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3 years ago

6

Mr. Fineman rolls a tennis ball off his desk. If his desk is 1 m tall, and the tennis ball is rolling off of his desk at a speed

of 1.6 m/s A.) How long will it take

1 answer:

Brums [2.3K]3 years ago

3 0

Answer:

The URL you requested has been blocked

The page you requested has been blocked because it contains a banned word.

URL: https://www.scribd.com/document/359448624/Solutions-Manual-pdf

User name: 22035460

Group name: HS-Students

Explanation:i really tryed but my school blocks us from learning but heres the link https://www.scribd.com/document/359448624/Solutions-Manual-pdf

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An object of unknown mass is initially at rest and dropped from a height h. It reaches the ground with a velocity v1 . The same

Hatshy [7]

Answer:

$v_2=\sqrt{2}v_1$

Explanation:

The velocity v₁ can be calculated with the kinematic formula:

$v_1^{2} =v_0^{2} +2gh$

Since the object is initially at rest, v₁ becomes:

$v_1=\sqrt{2gh}$

Where g is the acceleration due to gravity. Now, the velocity v₂ can be calculated with the same formula, but now the initial velocity is v₁:

$v_2^{2}=v_1^{2} +2gh$

Substituting v₁ in this expression and solving for v₂, we get:

$v_2^{2}=(\sqrt{2gh} )^{2} +2gh=4gh\\\\\implies v_2=\sqrt{4gh}=2\sqrt{gh}$

Now, dividing v₂ over v₁, we get the expression:

$\frac{v_2}{v_1}=\frac{2\sqrt{gh} }{\sqrt{2gh}}=\sqrt{2}\\ \\\implies v_2=\sqrt{2}v_1$

It means that v₂ is √2 times v₁.

4 0

4 years ago

how much time would it take for an airplane to reach its destination if it traveled at an average speed of 790 km/hr for a dista

NeTakaya

I think your question is incomplete because the distance between destination and departure point isn't given in the question

8 0

3 years ago

A solid sphere of mass 8.6 kg, made of metal whose density is 3,400 kg/, hangs by a cord. When the sphere is immersed in a liqui

creativ13 [48]

Answer:

A. $1900 kg/m^3$

Explanation:

We are given that

$m=8.6 kg$

Density, $\rho_s=3400 kg/m^3$

Tension,T=38 N

We have to find the density of liquid.

$T=mg-\rho_l Vg$

$g=9.8 m/s^2$

Volume,V= $\frac{m}{\rho_s}$

$38=8.6\times 9.8-\rho_l\times \frac{8.6}{3400}\times 9.8$

$\rho_l\times \frac{8.6}{3400}\times 9.8=8.6\times 9.8-38$

$\rho_l=\frac{(8.6\times 9.8-38)\times 3400}{8.6\times 9.8}$

$\rho_l=1867kg/m^3\approx 1900 kg/m^3$

Option A is true.

4 0

3 years ago

Which phrase best describes wave motion

White raven [17]

Answer:

I would describe wave motion as “propagating oscillations” as a general phrase.

For something like water waves, I would say “water waves are oscillations of the surface of a body of water that propagate in a given direction.”

For electromagnetic waves I would say “An electromagnetic wave is an oscillating disturbance of the electric and magnetic fields that propagates through space.”

Explanation:

PLZ MARK BRAINLIEST

4 0

3 years ago

Read 2 more answers

A mass of 1000 kg drops from a height of 10 m on a platform of negligible mass. It is desired to design a spring and dashpot on

Juliette [100K]

Answer:

k = $5\times 10^{4}\ N/m$

b = $0.707\times 10^{3}$

t = $7.1\times 10^{- 5}\ s$

Solution:

As per the question:

Mass of the block, m = 1000 kg

Height, h = 10 m

Equilibrium position, x = 0.2 m

Now,

The velocity when the mass falls from a height of 10 m is given by the third eqn of motion:

$v^{2} = u^{2} + 2gh$

where

u = initial velocity = 0

g = 10 $m/s^{2}$

Thus

$v = \sqrt{2\times 10\times 10} = 10\sqrt{2}\ m/s$

Force on the mass is given by:

F = mg = $1000\times 10 = 10000 N = 10\ kN$

Also, we know that the spring force is given by:

F = - kx

Thus

$k = \frac{F}{x} = \frac{10000}{0.2} = 5\times 10^{4}\ N/m$

Now, to find the damping constant b, we know that:

F = - bv

$b = \frac{F}{v} = \frac{10000}{10\sqrt{2}} = 0.707\times 10^{3}$

Now,

Time required for the platform to get settled to 1 mm or 0.001 m is given by:

$t = \frac{0.001}{v} = \frac{0.001}{10\sqrt{2}} = 7.1\times 10^{- 5}\ s$

4 0

3 years ago