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iragen [17]
3 years ago
7

6. Which of the following is the force needed to make a 1-kg object accelerate at 1 m/s??

Physics
2 answers:
Sever21 [200]3 years ago
8 0

Answer:

i think d. : 1 newton

Explanation:

because to accelerate something you need force and the unit for force is Newtons

blondinia [14]3 years ago
7 0
1 Newton I searched it up so I suppose it’s correct
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Suppose you observe that at night, the air just above the ground feels cooler than the air above it. then in the middle of a sun
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A car stops in 130 m. If it has an acceleration of -5 m/s2 what was the cars starting velocity?
Tatiana [17]

Answer:

<u>We are given:</u>

displacement (s) = 130 m

acceleration (a) = -5 m/s²

final velocity (v) = 0 m/s      [the cars 'stops' in 130 m]

initial velocity (u) = u m/s

<u>Solving for initial velocity:</u>

From the third equation of motion:

v² - u² = 2as

replacing the variables

(0)² - (u)² = 2(-5)(130)

-u² = -1300

u² = 1300

u = √1300

u = 36 m/s

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3 years ago
How does unequal solar heating lead to the Gulf Stream?
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Maple syrup, which comes from the sap of maple trees, contains water and natural sugars. It's a clear, brown liquid and the suga
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3 years ago
Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun
iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})     (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV

B. The energy of the emitted photon is given by the following formula:

E=h\frac{c}{\lambda}   (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J

Next, you use the equation (2) and solve for λ:

\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm

C. The radius of the orbit is given by:

r_n=n^2a_o   (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

r_5=5^2(2.380)=59.5

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0
3 years ago
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