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4 years ago

You have a graduated cylinder that you use to measure volume. The cylinder

Physics

2 answers:

Goshia [24]4 years ago

8 0

Answer:

35.5 ml

Explanation:

The other guy made me get the question wrong lol here ya go good luck

ad-work [718]4 years ago

5 0

Answer:

D. 21 ml

Explanation:

Since, the cylinder is marked and graduated in the intervals if 1 ml. Therefore, the values between two consecutive ml, such as between 30 ml and 31 ml can not be determined. Because, we do not have any scale in between the ml. So, the least count of this instrument is 1 ml. This graduated cylinder can give the answers to zero decimal places, accurately. And it can not determine any decimal value due to its graduating or the marking limitation. So, all the options given, contain a decimal value, except for the option D. In option D there is no decimal value, hence it is a correct answer.

D. <u>21 ml</u>

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A boy starting from rest, slides down a snowy hill on a sled. He travels 174.19 m to the bottom of the hill. He accelerates

AfilCa [17]

Answer:

11.6 s

Explanation:

We can use equation of motiom to solve this

$s = ut + \frac{1}{2} a {t}^{2 } \\ 174.19 = (0)t + \frac{1}{2} (2.61){t}^{2 } \\ t = 11.6s$

4 0

3 years ago

The position of an object is given by the equation x = 4.0t2 - 2.0 t - 4.5where x is in meters and t is in seconds. What is the

s344n2d4d5 [400]

Answer:

Explanation:

x = 4 t^2 - 2 t - 4.5

Position at t = 3 s

x = 4 (3)^2 - 2 (3) - 4.5 = 25.5 m

Velocity at t = 3 s

v = dx / dt = 8 t - 2

v ( t = 3 s) = 8 x 3 - 2 = 22 m/s

Acceleration at t = 3 s

a = dv / dt = 8

a ( t = 3 s ) = 8 m/s^2

When is the velocity = 0

v = 0

8 t - 2 = 0

t = 0.25 second

When is the position = 0

x = 0

4 t^2 - 2 t - 4.5 = 0

$t = \frac{2 \pm \sqrt{4 + 72}}{8}$

t = 1.4 second

8 0

3 years ago

An object moving north with an initial velocity of 14m/s accelerates 5m/s squared. What is the final velocity of the object?

OLEGan [10]

The final velocity of the object is 16m\s.
Hope this helps! :)

6 0

3 years ago

A wall has inner and outer surface temperatures of 25◦C and 8◦C, respectively. The interior and exterior air temperatures are 35

Angelina_Jolie [31]

Answer:

a) $\frac{\dot Q}{A} =60\ W.m^{-2}$

b) $\frac{\dot Q}{A} =110\ W.m^{-2}$

c) The wall may not be under steady because the two surfaces of the wall are exposed to the air at different temperatures and they have different convective coefficient.

Explanation:

Given:

temperature of the inner surface of the wall, $T_i=25^{\circ}C$

temperature of the outer surface of the wall, $T_o=8^{\circ}C$

temperature of the air outside, $T_{ao}=-3^{\circ}C$

temperature of the air inside, $T_{ai}=35^{\circ}C$

coefficient of heat convection on outside, $h_o=10\ W.m^{-2}.K^{-1}$

coefficient of heat convection on inside, $h_i=6\ W.m^{-2}.K^{-1}$

The heat flux between the interior air and the wall:

The convective heat transfer rate is given as,

$Q=h_i.A.\Delta T$

$\Rightarrow \frac{\dot Q}{A} =h_i\times (T_{ai}-T_i)$

$\frac{\dot Q}{A} =6\times (35-25)$

$\frac{\dot Q}{A} =60\ W.m^{-2}$

The heat flux between the exterior air and the wall:

$\Rightarrow \frac{\dot Q}{A} =h_o\times (T_{ao}-T_i)$

$\frac{\dot Q}{A}=10\times (8-(-3))$

$\frac{\dot Q}{A} =110\ W.m^{-2}$

The wall may not be under steady because the two surfaces of the wall are exposed to the air at different temperatures and they have different convective coefficient.

4 0

4 years ago

How much energy can be stored in a spring with k = 470 n/m if the maximum possible stretch is 17.0 cm ?

MrRissso [65]

The strain energy stored in a linear spring is
SE = (1/2)*k*x²
where
k = the spring constant
x = the extension (or compression) of the spring

Given:
k = 470 N/m
x = 17.0 cm = 0.17 m

Therefore
SE = 0.5*(470 N/m)*(0.17 m)² = 6.7915 J

Answer: 6.8 J (nearest tenth)

6 0

4 years ago