What is the mass of an object that has a weight of 52N?

Momentum is a measure of the<br> of an object.<br> Which term accurately completes the sentence?

RideAnS [48]

Answer:

Momentum is define as the product of the mass and velocity of a body. It is measured in Kgm/s.

Explanation:

Momentum is the product of mass and velocity of an object. When an object or a body of mass 'm' is moving with velocity 'v', then its momentum can be determined as;

momentum (P) = mass × velocity

i.e P = m × v

= mv

It is measured in Kgm/s.

The change in momentum of a body is referred to as its impulse (Ft).

ΔP = m(v - u) = Ft

Where: P is the momentum of the object, m is its mass, v is its final velocity, u is the initial velocity, F is the force and t is the time in which the force acts.

8 0

3 years ago

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A carnival game consists of a two masses on a curved frictionless track, as pictured below. The player pushes the larger object

Harman [31]

Answer:

v₁₀ = 1.90 m / s

Explanation:

In this exercise we are given the maximum height data, with energy we can know how fast the body came out

Final mechanical energy, maximum height

$Em_{f}$ = U = m g h

Initial mechanical energy, in the lower part of the track

Em₀ = K = ½ m v²

Em= $Em_{f}$

½ m v² = m g h

v = √ 2gh

Now we can use the moment to find the speed with which objects collide

The large object has a mass M = 5.41 kg a velocity starts v₁₀, the small object has a mass m = 1.68 kg an initial velocity of zero v₂₀ = 0 and final velocity v

Initial before the crash

p₀ = M v₁₀ + 0

Final after the crash

$p_{f}$ = M v1f + m v

p₀ = $p_{f}$

M v₁₀ = M $v_{1f}$ + m v

As the shock is elastic the kinetic energy is conserved

K₀ = $K_{f}$

½ M v₁₀² = ½ M $v_{1f}$ ² + ½ m v²

Let's write the system of equations

M v₁₀ = M $v_{1f}$ + m v

M v1₁₀² = M $v_{1f}$ ² + m v²

We cleared v1f in the first we replaced in the second

$v_{1f}$ = (M v₁₀ - mv) / M

M v₁₀² = M (M v₁₀ - mv)² / M² + m v²

M v₁₀² = 1 / M (M² v₁₀² - 2mM v v₁₀ + m² v²) +m v²

v₁₀² (M - M) + 2 m v v₁₀ - v² (m2 + m) / M = 0

2 m v₁₀ - v (m + 1) m/ M = 0

v₁₀ = v (m +1) / (2M)

Let's substitute the value of v

v1₁₀= √ (2gh) (m +1) / (2M)

Let's calculate

v₁₀ = √ (2 9.8 3) (1+ 1.68) / (2 5.41)

V₁₀ = 7.668 (2.68) / 10.82

v₁₀ = 1.90 m / s

5 0

3 years ago

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with an

Alona [7]

Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

$F=\dfrac{kQ^2}{r^2}$ ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

$F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}$ .....(2)

Dividing equation (1) and (2), we get :

$\dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}$