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Dmitry_Shevchenko [17]

3 years ago

13

A velocity vs. time graph is shown. A graph titled Velocity versus Time shows time in seconds on the x axis, numbered 0 to 5, ve

locity in meters per second on the y axis, numbered 0 to 20. A line starts at the origin and ends at (5, 20). What is the acceleration of the object?

2 answers:

masya89 [10]3 years ago

8 0

Answer:

a+4 m ...

Explanation:

Lisa [10]3 years ago

5 0

Answer:

$a = 4 \frac{m}{s {}^{2} }$

Explanation:

u for velocity

t for time

a for acceleration

$u = a \times t \\ 20 = a \times 5 \\ a = \frac{20}{5} \\ a = 4$

That's it! Hope I helped!

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If electrical current is moving through a horizontal wire toward your face, what direction is the induced magnetic field?

adoni [48]

Answer:(b)

Explanation:

Given

Electric current flowing through the horizontal wire towards the observer's face.

The direction of the magnetic field is given by the right-hand thumb rule, i.e. place the thumb in the direction of current and the wrapping of fingers will give the direction of the magnetic field

the direction of the magnetic field will be counterclockwise as observed by an observer.

4 0

2 years ago

How much total heat transfer is necessary to lower the temperature of 0.175 kg of steam from 125.5 °C to −19.5 °C, including the

GaryK [48]

Answer:

Explanation:

The heat required to change the temperature of steam from 125.5 °C to 100 °C is:

$Q_1 = ms_{steam} (125.5^0C - 100^0C) \\ \\ Q_1 = 0.175 \ kg ( 1520 \ J/kg.K ) (25.5^0 \ C) \\ \\ Q_1 = 6783 \ J$

The heat required to change the steam at 100°C to water at 100°C is;

$Q_2 = mL_v \\ \\ Q_2 = (0.175 \ kg) (2.25*10^6 \ J/kg ) \\ \\ Q_2 = 393750 \ J$

The heat required to change the temperature from 100°C to 0°C is

$Q_3 = ms_{water} (100^) \ C) \\ \\ Q_3 = (0.175\ kg)(4186 \ J/kg.K) (100 ^0c ) \\ \\ Q_3 = 73255 \ J$

The heat required to change the water at 0°C to ice at 0°C is:

$Q_4 = mL_f \\ \\ Q_4 = (0.175 \ kg)(3.34*10^5 \ J/kg) \\\\ Q_4 = 58450 \ J$

The heat required to change the temperature of ice from 0°Cto -19.5°C is:

$Q_5 = ms _{ice} (100^0 C) \\ \\ Q_5 = (0.175 \ kg)(2090 \ J/kg.K)(19.5^0C) \\ \\ Q_5 = 7132.125 \ J$

The total heat required to change the steam into ice is:

$Q = Q_1 + Q_2 + Q_3 + Q_4 +Q_5 \\ \\Q = (6788+393750+73255+58450+7132.125)J \\ \\ Q = 539325.125 \ J \\ \\ Q = 5.39*10^5 \ J$

b)

The time taken to convert steam from 125 °C to 100°C is:

$t_1 = \frac{Q_1}{P} = \frac{6738 \ J}{835 \ W} = 8.12 \ s$

The time taken to convert steam at 100°C to water at 100°C is:

$t_2 = \frac{Q_2}{P} =\frac{393750}{834} =471.56 \ s$

The time taken to convert water to 100° C to 0° C is:

$t_3 = \frac{Q_3}{P} =\frac{73255}{834} = 87.73 \ s$

The time taken to convert water at 0° to ice at 0° C is :

$t_4 = \frac{Q_4}{P} =\frac{58450}{834} = 70.08 \ s$

The time taken to convert ice from 0° C to -19.5° C is:

$t_5 = \frac{Q_5}{P} =\frac{7132.125}{834} = 8.55 \ s$

5 0

3 years ago

If element x has 94 protons how many electrons does it have

Igoryamba

94 electrons. protons and electrons are always the same, but neutrons are different.

8 0

2 years ago

Read 2 more answers

A disk rotates at constant angular acceleration, from angular position θ1 = 16.0 rad to angular position θ2 = 76.0 rad in 5.30 s

Oliga [24]

Answer:

(a) the angular velocity at θ1 is 11.64 rad/s

(b) the angular acceleration is 0.12 rad/ $s^{2}$

(c) the angular position was the disk initially at rest is - 428.27 rad

Explanation:

Given information :

θ1 = 16 rad

θ2 = 76 rad

ω2 = 11 rad/s

t = 5.3 s

(a) The angular velocity at θ1

First, we use the angular motion equation for constant acceleration

Δθ = (ω1+ω2)t/2

θ2 - θ1 = (ω1+ω2)t/2

ω1 + ω2 = 2 (θ2 - θ1) / t

ω1 = (2 (θ2 - θ1) / t ) - ω2

= (2 (76-16) / 5.3) - 11

= 11.64 rad/s

(b) the angular acceleration

ω2 = ω1 + α t

α t = ω2 - ω1

α = (ω2 - ω1)/t

= (11.64 - 11) / 5.3

= 0.12 rad/ $s^{2}$

(c) the angular position was the disk initially at rest, θ0

at rest ω0 = 0

ω2^2 = ω01 t + 2 α Δθ

2 α Δθ = ω2^2

θ2 - θ0 = ω2^2 / 2 α

θ0 = θ2 - (ω2^2) / 2 α

= 76 - ( $11^{2}$ / 2 x 0.12

= 76 - 504.16

= - 428.27 rad

4 0

3 years ago

State and prove bessel inequality

maria [59]

Statement :- We assume the orthagonal sequence ${{\{\phi\}}_{1}^{\infty}}$ in Hilbert space, now ${\forall \sf \:v\in \mathbb{V}}$ , the Fourier coefficients are given by:

${\quad \qquad \longrightarrow \sf a_{i}=(v,{\phi}_{i})}$

Then Bessel's inequality give us:

${\boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}$

Proof :- We assume the following equation is true

${\quad \qquad \longrightarrow \displaystyle \sf v_{n}=\sum_{i=1}^{n}a_{i}{\phi}_{i}}$

So that, ${\bf v_n}$ is projection of ${\bf v}$ onto the surface by the first ${\bf n}$ of the ${\bf \phi_{i}}$ . For any event, ${\sf (v-v_{n})\perp v_{n}}$

Now, by Pythagoras theorem:

${:\implies \quad \sf \Vert v\Vert^{2}=\Vert v-v_{n}\Vert^{2}+\Vert v_{n}\Vert^{2}}$

${:\implies \quad \displaystyle \sf ||v||^{2}=\Vert v-v_{n}\Vert^{2}+\sum_{i=1}^{n}\vert a_{i}\vert^{2}}$

Now, we can deduce that from the above equation that;

${:\implies \quad \displaystyle \sf \sum_{i=1}^{n}\vert a_{i} \vert^{2}\leqslant \Vert v\Vert^{2}}$

For ${\sf n\to \infty}$ , we have

${:\implies \quad \boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}$

Hence, Proved

5 0

2 years ago

Read 2 more answers