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2 years ago

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A force of constant magnitude F and fixed direction acts on an object of mass m that is initially at rest. If the force acts for

a time interval ∆t over a displacement ∆x , what is the magnitude of the resultant change in the linear momentum of the object?

1 answer:

ZanzabumX [31]2 years ago

5 0

Answer:

$\Delta P = F \Delta t$

Explanation:

As we know Newton's II law

$F$ = Rate of change in momentum

so we will have

$F = \frac{\Delta P}{\Delta t}$

now we will have

$\Delta P = F \Delta t$

so here we can say that change in momentum of the object is the product of force and interval of time for which the force is acting on it.

so we will have

$\Delta P = F \Delta t$

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A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest. If after the collision

TEA [102]

Answer:

D) No, since kinetic energy is not conserved.

Explanation:

Since momentum is always conserved in all collision

so in Y direction we can say

$0 = m(3.5 sin30) - mv_y$

$v_y = 1.75 m/s$

Now similarly in X direction we will have

$m(4.5) = m(3.5 cos30 ) + mv_x$

$v_x = 1.47 m/s$

now final kinetic energy of both puck after collision is given as

$KE_f = \frac{1}{2}m(3.5^2) + \frac{1}{2}m(1.75^2 + 1.47^2)$

$KE_f = 8.73 m$

initial kinetic energy of both pucks is given as

$KE_i = \frac{1}{2}m(4.5^2) + 0$

$KE_i = 10.125 m$

since KE is decreased here so it must be inelastic collision

D) No, since kinetic energy is not conserved.

5 0

2 years ago

Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of

Charra [1.4K]

To solve the problem it is necessary to apply the Torque equations and their respective definitions.

The Torque is defined as,

$\tau = I \alpha$

Where,

I=Inertial Moment

$\alpha =$ Angular acceleration

Also Torque with linear equation is defined as,

$\tau = F*d$

Where,

F = Force

d= distance

Our dates are given as,

R = 30 cm = 0.3m

m = 1.5 kg

F = 20 N

r = 4.0 cm = 0.04 m

t = 4.0s

Therefore matching two equation we have that,

$d*F = I\alpha$

For a wheel the moment inertia is defined as,

I= mR2, replacing we have

$d*F= \frac{mR^2a}{R}$

$d*F= mRa$

$a = \frac{rF}{ mR}$

$a = \frac{0.04*20}{1.5*0.3}$

$a=1.77 m/s^2$

Then the velocity of the wheel is

$V = a *t \\V=1.77*4 \\V=7.11 m/s$

Therefore the correct answer is D.

4 0

2 years ago

A man of mass m 1 5 70.0 kg is skating at v1 5 8.00 m/s behind his wife of mass m 2 5 50.0 kg, who is skating at v2 5 4.00 m/s.

ehidna [41]

Answer:

A. Kindly find attached free body diagram for your reference (smiles I guess I will make a terrible artist)

B. The collision is inelastic because both the husband and the wife moved together with same velocity as he grabs her on the waist

C. The general equation for conservation of momentum in terms of m 1, v 1, m 2, v 2, and final velocity vf

Say mass of husband is m1

Mass of the wife is m2

Velocity of the husband is v1

Velocity of the wife is v2

According to the conservation of momentum principle momentum before impact m1v1+m2v2 =momentum after impact Common velocity after impact (m1+m2)vf

The momentum equation is

m1v1+m2v2= (m1+m2)vf

D. To solve for vf we need to make it subject of formula

vf= {(m1v1) +(m2v2)}/(m1+m2)

E. Substituting our given data

vf=

{(1570*58)+(2550*54)}/(1570+2558)

vf=91060+137700/4120

vf=228760/4120

vf=55.52m/s

Their speed after collision is 55.52m/s

7 0

2 years ago

1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n

sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

$(1).\: x =v_0t$

$(2).\: y= 15m-\dfrac{1}{2}gt^2$

where $v_0$ is the initial velocity.

(a).

When the projectile hits the 50m mark, $y=0$ ; therefore,

$0=15-\dfrac{1}{2}gt^2$

solving for $t$ we get:

$t= 1.75s.$

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

$50m = v_0(1.75s)$

which gives

$\boxed{v_0 = 28.58m/s.}$

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

$v_x = 28.58m/s.$

the vertical component of the velocity is

$v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.$

which gives a speed $v$ of

$v = \sqrt{v_x^2+v_y^2}$

$\boxed{v =33.3m/s.}$

4 0

1 year ago

Microwave ovens emit microwave energy with a wavelength of 12.6 cm. what is the energy of exactly one photon of this microwave r

Helga [31]

We need the frequency of the photon, it is v = c/ λ

Where c is 3 x 10^8 ms^-1 and λ is the wave length

We also need the expression of connecting frequency to energy of photon

which is E = hv where h is Planck’s constant

Combining the two equations will give us:

E = h x c/λ

Inserting the values, we will have:

E = 6.626 x 10^-34 x 3 x 10^8 / 0.126

E = 1.578 x 10^ -24 J

7 0

2 years ago

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