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3 years ago

By what factor will the acceleration of an object change its mass stays constant but the force acting on it triples

Physics

1 answer:

Viktor [21]3 years ago

7 0

Answer:

lacceleration is given by

a=F/m

With force acting on the object remained constant ,the acceleration is inversly proportional to mass of the object i.e with increase in mass =>acceleration decreases and with decrease in mass => acceleration increases.

acceleration decreases

acceleration increase

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A ball tossed vertically upward from the ground next to a building passes the bottom of a window 1.8 s after being tossed and pa

VLD [36.1K]

The ball takes 0.2 sec to travel the height of the window, which is 2 m.

Then the speed of the ball at that time was

$v = \frac{d}{t}$

$v = \frac{2}{0.20}$

$v = 10\ \frac{m}{s}$

We know that:

$v = v_0 -gt$

Where $v_0$ is the initial velocity.

So:

$10 = v_0 -9.8(2)$

$10 + 9.8(2) = v_0$

$v_0 = 29.6\ m / s$

Then we have the equation for the position as a function of time.

$r = r_0 + v_0t - \frac{1}{2}gt^2$

Where

$r_0$ = initial position

r = position as a function of time

$v_0$ = initial velocity

g = acceleration of gravity

t = time.

If the ball is thrown from the ground then:

$r_0$ = 0 m

We want to find now the distance between the window and the ground.

When the ball reaches the bottom of the window t = 1.8s

So:

$r = 0 + 29.6(1.8) - 0.5(9.8)(1.8)^2$

$r = 37,404$ m

The window is 37,404 m high

Finally, to know how high the ball rises we must know at what moment the vertical velocity of the ball is zero.

$v = v_0 -gt\\\\0 = v_0 -gt\\\\gt = v_0\\\\t = \frac{29.6}{9.8}$

$t = 3.02\ s$

Now we replace t in the position equation

$r = 0 + 29.6(3.02) -0.5(9.8)(3.02)^2$

$r = 44.70\ m$

The ball reaches up to 44.70 m in height.

3 0

4 years ago

Two dogs are pulling a disk, but the disk is not moving. The disk does not move because the forces are (select one: unbalanced,

tiny-mole [99]

Answer:

balanced?

Explanation:

because if it wasn't moving that means they are pulling at a similar strength

8 0

3 years ago

Read 2 more answers

Consider an ideal monatomic gas of N particles with mass m in thermal equilibrium at a temperature T. The gas is contained in a

harina [27]

Answer:

$K.E.=\dfrac{3}{2}KT$

Explanation:

Given that

Number of particle =N

Equilibrium temperature= T

Side of cube = L

Gravitational acceleration =g

The kinetic energy of an atom given as

$K.E.=\dfrac{3}{2}KT$

Where

Equilibrium temperature= T

Boltzmann constant =K

K =1.380649×10−23 J/K

3 0

3 years ago

Un incendie s'est déclaré ce jeudi vers 18h30 dans un entrepôt de 2000m2 de l’usine Metalutex. Une entreprise de stockage de mat

dolphi86 [110]

Answer:

I am the bloopy blap blap man do u know english

3 0

3 years ago

You’re driving your car towards an intersection. A Porsche is stopped at the red light. You’re traveling at 36 km/h (10 m/s). As

34kurt

Your position at time $t$ , relative to the stop line:

$x_1=-15\,\mathrm m+\left(10\dfrac{\rm m}{\rm s}\right)t$

The Porsche's position:

$x_2=\dfrac12\left(3\dfrac{\rm m}{\mathrm s^2}\right)t^2$

a. You pass the Porsche immediately after the time it takes for $x_1=x_2$ :

$-15\,\mathrm m+\left(10\dfrac{\rm m}{\rm s}\right)t=\dfrac12\left(3\dfrac{\rm m}{\mathrm s^2}\right)t^2\implies t=2.3\,\rm s$

at which point you both will have traveled 7.8 m from the stop line.

b. The equation in part (a) has two solutions. The Porsche passes you at the second solution of about $t=4.4\,\rm s$ , at which point you both will have traveled 29 m.

c. At time $t$ , the Porsche is moving at velocity

$v=\left(3\dfrac{\rm m}{\mathrm s^2}\right)t$

so that at the moment it passes you, its speed is 13 m/s, which is about 46.8 km/h and below the speed limit, so neither of you will be pulled over.

6 0

4 years ago