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3 years ago

What is the time interval in which the average velocity is Negative?

Physics

1 answer:

expeople1 [14]3 years ago

5 0

Explanation:

Average velocity indicates direction and can be represented as a negative number when the displacement is in the negative direction. Average speed does not indicate direction and can only be positive or zero.

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Write any three differences between metals and nonmetals

Alisiya [41]

Metals Non-metals

They are solid at room temperature. They may be solid, liquid and gas at room temperature.

They have high melting and boiling points. They have low melting and boiling points.

They are good conductors of heat and electricity. They are bad conductors of heat and electricity.

4 0

3 years ago

A car speeds up from 10 m/s to 24 m/s. Change in velocity

JulsSmile [24]

Answer:

14m/s

Explanation:

subtract initial velocity by final velocity

5 0

2 years ago

You illuminate a slit with a width of 77.7 μm with a light of wavelength 721 nm and observe the resulting diffraction pattern on

SVETLANKA909090 [29]

Answer:

The width is $Z = 0.0424 \ m$

Explanation:

From the question we are told that

The width of the slit is $d = 77.7 \mu m = 77.7 *10^{-6} \ m$

The wavelength of the light is $\lambda = 721 \ nm$

The position of the screen is $D = 2.83 \ m$

Generally angle at which the first minimum of the interference pattern the light occurs is mathematically represented as

$\theta = sin ^{-1}[\frac{m \lambda}{d} ]$

Where m which is the order of the interference is 1

substituting values

$\theta = sin ^{-1}[\frac{1 *721*10^{-9}}{ 77.7*10^{-6}} ]$

$\theta = 0.5317 ^o$

Now the width of first minimum of the interference pattern is mathematically evaluated as

$Y = D sin \theta$

substituting values

$Y = 2.283 * sin (0.5317)$

$Y = 0.02 12 \ m$

Now the width of the pattern's central maximum is mathematically evaluated as

$Z = 2 * Y$

substituting values

$Z = 2 * 0.0212$

$Z = 0.0424 \ m$

3 0

4 years ago

A device that has the capacity to receive and store electrical energy is a(n)

Scilla [17]

Capacitor is, I think.

4 0

3 years ago

Newton’s Law of Cooling. Newton’s law of cooling states that the rate of change in the temperature T(t) of a body is proportiona

nexus9112 [7]

Answer:

After 30 minutes, the temperature of the body is: T₁₀ = 311.60 K

After 60 minutes, the temperature of the body is: T₂₀ = 298.18 K

Explanation:

Given that:

$\dfrac{dT}{dt}= K \bigg [ M(t) -T(t)\bigg]$

where;

K = 0.04

M(t) = 293

Then;

$\dfrac{dT}{dt}= 0.04 \bigg [ 293 -T\bigg]$

$\dfrac{dT}{dt}= 11.72 -0.04 \ T$

Using Euler's Formula;

$T_{n+1} = T_n + hf( t_n, T_n)$

where;

$f(t_n,T_n) = 11.72 - 0.04 T_n$

Then;

$T_{n+1} = T_n + 3.0 (11.72-0.04 \ T_n)$

$T_{n+1} = 0.88T_n + 35.16 \ ---(1)$

At initial state $t_0$ (0); $T_0$ = 360

At t₁ = 3.0 when T₀ = 360

$T_1= 0.88 T_o + 35.16$

$T_1= 0.88 (360) + 35.16$

$T_1 = 351.96 \ K$

At t₂ = 6.0 when T₂ = 0.88T₁ + 35.16

T₂ = 0.88(351.96) + 35.16

T₂ = 344.89 K

At t₃ = 9.0 when T₃ = 0.88T₂ + 35.16

T₃ = 0.88(344.89) + 35.16

T₃ =338.66 K

At t₄ = 12.0 when T₄ = 0.88T₃ + 35.16

T₄ = 0.88(338.66) + 35.16

T₄ = 333.18 K

At t₅ = 15.0 when T₅ = 0.88T₄ + 35.16

T₅ = 0.88(333.18) + 35.16

T₅ = 328.36 K

At t₆ = 18.0 when T₆ = 0.88T₅ + 35.16

T₆ = 0.88(328.36) + 35.16

T₆ = 324.12 K

At t₇ = 21.0 when T₇ = 0.88T₆ + 35.16

T₇ = 0.88(324.12) + 35.16

T₇ = 320.39 K

At t₈ = 24.0 when T₈ = 0.88T₇ + 35.16

T₈ = 0.88(320.29) + 35.16

T₈ = 317.02 K

At t₉ = 27.0 when T₉ = 0.88T₈ + 35.16

T₉ = 0.88(317.02) + 35.16

T₉ = 314.14 K

At t₁₀ = 30 when T₁₀ = 0.88T₉ + 35.16

T₁₀ = 0.88(314.14) + 35.16

T₁₀ = 311.60 K

At t₁₁ = 33.0 when T₁₁ = 0.88T₁₀ + 35.16

T₁₁ = 0.88(311.60) + 35.16

T₁₁ = 309.37 K

At t₁₂ = 36.0 when T₁₂ = 0.88T₁₁ + 35.16

T₁₂ = 0.88(309.37)+ 35.16

T₁₂ = 307.41 K

At t₁₃ = 39.0 when T₁₃ = 0.88T₁₂ + 35.16

T₁₃ = 0.88( 307.41) + 35.16

T₁₃ = 305.68 K

At t₁₄ = 42.0 when T₁₄ = 0.88T₁₃ + 35.16

T₁₄ = 0.88(305.68) + 35.16

T₁₄ = 304.16 K

At t₁₅ = 45.0 when T₁₅ = 0.88T₁₄ + 35.16

T₁₅ = 0.88(304.16) + 35.16

T₁₅ = 302.82 K

At t₁₆ = 48.0 when T₁₆ = 0.88T₁₅ + 35.16

T₁₆ = 0.88(302.82) + 35.16

T₁₆ = 301.64 K

At t₁₇ = 51.0 when T₁₇ = 0.88T₁₆ + 35.16

T₁₇ = 0.88(301.64) + 35.16

T₁₇ = 300.60 K

At t₁₈ = 54.0 when T₁₈ = 0.88T₁₇ + 35.16

T₁₈ = 0.88(300.60) + 35.16

T₁₈ = 299.69 K

At t₁₉ = 57.0 when T₁₉ = 0.88T₁₈ + 35.16

T₁₉ = 0.88(299.69) + 35.16

T₁₉ = 298.89 K

At t₂₀ = 60 when T₂₀ = 0.88T₁₉ + 35.16

T₂₀ = 0.88(298.89) + 35.16

T₂₀ = 298.18 K

4 0

3 years ago