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2 years ago

I neeeed heeeeeelllllllllppppppp

Physics

2 answers:

atroni [7]2 years ago

6 0

Answer:

the 2nd one

Explanation:

hope it helps

Gala2k [10]2 years ago

6 0

Answer: its the 2nd option

Explanation: when the net force of an object decreases, so does the acceleration

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A 1.15-kg mass oscillates according to the equation where x is in meters and in seconds. Determine (a) the amplitude, (b) the fr

ANEK [815]

The complete question is;

A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) the amplitude, (b) the frequency, (c) the total energy, and (d) the kinetic energy and potential energy when x = 0.360 m.

Answer:

A) Amplitude; A = 0.650 m

B) Frequency; f = 1.337 Hz

C) total energy = 17.142 J

D) Kinetic energy = 11.884 J

Potential Energy = 5.258 J

Explanation:

We are given;

Mass;m = 1.15 kg

Equation; x = 0.650 cos (8.40t)

(a) The standard form of a wave function is in the form y(x,t) = Asin(kx−ωt+ϕ)

So, comparing terms in our equation in the question to this, the amplitude is;

A = 0.650 m

(b) we know that formula for frequency is;

f = ω/2π

Again, comparing terms in the standard equation and our question, we can see that ω = 8.4

Thus;

f = 8.4/(2π)

f = 1.337 Hz

E = m•ω²•A²/2

Plugging in the relevant values, we have;

E = (1.15)(8.40)²(0.650)²/2

E = 17.142 J

(d) we want to find the kinetic energy and potential energy when x = 0.360 m.

The formula for kinetic energy in this case is given by;

K = (1/2)•m•ω²•(A² - x²)

Thus;

K = (1/2) × (1.15) × (8.40)² × ((0.650)² - (0.360)²)

K = 11.884 J

Also, the formula for the potential energy in this case is given by;

U = (1/2)•m•ω²•x²

Thus;

U = (1/2) × (1.15) × (8.40)² × (0.360)²

U = 5.258 J

3 0

2 years ago

Dave throws a 10 kg bowling ball straight up in the air. At the very tippy-top of its path, what is it's momentum?

svetoff [14.1K]

Answer:Zero

Explanation:

Given

mass of ball $m=10\ kg$

If the ball is thrown upward then at maximum point velocity of ball is zero because ball is no longer able to move upward

Momentum(P) of a particle is given by

$P=mass\times velocity$

$P=10\times 0$

$P=0$

Therefore at the highest point momentum is zero .

8 0

3 years ago

Kyle, a 95.0 kg football player, leaps straight up into the air (with no horizontal velocity) to catch a pass. He catches the 0.

babymother [125]

Answer:

The initial speed of the ball was 26.2 m/s

Explanation:

When the football player is in the air at his maximum height the vertical component of velocity is zero, To obtain the horizontal velocity when the player catches the ball we need to apply the linear momentum conservation theorem:

$m_1*v_{o1}+m_2*v_{o2}=m_t*v_f\\0.430kg*(v)+m_2*(0)=mt*vt$