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3 years ago

PLEASE HELP WILL GIVE BRAINLIEST!!! DUE SOON!! GIVE ME 2 REASONS WHY

Physics

2 answers:

TiliK225 [7]3 years ago

6 0

Answer:

★ For example, cold water has a higher density than warmer water. In Addition, water gets colder with depth because of the coldness, of the salty ocean water which sinks to the bottom of the ocean basins below the less dense water near the surface

Explanation:

Hope you have a great day :)

BartSMP [9]3 years ago

4 0

Cold water has a higher density than watm water. water gets colder with depth because of the cold, salty ocean water sinks to the bottom of the ocean basins below the less dense water near the surface

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If a force of 26 N is applied to an object with a mass of 8 kg, the object will accelerate at m/s2

malfutka [58]

False, the numbers do not add up

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3 years ago

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In which of these samples do the molecules mist likely have the least kinetic energy

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Provide the examples again, no samples are presents.

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3 years ago

If E=1/2Av^2+Bmx find the dimensions of A and B(Where E is energy,v,m and x are velocity,mass and distance respectively)

sp2606 [1]

Answer:

A = [kg]

B = [m/s²]

Explanation:

E = ½ Av² + Bmx

Substitute the units:

[J] = ½ A [m/s]² + B [kg] [m]

A Joule written in base units is:

1 J = 1 Nm = 1 kg m²/s²

Each term must have the same units.

[kg m²/s²] = A [m/s]²

[kg m²/s²] = A [m²/s²]

A = [kg]

[kg m²/s²] = B [kg] [m]

B = [m/s²]

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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

2 years ago

a car takes off from rest and covers a distance of 80m on a straight road in 10s.calculate the magnitude of its acceleration

olasank [31]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

Here's the solution :

Let's find the final velocity :

$\dfrac{displacement}{time}$

$\dfrac{80}{10}$

$8 \: \: ms {}^{ - 1}$

Initial velocity (u) = 0 (cuz it started from rest)

Final velocity (v) = 8 m/s

Time taken (t) = 10 sec

now, we know that :

$acceleration = \dfrac{v - u}{t}$

$\dfrac{8 - 0}{10}$

$\dfrac{8}{10}$

$0.8 \: \: m {s}^{ - 2}$

Acceleration = 0.8 m/s²

7 0

3 years ago