Pooping in my room and my room is upstairs and upstairs bathroom upstairs

An object moves in uniform circular motion what is true regarding the force on the object

PtichkaEL [24]

The force on the object has a constant strength, but its direction
keeps changing. The force is always directed from the object to
the center of the circle. It's called "centripetal force".

4 0

4 years ago

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Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 32.0°below the horizontal. If it st

scoray [572]

The figure of the problem is missing: find in attachment.

(a) 1.64 s

The ball follows a projectile motion path. The horizontal displacement is given by

$x(t) = v_0 cos \theta t$

where

$v_0$ is the initial speed

t is the time

$\theta=32.0^{\circ}$ is the angle below the horizontal

We can rewrite this equation as

$t=\frac{x(t)}{v_0 cos \theta}$ (1)

The vertical displacement instead is given by

$y(t) = -v_0 sin \theta t - \frac{1}{2}gt^2$ (2)

where

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

Substituting (1) into (2),

$y(t) = -x(t) tan \theta - \frac{1}{2}gt^2$

We know that for t = time of flight, the horizontal displacement is

$x(t) =50.8 m$

We also know that the vertical displacement is

$y(t) = -45 m$

Substituting everything into the equation, we can find the time of flight:

$\frac{1}{2}gt^2=-y -x tan \theta\\t=\sqrt{\frac{2(-y-xtan \theta)}{g}}=\sqrt{\frac{2(-(-45)-50.8 tan 32.0^{\circ})}{9.8}}=1.64 s$

(b) 36.5 m/s

We can now find the initial speed directly by using the equation for the horizontal displacement:

$x(t) = v_0 cos \theta t$

where we have

x = 50.8 m

$\theta=32.0^{\circ}$

Substituting the time of flight,

t = 1.64 s

We find:

$v_0 = \frac{x}{t cos \theta}=\frac{50.8}{(1.64)(cos 32.0^{\circ})}=36.5 m/s$

(c) 47.1 m/s at 48.8 degrees below the horizontal

As the ball follows a projectile motion, its horizontal velocity does not change, so its value remains equal to

$v_x = v_0 cos \theta = (36.5)(cos 32.0^{\circ})=31.0 m/s$

The initial vertical velocity is instead

$u_y = -v_0 sin \theta = -(36.5)(sin 32.0^{\circ})=-19.3 m/s$

And it changes according to the equation

$v_y = u_y -gt$

So at t = 1.64 s (when the ball hits the ground),

$v_y = -19.3 - (9.8)(1.64)=-35.4 m/s$

So the impact speed is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(31.0)^2+(-35.4)^2}=47.1 m/s$

While the direction is:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-35.4}{31.0})=-48.8^{\circ}$

8 0

3 years ago

Distinguish between atoms and molecules,

Lisa [10]

An atom is the smallest particle of an element which can take part in a chemical reaction

A molecule is the smallest particle of a substance that can normally exist alone and still retain the chemical properties of that substance, Be it an element or a compound.

Isotopic mass is the mass of an isotope of an element

Atomic mass is the mass of an atom.The relative atomic mass,A of an element is the number of times the average mass of one atom of that element is heavier than one-twelfth the mass of one atom of carbon-12

Molecular mass is the mass of a given molecule. The relative molecular mass,M, of an element or a compound is the number of times the average mass of one molecule of it is heavier than one-twelfth the mass of one atom of carbon-12

One mole of a substance is the amount containing as many elementary entities as the number of atoms in exactly 12 grams of Carbon-12 .

The molar mass of any substance is the mass of one mole of that substance expressed in grams

Hope I helped
Please mark brainliest

7 0

3 years ago

After the box comes to rest at position x₁, a person starts pushing the box, giving it a speed v₁. When the box reaches position

Vesna [10]

Answer:

$W_p = \frac{1}{2}mv_1^2$

Explanation:

As we know that box is initially at rest

So we will have

$v_i = 0$

now as it will be displaced from initial position to final position then final speed of the box is reached to

$v_f = v_1$

now we know by work energy theorem that work done by all the forces on the box will be equal to the change in kinetic energy

So here we will have

$W_p = \frac{1}{2}m(v_f^2 - v_i^2)$

$W_p = \frac{1}{2}m(v_1^2 - 0)$

$W_p = \frac{1}{2}mv_1^2$

8 0

4 years ago

Any force that causes an object to move in a circle is called a(n)

Vaselesa [24]

A). balanced force
b). unbalanced force
There's no such thing as either of these. A group of two or more forces can be balanced or unbalanced. A single force can't be.

c). gravitational force ... doesn't cause an object to move in a circle;
   Drop a stone from the roof of a tall building and watch it fall.
   It goes straight down, not in a circle.

d). centripetal force ... force directed toward the center of a circle,
   causes an object to move in a circle.

8 0

3 years ago

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