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3 years ago

A school bus moves at speed of 35 mi/hr for 20 miles. How long will it take the bus to get to school?

Physics

2 answers:

Irina18 [472]3 years ago

7 0

The time required for the bus to get to school (in hours) is

<em>(the distance from here to school) divided by (35 mi/hr) </em> .

In order to find the actual number, we would need to know the distance from here to school.

mafiozo [28]3 years ago

4 0

Answer:

Time, t = 0.57 hours

Explanation:

It is given that,

Speed of the school bus, v = 35 mi/hr

Distance covered by the bus, d = 20 miles

We need to find the time taken by the bus to get to school. Time taken by the bus is given by :

$t=\dfrac{d}{v}$

$t=\dfrac{20\ mi}{35\ mi/hr}$

t = 0.57 hours

So, the time taken by the bus to reach school is 0.57 hours. Hence, this is the required solution.

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Answer:

Pressure = ρgh

pressure (p) is measured in pascals (Pa)

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The fore of gravitational field strength (g) is measured in N/kg or m/s 2

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Answer = 235,200 Pa

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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

If you walk 30 meters forwards, and then turn around and walk 25 meters backwards, what is the distance that you walked? What di

xeze [42]

Given :

Walk in forward direction is 30 m .

Walk in backward direction is 25 m .

To Find :

The distance and displacement .

Solution :

We know , distance is total distance covered and displacement is distance between final and initial position .

So , distance travelled is :

D = 30 + 25 m = 55 m .

Now , we first move 30 m in forward direction and then 25 m in backward direction .

So , displacement is :

D = 30 - 25 m = 5 m .

Therefore , distance and displacement covered is 55 m and 5 m respectively .

Hence , this is the required solution .

5 0

3 years ago