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liubo4ka [24]
3 years ago
14

Jessica has a mass of 55kg she sleds down a hill that has a slope of 32 degrees. what is the component of her weight that is alo

ng her direction of motion ?

Physics
2 answers:
Snowcat [4.5K]3 years ago
8 0

Answer:

286 rounded, - A P E X

Explanation:

Arturiano [62]3 years ago
7 0

Answer:

W = 285.62 N

Explanation:

It is given that,

Mass of Jessica is 55 kg

Slope of the hill is 32 degrees

We need to find the component of her weight that is along her direction of motion.

The component along her direction of motion is shown in attached figure. It means

W_y=mg\sin\theta\\\\W_y=55\times 9.8\times \sin(32)\\\\W_y=285.62\ N

So, the component of her weight that is along her direction of motion is 285.62 N.

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An atom of this element would have a very easy time losing one electron to form an ionic bond with an atom of an element that wo
shepuryov [24]

Answer: potassium

Explanation: An atom of potassium would have a very easy time losing one electron to form an ionic bond with an atom of an element that would easily accept it.

8 0
3 years ago
What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?
adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, 3.03\times 10^{-19}J

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )

Where,

\lambda = Wavelength of radiation

R_H = Rydberg's Constant  = 10973731.6 m⁻¹

n_f = Higher energy level = 3

n_i= Lower energy level = 2

Putting the values, in above equation, we get:

\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )

\lambda=6.56\times 10^{-7}m

Now we have to calculate the energy.

E=\frac{hc}{\lambda}

where,

h = Planck's constant = 6.626\times 10^{-34}Js

c = speed of light = 3\times 10^8m/s

\lambda = wavelength = 6.56\times 10^{-7}m

Putting the values, in this formula, we get:

E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}

E=3.03\times 10^{-19}J

Therefore, the energy of one photon of hydrogen atom is, 3.03\times 10^{-19}J

3 0
3 years ago
What is the kinetic energy of a 200 kg satellite as it follows a circular orbit of radius 8x106m around the earth?
motikmotik
Given the equation for the Speed of a Satellite

v = SqRt{Gravitational Constant}{Mass of Earth} divided by the radius given in your problem

we have:


(square root whole term on right side)

v = G Me
———
r


so. (6.67x10^-11)(5.97x10^24)
___________________
(8.0x10^6)


v = 7055 m/s (which is reasonable)


so utilize the Kinetic Energy Formula


KE = 1/2mv^2


KE = 1/2(200)(7055)^2


KE = 4.977x10^9 J


4 0
3 years ago
Read 2 more answers
Determine the vector perpendicular to the plane of A= 31+ 6j - 2k and B=4i-j +3k
Sliva [168]

The vector perpendicular to the plane of A = 3i+ 6j - 2k and B = 4i-j +3k is 16 i - 17 j - 27 k

Let r be the vector perpendicular to A and B,

r = A * B

A = 3i + 6j - 2k

B = 4i - j + 3k

a1 = 3

a2 = 6

a3 = - 2

b1 = 4

b2 = - 1

b3 = 3

a * b = ( a2 b3 - b2 a3 ) i + ( a3 b1 - b3 a1 ) j + ( a1 b2 - b1 a2 ) k

a * b = [ ( 6 * 3 ) - ( - 1 * - 2 ) ] i + [ ( - 2 * 4 ) - ( 3 * 3 ) ] j + [ ( 3 * - 1 ) - ( 4 * 6 ) ] k

a * b = 16 i - 17 j - 27 k

The perpendicular vector, r = 16 i - 17 j - 27 k

Therefore, the vector perpendicular to the plane of A = 3i + 6j - 2k and B = 4i - j + 3k is 16 i - 17 j - 27 k

To know more about perpendicular vectors

brainly.com/question/14384780

#SPJ1

5 0
1 year ago
The photeselestric effect is observed when light of a sufficiently high frequency is focused onto a polished metal surface, emit
Helga [31]

Answer:

3.4\cdot 10^{-19} J

Explanation:

In order to convert the work function of cesium from electronvolts to Joules, we must use the following conversion factor:

1 eV = 1.6 \cdot 10^{-19} J

In our problem, the work function of cesium is

E=2.1 eV

so, we can convert it into Joules by using the following proportion:

1 eV : 1.6\cdot 10^{-19} J = 2.1 eV : x\\x=\frac{(1.6\cdot 10^{-19} J)(2.1 eV)}{1 eV}=3.4\cdot 10^{-19} J

8 0
3 years ago
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