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sammy [17]
2 years ago
15

The magnitude of the vertical velocity vector for an upwardly launched projectile _________. a stays constant b gets smaller and

then larger c decreases throughout the flight d increases throughout the flight
Physics
2 answers:
chubhunter [2.5K]2 years ago
7 0
B. Gets smaller and then larger
Masja [62]2 years ago
4 0

Answer:

changes by 9.8 m/s each second.

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If a marathon runner averages 9.50 mi/h, how many minutes does it take him or her to run a 26.22-mi marathon?
horsena [70]

Answer: 196 minutes

Explanation: 26.22/9.50 = 2.76

2 hours and 76 minutes equals 196 minutes

4 0
3 years ago
kenneth warmed up a 100- ml sample of air in and expandable container. he displayed the resluts in the following graph. the resl
shepuryov [24]

Answer:

300k

Explanation:

3 0
3 years ago
A technician services the carburetor, and then, performs a complete governor system adjustment. The governor system on the engin
madreJ [45]

Answer: Either of the two

Explanation:

Either of the two - the governed idle spring or the normal primary governor spring can be adjusted first.

The main function of the carburetor is to mix air with fuel.

5 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
The LR5 is the specialist submarine for underwater rescue. The average density of seawater is 1028 kg/ m3.
sladkih [1.3K]

Answer:

P = 7196 [kPa]

Explanation:

We can solve this problem using the expression that defines the pressure depending on the height of water column.

P = dens*g*h

where:

dens = 1028 [kg/m^3]

g = 10 [m/s^2]

h = 700 [m]

Therefore:

P = 1028*10*700

P = 7196000 [Pa]

P = 7196 [kPa]

5 0
4 years ago
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