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3 years ago

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1700 J of energy is lost from 0.14 kg object , the temperature decreases from 50°C to 45°C what is the specific heat of this obj

ect, amd what is the material ?

1 answer:

SCORPION-xisa [38]3 years ago

4 0

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Suppose a free-fall ride at an amusement park starts at rest and is in free fall. What is the velocity of the ride after 2.3 s?

horsena [70]

Answer:

V = a * t = 9.8 m/s^2 * 2.3 s = 22.5 m/s velocity after 2.3 s

S = 1/2 g t^2 since initial speed is zero

S = 1/2 * 9.8 m/s^2 * 5.29 s^2 = 25.9 m

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3 years ago

If an electron is accelerated from rest through a potential difference of 1.60 x 102V, what is its de Broglie wavelength

Ivahew [28]

Answer:

0.09 x10^-10m

Explanation:

Using wavelength=( 12.27 A)/√V

= 12.27 x 10^-10/ √1.6x10^2

= 0.09x10^-10m

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3 years ago

Zero field spot for opposite unequal charges

xz_007 [3.2K]

Answer:

The zero field location has to be on the line running between the two point charges because that's the only place where the field vectors could point in exactly opposite directions. It can't be between the two opposite charges because there the field vectors from both charges point toward the negative charge.

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3 years ago

water flows through a horizontal pipe with a cross-sectional area of 4m^2 at a speed of 5m/s with a pressure of 300,000pa at poi

LenKa [72]

The velocity at point B is 10 m/s with a pressure of 262500 Pa

<h3>Bernoulli equation</h3>

According to the continuity equation:

A₁V₁ = A₂V₂

Where A is the area and V is velocity, ρ = density of water = 1000 kg/m³

Hence:

4(5) = 2(V₂)

V₂ = 10 m/s

Using Bernoulli equation:

$P_1+\rho gh_1+\frac{1}{2}\rho V_1^2= P_2+\rho gh_2+\frac{1}{2}\rho V_2^2\\\\Hence:\\\\300000+\rho gh+(0.5)*1000*5^2=P_2+\rho gh+(0.5)*1000*10^2\\\\P_2=262500\ Pa\\$

The velocity at point B is 10 m/s with a pressure of 262500 Pa

Find out more on Bernoulli equation at: brainly.com/question/14082066

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3 years ago

The mass of the Moon is 7.35 x 1022 kg, while that of Earth is 5.98 x 1024 kg. The average distance from the center of the Moon

Kryger [21]

Answer:

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Explanation:

$m_e$ = Mass of the Earth = 5.98 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$r_1$ = Distance from the center of the Moon to the center of Earth = 6371000 m

$r_2$ = Distance from the center of the earth center to sun center

$m_m$ = Mass of moon = $7.35\times 10^{22}\ kg$

M = Mass of sun = $1.989\times 10^{30}\ kg$

$F_1=G\frac{m_em_m}{r_1^2}\\\Rightarrow F_1=6.67\times 10^{-11}\frac{5.98\times 10^{24}\times 7.35\times 10^{22}}{(384000000)^2}\\\Rightarrow F_1=1.988\times 10^{20}\ N$

$F_2=G\frac{Mm_e}{r_1^2}\\\Rightarrow F_2=6.67\times 10^{-11}\frac{5.98\times 10^{24}\times 1.989\times 10^{30}}{(149.6\times 10^9+6371000+695.51\times 10^6)^2}\\\Rightarrow F_2=3.511\times 10^{22} N$

$\frac{F_1}{F_2}=\frac{1.988\times 10^{20}}{3.511\times 10^{22}}\\\Rightarrow \frac{F_1}{F_2}=0.00566\\\Rightarrow F_1=F_20.00566$

Hence the force of moon on earth is 0.00566 times the force of earth on moon center to center

4 0

3 years ago