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3 years ago

help me pass the test please there is a couple of questions in my profile to answer i am leaving this one for you to help

Physics

1 answer:

kompoz [17]3 years ago

6 0

Answer:

0.5 m/s².

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 10 m/s

Time (t) = 20 s

Acceleration (a) =?

Acceleration can simply be defined as the rate of change of velocity with time. Mathematically, it is expressed as:

a = (v – u) / t

Where:

a is the acceleration.

v is the final velocity.

u is the initial velocity.

t is the time.

With the above formula, we can obtain the acceleration of the car as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 10 m/s

Time (t) = 20 s

Acceleration (a) =?

a = (v – u) / t

a = (10 – 0) / 20

a = 10/20

a = 0.5 m/s²

Therefore, the acceleration of the car is 0.5 m/s².

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Atoms with certain characteristics are radioactive. Which of those below will

Andrei [34K]

Answer:

D. all atoms with an atomic number over 83

8 0

2 years ago

Bill is farsighted and has a near point located 119 cm from his eyes. Anne is also farsighted, but her near point is 70.0 cm fro

Gennadij [26K]

Answer:

Explanation:

For Bill's glasses)

d_o=25 cm-2 cm=23.2 cm

d_i=129 cm-2 cm=127 cm

We know that

$1 / f = 1 / d_o + 1 / d_i
$

$f=(1 / d_o + 1 / d_i)^-1
$

so plug the numbers in

$f = (1 / (23) + 1 / (127))^-1
$

= 28.08 cm

For Anna's glasses)

d_o=25 cm-2 cm=23 cm

d_i=70.0 cm-2 cm=68 cm

Same thing

$(1 / f = 1 / d_o + 1 / d_i
)$

$f=(1 / do + 1 / di)^-1
$

so plug the numbers in

$f = (1 / (23) + 1 / (68))^-1
$

= 17.21 cm

5 0

3 years ago

A bicyclist travels due east 60.0 km in 3.5 hours. What is the cyclist average speed?

Nezavi [6.7K]

If it is average speed per hour, then you just have to divide 3.5 into 60
60/3.5 = 17.143 so around 17km/h

7 0

3 years ago

10. Amy is at basketball practice. She is about to shoot a free throw. When Amy releases the ball, A. the ball will have neither

Schach [20]

When she releases the ball, it has some of both. (C)

6 0

3 years ago

A 0.468 g sample of pentane, C 5H 12, was burned in a bomb calorimeter. The temperature of the calorimeter and 1.00 kg of water

Fittoniya [83]

Answer:

$Q_{lost}$ per mole of pentane = 3157.53 kJ/mol

Explanation:

Given:

Mass of pentane, m = 0.468 gram

Molar mass of pentane, M = 72.15

Now, mol of pentane, n = mass/M = 0.468/72.15 = 0.00648 mol of C5H12

Now,

ΔT = 23.65 - 20.45 = 3.2°C

Heat capacity of the calorimeter, C = 2.21 kJ/°C

Specific heat capacity of the water, Cp = 4.184 J/g.°C

Now,

the heat gained = the heat lost

$Q_{gained} = -Q_{lost}$

also,

$Q_{gained} = Q_{water} + Q_{calorimeter}$

$Q_{water} = m\times C\times(T_f-T_i)$

$Q_{water} = 1000\times4.184\times(23.65-20.45) = 13388.8\ J$

and

$Q_{calorimeter} = C\times\Delta T = 2.21\times1000\times3.2 = 7072\ J$

Now,

$Q_{total} = 13388.8 +7072 = 20460.8\ J$

we have,

$Q_{lost} = -Q_{gain} = - 20460.8\ J$ (Here negative sign depicts the release of the heat)

$Q_{lost}$ per mole of pentane =-20460.8/(0.00648 ) = 3157.53 kJ/mol

3 0

3 years ago

help me pass the test please there is a couple of questions in my profile to answer i am leaving this one for you to help ​

help me pass the test please there is a couple of questions in my profile to answer i am leaving this one for you to help