Please help with any of them!!!!!!

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

Brums [2.3K]

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

3 0

3 years ago

the tallest man made structure At present is theWarszawa radio massin Warsaw Poland.This radio mast rises 646m above ground near

PtichkaEL [24]

All of that fluff at the beginning is interesting, but completely irrelevant
to the question. The question is just asking for the mass of an object
that weighs 3.6N on Earth.

Weight = (mass) x (acceleration of gravity)

3.6N = (mass) x (9.8 m/s²)

Divide each side
by 9.8 m/s : Mass = 3.6N / 9.8 m/s² = <em>0.367 kilogram</em> (rounded)

7 0

3 years ago

If a certain shade of blue has a frequency of 7.33 x 10^14Hz what is the energy of exactly one photon of this light. Is this as

geniusboy [140]

In calculating the energy of a photon of light, we need the relationship for energy and the frequency which is expressed as:

E=hv

where h is the Planck's constant (6.626 x 10-34 J s)and v is the frequency.
E = 6.626 x 10-34 J s (<span>7.33 x 10^14 /s) = 4.857 x 10^-19 J</span>

6 0

4 years ago

Ayuda por favor cual es el objetivo del ultimate?

postnew [5]

Answer: i d k

Explanation:

3 0

3 years ago

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

3 years ago