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ankoles [38]
3 years ago
5

What is the average speed in mph for a car that travels for 5 hours and 20 minutes?

Physics
2 answers:
Ymorist [56]3 years ago
8 0

Answer: If you meant 5 miles in 20 minutes than it’s 1 mile in 5 minutes

Explanation:

s2008m [1.1K]3 years ago
5 0

The car's average speed is

0.1875 • (the number of miles it covered)

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4. A driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch,
stira [4]

Answer:

The driver's average velocity is 82.35 km/h.

Explanation:

Given:

The motion of the driver can be divided into 3 parts:

i. Displacement of the driver in 1.5 hours = 135 km

ii. Rest for 45 minutes.

iii. Displacement in next 2 hours = 215 km

The direction of motion remains same (east).

Now, total displacement of the driver is, D_{Total}=135+215=350 km.

Rest time is 45 minutes. Converting it to hours, we need to use the conversion factor 1\textrm{ min} = \frac{1}{60} hour.

So, 45 minutes in hours is equal to \frac{45}{60}=0.75 hours.

Now, total time taken for the complete journey is, \Delta t=1.5+\frac{45}{60}+2=1.5+0.75+2=4.25\textrm{ h}

Average velocity is given as:

v_{avg}=\frac{\textrm{Total displacement}}{Total time}=\frac{350}{4.25}=82.35\textrm{ km/h}

Therefore, the driver's average velocity is 82.35 km/h

4 0
3 years ago
In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10
sergij07 [2.7K]

Answer:

A)\Phi=83.84\times 10^{-9}

B)\Phi=0 Wb

C)emf=5.4090\times 10^{-4}V

Explanation:

Given that:

  • no. of turns i the coil, n=200
  • area of the coil, a=13.1 \times 10^{-4}\,m^2
  • time interval of rotation, t=3.1\times 10^{-2}\,s
  • intensity of magnetic field, B=6.4\times 10^{-5}\,T

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

\Phi=B.a\,cos \theta..................................(1)

\theta is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}

\Phi=83.84\times 10^{-9} Wb

(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

∴  \theta=90^{\circ}

From eq. (1)

\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}

\Phi=0 Wb

(C)

According to the Faraday's Law we have:

emf=n\frac{B.a}{t}

emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}

emf=5.4090\times 10^{-4}V

7 0
3 years ago
Anybody from India ?​
Lelu [443]

Answer:

No,why you say that

Explanation:

7 0
2 years ago
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When attempting to determine the coefficient of kinetic friction, why is it necessary to move the block with constant velocity
Readme [11.4K]

Answer:

Explanation:

In order to measure the coefficient of friction , we apply external force to move the body . When external force comes in motion , we adjust the external force so that it moves with zero acceleration or uniform velocity . In this case external force becomes equal to kinetic frictional force and then net force becomes zero because

net force = mass x acceleration = m x 0 = 0

Now frictional force = μ mg where μ is coefficient of kinetic friction

so F = μ mg where F is external force applied

μ = F / mg

Hence , to make external force equal to frictional force , it is necessary to make acceleration of body zero .

4 0
3 years ago
Ruthford's theory was identified by
8090 [49]

Answer:

your mom

Explanation:

6 0
3 years ago
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