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AlekseyPX
3 years ago
7

A car travels 60 miles per hour. how much time does it takes the car to travel 30 miles?

Physics
2 answers:
elena55 [62]3 years ago
7 0

1 hour= 60 mins

60 mins= 60 miles

30 miles= (60 miles÷60 minutes)×30 miles

             =30 mins


I am Lyosha [343]3 years ago
7 0
1 hr = 60 minutes 60 minutes / 60 miles = 1 minutes/miles 30 miles * 1 minute = 30 minutes 30 minutes
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1 mile. Is this a joke lol
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A conductor carrying a current I = 16.5 A is directed along the positive x axis and perpendicular to a uniform magnetic field. A
Jet001 [13]

To solve this problem we will apply the concepts related to the Magnetic Force, this is given by the product between the current, the body length, the magnetic field and the angle between the force and the magnetic field, mathematically that is,

F = ILBsin \theta

Here,

I = Current

L = Length

B = Magnetic Field

\theta = Angle between Force and Magnetic Field

But \theta = 90\°

F = ILB

Rearranging to find the Magnetic Field,

B = \frac{F}{IL}

Here the force per unit length,

B = \frac{1}{I}\frac{F}{L}

Replacing with our values,

B = \frac{0.130N/m}{16.5}

B = 0.0078T

Therefore the magnitude of the magnetic field in the region through which the current passes is 0.0078T

6 0
3 years ago
A car covers a distace of 20km in 20 seconds. Calculate its speed in km/h m/s.​
BartSMP [9]

Answer:

speed = 3600 kilometers per hour

= 3600 km/h

speed = 1000 meters per second

= 1000 m/s

Answer From Gauth Math

6 0
3 years ago
A parallel-plate capacitor is connected to a battery. What happens to the stored energy if the plate separation is doubled while
marta [7]

Answer:

(c)  As 'd' becomes doubled, energy decreases by the factor of 2

Explanation:

Energy stored in a parallel plate capacitor is given by:

U=\frac{1}{2}CV^2\\\\C=\frac{A\epsilon_{o}}{d}\\\\then\\\\U=\frac{1}{2}\frac{A\epsilon_{o}}{d}V^2--(1)\\\\

As capacitor remains connected to the battery so V remains constant. As can be seen from (1) that energy is inversely proportional to the separation between the plates so as 'd' becomes doubled, energy decreases by the factor of 2.

3 0
3 years ago
Read 2 more answers
A whistle of frequency 564 Hz moves in a circle of radius 71.2 cm at an angular speed of 17.1 rad/s. What are (a) the lowest and
trapecia [35]

Answer:

a) f'=544.66 \textup{Hz}

b) f'=584.75 \textup{Hz}

Explanation:

Given:

Frequency of the whistle, f = 564 Hz

Radius of the circle, r = 71.2 cm = 0.712 m

Angular speed, ω = 17.1 rad/s

speed of source, v_s = rω = 0.712 × 17.1 = 12.1752 m/s

speed of sound, v = 343 m/s

Now, applying the Doppler's effect formula, we have

f'=f\frac{v\pm v_d}{v\pm v_s}

where,

v_d = relative speed of the detector with respect to medium = 0

a) for lowest frequency, we have the formula as:

f'=f\frac{v}{v+v_s}

on substituting the values, we get

f'=564\times\frac{343}{343+12.1752}

or

f'=544.66 \textup{Hz}

b) for maximum frequency, we have the formula as:

f'=f\frac{v}{v-v_s}

on substituting the values, we get

f'=564\times\frac{343}{343-12.1752}

or

f'=584.75 \textup{Hz}

3 0
3 years ago
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