All categories

2 years ago

What is the kinetic energy of a 1 kilogram ball is thrown into the air with an initial velocity of 30 m/sec

1 answer:

5 0

To find this we should use the equation Ek=1/2mv^2 where Ek is the kinetic energy, m is the mass and v is the velocity. So by this Ek=1/2*1*30^2 which then makes Ek=450 joules

You might be interested in

Consider an old-fashion bicycle with a small wheel of radius 0.17 m and a large wheel of radius 0.92 m. Suppose the rider starts

Gala2k [10]

Answer:

10259.6 m

Explanation:

We are given that

Radius of small wheel,r=0.17 m

Radius of large wheel,r'=0.92 m

Initial velocity,u=0

Time,t=2.7 minutes=162 s

1 min=60 s

Velocity,v=10m/s

Time,t'=13.7 minutes=822 s

Time,t''=4.1 minutes=246 s

$v=u+at$

Substitute the values

$10=0+162a=162a$

$a=\frac{10}{162}=0.0617m/s^2$

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

Substitute the values

$s=\frac{1}{2}(0.0617)(162)^2=809.6 m$

$s'=vt'=10\times 822=8220 m$

$a'=\frac{v}{t''}=\frac{10}{246}$

$s''=\frac{1}{2}a't''^2=\frac{1}{2}\times \frac{10}{246}(246)^2=1230 m$

Total distance traveled by rider=s+s'+s''=809.6+8220+1230=10259.6 m

6 0

3 years ago

Read 2 more answers

Types of mechanical waves include

USPshnik [31]

I think surface waves

4 0

3 years ago

Heat is transfered from the heating elements to the pot

Varvara68 [4.7K]

Hear is transferred from the heating elements to the Pot by Conductivity

8 0

3 years ago

Describe the octet rule in terms of noble-gas configurations and potential energy

Wittaler [7]

Elements will gain or lose electrons to form a noble gas configuration. Atoms that meet the octet rule are stable because their valence electrons have a relatively low potential energy. Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

8 0

2 years ago

Read 2 more answers

The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is

stealth61 [152]

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

$\rho_m = 846lb/ft^3$