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Alik [6]
3 years ago
13

Atividade 1) A presença do campo magnético terrestre proporciona alguns efeitos interessantes. Dentre eles estão as auroras pola

res. Faça uma pesquisa e explique como esse fenômeno ocorre. Atividade 2 Utilizando a simulação mostrada na figura abaixo, aproxime o ímã da bússola em várias posições ao longo do plano. a) O que representam as pequenas agulhas ao redor do ímã? b) O que acontece com a agulha da bússola quando o ímã é movimentado ao seu redor? c) Marque a seleção "Mostrar medidor de campo". Movimente o medidor de campo O que você observa em relação ao campo magnético⃗ , quando o medidor de campo se aproxima do imã? c) Marque a caixa de seleção "planeta Terra". Como são mostradas as polaridades do ímã? Por que aparece dessa forma?
Physics
1 answer:
VLD [36.1K]3 years ago
7 0
The answer is 1 the first one is the only one that makes sense
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A plane stops from 250 mph in 25 seconds. Calculate the plane's acceleration.
Lelechka [254]

Explanation:

Given:

v₀ = 250 mph

v = 0 mph

t = 25 s

Find: a

v = at + v₀

(0 mph) = a (25 s) + (250 mph)

a = -10 mph/s

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Formula for weight on moon compared to weight on earth​
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Weight on moon = (0.16) • Earth weight

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A single-turn current loop, carrying a current of 3.50 a, is in the shape of a right triangle with sides 50.0, 120, and 130 cm.
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Thank your very much
3 0
3 years ago
A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for
stealth61 [152]

Answer:

(a) a_s=0.62\frac{m}{s^2}

(b) a_s=0.13\frac{m}{s^2}

(c) x_f=2.6m

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}

(c) Using the kinematics equation:

x_f=x_0+v_0t \pm  \frac{at^2}{2}

For the girl, we have x_0=0 and v_0=0. So:

x_f_g=\frac{a_gt^2}{2}(1)

For the sled, we have v_0=0. So:

x_f_s=x_0_s-\frac{a_st^2}{2}(2)

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s

Now, we solve (1) for x_f_g

x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m

5 0
3 years ago
An iron vat is 11 m long at room temperature (20°C). How much longer is it when it contains boiling water at 1 atm pressure?
RoseWind [281]

Answer:

The change in length is 0.01056 m.

Explanation:

Given that,

Length = 11 m

Temperature = 20°C

Pressure = 1 atm

Boiling temperature = 100°C

We need to calculate the length

Using formula of change in linear expansion

\dfrac{\Delta L}{L}=\alpha_{l}\Delta T

\Delta L=L\times\alpha_{l}\times(T_{f}-T_{i})

Put the value into the formula

\Delta L=11\times12\times10^{-6}\times(100-20)

\Delta L=11\times12\times10^{-6}\times80

\Delta L=0.01056\ m

Hence, The change in length is 0.01056 m.

5 0
3 years ago
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