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2 years ago

A person swims to the other end of a 20m long pol and back. What is their displacement?

1 answer:

4 0

Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. BUT Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position. so here distance is 40 m but displacement is obviously zero meter. so answer is 0 more information on: https://www.physicsclassroom.com/class/1DKin/Lesson-1/Distance-and-Displacement

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How deep can an object with 6360N hitting on a sponge get ???

borishaifa [10]

Answer:

The sponge must go $\dfrac{636}{m}\ \text{meter}$ deep

Explanation:

If F = 6360 N, then it is required to find how deep can an object with this force hitting on a sponge get.

We know that, F = mgh

m is mass

g is acceleration due to gravity

$h=\dfrac{F}{mg}\\\\h=\dfrac{6360}{10m}\\\\h=\dfrac{636}{m}\ \text{meter}$

So, the sponge must go $\dfrac{636}{m}\ \text{meter}$ deep.

5 0

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

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

$\Delta S=1.69J/K$

7 0

3 years ago

A child with a mass of 20 kg sits at a distance of 2 m from the pivot point of a seesaw. where should a 16-kg child sit to balan

Maksim231197 [3]

In the above problem, we need to find mass of the second child, so that the Center of Mass remains at the origin( pivot).

CM= m1r1+m2r2/m1+m2
0= 20*-2+16*r2/20+16
r2= 40/16
r2= +2.5 m