Ask question

Ask question

All categories

BaLLatris [955]

1 year ago

10

A ball having a mass of 0.20 kilograms is placed at a height of 3.25 meters. If it is dropped from this height, what will be the

kinetic energy of the ball when it reaches 1.5 meters above the ground?

2 answers:

asambeis [7]1 year ago

7 0

EC_1 + EP_1 = EC2 + EP_2

EC_2 = 0

EC_2 = EP_1 - EP_2

EC_2 = mg(H_1 - H_2) = 0.20 kg * 9.8 m/s^2 * (3.25 m - 1.5m) = 3.43 J

Pachacha [2.7K]1 year ago

6 0

Answer:

3.43 J

Explanation:

One form of energy can convert to another but cannot be created or destroyed.

A body has potential energy due to its position or configuration.

P.E. = m g h

A body has kinetic energy due to its motion.

K.E. = 0.5 mv²

Mechanical energy is the sum of potential energy and kinetic energy and in the absence of any external forces, the mechanical energy remains conserved.

Potential energy+ Kinetic energy (at the top) = Potential energy +kinetic energy (at 3.25 m)

⇒ m g h + 0 = m g h' + K.E.'

⇒K.E. ' = mg (h -h') = 0.20 kg × 9.81 m/s² × (3.25 m - 1.5 m) = 3.43 J

You might be interested in

A steel cable 1.25 in. in diameter and 50 ft long is to lift a 20-ton load without permanently deforming. What is the length of

Over [174]

Answer:

50.0543248872 ft

Explanation:

F = Load = 20 ton = $20\times 2000\ lb$

d = Diameter = 1.25 in

$L_1$ = Initial length = 50 ft

$L_2$ = Final length

A = Area = $\dfrac{\pi}{4}d^2$

Y = Young's modulus = $30\times 10^6\ psi$

Young's modulus is given by

$Y=\dfrac{FL}{A\Delta L}\\\Rightarrow Y=\dfrac{FL_1}{\dfrac{\pi}{4}d^2(L_2-L_1)}\\\Rightarrow L_2=\dfrac{4FL_1}{Y\pi d^2}+L_1\\\Rightarrow L_2=\dfrac{4\times 40000\times 50}{30\times 10^6\times \pi\times 1.25^2}+50\\\Rightarrow L_2=50.0543248872\ ft$

The length during the lift is 50.0543248872 ft

6 0

2 years ago

A stunt cyclist needs to make a calculation for an upcoming cycle jump. The cyclist is traveling 100 ft/sec toward an inclined r

Elena-2011 [213]

Answer: $A=23.57^{\circ}$

Explanation:

Given

velocity=100 ft/s

height of landing zone=10 ft

Equation of $x(t)=100 t\cos (A)$

$y(t)=-16t^2+100t\sin (A)+10$

Maximum height=35 feet

at maximum height

$\frac{\mathrm{d} Y}{\mathrm{d} t}=0$

$\frac{\mathrm{d} x}{\mathrm{d} t}=-32t+100\sin A$

$t=3.125\sin A$

At $t=3.125\sin A$

$Y(t)=35=-16\times (3.125\sin A)^2+100\times 3.125\sin A\times \sin A+10$

$312.5\sin^2 A-156.25\sin^2 A=25$

$sin^2 A=0.16$

$A=23.57^{\circ}$

8 0

2 years ago

A car drives 16 miles south and then 12 miles west. What is the magnitude of the car’s displacement? 4 miles 16 miles 20 miles 2

Ostrovityanka [42]

For this case, what we can do is use the Pythagorean theorem to find the magnitude of the displacement of the car.
We have then
$d ^ 2 = 16 ^ 2 + 12 ^ 2
$
From here, we clear the value of d.
We have then:
$d = \sqrt{16 ^ 2 + 12 ^ 2} 
$
Rewriting:
$d = \sqrt{256 + 144}$
$d = \sqrt{400}$
$d = 20 miles
$
Answer:
The magnitude of the car's displacement is:
d = 20 miles

7 0

2 years ago

Read 2 more answers

You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves along a frictionless track. however, af

djyliett [7]

I attached the missing picture.
We can figure this one out using the law of conservation of energy.
At point A the car would have potential energy and kinetic energy.
$A: mgh_1+\frac{mv_1^2}{2}$
Then, while the car is traveling down the track it loses some of its initial energy due to friction:
$W_f=F_f\cdot L$
So, we know that the car is approaching the point B with the following amount of energy:
$mgh_1+\frac{mv_1^2}{2}- F_fL$
The law of conservation of energy tells us that this energy must the same as the energy at point B.
The energy at point B is the sum of car's kinetic and potential energy:
$B: mgh_2+\frac{mv_2}{2}$
As said before this energy must be the same as the energy of a car approaching the loop:
$mgh_2+\frac{mv_2}{2}=mgh_1+\frac{mv_1^2}{2}- F_fL$
Now we solve the equation for $v_1$ :
$v_1^2=2g(h_2-h_1)+v_2^2+\frac{2F_fL}{m}\\
v_1^2=39.23\\
v_1=\sqrt{39.23}=6.26\frac{m}{s}$

4 0

1 year ago

Read 2 more answers

You just saw a magazine ad that shows several people having fun at a football game. They're all wearing a specific shoe brand. T

I am Lyosha [343]

The technique used is most likely Bandwagon.
So, B.

7 0

2 years ago

Read 2 more answers