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Luda [366]
1 year ago
9

Quando aquecemos água em nossas casas, ao nível do mar, utilizando um recipiente aberto, sua temperatura nunca ultrapassa os 100

°C. Isso ocorre porque:
a) ao atingir essa temperatura, a agua perde sua capacidade de absorver calor.
b) ao atingir essa temperatura, a agua passa a perder exatamente a mesma quantidade de calor que esta recebendo, mantendo assim a temperatura constante.
c) as mudanças de fase ocorrem à temperatura constante.
d) ao atingir essa temperatura, a agua começa a expelir o oxigenio e outros gases nela dissolvidos.
e) as mudanças de fases de agua sempre ocorrem em 100 graus Celsius
Physics
1 answer:
iren2701 [21]1 year ago
4 0

Answer:

i dont speak spanish

Explanation:

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A cart starts from rest and accelerates at 4.0 m/s2 for 5.0 s, then maintains that velocity for 10 s, and then decelerates at th
zhannawk [14.2K]

Answer:

Final speed of car = 12 m/s

Explanation:

We have equation of motion v = u + at, where v is final velocity, u is initial velocity, a is acceleration and t is time.

a) A cart starts from rest and accelerates at 4.0 m/s² for 5.0 s

        v = ?

        u = 0 m/s

        a = 4.0 m/s²

         t = 5 s

         v = u + at = 0 + 4 x 5 = 20 m/s

b) Then maintains that velocity for 10 s

        v = ?

        u = 20 m/s

        a = 0 m/s²

         t = 10 s

         v = u + at = 20 + 0 x 10 = 20 m/s

c) Then decelerates at the rate of 2.0 m/s² for 4.0 s

        v = ?

        u = 20 m/s

        a = -2.0 m/s²

         t = 4 s

         v = u + at = 20 + -2 x 4 = 12 m/s

Final speed of car = 12 m/s

3 0
2 years ago
The electric field near the earth's surface has magnitude of about 150n/c. what is the acceleration experienced by an electron n
qaws [65]
Felectric = q*E 
<span> Ftranslational = m*a 
</span><span> Felectric = Ftranslational
</span> <span>q*E = m*a 
</span><span> Solve for a 
</span><span> a = q/m*E </span>
<span> Our sign convention is "up is positive" 
</span><span> q = 1.6*10^-19 C 
</span><span> m = 1.67*10^-27 kg 
</span><span> E = -150 N/C (- because it is down and up is positive) 
</span> a =<span> -6,4*10^5</span><span> m/s^2 (downward) 
</span> answer
 a = -6,4*10^5 m/s^2 (downward) 
3 0
2 years ago
Calculate the amount of work done to draw a current of 8A from a point at 100V to a point at 120V in 2 seconds?
Morgarella [4.7K]
Given:
I=8A
t=2second
Potential difference,V=120-100=20volt
Workdone=V×i×t
=20×8×2
=320 joule.
3 0
2 years ago
a 75 kg man is standing at rest on ice while holding a 4kg ball. if the man throws the ball at a velocity of 3.50 m/s forward, w
AysviL [449]

Answer:

His resulting velocity will be 0.187 m/s backwards.

Explanation:

Given:

Mass of the man is, M=75\ kg

Mass of the ball is, m=4\ kg

Initial velocity of the man is, u_m=0\ m/s(rest)

Initial velocity of the ball is, u_b=0\ m/s(rest)

Final velocity of the ball is, v_b=3.50\ m/s

Final velocity of the man is, v_m=?\ m/s

In order to solve this problem, we apply law of conservation of momentum.

It states that sum of initial momentum is equal to the sum of final momentum.

Momentum is the product of mass and velocity.

Initial momentum = Initial momentum of man and ball

Initial momentum = Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm

Final momentum = Final momentum of man and ball

Final momentum = Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14

Now, initial momentum = final momentum

0=75v_m+14\\\\75v_m=-14\\\\v_m=\frac{-14}{75}\\\\v_m=-0.187\ m/s

The negative sign implies backward motion of the man.

Therefore, his resulting velocity is 0.187 m/s backwards.

3 0
2 years ago
Consider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 50 W/m · K and
tatuchka [14]

Answer:

solution:

dT/dx =T2-T1/L

&

q_x = -k*(dT/dx)

<u>Case (1)  </u>

dT/dx= (-20-50)/0.35==> -280 K/m

 q_x  =-50*(-280)*10^3==>14 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

 q_x  =-50*(80)*10^3==>-4 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

 q_x  =-50*(80)*10^3==>-4 kW

Case (3)

q_x  =-50*(160)*10^3==>-8 kW

T2=T1+dT/dx*L=70+160*0.25==> 110° C

Case (4)

q_x  =-50*(-80)*10^3==>4 kW

T1=T2-dT/dx*L=40+80*0.25==> 60° C

Case (5)

q_x  =-50*(200)*10^3==>-10 kW

T1=T2-dT/dx*L=30-200*0.25==> -20° C

note:

all graph are attached

6 0
2 years ago
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