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4 years ago

An object in motion will stay in motion unless acted upon by what type of force?

Physics

1 answer:

Travka [436]4 years ago

3 0

Answer:

Explanation:

The First Law of Motion states, "A body at rest will remain at rest, and a body in motion will remain in motion unless it is acted upon by an external force." This simply means that things cannot start, stop, or change direction all by themselves

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I drop an egg from a certain distance and it takes the egg 3.74 seconds to reach the ground. How high up was the egg?

Slav-nsk [51]

Answer:

The height from which the egg is dropped is <u>68.54 m</u>.

Explanation:

Given:

Initial velocity of egg is, $u=0\ m/s$ (Dropped means initial velocity is 0)

Time taken is, $t=3.74\ s$

Acceleration experienced by egg is due to gravity, $a=g=9.8\ m/s^2$

The height from which the egg is dropped is, $d=?$

Now, we use Newton's equation of motion that relates the distance, initial velocity, time and acceleration.

So, we have the following equation of motion:

$d=ut+\frac{1}{2}at^2$

Plug in all the given values and solve for 'd'. This gives,

$d=0+\frac{1}{2}\times 9.8\times (3.74)^2\\\\d=\frac{1\times 9.8\times 13.9876}{2}\\\\d=\frac{137.078}{2}\\\\d=68.54\ m$

Therefore, the height from which the egg is dropped is 68.54 m.

8 0

3 years ago

An airline employee tosses two suitcases in rapid succession with a horizontal velocity of 7.2 ft/s onto a 50-lb baggage carrier

zalisa [80]

Answer:

m₁ = 70 lb

Explanation:

Here we will use the law of conservation of momentum:

m₁u₁ + m₂u₂ + m₃u₃ = m₁v₁ + m₂v₂ + m₃v₃

where,

m₁ = mass of first suitcase = ?

m₂ = mass of second suitcase = 30 lb

m₃ = mass of baggage carrier = 50 lb

u₁ = initial speed of first suitcase = 7.2 ft/s

u₂ = initial speed of second suitcase = 7.2 ft/s

u₃ = initial speed of baggage carrier = 0 ft/s

v₁ = Final speed of first suitcase = 4.8 ft/s

v₂ = Final speed of second suitcase = 4.8 ft/s

v₃ = Final speed of baggage carrier = 4.8 ft/s

because after collision all three will have same speed

Therefore,

(m₁)(7.2 ft/s) + (30 lb)(7.2 ft/s) + (50 lb)(0 ft/s) = (m₁)(4.8 ft/s) + (30 lb)(4.8 ft/s) + (50 lb)(4.8 ft/s)

(m₁)(7.2 ft/s) + (216 lb ft/s) + (0 lb ft/s) = (m₁)(4.8 ft/s) + (144 lb ft/s) + (240 lb ft/s)

(m₁)(7.2 ft/s) - (m₁)(4.8 ft/s) = 168 lb ft/s

m₁ = (168 lb ft/s)/(2.4 ft/s)

6 0

3 years ago

A sphere of mass m" = 2 kg travels with a velocity of magnitude υ") = 8 m/s toward a sphere of mass m- = 3 kg initially at rest,

aleksklad [387]

a) 6.4 m/s

b) 2.1 m

c) $61.6^{\circ}$

d) 14.0 N

e) 4.6 m/s

f) 37.9 N

Explanation:

Since the system is isolated (no external forces on it), the total momentum of the system is conserved, so we can write:

$p_i = p_f\\m_1 u_1 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 2 kg$ is the mass of the 1st sphere

$m_2 = 3kg$ is the mass of the 2nd sphere

$u_1 = 8 m/s$ is the initial velocity of the 1st sphere

$v_1$ is the final velocity of the 1st sphere

$v_2$ is the final velocity of the 2nd sphere

Since the collision is elastic, the total kinetic energy is also conserved:

$E_i=E_k\\\frac{1}{2}m_1 u_1^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

Combining the two equations together, we can find the final velocity of the 2nd sphere:

$v_2=\frac{2m_1}{m_1+m_2}u_1=\frac{2(2)}{2+3}(8)=6.4 m/s$

Now we analyze the 2nd sphere from the moment it starts its motion till the moment it reaches the maximum height.

