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3 years ago

A spring toy jumps up from the floor and comes back down to the floor. Its initial speed is 12 m/s. What is the

Physics

1 answer:

dybincka [34]3 years ago

8 0

Answer:

$0\; \rm m \cdot s^{-1}$ .

Explanation:

In a projectile motion, the kinetic energy (KE) of the projectile is converted to gravitational potential energy (GPE) and then from GPE back to KE.

In this example, the spring toy is the projectile. Refer to the diagram attached.

The spring toy started with a velocity of $12\; \rm m\cdot s^{-1}$ , meaning that its initial KE is non-zero.
On the way up, the KE of this spring toy gets converted to GPE.
At the top of the trajectory, the GPE of this toy is maximized while its KE is minimized (zero).
As the toy returns to the ground, the GPE of this toy gets converted back to KE.
The GPE of this toy is zero when the toy is on the ground.

In other words, the kinetic energy (KE) of this toy would be $0$ when it is at the top of the trajectory and GPE is maximized.

Since the KE of this toy at the top of the trajectory would be $0$ , the velocity of this toy at that moment would also be $0\!$ .

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A car moves with an initial velocity of 19 m/s due north. (Part A) Find the velocity of the car after 5.6 s if its acceleration

galben [10]

Answer

Assuming

east is the positive x direction

north is the positive y direction

initial velocity , u = 19 j m/s

a) acceleration , a = 1.6 j m/s^2

Using first equation of motion

v = u + a × t

v = 19 + 5.6× 1.6

v = 28 j m/s

the velocity of the car after 5.6 s is 28 m/s north

acceleration , a = -1.5 j m/s^2

Using first equation of motion

v = u + a × t

v = 19 - 5.6 ×1.5

v = 10.6 j m/s

the velocity of the car after 5.6 s is 10.6 m/s north

5 0

3 years ago

An air balloon is moving upward at a constant speed of 3 m/s. Suddenly a passenger realizes that she left her camera on the grou

frosja888 [35]

Answer:t=0.3253 s

Explanation:

Given

speed of balloon is $u=3\ m/s$

speed of camera $u_1=20\ m/s$

Initial separation between camera and balloon is $d_o=5\ m$

Suppose after t sec of throw camera reach balloon then,

distance travel by balloon is

$s=ut$

$s=3\times t$

and distance travel by camera to reach balloon is

$s_1=ut+\frac{1}{2}at^2$

$s_1=20\times t-\frac{1}{2}gt^2$

Now

$\Rightarrow s_1=5+s$

$\Rightarrow 20\times t-\frac{1}{2}gt^2 =5+3t$

$\Rightarrow 5t^2-17t+5=0$

$\Rightarrow t=\dfrac{17\pm \sqrt{17^2-4(5)(5)}}{2\times 5}$

$\Rightarrow t=\dfrac{17\pm 13.747}{10}$

$\Rightarrow t=0.3253\ s\ \text{and}\ t=3.07\ s$

There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .

(b)When passenger catches the camera time is $t=0.3253\ s$

velocity is given by

$v=u+at$

$v=20-10\times 0.3253$

$v=16.747\ m/s$

and position of camera is same as of balloon so

Position is $=5+3\times 0.3253$

$=5.975\approx 6\ m$

8 0

3 years ago

A. Compare solids, liquids, and gases in terms of the way their volumes change (or don't change) when placed into different cont

svet-max [94.6K]

a) Solids keep shape, liquids take shape of containers but don't spill, gases take container's shape and spill out

b) if you heat gas, speed of its molecules will increase and they'll push the container's walls stronger, so the pressure will increase when the container heated

c) Heat flows from warmer to colder bodies

d) For monatomic gases it's U=1.5nRT only, molecular gas has bonds between atoms so total internal energy increases

e) Of gases

f) $v=\sqrt{\frac{3RT}{M}}=321$ m/s