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astra-53 [7]
2 years ago
11

Plz help I don’t know what to do

Physics
1 answer:
Musya8 [376]2 years ago
6 0
A. compression (the place on a wave where the lines are closest together)
b. rarefaction (the place on a wave where the lines are farthest apart)
c. longitudinal
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The net force experienced by an object is increased. What affect does this have on the acceleration of the object?
Evgesh-ka [11]

Answer:

The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.

Explanation:

hope it helps pls give me brainless

6 0
3 years ago
1.
rusak2 [61]

Answer:

9.6

Explanation:

to convert km to miles multiply by 1.609

7 0
3 years ago
A tuning fork generates sound waves with a frequency of 240 Hz. The waves travel in opposite directions along a hallway, are ref
Bingel [31]

Answer:

The phase difference between the reflected waves when they meet at the tuning fork is 159.29 rad.

Explanation:

Given that,

Frequency of sound wave = 240 Hz

Distance = 46.0 m

Distance of fork = 14 .0 m

We need to calculate the path difference

Using formula of path difference

\Delta x=2(L_{2}-L_{1})

Put the value into the formula

\Delta x =2((46.0-14.0)-14.0)

\Delta x=36\ m

We need to calculate the wavelength

Using formula of wavelength

\lambda=\dfrac{v}{f}

Put the value into the formula

\lambda=\dfrac{343}{240}

\lambda=1.42\ m

We need to calculate the phase difference

Using formula of the phase difference

\phi=\dfrac{2\pi}{\lambda}\times \delta x

Put the value into the formula

\phi=\dfrac{2\pi}{1.42}\times36

\phi=159.29\ rad

\phi\approx 68.2^{\circ}

Hence, The phase difference between the reflected waves when they meet at the tuning fork is 159.29 rad.

7 0
3 years ago
In an automobile collision, a 44-kilogram passenger moving at 15 meters per second is brought to rest by an air bag during a 0.1
const2013 [10]

Answer:

6,600N

Explanation:

According to second law of motion, Force = mass × acceleration

If acceleration = change in velocity/time = 15/0.10

Acceleration = 150m/s²

Given mass = 44kg

Force = 44× 150

Force = 6,600N

Magnitude of the average force exerted on the passenger during this time is 6,600N

8 0
3 years ago
Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces
Mandarinka [93]

Answer: 0.81\times 10^{16} beta particles

Explanation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass = 14.0 g

Molar mass = 137 g/mol

\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number 6.023\times 10^{23} of particles.

1 mole of cesium contains atoms =  6.023\times 10^{23}

0.102 moles of cesium contains atoms =  \frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}

The relation of atoms with time for radioactivbe decay is:

N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}

Where N_t =atoms left undecayed

N_0 = initial atoms

t = time taken for decay = 3 minutes

{t_{\frac{1}{2}}} = half life = 30.0 years = 1.577\times 10^7 minutes

The fraction that decays  :  1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}

Amount of particles that decay is  = 0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}

Thus 0.81\times 10^{16} beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0
2 years ago
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