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EastWind [94]
3 years ago
14

A cyclist moving with a constant velocity of 6.0 m/s forward passes a car that is just starting. If the car has a constant accel

eration of 2.0 m/s2, when will the car overtake the cyclist?
Physics
1 answer:
sergiy2304 [10]3 years ago
6 0

After 6 seconds, the car will surpass the cyclist.

<h3><u>Explanation:</u></h3>

The speed of the cyclist = 6 m/s.

Let after time t sec, the car will overtake the cyclist.

So, distance covered by the cyclist in t sec = 6t m

Initial velocity of the car is 0 m/s, because the car is just starting.

Acceleration of the car =2 m/s^2.

Final velocity of the car =6 m/s.

So to cover the distance 6t, the time required by the car = \frac{1}{2} \times a \times t^2 = \frac{1}{2} \times 2 \times t^2

6t = \frac{1}{2} \times 2 \times t^2\\6t = t^2

t =6 sec

So, after 6 seconds, the car will surpass the cycle.  

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Are the statements of the Flat Earth Society an example of science or pseudoscience?
Y_Kistochka [10]

Answer:

Pseudoscience

Explanation:

Pseudoscience is defined as set of ideas that present themselves as a science, but they do not fulfill the criteria to be called science. In the same way, in the middle of 20th century some people called earth is flat. They challenged the spherical shape of earth. But their hypothesis does not fulfill the criteria of science. Therefore, the statements of flat earth society are pseudoscience.

5 0
3 years ago
Read 2 more answers
On what factors does critical velocity depend on
ASHA 777 [7]

Explanation:

The critical velocity is that velocity of liquid flow, up to which its flow is streamlined (laminar)& above which its flow becomes turbulent. It's denoted by Vc & it depends upon: Coefficient of viscosity of liquid (η) Density of liquid. Radius of the tube.

6 0
3 years ago
the electron are accelrated to a speed of 2.40*10^7 in 1.8*10^-9, the force experinced by an electron ?
dalvyx [7]
Force = mass × acceleration

To find acceleration, we can divide the speed by the time it took:

acceleration = 2.40×10^7 / 1.8×10^-9

acceleration = 1.33×10^16

the mass is equal to the mass of an electron

force = (9.11×10^-31)(1.33×10^16)

force = 1.21×10^-14 N
3 0
3 years ago
A merry-go-round is spinning at a rate of 4.04.0 revolutions per minute. Cora is sitting 0.50.5 m from the center of the merry-g
dsp73

Answer:

angular speed of both the children will be same

Explanation:

Rate of revolution of the merry go round is given as

f = 4.04 rev/min

so here we have

f = \frac{4.04}{60} =0.067 rev/s

here we know that angular frequency is given as

\omega = 2\pi f

\omega = 2\pi(0.067)

\omega = 0.42 rad/s

now this is the angular speed of the disc and this speed will remain same for all points lying on the disc

Angular speed do not depends on the distance from the center but it will be same for all positions of the disc

7 0
3 years ago
A 44-turn rectangular coil with length ℓ = 17.0 cm and width w = 8.10 cm is in a region with its axis initially aligned to a hor
Mumz [18]

Answer:

The maximum induced emf in the rotating coil  = 29.66V

The induced emf in the rotating coil when (t = 1.00 s) = 26.66V

The maximum rate of change of the magnetic flux through the rotating coil = 0.674Wb/s

Explanation:

Lets state the parameters we are being given right from the question:

Number of rectangular coil, (N) = 44

Length of Coil, l =17cm in meters we have; (l) = 17 × 10⁻² m

Width of Coil, w =8.10cm in meters we have; (w) = 8.10 × 10⁻² m

Magnitude of Uniform Magnetic Field (B) = 767mT= 765 × 10⁻³ T

Angular Speed of Coil, (ω) = 64 rad/s

(a)

To calculate the induced emf in the rotating cell,we can use the formula:

emf = NBAωsin(ωt)

For maximum induced emf, the value of sin(ωt) will be 1

emf_max = NBAω ; if (A = l × w) , we have:

emf_max  = NB(l × w)ω

subsitituting the parameters into the above equation; we have:

emf_max  = 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64

= 29.66V

(b)

At t = 1s, the induced emf is calculated as:

emf = NBAωsin(ωt)

substituting the parameters into the equation, we have:

emf =   44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64 × sin (64 × 1)

=26.66V

(c)

To calculate the maximum rate of change of the magnetic flux through the rotating coil; we need to reflect on the equation for the maximum induced emf in terms of magnetic flux.

i.e emf_max = N\frac{d∅}{dt}

since emf_max = 29.66 and N = 44; we have:

29.66 =  44\frac{d∅}{dt}

\frac{d∅}{dt} = \frac{29.66}{44}

= 0.674 Wb/s

5 0
3 years ago
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