All categories

4 years ago

Have any one loved someone?

Physics

2 answers:

Elena-2011 [213]4 years ago

5 0

Yes many different people have loved someone

fredd [130]4 years ago

3 0

yes and owwwwwwwww did i love it sooo much till it was gone...............................................................PIZZA

You might be interested in

2. A 2kg body attached to a spring undergoes a SHM of amplitude 0.4m and period

Rudik [331]

Answer:

a) 12.8 N

b) 3.2 m/s²

Explanation:

I'm guessing the period is 0.5π s.

Period of a spring in simple harmonic motion is:

T = 2π √(m/k)

Given T = 0.5π and m = 2 kg:

0.5π = 2π √(2/k)

0.25 = √(2/k)

0.0625 = 2/k

k = 32

The spring constant is 32 N/m, and the maximum displacement is 0.4 m. The maximum force can be found with Hooke's law:

F = kx

F = (32 N/m) (0.4 m)

F = 12.8 N

The acceleration can be found with Newton's second law:

∑F = ma

kx = ma

(32 N/m) (0.2 m) = (2 kg) a

a = 3.2 m/s²

8 0

4 years ago

The elements in Group 2 have similar chemical properties because each atom of these elements has the same

4vir4ik [10]

Number of electrons in the outer energy shell

4 0

3 years ago

Calculate the force of gravitation due to Earth on A ball of 1 kg mass lying on the ground

sergejj [24]

Mass=1kg
Acceleration due to gravity=9.8m/s^2

Using newtons second law

$\\ \sf\longmapsto Force=Mass\times Acceleration$

$\\ \sf\longmapsto Force=1(9.8)$

$\\ \sf\longmapsto Force=9.8N$

3 0

3 years ago

When a cell divides to help repair an organism, what type of cell division is it?

marin [14]

Answer:

Mitosis

Explanation:

Mitosis is the process of making new body cells

4 0

4 years ago

The following questions present a twist on the scenario above to test your understanding. Suppose another stone is thrown horizo

Ipatiy [6.2K]

The first part of the text is missing, you can find on google:

"A ball is thrown horizontally from the roof of a building 45 m. If it strikes the ground 56 m away, find the following values."

Let's now solve the different parts.

(a) 3.03 s

The time of flight can be found by analyzing the vertical motion only. The vertical displacement at time t is given by

$y(t) = h -\frac{1}{2}gt^2$

where

h = 45 m is the initial height

g = 9.8 m/s^2 is the acceleration of gravity

When y=0, the ball reaches the ground, so the time taken for this to happen can be found by substituting y=0 and solving for the time:

$0=h-\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(45)}{9.8}}=3.03 s$

(b) 18.5 m/s

For this part, we need to analyze the horizontal motion only, which is a uniform motion at constant speed.

The horizontal position is given by

$x=v_x t$

where

$v_x$ is the horizontal speed, which is constant

t is the time

At t = 3.03 s (time of flight), we know that the horizontal position is x = 56 m. By substituting these numbers and solving for vx, we find the horizontal speed:

$v_x = \frac{x}{t}=\frac{56}{3.03}=18.5 m/s$

The ball was thrown horizontally: this means that its initial vertical speed was zero, so 18.5 m/s was also its initial overall speed.

(c) 35.0 m/s at 58.1 degrees below the horizontal

At the impact, we know that the horizontal speed is still the same:

$v_x = 18.5 m/s$

we need to find the vertical velocity. This can be done by using the equation

$v_y = u_y -gt$

where

$u_y =0$ is the initial vertical velocity

g is the acceleration of gravity

t is the time

Substituting t = 3.03 s, we find the vertical velocity at the time of impact:

$v_y = -(9.8)(3.03)=-29.7 m/s$

So the magnitude of the velocity at the impact (so, the speed at the impact) is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.5^2+(-29.7)^2}=35.0 m/s$

The angle instead can be found as:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-29.7}{18.5})=-58.1^{\circ}$

so, 58.1 degrees below the horizontal.

4 0

3 years ago