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Sergeu [11.5K]
2 years ago
15

If the soccer player runs with a speed of 4.6m/s, how long does it take him to run 60m?

Physics
2 answers:
Diano4ka-milaya [45]2 years ago
7 0
About 13 s (60/4.6)
satela [25.4K]2 years ago
5 0

Explanation:

Hello there!

Just divide 60 by 4.6:

60/4.6=13.04 seconds.

;)

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Large-scale environmental catastrophes _______.
VLD [36.1K]
I'll give just one. The eruption of Mount Tambora in 1815 was one of the volcanic eruptions with widespread damage. Mount Tambora erupted, bringing thermal waves and tsunamis that killed 38,000 people (estimated). It is not just the tsunami and thermal waves, but also the ash plume that caused a cool down on the earth's atmosphere by 5°C causing "The Year Without Summer" and caused crops worldwide to fail, or degrade causing famine. That is all I know :) have a good day.
5 0
1 year ago
A brick of mass 2 kg is dropped from a rest position 5 m above the ground. what is its velocity at a height of 3 m above the gro
Rina8888 [55]
We can solve the problem by using the law of conservation of energy.

Using the ground as reference point, the mechanical energy of the brick when it is at 5 m from the ground is just potential energy (because the brick is initially at rest, so it doesn't have kinetic energy):
E= U = mgh=(2 kg)((9.81 m/s^2)(5 m)=98.1 J

when the brick is at h'=3 m from the ground, its mechanical energy is now sum of kinetic energy and potential energy:
E= K+U= \frac{1}{2} mv^2 + mgh'

where v is the velocity of the brick. Since E is conserved, it must be equal to the initial energy (98.1 J), so we can solve this equation to find v:
v= \sqrt{ \frac{2(E-mgh')}{m} }=6.3 m/s
8 0
1 year ago
Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance
scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

x_{1}=\dfrac{\lambda D}{2d}

Second minima from center

x_{2}=\dfrac{3\lambda D}{2d}

The distance between the places where the intensity is zero due to the double slit effect

\Delta x_{d}=x_{2}-x_{1}

\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}

\Delta x_{d}=\dfrac{\lambda D}{d}

Put the value into the formula

\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}

\Delta x_{d}=0.015 =15\times10^{-3}\ m

\Delta x_{d}=15\ mm

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0
1 year ago
The three point charges +4.0 μC, -5.0 μC, and -9.0 μC are placed on the x-axis at the points x = 0 cm, x = 40 cm, and x = 120 cm
ale4655 [162]

Answer:

 

Explanation:

4μC will attract -9μC towards the centre and -5μC will repel it away from the centre.  Both these forces are opposite to each other.

Force due to 4μC on -9μC towards the centre

= k x Q₁ Q₂/R² = 9 X 10⁹ X 4 X 10⁻⁶ X 9 X 10⁻⁶ / (1.2)² = 225 X 10⁻³ N/C

Force due to -5μC on -9μC  away from the centre

= 9 x 10⁹ x 5 x 10⁻⁶x 9 x 10⁻⁶/( 0.8)² = 632.8 x 10⁻³ .N/C

Ner field =407.8 N/C.

6 0
2 years ago
Read 2 more answers
Derive an expression for the gravitational potential energy of a system consisting of Earth and a brick of mass m placed at Eart
Arlecino [84]

Answer:

The gravitational potential energy of a system is -3/2 (GmE)(m)/RE

Explanation:

Given

mE = Mass of Earth

RE = Radius of Earth

G = Gravitational Constant

Let p = The mass density of the earth is

p = M/(4/3πRE³)

p = 3M/4πRE³

Taking for instance,a very thin spherical shell in the earth;

Let r = radius

dr = thickness

Its volume is given by;

dV = 4πr²dr

Since mass = density* volume;

It's mass would be

dm = p * 4πr²dr

The gravitational potential at the center due would equal;

dV = -Gdm/r

Substitute (p * 4πr²dr) for dm

dV = -G(p * 4πr²dr)/r

dV = -G(p * 4πrdr)

The gravitational potential at the center of the earth would equal;

V = ∫dV

V = ∫ -G(p * 4πrdr) {RE,0}

V = -4πGp∫rdr {RE,0}

V = -4πGp (r²/2) {RE,0}

V = -4πGp{RE²/2)

V = -4Gπ * 3M/4πRE³ * RE²/2

V = -3/2 GmE/RE

The gravitational potential energy of the system of the earth and the brick at the center equals

U = Vm

U = -3/2 GmE/RE * m

U = -3/2 (GmE)(m)/RE

5 0
1 year ago
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