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Svetradugi [14.3K]
3 years ago
14

A large heavy rock, a medium size stone and a small marble are dropped at the same time from the roof of a building. Neglecting

air resistance, which object will strike the ground first?
Physics
2 answers:
KengaRu [80]3 years ago
6 0
They would all strike the ground at the same time. If air resistance is neglected for example in an airless tube, then gravity is the only force acting on them and it has the same effect on them dropping so the drop at the same speed.
lutik1710 [3]3 years ago
4 0
I believe that they would all hit at the same time due to the force of gravity actuating on them. All things fall at the same speed I believe.
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A 9.4 kg dog takes a nap in a canoe and wakes up to find the canoe has drifted out onto the lake but now is stationary. He walks
bekas [8.4K]

Answer:

28.2 kg

Explanation:

Data provided in the question:

Mass of dog, m = 9.4 kg

Velocity of the dog with respect to water, v = 0.48 m/s

Velocity of the canoe, V = - 0.16 m/s          [ - sign means velocity in opposite direction]

Now,

Let the mass of canoe be 'M'

Therefore,

From the conservation of momentum, we have

⇒ mv + MV = 0

or

⇒ 9.4 × 0.48 + M(- 0.16) = 0

or

⇒ 4.512 - 0.16M = 0

or

⇒ 0.16M = 4.512

or

⇒ M = 28.2 kg

7 0
3 years ago
A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow that exerts a fricti
Ugo [173]

Use Newton's second law to determine the acceleration being applied to the sled. There are three forces at work on the sled (its weight, the force normal to the ground, and friction) but two of them cancel, leaving friction as the only effective force. This vector is pointed in the opposite direction of the sled's movement, so if we take the direction of its movement to be the positive axis, we would find the acceleration due to the friction to be

\vec F_G+\vec F_N+\vec F_F=m\vec a\iff-12\,\mathrm N=(20\,\mathrm{kg})a\implies a=-0.6\,\dfrac{\rm m}{\mathrm s^2}

Now we use the formula

{v_f}^2-{v_i}^2=2a(x_f-x_i)

to find the distance it travels. The sled comes to a rest, so v_f=0, and let's take the starting position x_i=0 to be the origin. Then the distance traveled x_f-x_i=x_f is

-\left(4.5\,\dfrac{\rm m}{\rm s}\right)^2=2\left(-0.6\,\dfrac{\rm m}{\mathrm s^2}\right)x_f\implies x_f\approx17\,\mathrm m

6 0
3 years ago
A light spring of constant 179 N/m rests vertically on the bottom of a large beaker of water. A 5.32 kg block of wood of density
Digiron [165]

Answer:

Compression of the spring: 0.18 m (downward)

Explanation:

The forces acting on the block of wood are:

- The force of gravity, acting downward, of magnitude mg, where m = 5.32 kg is the mass of the block and g=9.8 m/s^2 is the acceleration due to gravity

- The force exerted by the spring, downward, of magnitude kx, where k=179N/m is the spring constant and x is the elongation of the spring

- The buoyant force, upward, of magnitude \rho V g, where \rho=1000 kg/m^3 is the water density and V the volume of the block

Since the block is in equilibrium, the net force is zero, so we can write

mg+kx-\rho V g=0 (1)

We have to find the volume of the block first. We have:

m = 5.32 kg (mass)

\rho_w = 622 kg/m^3 (wood density)

So, the volume is

V=\frac{m}{\rho_w}=\frac{5.32}{622}=0.0086 m^3

So now we can re-arrange eq.(1) to find the elongation of the spring, x:

x=\frac{-mg+\rho Vg}{k}=\frac{-(5.32)(9.8)+(1000)(0.0086)(9.8)}{179}=0.18 m

So, the spring is compressed by 0.18 m.

7 0
3 years ago
Why do you think ionic bonds are generally weaker than covalent bonds?
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5 0
3 years ago
Is aluminum foil reflecting onto something conduction, convection, or radiation?
mojhsa [17]
I had the SAME problem, put down Radiation and it’s thermal/light.
4 0
3 years ago
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