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3 years ago

during a _____, a substance mixes with another substance and changes into completely new substances with different properties.

Physics

1 answer:

Tatiana [17]3 years ago

7 0

chemical change or chemical reaction

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A boat moves up and down as water waves pass under the boat. If the amplitude of the wave gets bigger, then

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The anwser is A or D

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3 years ago

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A delivery truck leaves a warehouse and travels 3.20 km east. The truck makes a right turn and travels 2.45 km south to arrive a

boyakko [2]

Explanation:

It is given that,

Displacement of the delivery truck, $d_1=3.2\ km$ (due east)

Then the truck moves, $d_2=2.45\ km$ (due south)

Let d is the magnitude of the truck’s displacement from the warehouse. The net displacement is given by :

$d=\sqrt{d_1^2+d_2^2}$

$d=\sqrt{3.2^2+2.45^2}$

d = 4.03 km

Let $\theta$ is the direction of the truck’s displacement from the warehouse from south of east.

$\theta=tan^{-1}(\dfrac{2.45}{3.2})$

$\theta=37.43^{\circ}$

So, the magnitude and direction of the truck’s displacement from the warehouse is 4.03 km, 37.4° south of east.

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3 years ago

What is the total entropy change if a 0.280 efficiency engine takes 3.78E3 J of heat from a 3.50E2 degC reservoir and exhausts i

babymother [125]

Answer:

$\Delta S=1.69J/K$

Explanation:

We know,

$\eta=1-\frac{T_2}{T_1}=1-\frac{Q_2}{Q_1}$ ..............(1)

where,

η = Efficiency of the engine

T₁ = Initial Temperature

T₂ = Final Temperature

Q₁ = Heat available initially

Q₂ = Heat after reaching the temperature T₂

Given:

η =0.280

T₁ = 3.50×10² °C = 350°C = 350+273 = 623K

Q₁ = 3.78 × 10³ J

Substituting the values in the equation (1) we get

$0.28=1-\frac{Q_2}{3.78\times 10^{3}}$

$\frac{Q_2}{3.78\times 10^3}=0.72$

$Q_2=3.78\times 10^3\times0.72$

⇒ $Q_2 =2.721\times 10^3 J$

Now,

The entropy change ( $\Delta S$ ) is given as:

$\Delta S=\frac{\Delta Q}{T_1}$

$\Delta S=\frac{Q_1-Q_2}{T_1}$

substituting the values in the above equation we get

$\Delta S=\frac{3.78\times 10^{3}-2.721\times 10^3 J}{623K}$

$\Delta S=1.69J/K$

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3 years ago

A projectile is launched upward at an angle of 70⁰ from the horizontal and strikes the ground a certain distance down range. For

OLga [1]

Answer:20°

Explanation:

Recall

Range R of a projectile is given by U^2sin2A/g

We're U = velocity,A= angle of projection and g is acceleration due to gravity

From the question the range R are the same

Hence R1=R2

U1^2sin2A/g=U2^2sin2B/g

But U1=U2 and g=g

Hence sin2A=sin 2B

Sin 2*70= sin2*B

0.6427=sin2B

B=sin inverse(0.6427)=40/2=20°

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3 years ago

Can we fly if we were falling for two days

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...................no

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2 years ago

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