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bixtya [17]
3 years ago
11

A car moving at constant speed in a straight line with the engine providing a driving force equal to the resistive force F.

Physics
2 answers:
Sergeu [11.5K]3 years ago
7 0

Answer:

Option (b)

Explanation:

Let the initial speed of the car is u and the final speed is zero. Let mass of the car is m.

Case I

Acceleartion of the car, a = - F/m

Use III equation of motion,

v^{2} = u^{2} + 2 a s

0^{2} = u^{2} - 2 \frac{F}{m} \times 100

u^{2} = \frac{200 F}{m}     ...... (1)

Case II

Acceleration of the car, a = - 0.8 F/m

Use III equation of motion,

v^{2} = u^{2} + 2 a s

0^{2} = u^{2} - 2 \frac{0.8F}{m} \times s

0^{2} = u^{2} - 2 \frac{0.8F}{m} \times s

u^{2} = \frac{1.6F}{m} \times s      ..... (2)

Dividing equation (2) by equation (1), we get

s = 125 m

Lunna [17]3 years ago
6 0

125 b

simultaneous kinematic equations two variables are F and stopping distance

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slamgirl [31]

Answer:

The charge is 2.75\times10^{-13}\ C

Explanation:

Given that,

Distance = 2.5 mm

Electric field = 800 NC

Length L=5.0\times5.0\times10^{-4}\ m

We need to calculate the linear charge density

Using formula of linear charge density

E=\dfrac{2k\lambda}{r}

\lambda=\dfrac{Er}{2k}

Put the value into the formula

\lambda=\dfrac{800\times2.5\times10^{-3}}{2\times9\times10^{9}}

\lambda=1.1\times10^{-10}\ C/m

We need to calculate the charge

Using formula of charge

Q=\lambda\timesL

Put the value into the formula

Q=1.1\times10^{-10}\times(5.0\times5.0\times10^{-4})

Q=2.75\times10^{-13}\ C

Hence, The charge is 2.75\times10^{-13}\ C

5 0
2 years ago
The edge of a flying disc with a radius of 0.13 m spins with a tangential speed of 3.3 m/s. The centripetal acceleration of the
Marat540 [252]

Answer:

Centripetal acceleration = 83.77m/s²

Explanation:

<u>Given the following data;</u>

Radius, r = 0.13m

Velocity, v = 3.3m/s

To find centripetal acceleration;

Centripetal acceleration is given by the formula;

Acceleration, a = \frac {v^{2}}{r}

Substituting into the equation, we have;

Centripetal \; acceleration, a = \frac {3.3^{2}}{0.13}

Centripetal \; acceleration, a = \frac {10.89}{0.13}

<em>Centripetal acceleration = 83.77m/s²</em>

<em>Therefore, the centripetal acceleration of the edge of the disc is 83.77 m/s². </em>

5 0
3 years ago
Read 2 more answers
A car goes East from 2 m/s to 16m/s in 3.5s. What is the Car's acceleration?
BaLLatris [955]
A=v1+v2 / t
2+16/3.5=?
8 0
2 years ago
Grace rides her bike at a constant speed of 6 miles per hour. How far can she travel in 2 1/2 hours?
aliina [53]

Answer:

15 miles

Explanation:

6 miles per hour

2 1/2 hours

6 x 2 = 12

6 x 1/2 = 3

12 + 3 = 15

7 0
3 years ago
A diffraction grating has 500 slits/mm. What is the longest wavelength of light for which there will be a third-order maximum?
Alexxandr [17]

Answer:

The longest wavelength of light  is 666.7 nm

Explanation:

The general form of the grating equation is

mλ = d(sinθi + sinθr)

where;

m is third-order maximum = 3

λ is the wavelength,

d is the slit spacing (m/slit)

θi  is the incident angle

θr is the diffracted angle

Note: at longest wavelength, sinθi + sinθr = 1

λ = d/m

d = 1/500 slits/mm

λ = 1 mm/(500 *3) = 1mm/1500 = 666.7 X 10⁻⁶ mm = 666.7 nm

Therefore, the longest wavelength of light  is 666.7 nm

8 0
3 years ago
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