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Amiraneli [1.4K]
2 years ago
5

Calculate the speed of a car that travels 355km in 15 hours , a good answer please

Physics
2 answers:
Ludmilka [50]2 years ago
8 0

Answer:

23.67 km / hr

Explanation:

car travels 355 km (d)

duration = 15 hrs  (t)

average speed formula = v = d / t

v = 355 km / 15 hr

v = 23.67 km / hr

Zigmanuir [339]2 years ago
8 0

Explanation:

<em>Hello</em><em>,</em><em> </em><em>there</em><em>!</em><em>!</em><em>!</em>

<em>Here</em><em>,</em><em> </em><em>your</em><em> </em><em>question</em><em> </em><em>is</em><em> </em><em>asking</em><em> </em><em>about</em><em> </em><em>the</em><em> </em><em>speed</em><em> </em><em>and</em><em> </em><em>the</em><em> </em><em>given</em><em> </em><em>are</em><em>:</em>

<em>distance</em><em> </em><em>(</em><em>s</em><em>)</em><em>=</em><em> </em><em>3</em><em>5</em><em>5</em><em>km</em>

<em>time</em><em> </em><em>(</em><em>t</em><em>)</em><em> </em><em>=</em><em> </em><em>1</em><em>5</em><em>hrs</em><em>.</em>

<em>speed</em><em> </em><em>(</em><em>v</em><em>)</em><em>=</em><em>?</em>

<em>now</em><em>,</em><em> </em>

<em>we</em><em> </em><em>have</em><em> </em><em>the</em><em> </em><em>formula</em><em>:</em>

<em> </em>

<em>v</em><em>=</em><em> </em><em>s</em><em>/</em><em>t</em>

<em>or</em><em>,</em><em> </em><em>v</em><em> </em><em>=</em><em> </em><em>3</em><em>3</em><em>5</em><em>km</em><em> </em><em>/</em><em>1</em><em>5</em><em> </em><em>hrs</em><em>.</em>

<em>Therefore</em><em>,</em><em> </em><em>the</em><em> </em><em>speed</em><em> </em><em>of</em><em> </em><em>car</em><em> </em><em>was</em><em> </em><em>2</em><em>3</em><em>.</em><em>6</em><em>7</em><em>km</em><em>/</em><em>hr</em><em>.</em>

<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em>

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The bicyclist accelerates with magnitude <em>a</em> such that

25.0 m = 1/2 <em>a</em> (4.90 s)²

Solve for <em>a</em> :

<em>a</em> = (25.0 m) / (1/2 (4.90 s)²) ≈ 2.08 m/s²

Then her final speed is <em>v</em> such that

<em>v</em> ² - 0² = 2<em>a</em> (25.0 m)

Solve for <em>v</em> :

<em>v</em> = √(2 (2.08 m/s²) / (25.0 m)) ≈ 10.2 m/s

Convert to mph. If you know that 1 m ≈ 3.28 ft, then

(10.2 m/s) • (3.28 ft/m) • (1/5280 mi/ft) • (3600 s/h) ≈ 22.8 mi/h

8 0
2 years ago
Suppose there is a uniform magnetic field, B, pointing into the page (so your index finger will point into the page). If the vel
Naily [24]

Answer:

Explanation:

We shall show all given data in vector form and calculate the direction of force with the help of following formula

force F = q ( v x B )

q is charge , v is velocity and B is magnetic field.

Given B = - Bk ( i is  right  , j is  upwards  and k is straight up the page  )

v = v j

F = q ( vj x - Bk )

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The direction is towards left .

a ) If velocity is down

v = - v j

F = q ( - vj x - bk )

= qvB i

Direction is right .

b ) v = v i

F = q ( vi x - Bk )

= qvB j

force is upwards

c ) v = - vi

F = q ( -vi x - Bk )

= -qvBj

force is downwards

d ) v = - v k

F = q( - vk x -Bk  )

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No force will be created

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F = q(  vk x -Bk  )

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No force will be created  

3 0
3 years ago
A 2 kg ball is moving 3 m/s when it starts rolling up a hill.
AURORKA [14]

Answer:

the height reached is = 0.458 [m]

Explanation:

We need to make a sketch of the ball and see the location of the reference point where the potential energy is zero. But the kinetic energy will be defined by the following expression:

Ek=\frac{1}{2} *m*v^{2} \\where:Ek= kinetic energy [J]\\m = mass of the ball [kg]\\v = velocity of the ball [m/s]

Replacing the values on the equation we have:

Ek=\frac{1}{2}*(2)*(3^{2} )\\ Ek=9[J]\\

This kinetic energy will be transformed in potential energy in the moment when the ball starts to rolling up. Therefore the maximum height reached by the ball depends of the initial velocity given to the ball.

Ek=Ep\\where\\Ep=potential energy [J]\\Ep=m*g*h\\where\\g=gravity = 9.81[m/s^2]\\h=height reached [m]\\

Now we have:

h=\frac{Ep}{m*g} \\h=\frac{9}{2*9.81} \\\\h=0.45 [m]

In that moment when the ball reach the 0.45 [m] the potencial energy will be maximum and equal to the kinetic energy when the ball has a velocity of 3[m/s]

6 0
3 years ago
What's 5 and 6? Physics
andre [41]
5) 204 meters 
6)
A) 150 miles
B)241 km
4 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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