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2 years ago

1. What is the momentum of a golf ball with a mass of 62 g moving at 73 m/s?

Physics

1 answer:

Anit [1.1K]2 years ago

4 0

Answer:

<h3>The answer is 4.53 kgm/s</h3>

Explanation:

The momentum of an object can be found by using the formula

<h3>momentum = mass × velocity</h3>

From the question

mass = 62 g = 0.062 kg

velocity = 73 m/s

We have

momentum = 0.062 × 73 = 4.526

We have the final answer as

Hope this helps you

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A spring stores 10. joules of elastic potential

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The answer would be . Since we are looking for the spring constant you would need to use the formula
$PEs = \frac{1}{2} k {x}^{2}$
. Then you'd substitute, for PEs and x.
$10j=1/2k(.20m^2).$
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A person who weighs 625 N is riding a 98-N mountain bike. Suppose that the entire weight of the rider and bike is supported equa

kozerog [31]

Answer:

A = 4.76 x 10⁻⁴ m²

Explanation:

given,

weight of the person = 625 N

weight of the bike = 98 N

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Area of contact on each Tyre = ?

total weight of the system = 625 + 98

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Let F be the force on both the Tyre

F + F = W

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F = 361.5 N

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$A = \dfrac{F}{P}$

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2 years ago

un serbatoio d'acqua di forma cilindrica è riempito con acqua dolce fino a un livello di 6,40 m calcola la pressione dell'acqua

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Per calcolare la pressione a una certa profondità, devi considerare la legge di Stevino:
<span>p = ρ · g · h

Tenendo conto che:
g = 9,81 m/s²
ρ = 1000 kg/m³

Troviamo:
p(h</span>₁) = ρ · g · h₁ = 1000 · 9,81 · 4,50 = 44145 Pa
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2 years ago

A child pushes her toy across a level floor at a steady velocity of 0.50 m/s using an applied force of 2.0 N. If the weight of t

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Since toy is moving at constant speed that means that force that child is applying on toy is equal to force of friction.

Rate of speed that toy is moving is irelevant.

childs force is:
Fc = 2N
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Ff = a*Q

where Q is weight of the toy and a is friction

if we express a we get
a = F/Q = 2/8 = 0.25

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2 years ago

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In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

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2 years ago