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3 years ago

5

A horse starts from rest and speeds uo with constant acceleration. If it has gone a distance of 100 m at the point when it reach

es a speed of 12 m/s, what is the acceleration?

1 answer:

makkiz [27]3 years ago

3 0

Answer:

yee yee

Explanation:

You might be interested in

You are conducting an experiment inside a train car that may move along level rail tracks. A load is hung from the ceiling on a

allochka39001 [22]

Answer:

b.

c.

e.

h.

Explanation:

If we consider the ball,the tension is not balance by weight then the force acting will be non zero.It means that train car in not inertial frame of reference.And the same time the train car must accelerated with respected to the earth.It is not moving straight path.When train is in rest position but the string will be at the constant angle.

So the following option are correct:

b.

c.

e.

h.

6 0

3 years ago

A 30 N rock falls from a 40 m cliff. At what point during its fall are its

vaieri [72.5K]

Answer:

<em>Both energies are equal when the rock has fallen 20 m or equivalently when it is at a height of 20 m.</em>

Explanation:

<u>Potential and Kinetic Energy</u>

The gravitational potential energy is the energy an object has due to its height above the ground. The formula is

$U=mgh$

Where:

m = mass of the object

g = acceleration of gravity (9.8~m/s^2)

h = height

Note we can also use the object's weight W=mg into the formula:

$U=Wh$

The kinetic energy is the energy an object has due to its speed:

$\displaystyle K=\frac{1}{2}mv^2$

Where v is the object's speed.

Initially, the object has no kinetic energy because it's assumed at rest.

The W=30 N rock falls from a height of h=40 m, thus:

$U=30*40=1,200 J$

Since the sum of the kinetic and potential energies is constant:

U' + K' = 1,200 J

Here, U' and K' are the energies at any point of the motion. Since both must be the same:

U' = K' = 600 J

U'=Wh'=600

Solving for h':

$\displaystyle h'=\frac{600}{W}=\frac{600}{30}=20~m$

Both energies are equal when the rock has fallen 20 m or equivalently when it is at a height of 20 m.

3 0

3 years ago

The gas sample has a volume of 45.1 μL at 24.7 °C. What is the volume of the gas after the sample is heated to 37.2 °C at consta

likoan [24]

Answer:

V₂ = 46.99 μL.

Explanation:

Given that

V₁ = 45.1 μL

T₁ = 24.7°C = 273 + 24.7 = 297.7 K

T₂ = 37.2°C = 273+37.2=310.2 K

Lets take ,The final volume = V₂

We know that ,the ideal gas equation

If the pressure of the gas is constant ,then we can say that

$\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}$

${V_2}=V_1\times \dfrac{T_2}{T_1}$

Now by putting the values in the above equation we get

${V_2}=45.1\times \dfrac{310.2}{297.7}\ \mu \ L$

V₂ = 46.99 μL.

The final volume will be 46.99 μL.

7 0

3 years ago

A cannon shoots a ball at an angle θ above the horizontal ground. (a) Neglecting air resistance, use Newton's second law to find

nikitadnepr [17]

Answer:

a) x = v₀ₓ t , y = $v_{oy}$ t - ½ g t²

Explanation:

This is a projectile launch problem, let's use Newton's second law on each axis

X axis

F = m a

Since there is no acceleration on the x axis, the force on this axis is zero

Y Axis

-W = m $a_{y}$

-m g = m $a_{y}$

$a_{y}$ = -g

In this axis the acceleration is the acceleration of gravity

Now we can use science to find the position of the body on each axis

X axis

x = v₀ₓ t + ½ a t²

As the acceleration on this axis is zero

x = v₀ₓ t

Y Axis

y = $v_{oy}$ t + ½ $a_{y}$ t²

The acceleration on this axis is –g

y = $v_{oy}$ t - ½ g t²

B) to find the maximum value of distance r

r =√ x² + y²

r = √( v₀ₓ² t² + ( $v_{oy}$ t + ½ g t²)²

We can find the maximum value of r using time respect derivatives

dr / dt = 0

0 = ½ 1/√( v₀ₓ² t² + ( $v_{oy}$ t + ½ g t²)² (v₀ₓ² 2t + 2 ( $v_{oy}$ t - ½ g t²)( $v_{oy}$ - ½ g2t)

We simplify this expression

0 = v₀ₓ² 2t + 2 ( $v_{oy}$ t - ½ g t²) ( $v_{oy}$ - ½ g2t)

-v₀ₓ² t =( $v_{oy}$ t - ½ g t²) ( $v_{oy}$ - ½ g2t)

-v₀ₓ² t = $v_{oy}$ ² t -3/2 gt² $v_{oy}$ + 1/2 g² t³

½ g² t²- 3/2 g $v_{oy}$ t = $v_{oy}$ ² + v₀ₓ²

Let's use trigonometry to find go and vox

sin θ = $v_{oy}$ / v₀

cos θ = vox / v₀

$v_{oy}$ = v₀ sin θ

v₀ₓ = v₀ cos θ

We replace

½ g² t² -3/2 g $v_{oy}$ t = v₀ (sin² θ + cos² θ)

g t² - 3v₀ sin θ t = 2 v₀/g

The time is maximum for the angle is zero

8 0

3 years ago

What causes an aurora?

geniusboy [140]

Answer:

When charged particles from the sun strike atoms in Earth's atmosphere, they cause electrons in the atoms to move to a higher-energy state. When the electrons drop back to a lower energy state, they release a photon: light. This process creates the beautiful aurora, or northern lights.

Explanation:

6 0

4 years ago