle ear is induced. These ossicles transmit their vibration to the oval window, which is a membranous structure that communicates the middle ear with the cochlea of the inner ear. When the oval membrane moves, it moves the liquid (perilymph) that fills one of the three cavities of the cochlea generating waves in it. These waves mechanically stimulate the sensory cells (hair cells) located in the organ of Corti, within the cochlea in the central cavity, the middle ramp. This cavity is filled with a liquid rich in K +, endolymph. The cells embedded in the endolymph, change their permeability to K + due to the movement of the cilia and respond by releasing a neurotransmitter that excites the nerve terminals, which initiate the auditory sensory pathway.

3 0

3 years ago

A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule

ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

$p_i = 0$

The final total momentum is instead:

$p_f = m_a v_a + m_c v_c$

where

$m_a = 125 kg$ is the mass of the astronaut

$v_a = 2.50 m/s$ is the velocity of the astronaut

$m_c = 1900 kg$ is the mass of the capsule

$v_c$ is the velocity of the capsule

Since the total momentum must be conserved, we have

$p_i = p_f = 0$

$m_a v_a + m_c v_c=0$

Solving the equation for $v_c$ , we find

$v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s$

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

$F \Delta t = \Delta p$