Since its total mechanical energy is conserved, its initial kinetic energy is entirely converted into gravitational potential energy at the highest point.

So we can write:

$KE_i = PE_f$

$\frac{1}{2}mv^2 = mgh$

where

m = 3 kg is the mass of the sphere

v = 6.4 m/s is the initial speed of the sphere

$g=9.8 m/s^2$ is the acceleration due to gravity

h is the maximum height reached

Solving for h, we find

$h=\frac{v^2}{2g}=\frac{(6.4)^2}{2(9.8)}=2.1 m$

Here the 2nd sphere is tied to a rope of length

L = 4 m

We know that the maximum height reached by the sphere in its motion is

h = 2.1 m

Calling $\theta$ the angle that the rope makes with the vertical, we can write

$h = L-Lcos \theta$

Which can be rewritten as

$h=L(1-cos \theta)$

Solving for $\theta$ , we can find the angle between the rope and the vertical:

$cos \theta = 1-\frac{h}{L}=1-\frac{2.1}{4}=0.475\\\theta=cos^{-1}(0.475)=61.6^{\circ}$

The motion of the sphere is part of a circular motion. The forces acting along the centripetal direction are:

- The tension in the rope, T, inward

- The component of the weight along the radial direction, $mg cos \theta$ , outward

Their resultant must be equal to the centripetal force, so we can write:

$T-mg cos \theta = m\frac{v^2}{r}$

where r = L (the radius of the circle is the length of the rope).

However, when the sphere is at the highest point, it is at rest, so

v = 0

Therefore we have

$T-mg cos \theta=0$

So we can find the tension:

$T=mg cos \theta=(3)(9.8)(cos 61.6^{\circ})=14.0 N$

We can solve this part by applying again the law of conservation of energy.

In fact, when the sphere is at a height of h = 1 m, it has both kinetic and potential energy. So we can write:

$KE_i = KE_f + PE_f\\\frac{1}{2}mv^2 = \frac{1}{2}mv'^2 + mgh'$

where:

$KE_i$ is the initial kinetic energy

$KE_f$ is the kinetic energy at 1 m

$PE_f$ is the final potential energy

v = 6.4 m/s is the speed at the bottom

v' is the speed at a height of 1 m

h' = 1 m is the height

m = 3 kg is the mass of the sphere

And solving for v', we find:

$v'=\sqrt{v^2-2gh'}=\sqrt{6.4^2-2(9.8)(1)}=4.6 m/s$

Again, since the sphere is in circular motion, the equation of the forces along the radial direction is

$T-mg cos \theta = m\frac{v^2}{r}$

where

T is the tension in the string

$mg cos \theta$ is the component of the weight in the radial direction

$m\frac{v^2}{r}$ is the centripetal force

In this situation we have

v = 4.6 m/s is the speed of the sphere

$cos \theta$ can be rewritten as (see part c)

$cos \theta = 1-\frac{h'}{L}$

where in this case,

h' = 1 m

L = 4 m

And $r=L=4 m$ is the radius of the circle

Substituting and solving for T, we find:

$T=mg cos \theta + m\frac{v^2}{r}=mg(1-\frac{h'}{L})+m\frac{v^2}{L}=\\=(3)(9.8)(1-\frac{1}{4})+(3)\frac{4.6^2}{4}=37.9 N$

4 0

3 years ago

What is the process when beach sediment moves down the beach with the current?

Solnce55 [7]

The answer is:
________________________________________
"longshore drift" .
___________________________________________

5 0

4 years ago

Characteristics

MAXImum [283]

The answer is B. Comet

6 0

4 years ago

An object in motion will stay in motion unless acted upon by what type of force?​

An object in motion will stay in motion unless acted upon by what type of force?