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Verdich [7]
4 years ago
11

A ball is launched with initial speed v from ground level up a frictionless slope (This means the ball slides up the slope witho

ut rolling). The slope makes an angle θ with the horizontal. Using conservation of energy, find the maximum vertical height hmax to which the ball will climb.
Physics
1 answer:
amid [387]4 years ago
4 0

Answer:

hmax = 1/2 · v²/g

Explanation:

Hi there!

Due to the conservation of energy and since there is no dissipative force (like friction) all the kinetic energy (KE) of the ball has to be converted into gravitational potential energy (PE) when the ball comes to stop.

KE = PE

Where KE is the initial kinetic energy and PE is the final potential energy.

The kinetic energy of the ball is calculated as follows:

KE = 1/2 · m · v²

Where:

m = mass of the ball

v = velocity.

The potential energy is calculated as follows:

PE = m · g · h

Where:

m = mass of the ball.

g = acceleration due to gravity (known value: 9.81 m/s²).

h = height.

At  the maximum height, the potential energy is equal to the initial kinetic energy because the energy is conserved, i.e, all the kinetic energy was converted into potential energy (there was no energy dissipation as heat because there was no friction). Then:

PE = KE

m · g · hmax = 1/2 · m · v²

Solving  for hmax:

hmax = 1/2 · v² / g

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Lapatulllka [165]

To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.

The altitude is,

z = 20km

And the velocity can be written as,

V = 3530km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

V = 980.55m/s

From the properties of standard atmosphere at altitude z = 20km temperature is

T = 216.66K

k = 1.4

R = 287 J/kg

Velocity of sound at this altitude is

a = \sqrt{kRT}

a = \sqrt{(1.4)(287)(216.66)}

a = 295.049m/s

Then the Mach number

Ma = \frac{V}{a}

Ma = \frac{980.55}{296.049}

Ma = 3.312

So front stagnation temperature

T_0 = T(1+\frac{k-1}{2}Ma^2)

T_0 = (216.66)(1+\frac{1.4-1}{2}*3.312^2)

T_0 = 689.87K

Therefore the temperature at its front stagnation point is 689.87K

6 0
4 years ago
A 26.0 g ball is fired horizontally with initial speed v0 toward a 110 g ball that is hanging motionless from a 1.10 m -long str
mart [117]

Answer:

7.3 ms^{-1}

Explanation:

Consider the motion of the ball attached to string.

In triangle ABD

Cos50 = \frac{AB}{AD} \\Cos50 = \frac{AB}{L}\\AB = L Cos50

height gained by the ball is given as

h = BC = AC - AD \\h = L - L Cos50\\h = 1.10 - 1.10 Cos50\\h = 0.393 m

M  = mass of the ball attached to string = 110 g

V = speed of the ball attached to string just after collision

Using conservation of energy

Potential energy gained = Kinetic energy lost

Mgh = (0.5) M V^{2} \\V = sqrt(2gh)\\V = sqrt(2(9.8)(0.393))\\V = 2.8 ms^{-1}

Consider the collision between the two balls

m  = mass of the ball fired = 26 g

v_{o} = initial velocity of ball fired before collision = ?

v_{f} = final velocity of ball fired after collision = ?

using conservation of momentum

m v_{o} = MV + m v_{f}\\26 v_{o} = (110)(2.8) + 26 v_{f}\\v_{f} = v_{o} - 11.85

Using conservation of kinetic energy

m v_{o}^{2} = MV^{2} + m v_{f}^{2} \\26 v_{o}^{2} = 110 (2.8)^{2} + 26 (v_{o} - 11.85)^{2} \\v_{o} = 7.3 ms^{-1}

3 0
3 years ago
The electric field is strongest where equipotential curves are:_______.A) closest togetherB) farthest apartC) most nearly straig
cluponka [151]

Answer:

A closest

Explanation:

This is because the electric field will be strongest or largestwhen the equipotential curves are closest together

We know that the field is

E= V/d

Where is distance and we see that d being the denominator will only make E bigger if it becomes smaller that is the curves closest

3 0
3 years ago
Astronomers observe two separate solar systems each consisting of a planet orbiting a sun. The two orbits are circular and have
Oxana [17]

Answer:

(D) 3

Explanation:

The angular momentum is given by:

\vec{L}=\vec{r}\ X \ \vec{p}

Thus, the magnitude of the angular momenta of both solar systems are given by:

L_1=Rm_1v_1=Rm_1(\omega R)=R^2m_1(\frac{2\pi}{T_1})=2\pi R^2\frac{m_1}{T_1}\\\\L_2=Rm_2v_2=2\pi R^2\frac{m_2}{T_2}

where we have taken that both systems has the same radius.

By taking into account that T1=3T2, we have

L_1=2\pi R^2\frac{m_1}{3T_2}=\frac{1}{3}2\pi R^2\frac{1}{T_2}m_1=\frac{1}{3}\frac{L_2}{m_2}m_1

but L1=L2=L:

L=\frac{1}{3}L\frac{m_1}{m_2}\\\\\frac{m_1}{m_2}=3

Hence, the answer is (D) 3

HOPE THIS HELPS!!

3 0
3 years ago
Why does the Earth have more gravitation pull than the moon?
Nookie1986 [14]
As accurately described by Einstein's theory of relativity, gravity is not necessarily a force, but a consequence of the curvature of space time that is caused by the uneven distribution of mass. But this could be understood more easily through Newton's Law of Universal Motion. The equation is shown below:

F = G(m₁m₂/d²), where
F is the gravitational force
G is called Newton's universal gravitation constant equal to 6.673×10⁻¹¹ N m² kg⁻²
m is the mass of the objects 1
d is the distance between the objects
 
Basing on the equation, the gravitational force depends on the mass the distance between the objects. So, when you compare the gravitational pull between Earth and moon. you do not need to include the effect of distance because, together. they have the same amount of d. So, it mainly depends on the masses. Since F is directly proportional to m, the greater the mass, the greater is the pull.

So, the answer is: <span>The Earth has more mass than the moon.</span> 
5 0
4 years ago
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