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3 years ago

Acording to Newton’s second law of motion if a rigid body of unchanging mass is observed accelerating what must be happening

Physics

2 answers:

blondinia [14]3 years ago

6 0

A force is being applied to the mass.

Kitty [74]3 years ago

5 0

The net force on the object can't be zero.

The forces acting on it are not balanced.

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Two equal forces act on two different objects, one of which has a mass ten times as large as the other. The larger object will h

elena-s [515]

Answer:

The larger object will have <u>smaller</u> acceleration that the less massive object.

Explanation:

Generally force is mathematically represented as

$F = ma$

=> $m = \frac{F}{a }$

at constant force we have

$m \ \alpha \ \frac{1}{a}$

So if m is increasing a will be decreasing which means the object with the larger mass will have less acceleration

4 0

3 years ago

Which type of energy is emitted from earth to the atmosphere? *?

notsponge [240]

Infrared radiation - Greenhouse gases in the atmosphere absorb most of the earth's emitted infrared radiation which heats the lower atmosphere

7 0

3 years ago

A particle with mass 1.09 kg oscillates horizontally at the end of a horizontal spring. A student measures an amplitude of 0.985

e-lub [12.9K]

Answer:

a) $f = 0.598\,hz$ , b) $v_{max} = 3.701\,\frac{m}{s}$ , c) $k = 15.385\,\frac{N}{m}$ , d) $U = 1.081\,J$ , e) $K = 6.382\,J$ , f) $v\approx 3.422\,\frac{m}{s}$

Explanation:

a) The frequency of oscillation is:

$f = \frac{76}{127\,hz}$

$f = 0.598\,hz$

b) The angular frequency is:

$\omega = 2\pi \cdot f$

$\omega = 2\pi \cdot (0.598\,hz)$

$\omega = 3.757\,\frac{rad}{s}$

Lastly, the speed at the equilibrium position is:

$v_{max} = \omega \cdot A$

$v_{max} = (3.757\,\frac{rad}{s} )\cdot (0.985\,m)$

$v_{max} = 3.701\,\frac{m}{s}$

c) The spring constant is:

$\omega = \sqrt{\frac{k}{m}}$

$k = \omega^{2}\cdot m$

$k = (3.757\,\frac{rad}{s} )^{2}\cdot (1.09\,kg)$

$k = 15.385\,\frac{N}{m}$

d) The potential energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$U = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.375\,m)^{2}$

$U = 1.081\,J$

e) The maximum potential energy is:

$U_{max} = \frac{1}{2}\cdot (15.385\,\frac{N}{m} )\cdot (0.985\,m)^{2}$

$U_{max} = 7.463\,J$

The kinetic energy when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$K = U_{max} - U$

$K = 7.463\,J - 1.081\,J$

$K = 6.382\,J$

f) The speed when the particle is located 38.1 % of the amplitude away from the equilibrium position is:

$K = \frac{1}{2}\cdot m \cdot v^{2}$

$v = \sqrt{\frac{2\cdot K}{m} }$

$v = \sqrt{\frac{2\cdot (6.382\,J)}{1.09\,kg} }$

$v\approx 3.422\,\frac{m}{s}$

4 0

3 years ago

A ferris wheel is 35 meters in diameter and boarded from a platform that is 2 meters above the ground. The six o'clock position

Drupady [299]

Answer:

$h=f(t)= -17.5cos(\pi /3)+20.5$

Explanation:

Amplitude is 35/2=17.5

Midline= Distance from ground + Amplitude = 17.5+3= 20.5

Period is time taken to finish 6 minutes

2π/b=T

2π/b=6

b=π/3

$h=f(t)= -17.5cos(\pi /3)+20.5$

6 0

3 years ago

Read 2 more answers

A projectile of mass 2.0 kg is fired in the air at an angle of 40.0 ° to the horizon at a speed of 50.0 m/s. At the highest poin

tekilochka [14]

Answer:

a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

0.7 kg is 109.4 ms on the x axis

b) Y = 109.3 m

Explanation:

This is a moment and projectile launch exercise.

a) Let's start by finding the initial velocity of the projectile

sin 40 = voy / v₀

$v_{oy}$ = v₀ sin 40

$v_{oy}$ = 50.0 sin40

$v_{oy}$ = 32.14 m / s

cos 40 = v₀ₓ / V₀

v₀ₓ = v₀ cos 40

v₀ₓ = 50.0 cos 40

v₀ₓ = 38.3 m / s

Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis

Let's write the amounts

Initial mass of the projectile M = 2.0 kg

Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and $v_{y}$ = -10.0 m / s

Fragment mass 2 m₂ = 0.7 kg moves in the x direction

Fragment mass 3 m₃ = 0.3 kg moves up (y axis)

Moment before the break

X axis

p₀ₓ = m v₀ₓ

Y Axis y

$p_{oy}$ = 0

After the break

X axis

$p_{fx}$ = m₂ v₂

Axis y

$p_{fy}$ = m₁ v₁ + m₃ v₃

Let's write the conservation of the moment and calculate

Y Axis

0 = m₁ v₁ + m₃ v₃

Let's clear the speed of fragment 3

v₃ = - m₁ v₁ / m₃

v₃ = - (-10) 1 / 0.3

v₃ = 33.3 m / s

X axis

M v₀ₓ = m₂ v₂

v₂ = v₀ₓ M / m₂

v₂ = 38.3 2 / 0.7

v₂ = 109.4 m / s

The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

0.7 kg is 109.4 ms on the x axis

b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics

$v_{fy}$ ² = $v_{oy}$ ² - 2 gy

0 = $v_{oy}$ ²-2gy

y = $v_{oy}$ ² / 2g

y = 32.14² / 2 9.8

y = 52.7 m

This is the height where the break occurs, which is the initial height for body movement of 0.3 kg

$v_{f}$ ² = $v_{y}$ ² - 2 g y₂

0 = $v_{y}$ ² - 2 g y₂

y₂ = $v_{y}$ ² / 2g

y₂ = 33.3²/2 9.8

y₂ = 56.58 m

Total body height is

Y = y + y₂

Y = 52.7 + 56.58

Y = 109.3 m

8 0

2 years ago