3 years ago

In which group is hydrogen located on the periodic table?

Physics

1 answer:

Lelu [443]3 years ago

3 0

Hydrogen is located in Group 1

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Salma takes 15 minutes from her house to reach her school on a bicycle. If the bicycle has a speed of 2 m/s, calculate the dista

Sidana [21]

Distance = (15 minutes) x (60 sec/min) x (2 meter/sec)

Distance = (15 x 60 x 2) (minute-sec-meter / min-sec)

Distance = 1,800 meters

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3 years ago

What is the speed of a wave that has a frequency of 45 Hz and a wavelength of 0.1 meters?

Yuri [45]

λν=c
(wavelength x frequency = speed)

speed = 45 x 0.1
= 4.5 m/s

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4 years ago

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Problem 21-40a:

Greeley [361]

Answer:

Part a)

$\frac{F_e}{F_g} = 2.74 \times 10^{12}$

Part b)

$q = 3.37 \times 10^{-4} C$

Explanation:

As we know that electric force on electric charge is given as

$F = qE$

here we have

$q = 1.6 \times 10^{-19}C$

E = 153 N/C

now force is given as

$F = (1.6 \times 10^{-19})(153) = 2.45 \times 10^{-17} N$

Gravitational force on electric charge near surface of earth is given as

$F_g = mg$

$F_g = (9.1 \times 10^{-31})(9.81) = 8.93 \times 10^[-30} N$

now the ratio of two forces is given as

$\frac{F_e}{F_g} = \frac{2.45 \times 10^{-17}}{8.93 \times 10^{-30}}$

$\frac{F_e}{F_g} = 2.74 \times 10^{12}$

Part b)

Now the ball is balanced by the electric force and the force of gravity on it

so here we have

$F_g = qE$

$mg = qE$

$(5.25 \times 10^{-3})(9.81) = q(153)$

here we have

$q = 3.37 \times 10^{-4} C$

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4 years ago

Question 9 of 10 How many carbon (C) atoms are in a molecule of CH,O?

Juli2301 [7.4K]

Answer:

1 molecule not shure

Explanation:

6 0

3 years ago

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A person holding a lunch bag is moving upward in a hot air balloon at a constant speed of 7.3 m/s . When the balloon is 24 m abo

Kitty [74]

Explanation:

Given that,

Initial speed of the bag, u = 7.3 m/s

Height above ground, s = 24 m

We need to find the speed of the bag just before it reaches the ground. It can be calculated using third equation of motion as :

$v^2=u^2+2as$

$v^2=(7.3)^2+2\times 9.8\times 24$

$v=\sqrt{523.69}$

v = 22.88 m/s

So, the speed of the bag just before it reaches the ground is 22.38 m/s. Hence, this is the required solution.

8 0

3 years ago