Physics, 13.04.2021 05:20 AmityHeart

What is the force holding you down?

Answers: 2

Show answers

Answers

Answer from: quynhnguyen7454

The force holding you down is gravity.

Explanation:

Gravity is a force between two objects with mass. It pulls things together. You have mass, and the Earth has mass, so gravity tries to pull you and the Earth together. The gravitational force is much bigger for more massive objects.

Answer from: haleybain6353

Gravitational Force

Explanation:

Gravitational Force holding everything down

-TheUnknownScientist

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

Answer from: Quest

Initial velocity, u = 4.9 m/sfinal velocity = v m/sinitial height, h₁ = 245 m final height, h₂ = 0a = -9.8 m/s²time taken = t secondss = ut + 0.5at²⇒ (h₂-h₁) = ut + 0.5at²⇒ 0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get t = 7.59sv = u + at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is 69.48 m/s downward

Another question on Physics

Physics, 21.06.2019 21:40

Avery hard rubber ball (m = 0.5 kg) is falling vertically at 4 m/s just before it bounces on the floor. the ball rebounds back at essentially the same speed. if the collision with the floor lasts 0.05 s, what is the average force exerted by the floor on the ball?

Answers: 3

Answer

Physics, 22.06.2019 10:30

Avery long solenoid of inner radius 2.75 cm creates an oscillating magnetic field of the form = for this solenoid, =0.00425 t and =319 rad/s. (a)what is the maximum value of the induced electric field at a perpendicular distance 1.45 cm from the axis of the solenoid? (b) what is the maximum value of the inducted electric field at a point 5.85 cm perpendicular from the axis of the solenoid?

Answers: 1

Answer

Physics, 22.06.2019 16:00

At what time will the interfering waves have an amplitudeof zero?

Answers: 1

Answer

Physics, 22.06.2019 18:00