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IrinaVladis [17]
3 years ago
11

which of the following is not an example of force? A. magnetism B. Electric force C. energy D. friction​

Physics
1 answer:
Tcecarenko [31]3 years ago
5 0

Answer:

Friction​

Explanation:

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Two horses are side by side on a carousel. Which has a greater tangential speed the one closer to the center or the one farther
photoshop1234 [79]

Answer:

The horse father from the center has a greater tangential speed. Although both horses complete one circle in the same time period, the one farther from the center covers a greater distance during that same period.

Explanation:

8 0
3 years ago
If the mass of the book is 50 sliding with acceleration 1.2 m/s ^ 2 then the friction force is
Tatiana [17]

Answer:

73N

Explanation:Just multiply 1.2^2 by 50

3 0
3 years ago
calculate the distance a plane flies on a 7.95 hour flight from Chicago to London. Assume a constant speed of 800.0 km/h
soldier1979 [14.2K]

Answer:

6360 km

Explanation:

Use the kinematics equation x=v_ot+\frac{1}{2}at^2.  We are given t = 7.95 hours and a = 0 m/s^2 (constant speed means there is no acceleration).  Solve for x.

x=(800)(7.95)+\frac{1}{2}(0)(7.95^2)\\x=6360 \ km

4 0
3 years ago
a bus starting from rest moves with a uniform acceleration of 0.1 metre per second square for 2 minutes find the speed acquired
blondinia [14]
S ?
U 0m/s
V ?
A 0.1m/s^2
T 2min (120 sec)

S=ut+0.5at^2
S=0(120 sec)+0.5(0.1m/s^2)(120 sec)^2
S=720m

Distance double 720m*2=1440m

V^2=u^2+2as
V^2=(0)^2+2(0.1 m/s^2)(1440m)
V^2=288
V= square root of 288=12 root 2=16.97 to 2 decimal places
6 0
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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