1. What is evaporation and how does it affect weather?

What effect does a catalyst have on the energy of a reaction system? HELP ASAP PLZ!!

Lubov Fominskaja [6]

Answer:

I think option A is correct

=> it increases the initial energy of the reactants

hope it helps

5 0

3 years ago

Sarah, whose mass is 40 kg, is on her way to school after a winter storm when she accidentally slips on a patch of ice whose coe

RideAnS [48]

Sarah's acceleration is $-0.49 m/s^2$

Explanation:

The force of kinetic friction acting on Sarah has a magnitude which is given by:

$F_f = \mu mg$

where

$\mu$ is the coefficient of kinetic friction

m is Sarah's mass

g is the acceleration of gravity

Moreover, according to Newton's second law of motion, we know that the net force on Sarah is equal to its mass times its acceleration:

$F=ma$

where a is the acceleration

Since the force of friction is the only force acting on Sarah, we can say that the net force is equal to the force of friction, therefore:

$F=-\mu mg = ma$

where the negative sign is due to the fact that the force of friction has a direction opposite to the motion of Sarah. Solving for a, we find

$a=-\mu g$

And substituting the following values:

$\mu = 0.05$ (coefficient of friction)

$g=9.81 m/s^2$ (acceleration of gravity)

we find:

$a=-(0.05)(9.81)=-0.49 m/s^2$

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

4 0

3 years ago

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af

vekshin1

Answer:

$T=51.64^\circ F$

$t=180.10s$

Explanation:

The Newton's law in this case is:

$T(t)=T_m+Ce^{kt}$

Here, $T_m$ is the air temperture, C and k are constants.

We have

$70^\circ F$ in $t=0$

So:

$T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F$

And we have $60^\circ F$ in $t=30 s$ , So:

$T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061$

Now, we have:

$T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)$

Applying (1) for $t=1 min=60s$ :

$T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F$

Applying (1) for $T=30^\circ F$ :

$30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s$

8 0

3 years ago

What is the equation for an inelastic collision

abruzzese [7]

M1 v1 = (m1 + m2)v2.

All of the exponents should be lowered to the bottom right of the letters.

7 0

3 years ago

A piece of clay sits 0.10 m from the center of a potter’s wheel. If the potter spins the wheel at an angular speed of 15.5 rad/s

finlep [7]

Answer:

$a=24.025\ m/s^2$

Explanation:

Given that

Distance from the center ,r= 0.1 m

The angular speed ,ω = 15.5 rad/s

We know that centripetal acceleration is given as

a=ω² r

a=Acceleration

r=Radius

ω=angular speed

a=ω² r

Now by putting the values in the above equation we get

$a=15.5^2\times 0.1\ m/s^2$

$a=24.025\ m/s^2$

Therefore the acceleration of the clay will be $a=24.025\ m/s^2$ .

4 0

3 years ago