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liraira [26]
2 years ago
15

Calculate the average time it took the car to travel 0. 25 and 0. 50 meters with three washers attached to the pulley. Record th

e averages, rounded to two decimal places, in Table C of your Student Guide. What is the average time it took the car to travel 0. 25 meters? seconds What is the average time it took the car to travel 0. 50 meters? seconds.
Physics
1 answer:
EastWind [94]2 years ago
6 0

The Speed of a moving vehicle can be measured. The average time it will taken too cover both distance are shown below;

The average travel times traveled are :

0.25 meters = 2.23 seconds

0.50 meters = 3.13 seconds

Therefore, the Average travel time to cover 0.25 meters is:

(Trial 1 + Trial 2 + Trial 3) ÷ 3

(2.24 + 2.21 + 2.23) / 3 = 6.68 = 2.2266 = 2.23 seconds ( approximated to 2 decimal places).

Therefore, the Average travel time to cover 0.50 meters is

(Trial 1 + Trial 2 + Trial 3) ÷ 3

(3.16 + 3.08 + 3.15) / 3 = 9.39 ÷ 3 = 3.13 seconds

Conclusively, the average time taken to cover 0.25 and 0.50 meters is said to be 2.23 and 3.13 seconds for the both..

Learn more from

brainly.com/question/25738530

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A car slows down uniformly from a speed of 21.0m/s to rest in 6.00s. How far did it travel in that time?
ss7ja [257]

(sorry something wrong w my keyboard so write each line for the explnation!)

63.0 m

Explanation:

Acceleration of car

=

v

−

u

t

=

0 ms

−

1

−

21.0 ms

−

1

6.00 s

=

−

3.50 ms

−

2

S

=

v

2

−

u

2

2a

S

=

(

0 ms

−

1

)

2

−

(

21.0 ms

−

1

)

2

2

×

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3.50 ms

−

2

S

=

63.0

m

4 0
3 years ago
Use the graph to answer the question.
Setler79 [48]

Answer:175

Explanation:

5 0
3 years ago
Read 2 more answers
Juliette is driving her car when she sees a cat run across the road. If she is able to stop the car over a distance of 0.025km i
jek_recluse [69]
velocity=\frac{distance}{time}=\frac{0,025km}{2,5s}=\frac{25m}{2,5s}=100\frac{m}{s}\\\\
acceleration=\frac{velocity}{time}=\frac{-10\frac{m}{s}}{2,5s}=-4\frac{m}{s^2}\\\\Solution\ is\ D.
6 0
3 years ago
A 0.12 g honeybee acquires a charge of +24pC while flying. The earth's electric field near the surface is typically 100 N/C, dow
shusha [124]

Answer:

150000000

\dfrac{F_e}{F_g}=0.00000203873598369

49050000 N/C

Explanation:

q = Charge = 24 pC

m = Mass of honeybee = 0.12 g

E = Electric field = 100 N/C

g = Acceleration due to gravity = 9.81 m/s²

1\ C=6.25\times 10^{18}\ electrons

Number electrons is

n=24\times 10^{-12}\times 6.25\times 10^{18}\\\Rightarrow n=150000000

The number of electrons added or removed was 150000000

Force is given by

F_e=Eq\\\Rightarrow F_e=100\times 24\times 10^{-12}\\\Rightarrow F_e=2.4\times 10^{-9}\ N

The ratio is

\dfrac{F_e}{F_g}=\dfrac{2.4\times 10^{-9}}{0.12\times 10^{-3}\times 9.81}\\\Rightarrow \dfrac{F_e}{F_g}=0.00000203873598369

The ratio is \dfrac{F_e}{F_g}=0.00000203873598369

Balancing the forces we get

Eq=mg\\\Rightarrow E=\dfrac{mg}{q}\\\Rightarrow E=\dfrac{0.12\times 10^{-3}\times 9.81}{24\times 10^{-12}}\\\Rightarrow E=49050000\ N/C

The electric field required is 49050000 N/C

4 0
3 years ago
Two satellites A and B orbit the Earth in the same plane. Their masses are 5 m and 7 m, respectively, and their radii 4 r and 7
Dmitry [639]

Answer:

The ratio of their orbital speeds are 5:4.

Explanation:

Given that,

Mass of A = 5 m

Mass of B = 7 m

Radius of A = 4 r

Radius of B = 7 r

The orbital speed of satellite A,

v_{A}=\sqrt{\dfrac{GM_{A}}{R_{A}}}......(I)

The orbital speed of satellite B,

v_{B}=\sqrt{\dfrac{GM_{B}}{R_{B}}}......(I)

We need to calculate the ratio of their orbital speeds

Using equation (I) and (II)

\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{\dfrac{GM_{A}}{R_{A}}}{\dfrac{GM_{B}}{R_{B}}}}

Put the value into the formula

\dfrac{v_{A}}{v_{B}}=\sqrt{\dfrac{G\times5m\times7r}{G\times7m\times4r}}

\dfrac{v_{A}}{v_{B}}=\dfrac{5}{4}

Hence, The ratio of their orbital speeds are 5:4.

8 0
3 years ago
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