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finlep [7]
3 years ago
14

Two golf balls are hit from the same point on a flat field. Both are hit at an angle of 20° above the horizontal. Ball 2 has twi

ce the initial speed of ball 1. If ball 1 lands a distance d1 from the initial point, at what distance d2 does ball 2 land from the initial point? (Neglect any effects due to air resistance.)
A) dd_{2} = 2d_{_{1}}

B) d_{2} = 0.5d_{_{1}}

C)d_{2} = 8d_{_{1}}

D) d_{2} = d_{_{1}}

E) d_{2} = 4d_{_{1}}
Physics
1 answer:
Nutka1998 [239]3 years ago
5 0

Answer:

E) d_{2} = 4d_{_{1}}

Explanation:

We know that the range of a projectile is given by:

R=\frac{u^2.sin\ \2\theta}{g}

where:

u = initial velocity of the projectile

\theta = angle of projection of projectile

<u>Now, range for ball 1 :</u>

d_1=\frac{u^2.sin\ \2\theta}{g}

<u>Now, range for ball 2 with double velocity and all the other parameters being the same:</u>

d_2=\frac{(2u)^2.sin\ \2\theta}{g}

d_2=4\times \frac{u^2.sin\ \2\theta}{g}

d_2=4\times d_1

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Consider the same roller coaster. It starts at a height of 40.0 m but once released, it can only reach a height of 25.0 m above
poizon [28]

Answer:

The magnitude of the frictional force between the car and the track is 367.763 N.

Explanation:

The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:

U_{g,1} = U_{g,2} + W_{dis}

Where:

U_{g,1}, U_{g,2} - Initial and final gravitational potential energy, measured in joules.

W_{dis} - Dissipated work due to friction, measured in joules.

Gravitational potential energy is described by the following formula:

U = m \cdot g \cdot y

Where:

m - Mass, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

y - Height with respect to reference point, measured in meters.

In addition, dissipated work due to friction is:

W_{dis} = f \cdot \Delta s

Where:

f - Friction force, measured in newtons.

\Delta s - Travelled distance, measured in meters.

Now, the energy equation is expanded and frictional force is cleared:

m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s

f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}

If m = 1000\,kg, g = 9.807\,\frac{m}{s^{2}}, y_{1} = 40\,m, y_{2} = 25\,m and \Delta s = 400\,m, then:

f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}

f = 367.763\,N

The magnitude of the frictional force between the car and the track is 367.763 N.

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4 years ago
A cubical box, 5.00 cm on each side, is immersed in a fluid. The gauge pressure at the top surface of the box is 594 Pa and the
Zolol [24]

Answer:

The density of the fluid is 1100 kg/m³.

Explanation:

Given that,

Height = 5.00 cm

Pressure at top =594 Pa

Pressure at bottom = 1133 Pa

We need to calculate the change in pressure

Using formula of change in pressure

\Delta P=P_{b}-P_{t}

Where, P_{b} = Pressure at bottom

P_{t} = Pressure at top

put the value into the formula

\Delta P=1133-594

\Delta P=539\ Pa

Using formula of pressure for density

\Delta P = \rho g h

\rho =\dfrac{\Delta P}{gh}

Where, \rho = density

P = pressure

h = height

Put the value in to the formula

\rho =\dfrac{539}{5.00\times10^{-2}\times9.8}

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4 0
4 years ago
When two point charges are a distance dd part, the electric force that each one feels from the other has magnitude F.F . In orde
Harman [31]

Answer:

In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. r' = r/2.

Explanation :

The electric force between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between charges. It is given by :

F=\dfrac{kq_1q_2}{r^2}

r is the separation between charges  

F\propto \dfrac{1}{r^2}

r=\sqrt{\dfrac{1}{F}}

If F'= 2F

r'=\dfrac{1}{\sqrt{2F} }

In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. r'=\dfrac{1}{\sqrt{2F} }. Hence, this is the required solution.                                                                                    

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