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Illusion [34]
4 years ago
12

a car can go from 0 - 45m/s in 3.4 seconds A. how much is its acceleration in m/s B. how much longer would it take the car to ge

t up to 70m/s
Physics
1 answer:
Nimfa-mama [501]4 years ago
3 0

Answer:

13.2m/s²

1.89s

Explanation:

Given parameters:

Initial velocity  = 0m/s

Final velocity 1   = 45m/s

Final velocity 2  = 70m/s

Unknown:

Time taken  = 3.4s

Solution:

Acceleration is the rate of change of velocity with time.

   Acceleration  = \frac{Final velocity 1 - initial velocity }{3.4}    

  Acceleration  = \frac{45 - 0}{3.4}   = 13.2m/s²

B. How much longer would it take the car to get up to 70m/s

Maintaining the is acceleration;

       13.2 = \frac{70 - 45}{t}  

        25  = 13.2t

          t  = 1.89s

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What is kinematics?? koi Hindustani online h kya?​
AlexFokin [52]

Answer:

the branch of mechanics concerned with the motion of objects without reference to the forces which cause the motion.

3 0
3 years ago
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Suppose you are holding a basketball while standing still on a skateboard. You and the skateboard have a mass of 50kg. You throw
Nutka1998 [239]
Your acceleration before you throw the ball is zero. After the throw, you use the equation 

<span>F = m * a </span>

<span>To solve for acceleration </span>

<span>a = F/m </span>

<span>a = (10 N) / (50 kg) </span>

<span>a = 0.2 m/s^2</span>
4 0
4 years ago
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For a galvanic cell to generate an electric current flowing from anode to cathode, what must be true
photoshop1234 [79]

Complete question:

For a galvanic cell to generate an electric current flowing from anode to cathode, what must be true?

(a) Electrons flow from the anode to the cathode

(b) Electrons flow from the more negatively charged electrode to the more positively charged electrode

(c) Electrons flow from higher potential energy to lower potential energy

(d) All of the above are true.

Answer:

(d) All of the above are true.

Explanation:

A galvanic or Voltaic cell is a primary type of electrochemical cell that is used to generate electrical energy from the chemical reactions that take place in it.

It consists of a positive electrode (cathode) and a negative electrode (anode) for the movement of charges.

(a) Electrons flow from the anode to the cathode. TRUE

Anode is the negative electrode and for electron current, electrons flow from negative electrode to positive electrode.

(b) Electrons flow from the more negatively charged electrode to the more positively charged electrode. TRUE

Based on electron current flow.

(c) Electrons flow from higher potential energy to lower potential energy. TRUE

The driving force of the electron flow is the potential difference. Electrons must flow from higher potential to lower potential.

All the options are correct, so we select option "D"

8 0
4 years ago
Light of wavelength 550 nm falls on a
Brums [2.3K]

Answer:

The first diffraction maximum fringe will be at approximately 2.7 meters from the central maximum.

Explanation:

We can describe single slit diffraction phenomenon with the equation:

a\sin\theta=m\lambda (1)

with θ the angular position of the minimum of order m respect the central maximum, a the slit width and λ the wavelength of the incident light. Because the distances between the first minima and the central maximum (y_{m}) are small compared to the distance between the screen and the slit (x), we can approximate \sin\theta\approx\tan\theta=\frac{y}{x}, using this on (1):

a\frac{y_{m}}{x}=m\lambda

solving for y

y_{m}= \frac{mx\lambda}{a}

Note that y_{m}is the distance between a minimum and the central maximum but we need the position of a maximum not a minimum, here we can use the fact that a maximum is approximately between two minima, so the first diffraction maximum fringe is between the minima of order 1 and 2, so we should find y_{1}, y_{2} add them and divide by two:

y_{1}= \frac{(1)(10.0m)(550\times10^{-9}\,m)}{3.00\times10^{-6}\,m}

y_{1}= 1.8 m

y_{2}= \frac{(2)(10.0m)(550\times10^{-9}\,m)}{3.00\times10^{-6}\,m}

y_{1}= 3.6 m

maximum = \frac{1.8+3.6}{2}=2.7m

7 0
3 years ago
You throw a baseball directly upward at time t = 0 at an initial speed of 12.3 m/s. What is the maximum height the ball reaches
Semmy [17]

Explanation:

At the maximum height, the ball's velocity is 0.

v² = v₀² + 2a(x - x₀)

(0 m/s)² = (12.3 m/s)² + 2(-9.80 m/s²)(x - 0 m)

x = 7.72 m

The ball reaches a maximum height of 7.72 m.

The times where the ball passes through half that height is:

x = x₀ + v₀ t + ½ at²

(7.72 m / 2) = (0 m) + (12.3 m/s) t + ½ (-9.8 m/s²) t²

3.86 = 12.3 t - 4.9 t²

4.9 t² - 12.3 t + 3.86 = 0

Using quadratic formula:

t = [ -b ± √(b² - 4ac) ] / 2a

t = [ 12.3 ± √(12.3² - 4(4.9)(3.86)) ] / 9.8

t = 0.368, 2.14

The ball reaches half the maximum height after 0.368 seconds and after 2.14 seconds.

5 0
4 years ago
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