A car traveling comes to a stop at a red light.How many forces are taking place?

A rope with a mass density of 1 kg/m has one end tied to a vertical support. You hold the other end so that the rope is horizont

PIT_PIT [208]

Answer:

$v' = 2.83 m/s$

Explanation:

Velocity of wave in stretched string is given by the formula

$v = \sqrt{\frac{T}{\mu}}$

here we know that

T = 4 N

also we know that linear mass density is given as

$\mu = 1 kg/m$

so we have

$v = \sqrt{\frac{4}{1}} = 2 m/s$

now the tension in the string is double

so the velocity is given as

$v' = \sqrt{\frac{8}{1}} = 2\sqrt2 m/s$

$v' = 2.83 m/s$

4 0

2 years ago

Two particles oscillate in simple harmonic motion along a common straight-line segment of length 1.0 m. Each particle has a peri

igor_vitrenko [27]

Answer:

a) the particles are 0.217 m apart

b) the particles are moving in the same direction.

Explanation:

a) The amplitude of the oscillations is A/2 and the period of each particle is

T = 1.5 s however, they differ by a phase of π/6 rad. Let the phase of the first particle be zero so that the phase of the second particle is π/6. So we can write the coordinates of each of the particles as,

x₁ = A/2 cos(ωt)

x₂ = A/2 cos(ωt + π/6)

we can write the angular frequency ω, as

ω = 2π / T

so,

x₁ = A/2 cos(2π / T)

x₂ = A/2 cos(2π / T + π/6)

Thus, the coordinates of the particles at t = 0.45 s are,

x₁ = A/2 cos((2π × 0.45) / 1.5)) = -0.155 A

x₂ = A/2 cos((2π × 0.45) / 1.5) + π/6) = -0.372 A

Their separation at that time is, therefore,

Δx = x₁ - x₂

= -0.155 A + 0.372 A

= 0.217 A

since A = 1 m

Thus,

Δx = 0.217 m

b) In order to find their directions, we must take the derivatives at t = 0.45 s.

Therefore,

v₁ = dx₁ / dt

= (-πA / T) sin(2πt / T)

= -(π(1) / 1.5) sin(2π(0.45) / 1.5)

= -1.99

and,

v₂ = dx₂ / dt

= (-πA / T) sin((2πt / T) + π/6)

= -(π(1) / 1.5) sin((2π(0.45) / 1.5) + π/6)

= -1.40

Since both v₁ and v₂ are negative, this shows that the particles are moving in the same direction.

6 0

2 years ago

A force of 85 N is used to push a box along the floor a distance of 15 m. How much work was done?

denpristay [2]

Answer:

1275J

Explanation:

Given parameters:

Force on box = 85N

Distance moved = 15m

Unknown:

Work done = ?

Solution:

Work done is the amount of force applied on a body to move it through a specific distance.

Work done = Force x distance

Now insert the parameters and solve;

Work done = 85 x 15 = 1275J

4 0

2 years ago

Read 2 more answers

the gravitational force between two objects is 1600 and what will be the gravitational force between the objects if the distance

Xelga [282]

I believe this is what you have to do:

The force between a mass M and a point mass m is represented by

$F = G\frac{Mm}{r^{2} }$

So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁

So F₁ = G(Mm/r^2)

Now the distance has doubled so lets account for this in F₂:

F₂ = G(Mm/(2r)^2)

Now square the 2 that gives you four and we can pull that out in front to give

F₂ = $\frac{1}{4}$ G(Mm/r^2)

Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations

now we see that:

F₂ = $\frac{1}{4}$ F₁

So the second force will be 0.25 (1/4) x 1600 or 400 N.

6 0

2 years ago

what volume of alcohol will have the same mass as 4.2m^3 of petrol? (density of alcohol 0.4kg/m^3 and petrol is 0.3kg/m^3)

Alexxx [7]

Answer:

3.15m³

Explanation:

To solve this problem, let us first find the mass of the petrol from the given dimension.

Mass = density x volume

Volume of petrol = 4.2m³

Density of petrol = 0.3kgm⁻³

Mass of petrol = 4.2 x 0.3 = 1.26kg

So;

We can now find the volume of the alcohol

Volume of alcohol = $\frac{mass}{density}$

Mass of alcohol = 1.26kg

Density of alcohol = 0.4kgm⁻³

Volume of alcohol = $\frac{1.26}{0.4}$ = 3.15m³

7 0

2 years ago