All categories

3 years ago

Help please! Asap!! (I selected some yes/no but I don't know the answer) (Either yes or no)

Physics

2 answers:

vlabodo [156]3 years ago

5 0

Mass ... no, no. ... Drop height ... yes, yes. ... Object size ... no, no.

Yuki888 [10]3 years ago

4 0

Mass - Yes - Yes

Drop height - Yes - Yes

Initial velocity - No - No

Acceleration due to gravity - Yes - Yes

Object size - No - No

Hope this helps!

You might be interested in

In which situation is the gravitational force between two objects hard to detect? (Options)

svlad2 [7]

A - the objects are too small

GRAVITATIONAL FORCE IS EXPERIENCED BY ALL OBJECTS IN THE UNIVERSE ALL THE TIME. BUT THE ORDINARY OBJECTS YOU SEE EVERY DAY HAVE MASSES SO SMALL THAT THEIR ATTRACTION TOWARD EACH OTHER IS HARD TO DETECT. -https://www.ftsd.org/cms/lib6/MT01001165/Centricity/ModuleInstance/630/CHAPTER_2_NOTES_FOR_EIGHTH_GRADE_PHYSICAL_SCIENCE.pdf

5 0

3 years ago

Read 2 more answers

What is the most important factor for the formation of our planets

ohaa [14]

The gravitational pull of the Sun the interstellar dust attracting heat away from the protosun the process of nuclear fusion the nebular cloud condensing.

7 0

3 years ago

A certain type of laser emits light that has a frequency of 4.2 × 1014 Hz. The light, however, occurs as a series of short pulse

bogdanovich [222]

Explanation:

It is given that,

Frequency of the laser light, $f=4.2\times 10^{14}\ Hz$

Time, $t=3.2\times 10^{-11}\ s$

(a) Let $\lambda$ is the wavelength of this light. It can be calculated as :

$\lambda=\dfrac{c}{f}$

$\lambda=\dfrac{3\times 10^8}{4.2\times 10^{14}}$

$\lambda=7.14\times 10^{-7}\ m$

$\lambda=714\ nm$

(b) Let n is the number of the wavelengths in one pulse. It can be calculated as :

$n=f\times t$

$n=4.2\times 10^{14}\times 3.2\times 10^{-11}$

n = 13440

Hence, this is the required solution.

8 0

3 years ago

Force → F = ( − 8.0 N ) ˆ i + ( 6.0 N ) ˆ j acts on a particle with position vector → r = ( 3.0 m ) ˆ i + ( 4.0 m ) ˆ j . What a

andrey2020 [161]

Explanation:

It is given that,

Force, $F=(-8\ N)i+(6\ N)j$

Position vector, $r=(3i+4j)\ m$

(a) The torque on the particle about the origin is given by :

$\tau=F\times r\\\\\tau=(-8i+6j)\times (3i+4j)\\\\\tau=(-50k)\ N-m$

(b) To find the angle between r and F use dot product formula as :

$F{\cdot} r=|F||r|\ \cos\theta\\\\\cos\theta=\dfrac{F{\cdot} r}{|F| |r|}\\\\\cos\theta=\dfrac{(-8i+6j){\cdot} (3i+4j)}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=\dfrac{-24+24}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=0\\\\\theta=90^{\circ}$

Hence, this is the required solution.

8 0

3 years ago

Read 2 more answers

I just need number 2

il63 [147K]

We will apply the conservation of linear momentum to answer this question.

Whenever there is an interaction between any number of objects, the total momentum before is the same as the total momentum after. For simplicity's sake we mostly use this equation to keep track of the momenta of two objects before and after a collision:

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

Note that v₁ and v₁' is the velocity of m₁ before and after the collision.

Let's choose m₁ and v₁ to represent the bullet's mass and velocity.

m₂ and v₂ represents the wood block's mass and velocity.

The bullet and wood will stick together after the collision, so their final velocities will be the same. v₁' = v₂'. We can simplify the equation by replacing these terms with a single term v'

m₁v₁ + m₂v₂ = m₁v' + m₂v'

m₁v₁ + m₂v₂ = (m₁+m₂)v'

Let's assume the wood block is initially at rest, so v₂ is 0. We can use this to further simplify the equation.

m₁v₁ = (m₁+m₂)v'

Here are the given values:

m₁ = 0.005kg

v₁ = 500m/s

m₂ = 5kg

Plug in the values and solve for v'

0.005×500 = (0.005+5)v'

v' = 0.4995m/s

v' ≅ 0.5m/s

4 0

3 years ago