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4 years ago

14

The speed of a bus increases uniformly from 15meter per seconds to 60 meter per seconds in 20second.Calculate, the acceleration.

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1 answer:

Anarel [89]4 years ago

6 0

Answer:

2.25 m/s²

Explanation:

Acceleration = change in velocity / time taken

= (60-15)/20

= 2.25 m/s²

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The Achilles tendon connects the muscles in your calf to the back of your foot. When you are sprinting, your Achilles tendon alt

Andrews [41]

Explanation:

It is given that,

Mass of the runner, m = 70 kg

Length of the tendon, l = 15 cm = 0.15 m

Area of cross section, $A=110\ mm^2=0.00011\ m^2$

Part A,

Let the runner's Achilles tendon stretch if the force on it is 8.0 times his weight, F = 8 mg

Young's modulus for tendon is, $Y=0.15\times 10^{10}\ N/m^2$

The formula of the Young modulus is given by :

$Y=\dfrac{F/A}{\dfrac{\Delta L}{L}}$

$0.15\times 10^{10}=\dfrac{8\times 70\times 9.8/0.00011}{\dfrac{\Delta L}{0.15}}$

$\Delta L=0.0049\ m$

Part B,

The fraction of the tendon's length does this correspond is given by :

$\dfrac{\Delta L}{L}=\dfrac{0.0049}{0.15}$

$\dfrac{\Delta L}{L}=0.0326$

Hence, this is the required solution.

6 0

3 years ago

Light from a helium-neon laser (λ=633nm) passes through a circular aperture and is observed on a screen 5.80 mm behind the apert

Helga [31]

Explanation:

Whenever the light passes through hole or slit then it tends to bend that is actually a diffraction. It will then made the interference pattern of light and dark bands that due to constructive and destructive interference.

So by using the equation of diffraction,

dsinA = nL

SinA is a geometric component it can be written as,

$d\frac{y}{x} = nL$

∵ x is the distance from screen.

∵ y is the half of the width of central maximum.

now by putting the values in mm,

$d\frac{17}{5.80} = 0.000633$

d = 2.15× $10^{-4}$ mm

7 0

3 years ago

Beating a carpet with a carpet beater.

prisoha [69]

I would say a short person with muscles considering they are closer to the ground, but they may not be able to build up as much force in such a short time compared to the tall person.

8 0

3 years ago

Atomic hydrogen produces a well-known series of spectral lines in several regions of the electromagnetic spectrum. Each series f

navik [9.2K]

Answer:

n₁ = 3

Explanation:

The energy of the states in the hydrogen atom is explained by the Bohr model, the transitions heal when an electron passes from a state of higher energy to another of lower energy,

ΔE = $E_{nf}$ - E₀ = - k²e² / 2m (1 / $n_{f}$ ²2 - 1 / n₀²)

The energy of this transition is given by the Planck equation

E = h f = h c / λ

h c / λ = -k²e² / 2m (1 / no ²- 1 / no²)

1 / λ = Ry (1/ $n_{f}$ ² - 1 / n₀²)

Let's apply these equations to our case

λ = 821 nm = 821 10⁻⁹ m

E = h c / λ

E = 6.63 10⁻³⁴ 3 10⁸/821 10⁻⁹

E = 2.423 10⁻¹⁹ J

Now we can use the Bohr equation

Let's reduce to eV

E = 2,423 10⁻¹⁹ J (1eV / 1.6 10⁻¹⁹) = 1,514 eV

$E_{nf}$ - E₀ = -13.606 (1 / $n_{f}$ ² - 1 / n₀²) [eV]

Let's look for the energy of some levels

n $E_{n}$ (eV) $E_{nf}$ - E $E_{ni}$ (eV)

1 -13,606 E₂-E₁ = 10.20

2 -3.4015 E₃-E₂ = 1.89

3 -1.512 E₄- E₃ = 0.662

4 -0.850375

We see the lines of greatest energy for each possible series, the closest to our transition is n₁ = 3 in which a transition from infinity (n = inf) to this level has an energy of 1,512 eV that is very close to the given value

