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melomori [17]
3 years ago
5

The mass of a golf ball is 45.9

Physics
1 answer:
Tema [17]3 years ago
8 0
The equation for the de Broglie wavelength is: 
<span>λ = (h/mv) √[1-(v²/c²)], </span>
<span>where h is Plank's Constant, m is the rest mass, v is velocity, and c is the velocity of light in vacuum. However, if c>>v (and it is, in this case) then the expression under the radical sign approaches 1, and the equation simplifies to: </span>
<span>λ = h/mv. </span>
<span>Substituting, (remember to convert the mass to kg, since 1 J = 1 kg·m²/s²): </span>
<span>λ = (6.63x10^-34 J·s) / (0.0459 kg) (72.0 m/s) = 2.00x10^-34 m.</span>
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You would use d. table

6 0
3 years ago
What substance is used by plants as a source of food
harkovskaia [24]

Answer:

dart

Explanation:

dart and sun and water so that the plant be okay

7 0
3 years ago
13.<br>14. Why does a person feel weightlessness during freefall?​
Sveta_85 [38]

Answer:

When in free fall, the only force acting upon your body is the force of gravity - a non-contact force. Since the force of gravity cannot be felt without any other opposing forces, you would have no sensation of it. You would feel weightless when in a state of free fall.

6 0
3 years ago
The speed of a 500 g cricket ball changes from 10m/s to 30m/s in just 7 seconds. What is the force acting on the cart?
9966 [12]

The force acting on the cart is 1.43 N.

<h3>What is force?</h3>

Force can be defined as the product of mass and acceleration.

To calculate the force acting on the cart, we use the formula below.

Formula:

  • F = m(v-u)/t................. Equation 1

Where:

  • F = Force acting on the cart
  • m = mass of the cart
  • v = Final velocity
  • u = initial velocity
  • t = time

From the question,

Given:

  • m = 500 g = 0.5 kg
  • v = 30 m/s
  • u = 10 m/s
  • t = 7 seconds

Substitute these values into equation 1

  • F = 0.5(30-10)/7
  • F = 10/7
  • F = 1.43 N.

Hence, the force acting on the cart is 1.43 N.

Learn more about force here: brainly.com/question/13370981

7 0
2 years ago
(a) In what direction would the ship in Exercise 3.57 have to travel in order to have a velocity straight north relative to the
shtirl [24]

Answer:

Direction of ship: 9.45° West of North

Ship's relative speed: 7.87m/s

Explanation:

A. Direction of ship: since horizontal of the velocity of boat relative to the ground is 0

Vx=0

Therefore, -VsSin∅+VcCos∅40°

Sin∅ = Vc/Vs × Cos 40°

Sin∅ = 1.5/7 ×Cos40°

Sin∅= 0.164

∅= Sin-¹ (0.164)

∅= 9.45° W of N

B. Ship's relative speed:

Vy= VsCos∅ + Vcsin40°

= 7Cos9.45° + 1.5sin40°

= 7×0.986 + 1.5×0.642

= 7.865

= 7.87m/s

4 0
3 years ago
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