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n200080 [17]
2 years ago
11

What constant acceleration is required to increase the speed of a car from 26 mi/h to 51 mi/h in 5 seconds

Physics
1 answer:
mr_godi [17]2 years ago
8 0

2.25 m/s² of acceleration is required to increase the speed of a car from 26 mi/h to 51 mi/h in 5 seconds.

To find the answer, we need to know about the acceleration.

<h3>What is acceleration?</h3>
  • Acceleration is given as the ratio of velocity to time.
  • Mathematically, acceleration= velocity/time.
<h3>What is the acceleration required to increase the speed of a car from 26 mi/h to 51 mi/h in 5 seconds?</h3>
  • Here change in velocity of the car is 51-26= 25 mi/h.
  • As 1 mi/h = 0.45 m/s. So 25mi/h = 11.25 m/s.
  • Acceleration= (11.25m/s)/5s = 2.25 m/s².

Thus, we can conclude that the constant acceleration is 2.25 m/s².

Learn more about the acceleration here:

brainly.com/question/460763

#SPJ4

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Figure one, voltmeters
slava [35]

Answer:

(i) Half

(ii) 3 V

(iii) V₁

Explanation:

(i) The given parameters are;

The circuits have identical resistances

The number of resistors in circuit 1 = 1 resistor

The number of resistors in circuit 2 = 2 resistors

Let 'R' represent the value of each resistor, we have;

The total resistance of circuit 1 = R Ohm

The total resistance of circuit 3 = 2·R Ohm

∴ The total resistance of circuit 1  = (1/2) × The total resistance of circuit 3

∴ The resistance of circuit 1 is <u>half</u> the resistance of circuit 3

(ii) The potential difference of each cell, V = 1.5 volts

The number of cells in circuit 2 = 2 cells

The total potential difference of the cells of circuit 2 = 2 × 1.5 volts = 3 × volts = 3 V.

The voltmeter reading = The potential difference across the cell or cells it is applied

∴ The voltmeter reading on voltmeter, V₂, applied across the cells of circuit 2 = 3 V

(iii) The voltmeter reading V₁ = 1.5 V

The voltmeter reading V₂ = 3 V

The voltmeter reading V₃ = 4.5/(2·R) × R = 2.25 V

Therefore, the voltmeter reading with the smallest volt, is V₁ = 1.5 V

6 0
3 years ago
Dave rows a boat across a river at 4.0 m/s. the river flows at 6.0 m/s and is 360 m across.
Ad libitum [116K]

a) Let's call x the direction parallel to the river and y the direction perpendicular to the river.

Dave's velocity of 4.0 m/s corresponds to the velocity along y (across the river), while 6.0 m/s corresponds to the velocity of the boat along x. Therefore, the drection of Dave's boat is given by:

\theta= arctan(\frac{v_y}{v_x})=arctan(\frac{4.0 m/s}{6.0 m/s})=arctan(0.67)=33.7^{\circ}

relative to the direction of the river.


b) The distance Dave has to travel it S=360 m, along the y direction. Since the velocity along y is constant (4.0 m/s), this is a uniform motion, so the time taken to cross the river is given by

t=\frac{S_y}{v_y}=\frac{360 m}{4.0 m/s}=90 s


c) The boat takes 90 s in total to cross the river. The displacement along the y-direction, during this time, is 360 m. The displacement along the x-direction is

S_x = v_x t =(6.0 m/s)(90 s)=540 m

so, Dave's landing point is 540 m downstream.


d) If there were no current, Dave would still take 90 seconds to cross the river, because its velocity on the y-axis (4.0 m/s) does not change, so the problem would be solved exactly as done at point b).

7 0
4 years ago
I need help asap. #3, 12points
WINSTONCH [101]

Answer:

-10C= Solid

10C= Liquid

50C= Liquid

90C= Liquid

110C= Gas

120C= Gas

Explanation:

Below 0 degree C (Celsius), water is frozen means it is in the form of ice. After 0 degree, once we keep it in room temperature, the ice starts becoming liquid (water), and once we heat water, after 100 degree C (Celsius) water starts boiling and thus starts entering gaseous state.

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What is the speed of the fastest baseball pitch ever thrown?
Annette [7]

The fastest pitch ever thrown was 105 MPH. Hope this helps!

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Where is cloud data stored?
Annette [7]

Answer: i think its local drive

Explanation:

5 0
3 years ago
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