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2 years ago

An exothermic reaction ___ energy.

1 answer:

5 0

Well, an exothermic reaction is a reaction, that involves the release of heat. So your answer would be A, "releases".

Good luck!

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Who expressed particles by wave equations?

ehidna [41]

I believe it was famous quantum physicist Erwin Schrödinger

7 0

2 years ago

The line spectrum of lithium has a red line at 670.8 nm. Calculate the energy of a photon with this wavelength.

Vadim26 [7]

First of all, we can calculate the frequency of the photon, which is related to the wavelength by the following equation:
$f= \frac{c}{\lambda}$
where c is the speed of light and $\lambda=670.8 nm=670.8 \cdot 10^{-9} m$ is the photon wavelength. Substituting into the formula, we find the frequency
$f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{670.8 \cdot 10^{-9} m}=4.47 \cdot 10^{14} Hz$

Now we can find the energy of the photon with the following equation:
$E=hf$
where h is the Planck constant. Substituting numbers, we find
$E=hf=(6.6 \cdot 10^{-34}Js)(4.47 \cdot 10^{14} Hz)=2.95 \cdot 10^{-19} J$

5 0

2 years ago

The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0° above the horizo

Jobisdone [24]

(a) 4.0 m/s

We can solve this part just by analyzing the vertical motion of the froghopper.

The initial vertical velocity of the froghopper as it jumps from the ground is given by

$u_y = u_0 sin \theta$ (1)

where

$u_0$ is the takeoff speed

$\theta=58.0^{\circ}$ is the angle of takeoff

The maximum height reached by the froghopper is

h = 58.7 cm = 0.587 m

We know that at the point of maximum height, the vertical velocity is zero:

$v_y = 0$

Since the vertical motion is an accelerated motion with constant (de)celeration $g=-9.8 m/s^2$ , we can use the following SUVAT equation:

$v_y^2 - u_y^2 = 2gh$

Solving for $u_y$ ,

$u_y = \sqrt{v_y^2-2gh}=\sqrt{-2(-9.8)(0.587)}=3.4 m/s$

And using eq.(1), we can now find the initial takeoff speed:

$u_0 = \frac{u_y}{sin \theta}=\frac{3.4}{sin 58.0^{\circ}}=4.0 m/s$

(b) 1.47 m

For this part, we have to analyze the horizontal motion of the froghopper.

The horizontal velocity of the froghopper is

$u_x = u_0 cos \theta = (4.0) cos 58.0^{\circ} =2.1 m/s$

And this horizontal velocity is constant during the entire motion.

We now have to calculate the time the froghopper takes to reach the ground: this is equal to twice the time it takes to reach the maximum height.

The time needed to reach the maximum height can be found through the equation

$v_y = u_y + gt$

Solving for t,

$t=-\frac{u_y}{g}=-\frac{3.4}{9.8}=0.35 s$

So the time the froghopper takes to reach the ground is

$T=2t=2(0.35)=0.70 s$

And since the horizontal motion is a uniform motion, we can now find the horizontal distance covered:

$d=u_x T = (2.1)(0.70)=1.47 m$

7 0

3 years ago

A student look at ocean waves coming into the beach. An oceanwavd with more energy will

pshichka [43]

C. Travel toward the beach faster

5 0

3 years ago

Read 2 more answers

Difference between hair dryer and heat gun

Bumek [7]

The heat gun obviously wins this round. Master Appliance heat guns can reach temperatures of up to 1,000 Fahrenheit. A handheld blow dryer might reach 131 degrees Fahrenheit. A hair dryer gets hot enough to burn skin, but not hot enough to complete serious tasks like striping paint and removing serious. By the way I got this from google.

5 0

2 years ago