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4 years ago

The mass of a golf ball is 45.9 g . if it leaves the tee with a speed of 62.0 m/s , what is its corresponding wavelength?

Physics

1 answer:

inn [45]4 years ago

4 0

The wavelength of the golf ball is <u>2.328×10⁻³⁴m.</u>

All moving particles with mass have a matter wave associated with it. These matter waves are called deBroglie waves.

The deBroglie wavelength λ of a particle is given by,

$\lambda=\frac{h}{mv}$

Here, h is the Planck's constant, m is the mass of the ball and v is its velocity.

Calculate the deBroglie wavelength of the moving golf ball by substituting 6.626×10⁻³⁴J s for h, 45.9×10⁻³kg for m and 62.0 m/s for v.

$\lambda=\frac{h}{mv}\\ =\frac{6.626*10^-^3^4Js}{(45.9*10^-^3kg)(62.0m/s)} \\ =2.328*10^-^3^4m$

The wavelength of the golf ball is <u>2.328×10⁻³⁴m.</u>

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A. True
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Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.

Artist 52 [7]

Answer:

a) C.M $=(\bar x, \bar y)=(0.767,0.7)m$

b) $(x_4,y_4)=(-1.917,-1.75)m$

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

$\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}$

$\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}$

Where M represent the sum of all the masses on the system.

And the center of mass C.M $=(\bar x, \bar y)$

Part a

$m_1= 3 kg, m_2=5kg,m_3=7kg$ represent the masses.

$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

C.M $=(\bar x, \bar y)=(0.767,0.7)m$

Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m$

If we solve for a we got:

$(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0$

$a=-\frac{(5kg*2.3m)}{6kg}=-1.917m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m$

$(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0$

And solving for b we got:

$b=-\frac{(7kg*1.5m)}{6kg}=-1.75m$

So the coordinates for this new particle are:

$(x_4,y_4)=(-1.917,-1.75)m$

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3 years ago

A 5.0 V battery contains 775 C of charge. How much electricity energy can it produce?

KatRina [158]

Answer:

3875J

Explanation:

Energy is defined as the power × time

And it's defined as

Power = IV - I- current and V- voltage

Now quantity of electricity; Q = I × t

Where I is current and t is time

Now Energy = I ×V×t = V× I×t = V× Q;

where Q is quantity of electricity 775C and V is 5.0volt

Hence 775 × (5) =3875J

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3 years ago

Pam, wearing a rocket pack, stands on frictionless ice. She has a mass of 49 kg. The rocket supplies a constant force for 22.0 m

ss7ja [257]

Answer:

Magnitude of the force is 4350N

Explanation:

As the woman accelerates at a distance of 22 m to go from rest to 62.5 m / s, we can use the kinematics to find the acceleration

v² = v₀² + 2 a x

v₀ = 0

a = v² / 2x

a = 62.5²/(2 × 22)

a = 88.78m/s²

the time you need to get this speed

v = v₀ + a t

t = v / a

t = 62.5 / 88.78

t = 0.704s

Let's caculate the magnitude of the force

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= 49 × 88.78

= 4350.22

≅ 4350N

Magnitude of the force is 4350N

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W = mg + ma

Where mg represents the weight reading without external acceleration and ma as the weight reading due to the external acceleration.

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