The Sun is divided into three regions.<br> True оr False?

A 4 kg toy car moves horizontally on a rough road with coefficient of kinetic friction 0.2. It accelerates from rest to 20 m/s i

attashe74 [19]

The total work done on the car is 784Joule.

<h3>What's the acceleration of the car?</h3>

As per Newton's equation of motion, V= U+at
U= initial velocity= 0 m/s

V= vinal velocity= 20m/s

t= time = 10s

a= acceleration

So, 20= 0+ 10a

=> a= 20/10= 2m/s²

<h3>What's the distance covered by the car in 10 seconds?</h3>

As per Newton's equation of motion,

V²-U² = 2aS

S= distance covered by the car
So, 20²-0=2×2×S=4S

=> 400= 4S

=> S= 400/4= 100m

<h3>What's the work done on the car due to frictional force?</h3>

Work done by frictional force= frictional force × distance

= (0.2×4×9.8)×100

= 784Joule

Thus, we can conclude that the work done on the car is 784Joule.

Learn more about the work done here:

brainly.com/question/25573309

#SPJ1

4 0

1 year ago

A crucial characteristic of turbines in tidal generators is that _____.

Mila [183]

<span>The blades should turn in two directions.</span>

4 0

3 years ago

Read 2 more answers

If the 50 kg boy were in a spacecraft 5.0r from the center of the earth, what would his weight be?

Lady_Fox [76]

·The acceleration of gravity is proportional to

   1 / (the square of the distance from the center) .

When we're on the surface, we're 1 radius from the center of the Earth,
and the acceleration of gravity is 9.8 m/s² .

The boy's weight = (mass) · (gravity) = (50kg) · (9.8 m/s²)

   = 490 newtons .

At the distance of 5 radii from the center (4 radii altitude from the surface),
the acceleration of gravity is

   (9.8 m/s²) · (1/5)² = 0.39 m/s² .

The boy's weight is (mass) · (gravity) = (50kg) · (0.39 m/s²)

= 19.6 newtons .

Just as we expected, his weight at that distance is

   (19.6 / 490) = 0.04 = 1/25 = 1/5² of his weight on the surface.

7 0

3 years ago

A satellite was in two separate crashes. In both crashes, the satellite had the same mass. Engineers want to know about the spee

Komok [63]

Answer:

The fastest satellite must change orbit

The most massive body (m₁) transfers more momentum to the satellite,

Explanation:

For this problem we consider a system formed by the satellite and each of the bodies with which it collides, in this system the forces during the collision are internal, the amount of movement must be conserved. Let's write the momentum is two instants

Most massive body (m1)

initial. Before the crash

p₀₁ = M v + m₁ v₁

after the crash

$p_{f1}$ = M v´ + m₁ v₁´

how momentum is conserved

p₀ = p_{f}

Lighter body (m2)

p₀₂ = M v + m₂ v₂

p_{f2} = M v´ + m₂ v₂´

Let's clarify that the speed of the satellite and the object do not have the same direction, in general these shocks are elastic.

We can see that p₀₁> p₀₂

Let us analyze the two cases when the body collides, The most massive body (m₁) transfers more momentum to the satellite, therefore there must be a greater change in its momentum and velocity.

The fastest satellite must change orbit, thus rotating at a different distance from Earth

5 0

3 years ago

2. A pair of narrow, parallel slits sep by 0.25 mm is illuminated by 546 nm green light. The interference pattern is observed on

Agata [3.3K]

Answer:

for the first interference m = 1 y = 2,839 10-3 m

for the second interference m = 2 y = 5,678 10-3 m

Explanation:

The double slit interference phenomenon, for constructive interference is described by the expression

d sin θ = m λ

where d is the separation between the slits, λ the wavelength and m an integer that corresponds to the interference we see.

In these experiments in general the observation screen is L >> d, let's use trigonometry to find the angles

tan θ = y / L

with the angle it is small,

tan θ = sin θ / cos θ = sin θ

we substitute

sin θ = y / L

d y / L = m λ

the distance between the central maximum and an interference line is

y = m λ L / d

let's reduce the magnitudes to the SI system

λ = 546 nm = 546 10⁻⁹ m

d = 0.25 mm = 0.25 10⁻³ m

let's substitute the values

y = m 546 10⁻⁹ 1.3 / 0.25 10⁻³

y = m 2,839 10⁻³

the explicit value for a line depends on the value of the integer m, for example

for the first interference m = 1

the distance from the central maximum to the first line is y = 2,839 10-3 m

for the second interference m = 2

the distance from the central maximum to the second line is y = 5,678 10-3 m

3 0

3 years ago

The Sun is divided into three regions. True оr False?