8 0

3 years ago

A worker pushes a 50 kg crate a distance of 7.5 m across a level floor. He

Taya2010 [7]

a) 73.5 N

b) 551.3 J

c) -551.3 J

d) 0 J

e) 0 J

f) 0 J

g) 0 J

Explanation:

a)

There are two forces acting on the crate:

- The push of the worker, F, in the forward direction

- The frictional force, $F_f=\mu mg$ , in the backward direction, where

$\mu=0.15$ is the coefficient of friction

m = 50 kg is the mass of the crate

$g=9.8 m/s^2$ is the acceleration due to gravity

According to Newton's second law of motion, the net force on the crate must be equal to the product of mass and acceleration, so:

$F-F_f=ma$

However, the crate here is moving with constant velocity, so its acceleration is zero:

$a=0$

So the previous equation becomes:

$F-F_f=0$

And we can find the magnitude of the applied force:

$F=F_f=\mu mg=(0.15)(50)(9.8)=73.5 N$

b)

The work done by the applied force on the crate is

$W_F=Fd cos \theta$

where:

F is the magnitude of the force

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here we have:

F = 73.5 N

d = 7.5 m

$\theta=0^{\circ}$ (the force is applied in the same direction as the displacement)

Therefore,

$W_F=(73.5)(7.5)(cos 0^{\circ})=551.3 J$

c)

The work done by friction on the crate is:

$W_{F_f}=F_f d cos \theta$

where in this case:

$F_f=73.5 N$ is the magnitude of the force of friction

d = 7.5 m is the displacement of the crate

$\theta=180^{\circ}$ , because the displacement is forward and the force of friction is backward, so they are in opposite direction

Therefore, the work done by the force of friction is:

$W_{F_f}=(73.5)(7.5)(cos 180^{\circ})=-551.3 J$

d)

To find the normal force, we analyze the situation of the force along the vertical direction.

We have two forces on the vertical direction:

- The normal force, N, upward

- The force of gravity, $mg$ , downward, where

m = 50 kg is the mass of the crate

$g=9.8 m/s^2$ is the acceleration due to gravity

Since the crate is in equilibrium in this direction, the vertical acceleration is zero, so the two forces balance each other:

$N-mg=0\\N=mg=(50)(9.8)=490 N$

The work done by the normal force is:

$W_N=Nd cos \theta$

In this case, $\theta=90^{\circ}$ , since the normal force is perpendicular to the displacement of the crate; therefore, the work done is

$W_N=(490)(7.5)(cos 90^{\circ})=0$

e)

The work done by the gravitational force is:

$W_g=F_g d cos \theta$

where:

$F_g=mg=(50)(9.8)=490 N$ is the gravitational force

d = 7.5 m is the displacement of the crate

$\theta=90^{\circ}$ is the angle between the direction of the gravitational force (downward) and the displacement (forward)

Therefore, the work done by gravity is

$W_g=(490)(7.5)(cos 90^{\circ})=0 J$

f)

The total work done on the crate can be calculated by adding the work done by each force:

$W=W_F+W_{F_f}+W_N+W_g$

where we have:

$W_F=+551.3 J$ is the work done by the applied force

$W_{F_f}=-551.3 J$ is the work done by the frictional force

$W_N=0$ is the work done by the normal force

$W_g=0$ is the work done by the force of gravity

Substituting,

$W=+551.3+(-551.3)+0+0=0 J$

So, the total work is 0 J.

g)

According to the work-energy theorem, the change in kinetic energy of the crate is equal to the work done on it, therefore:

$W=\Delta E_K$

where

W is the work done on the crate

$\Delta E_K$ is the change in kinetic energy of the crate

In this problem, we have:

$W=0$ (total work done on the crate is zero)

Therefore, the change in kinetic energy of the crate is:

$\Delta E_K = W = 0$

5 0

4 years ago