10. John does 176 J of work lifting himself a distance of 0.40 m. How

Tyson throws a shot put ball weighing 7.26 kg. At a height of 2.1 m above the ground, the mechanical energy of the ball is 172.1

max2010maxim [7]

Answer:

2.5 m/s

Explanation:

Mechanical energy is the sum of the potential and kinetic energy.

E = PE + KE

E = mgh + ½mv²

172.1 J = (7.26 kg) (9.8 m/s²) (2.1 m) + ½ (7.26 kg) v²

v = 2.5 m/s

7 0

3 years ago

Read 2 more answers

Aristotle was a true scientist. True or false?

leonid [27]

No, because he was a philosopher

4 0

3 years ago

Read 2 more answers

0.250 moles of gas are in a piston.The gas does 109 J of work while 240 J of heat are added. What is the change in internal ener

irina [24]

Answer:

+131Joules

Explanation:

Energy can be expressed using below expresion.

ΔE = (q + w).........eqn(1)

q will be + be if heat is gained hence, q= 240 J

work "w" will be - ve if work is done by the system, hence w= -109 J

Then substitute into eqn(1)

Change in Internal energy=

= (240 -109 )

= +131J

6 0

3 years ago

Two cars are moving towards each other and sound emitted by first car with real frequency of 3000 hertz is detected by a person

sertanlavr [38]

Answer:

v ’= 21.44 m / s

Explanation:

This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s

f ’= f (v + v₀) / (v- $v_{s}$ )

where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer

in this exercise both the source and the observer are moving, we will assume that both have the same speed,

v₀ = v_{s} = v ’

we substitute

f ’= f (v + v’) / (v - v ’)

f ’/ f (v-v’) = v + v ’

v (f ’/ f -1) = v’ (1 + f ’/ f)

v ’= (f’ / f-1) / (1 + f ’/ f) v

v ’= (f’-f) / (f + f’) v

let's calculate

v ’= (3400 -3000) / (3000 +3400) 343

v ’= 400/6400 343

v ’= 21.44 m / s

3 0

3 years ago

An astronaut holds a rock 100 m above surface of Planet X. The rock is then thrown upwards with a sleek of 15m/s. The rock reach

Gelneren [198K]

Answer: $5 m/s^{2}$

Explanation:

This problem is related to vertical motion, and the equation that models it is:

$y=y_{o}+V_{o}sin\theta t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0m$ is the rock's final height

$y_{o}=100 m$ is the rock's initial height

$V_{o}=15 m/s$ is the rock's initial velocity

$\theta=90\°$ is the angle at which the rock was thrown (directly upwards)

$t=10 s$ is the time

$g$ is the acceleration due gravity in Planet X

Isolating $g$ and taking into account $sin(90\°)=1$ :

$g=(-\frac{2}{t^{2}})(y-y_{o}-V_{o}t)$ (2)

$g=(-\frac{2}{(10 s)^{2}})(0 m-100 m-(15 m/s)(10 s))$ (3)

$g=5 m/s^{2}$ (4) This is the acceleration due gravity in Planet X

5 0

3 years ago