All categories

3 years ago

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

Physics

2 answers:

svlad2 [7]3 years ago

5 0

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

Anit [1.1K]3 years ago

4 0

Answer:

Hence, the correct option is (A) "If the mass of each object is very small".

Explanation:

The gravitational force between two objects is directly proportional to the product of masses and inversely proportional to the square of the distance between them.

The mathematical expression is given by :

$F=G\dfrac{m_1m_2}{d^2}$

d is the distance between masses

Gravitational force is one of the fundamental force in the universe. If the masses of objects is too small, then their interaction with each other is hard to detect.

Hence, the correct option is (A) "If the mass of each object is very small".

You might be interested in

True or False: The arrows on a motion map should point in the directions of motion.

slavikrds [6]

Answer:

true,true,false

Explanation:

its false because if it is equal it would show an arrow pointing left and a 20 and the same for the right

3 0

2 years ago

Find the acceleration that can result from a net force of 13 N exerted on a 3.6-kg cart. (Note: The unit N/kg is equivalent to m

STatiana [176]

Answer:

a = 3.61[m/s^2]

Explanation:

To find this acceleration we must remember newton's second law which tells us that the total sum of forces is equal to the product of mass by acceleration.

In this case we have:

$F = m*a\\\\m=mass = 3.6[kg]\\F = force = 13[N]\\13 = 3.6*a\\a = 3.61[m/s^2]$

7 0

2 years ago

If the forces acting on an object are balanced, the object may change speed, change direction, or do both. Please select the bes

Leviafan [203]

False is ur answer sorry if i am too late

8 0

3 years ago

Read 2 more answers

The pupil of a cat's eye narrows to a slit width of 0.5 mm in daylight. What is the angular resolution of the cat's eye in dayli

Elenna [48]

Answer:

C. 10⁻³ rads

Explanation:

Here, we shall use Rayleigh's Criterion to find out the angular resolution of Cat's eye during day light. Rayleigh's Criterion is written as follows:

θ = λ/a

where,

θ = angular resolution of Cat's eye = ?

λ = wavelength = 500 nm = 5 x 10⁻⁷ m

a = slit width of eye = 0.5 mm = 5 x 10⁻⁴ m

Therefore,

θ = (5 x 10⁻⁷ m/5 x 10⁻⁴ m)

Therefore,

θ = 0.001

θ = Sin⁻¹(0.001)

θ = 0.001 rad = 1 x 10⁻³ rad

Hence, the correct answer is:

4 0

2 years ago

A tennis ball is dropped from a height of 1.21m above the ground. Calculate its velocity when it is 0.27m from the ground.

wlad13 [49]

9.81m/s^2 x .27 = 2.6487m/s

8 0

2 years ago