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vaieri [72.5K]
4 years ago
14

Consider a small child sliding down two differently shaped slides that have the same coefficient of kinetic friction. How must t

he second slide be shaped to provide the same change in gravitational potential energy of the child-Earth system but more change in kinetic energy for the child while they slide down
Physics
1 answer:
timama [110]4 years ago
6 0

Answer:

Having a bigger angle above the horizontal

Explanation:

Applying the energy conservation theorem:

K_1+U_1+W_{ext}=K_2+U_2

The kinetic energy is reduced because of the work done by the friction force.

The friction force is given by:

F_f=F_N*\µ

so the friction force depends on the Normal force, because the slide has an angle the normal force is given by:

F_N=m.g*cos(\theta)

So when the angle of the slide is bigger, the friction force decreases, for example:

for 45 degrees:

F_N=m.g*cos(45)\\F_N=0.70(m.g)\\

for 75 degrees:

F_N=m.g*cos(75)\\F_N=0.26(m.g)\\

as you can see if the angle is bigger above the horizontal, the friction force is reduced and so the work done by that force. We didn't have to change the height of the slide, so the potential gravitational energy remains the same.

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Em muitas situações do dia a dia, temos que solucionar questões que exigem conhecimento das relações existentes entre as grandez
VladimirAG [237]

Answer:

R\$37,96

Explanation:

1) Olá, vamos calcular primeiramente o Volume da Piscina. Ela é uma piscina com uma forma tradicional, retangular. Temos a largura, o comprimento e a profundidade.

Volume=a.b.c\\\\Volume=largura*comprimento*altura\\Volume= 10*15*1,5\\Volume=225m^{3}

2) Precisamos fazer uma correlação entre a medida de capacidade (litro) e de volume, já que o custo é por litro d'água.

1 litro = 1dm^3\\1 m^{3}=1000 \:dm^{3}=1000 \:litros\\225\:m^{3}=225000\: litros

3) Observe que Francisco gasta 0,005 g de NaClO por semana.

Considerando que o mês tenha 4,5 semanas. E que quanto mais se trata a piscina mais se gasta Hipoclorito de Sódio. Temos uma relação diretamente proporcional.

Finalmente,

3.1) Rendimento do hipoclorito de produto em gramas por semana:

1 semana   \rightarrow      0,005g\\4,5 semanas  \rightarrow 0,0225g

3.2) Consumo mensal (y), considerando 4,5 semanas:

0,005g\rightarrow1 litro\\y\rightarrow225000\\y=1125 gramas \: para\: 225000 litros.\\\\1125gramas \:por\:litro\rightarrow 1 semana\\q \rightarrow 4,5\\q=5062,5 \:gramas=\:5,0625 Kg \:de \:NaClO\\\\

3.3)  O custo da manutenção, já que 18 kg de NaClO  custa R$ 135 e são grandezas diretamente proporcionais:

18 kg\rightarrow \$135\\5,0625  \:kg \:de \:NaOCl\rightarrow p\\p=\frac{5,0625*135}{18}=\frac{683.4375}{18} \therefore p=R\$37,96

8 0
3 years ago
Explain what matter is
djverab [1.8K]

Answer:

it doesnt matter

Explanation:

matter is what makes up everything around us; anything takin up space

8 0
3 years ago
Read 2 more answers
Matthew is waterskiing. As the boat starts moving, he is at an angle of 8.0° to the right of the boat. The boat applies 250 newt
Citrus2011 [14]

The work done is B. 1.2\cdot 10^4 J

Explanation:

The work done by a force on an object is given by

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and of the displacement

For the boat in this problem, we have:

F = 250 N (force applied)

d = 50 m (displacement)

\theta=8.0^{\circ}

Substituting, we find the work done:

W=(250)(50)(cos 8^{\circ})=1.2\cdot 10^4 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

5 0
3 years ago
Samuel is running in a parks nature trail.at first checkpoint he ran 925 m in 10 minutes.at second checkpoint he slowed down and
geniusboy [140]

Answer:

hey did you complete the whole test?

Explanation:

7 0
3 years ago
It is 2058 and you are taking your grandchildren to Mars. At an elevation of 34.7 km above the surface of Mars, your spacecraft
Paul [167]

Answer: 1.23\ m/s^2

Explanation:

Given

At an elevation of y=34.7\ km, spacecraft is dropping vertically at a speed of u=293\ m/s

Final velocity of the spacecraft is v=0

using equation of motion i.e. v^2-u^2=2as

Insert the values

\Rightarrow 0-(293)^2=2\times a\times (34.7\times 10^3)\\\\\Rightarrow a=-\dfrac{293^2}{2\times 34.7\times 10^3}\\\\\Rightarrow a=-1.23\ m/s^2

Therefore, magnitude of acceleration is 1.23\ m/s^2.

8 0
3 years ago
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