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3 years ago

5

What is the force the when does when Gravity pushes you

2 answers:

BARSIC [14]3 years ago

6 0

Answer:

Image result for what is the force the when does when Gravity pushes you

The important thing to remember is that gravity is neither a push nor a pull; what we interpret as a “force” or the acceleration due to gravity is actually the curvature of space and time — the path itself stoops downward.

Explanation:

Image result for what is the force the when does when Gravity pushes you

The important thing to remember is that gravity is neither a push nor a pull; what we interpret as a “force” or the acceleration due to gravity is actually the curvature of space and time — the path itself stoops downward.

Nezavi [6.7K]3 years ago

5 0

Answer:

It is called gravitational force.

Explanation:

${ \boxed{ \rm{F = \frac{GM_{a} m _{a}}{ {r}^{2} } }}}$

F is the gravitational force
Ma is the mass of a planetary body
ma is mass lf body being attracted
r is the separation distance
G is the universal gravitational constant

→ Assume you are in space and you have a mass of ma, and you are r metres from the Earth of mass Ma. The force F will be experienced and you will tend to be pulled towards the Earth when you are in its gravitational field.

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A spring suspended vertically is 11.9 cm long. When you suspend a 37 g weight from the spring, at rest, the spring is 21.5 cm lo

Naddika [18.5K]

Answer:

Time period of oscillations is 0.62 s

Explanation:

Due to suspension of weight the change in the length of the spring is given as

$\Delta L = L_f - L_i$

$\Delta L = 21.5 - 11.9 = 9.6 cm$

now we know that spring is stretched due to its weight so at equilibrium the force due to weight is counter balanced by the spring force

$mg = kx$

$0.037 (9.81) = k(0.096)$

$k = 3.78 N/m$

Now the period of oscillation of spring is given as

$T = 2\pi \sqrt{\frac{m}{k}}$

Now plug in all values in it

$T = 2\pi \sqrt{\frac{0.037}{3.78}}$

$T = 0.62 s$

5 0

3 years ago

g You shine orange laser light that has a wavelength of 600 nm through a narrow slit. The slit forms a diffraction pattern on a

zimovet [89]

Answer:

λ = 3 10⁻⁷ m, UV laser

Explanation:

The diffraction phenomenon is described by the expression

a sin θ = m λ

let's use trigonometry

tan θ = y / L

as in this phenomenon the angles are small

tan θ = $\frac{sin \ \theta}{cos \ \theta}$ = sin θ

sin θ = y / L

we substitute

a y / L = m λ

let's apply this equation to the initial data

a 0.04 / L = 1 600 10⁻⁹

a / L = 1.5 10⁻⁵

now they tell us that we change the laser and we have y = 0.04 m for m = 2

a 0.04 / L = 2 λ

a / L = 50 λ

we solve the two expression is

1.5 10⁻⁵ = 50 λ

λ = 1.5 10⁻⁵ / 50

λ = 3 10⁻⁷ m

UV laser

3 0

3 years ago

Two people are trying to pull a 5-kilogram block in opposite directions. The first person pulls to the right with a force of 25

klasskru [66]

TEST TEST TEST TEST TEST TEST TEST

7 0

3 years ago

A jogger hears a car alarm and decides to investigate. While running toward the car, she hears an alarm frequency of 872.10 Hz.

Nata [24]

Answer:

v = 4.18 m/s

Explanation:

given,

frequency of the alarm = 872.10 Hz

after passing car frequency she hear = 851.10 Hz

Speed of sound = 343 m/s

speed of the jogger = ?

speed of the

$v_f = \dfrac{872.10-851.10}{2}$

$v_f =10.5\ Hz$

v_o = 872.1 - 10.5

$V_0 = 861.6\ Hz$

The speed of jogger

$v = \dfrac{v_1 \times 343}{v_0}-343$

$v = \dfrac{872.1 \times 343}{861.6}-343$

v = 4.18 m/s

5 0

3 years ago

Readme [11.4K]

Answer:

I got you.. i'm in middle school and had that same question.

Explanation:

Refer to the diagram shown below.

The vertical distance traveled is

s = 25 m

The initial vertical launch velocity is zero.

Therefore

s = (1/2)*g*t²

where g = 9.8 m/s²

t = the time of flight, s

That is,

0.5*9.8*t² = 25

t² = 25/4.9 = 5.102

t = 2.26 s

Answer: 2.26 s

3 0

3 years ago