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serious [3.7K]
3 years ago
9

Two golf balls are hit from the same point on a flat field. Both are hit at an angle of 55 degree above the horizontal. Ball 2 h

as twice the initial speed of ball 1. If ball 1 lands a distance d_1 from the initial point, at what distance d_2 does ball 2 land from the initial point? (Neglect any effects due to air resistance.)
d_2 = d_1
d_2 = 4d_1
d_2 = 8d_1
d_2 = 0.5d_1
d_2 = 2d_1
Physics
1 answer:
juin [17]3 years ago
4 0

Answer:

d_2 = 4d_1

Explanation:

The range or horizontal distance covered by a projectile projected with a velocity U at an angel of θ to the horizontal is given by

R = U²sin2θ/g

Let the range or horizontal distance of ball 1 with initial velocity U projected at an angle θ = 55° be

d_1 = U²sin2θ/g

Let the range or horizontal distance of ball 2 with initial velocity V = 2U projected at an angle θ = 55° be

d_2 = V²sin2θ/g

= (2U)²sin2θ/g

= 4U²sin2θ/g

= 4d_1   (since d_1 = U²sin2θ/g)

So, the ball 2 lands a distance d_2 = 4d_1 from the initial point.

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A person jumps out a fourth-story window 14 m above a firefighter safety net. The survivor stretches the net 1.8 m before coming
Monica [59]

Answer:

The deceleration is  a =  - 76.27 m/s^2

Explanation:

From the question we are told that

   The height above  firefighter safety net is H  = 14 \ m

   The length by which the net is stretched is s =  1.8 \ m

   

From the law of energy conservation

    KE_T + PE_T =  KE_B + PE_B

 Where KE_T is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )

   and  PE_T is the potential energy of the before jumping  which is mathematically represented at

          PE_T  = mg H

and  KE_B is the kinetic energy of the person just before landing on the safety net  which is mathematically represented at

        KE_B = \frac{1}{2} m v^2

and  PE_B is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )

   So the above equation becomes

          mgH =  \frac{1}{2} m v^2

=>           v =  \sqrt{2 gH }

    substituting values

                v =  16.57 m/s

Applying the equation o motion

             v_f =  v  + 2 a s

Now the final velocity is zero because the person comes to rest

      So

         0 = 16.57 + 2 * a * 1.8

            a =  - \frac{16.57^2 }{2 * 1.8}

            a =  - 76.27 m/s^2

         

         

4 0
3 years ago
What type of circuit is it please give with explanation !!!!
Archy [21]
The circuit is a parallel circuit.

Because even one light bulb is broken, the others are still lit. Therefore it is not a series circuit
7 0
3 years ago
If Rebecca is traveling at 7 m/s east for 1 hour how much is Rebecca displaced
Mekhanik [1.2K]

Answer:

25.2 km east

Explanation:

The relationship between velocity and displacement is given by:

v=\frac{d}{t}

where

v is the velocity

d is the displacement

t is the time

In this problem, we have:

v=7 m/s east is Rebecca's velocity

t=1 h=60 min=3600 s is the time taken

Re-arranging the equation, we can find Rebecca's displacement:

d=vt=(7 m/s)(3600 s)=25,200 m=25.2 km

and the direction is same as velocity (east)

3 0
3 years ago
A ball is rolling across the floor at a constant velocity. What is the value of the change in its kinetic energy, ΔEk?
Katyanochek1 [597]

B. When the ball is rolling across the floor at a constant velocity, the change in its kinetic energy is zero.

<h3>What is change in kinetic energy?</h3>

The change in kinetic energy of an object is the dereference between the final kinetic energy and the initial kinetic energy.

ΔK.E = K.Ef - K.Ei

ΔK.E = 0.5m(vf² - vi²)

where;

  • K.Ef is the final kinetic energy
  • K.Ei is the initial kinetic energy
  • vf is final velocity
  • vi is initial velocity

At constant velocity, the initial velocity and final velocity are equal.

ΔK.E = 0.5m(0) = 0

Thus, when the ball is rolling across the floor at a constant velocity, the change in its kinetic energy is zero.

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

5 0
2 years ago
By what factor does the peak frequency change if the celsius temperature of an object is doubled from 20.0 ∘c to 40.0 ∘c?
mart [117]

Answer:

it increases by a factor 1.07

Explanation:

The peak wavelength of an object is given by Wien's displacement law:

\lambda=\frac{b}{T} (1)

where

b is the Wien's displacement constant

T is the temperature (in Kelvins) of the object

given the relationship between frequency and wavelength of an electromagnetic wave:

f=\frac{c}{\lambda}

where c is the speed of light, we can rewrite (1) as

\frac{c}{f}=\frac{b}{T}\\f=\frac{Tc}{b}

So the peak frequency is directly proportional to the temperature in Kelvin.

In this problem, the temperature of the object changes from

T_1 = 20.0^{\circ}+273=293 K

to

T_2 = 40.0^{\circ}+273 = 313 K

so the peak frequency changes by a factor

\frac{f_2}{f_1} \propto \frac{T_2}{T_1}=\frac{313 K}{293 K}=1.07

8 0
3 years ago
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