If an object has a mass of 8 kg, what is its approximate weight on Earth?

In an inelastic collision a 2.5 kg ball moving at 7.5 m/s is caught by a 70kg man while the man is standing on ice. What is the

MrRa [10]

The velocity of the ball and the man is 0.259 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, in an isolated system, the total momentum before and after the collision must be conserved. Therefore, for the ball-man system, we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$

where:

$m_1 = 2.5 kg$ is the mass of the ball

$u_1 = 7.5 m/s$ is the initial velocity of the ball

$m_2 = 70 kg$ is the mass of the man

$u_2 = 0$ is the initial velocity of the man

$v$ is the final velocity of the man and the ball after the collision

Re-arranging the equation and substituting the values, we find the final velocity:

$v=\frac{m_1 u_1}{m_1+m_2}=\frac{(2.5)(7.5)}{2.5+70}=0.259 m/s$

So, the man and the ball slides on the ice at 0.259 m/s.

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3 0

3 years ago

The magnitude of the angular momentum of the two-satellite system is best represented by

zloy xaker [14]

The magnitude of the angular momentum of the two-satellite system is best represented as, L=m₁v₁r₁-m₂v₂r₂.

<h3>What is angular momentum.?</h3>

The rotational analog of linear momentum is angular momentum also known as moment of momentum or rotational momentum.

It is significant in physics because it is a conserved quantity. the total angular momentum of a closed system remains constant. Both the direction and magnitude of angular momentum are conserved.

The magnitude of the angular momentum of the two-satellite system is best represented as;

L=∑mvr

L=m₁v₁r₁-m₂v₂r₂

Hence, the magnitude of the angular momentum of the two-satellite system is best represented as, L=m₁v₁r₁-m₂v₂r₂.

To learn more about the angular momentum, refer to the link;

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3 0

2 years ago

In the Bohr model of the hydrogen atom, an electron({rm mass};m=9.1; times 10^{ - 31;}{rm kg}) orbits a proton at a distance of

max2010maxim [7]

Answer:

n=6.56×10¹⁵Hz

Explanation:

Given Data

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

To find

Revolutions per second

Solution

Let F be the force of attraction

let n be the number of revolutions per sec made by the electron around the nucleus then the centripetal force is given by

F=mω²r......................where ω=2π n

F=m4π²n²r...............eq(i)

as the values given where

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

we have to find n from eq(i)

n²=F/(m4π²r)

$n^{2} =\frac{8.2*10^{-8} }{9.11*10^{-31}* 4\pi^{2} *5.3*10^{-11} }\\ n^{2}=4.31*10^{31}\\ n=\sqrt{4.31*10^{31}}\\ n=6.56*10^{15}Hz$

8 0

3 years ago

The position of an object in simple harmonic motion is defined by the function y = (0.50 m) sin (πt/2). Determine the maximum sp

Gnoma [55]

The maximum speed of the object under simple harmonic motion is 0.786 m/s.

The given parameters:

Position of the particle, y = 0.5m sin(πt/2)

<h3>Wave equation for simple harmonic motion;</h3>

y = A sin(ωt + Ф)

where;

A is the amplitude = 0.5 m
ω is the angular speed = π/2

The maximum speed of the object is calculated as follows;

$V_{max} = A \omega\\\\V_{max} = 0.5 \times \frac{\pi}{2} = \frac{\pi}{4} \ m/s = 0.786 \ m/s$

Thus, the maximum speed of the object under simple harmonic motion is 0.786 m/s.

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5 0

2 years ago

Consider an ideal gas at 27.0 degrees Celsius and 1.00 atmosphere pressure. Imagine the molecules to be uniformly spaced, with e

My name is Ann [436]

To solve the exercise it is necessary to keep in mind the concepts about the ideal gas equation and the volume in the cube.

However, for this case the Boyle equation will not be used, but the one that corresponds to the Boltzmann equation for ideal gas, in this way it is understood that

$PV =NkT$