1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Darina [25.2K]
3 years ago
7

A 59.3 kg diver jumps off a board

Physics
1 answer:
ser-zykov [4K]3 years ago
6 0

The board is 2.50m high.

Why?

We can calculate how high was the board applying the Law of Conservation of Mechanical Energy. This Law states that the mechanical energy (kinematic and potential) will be conserved during the motion.

It can be described with the following formula:

E_{M_{1}}=E_{M_{2}}\\\\PE_{1}+KE_{1}=PE_{2}+KE_{2}

PE=m*g*h\\KE=\frac{1}{2}m*v^{2}

At the top of the boar, the kinetic energy is equal to 0.

At the water, the potential energy is equal to 0.

So,

PE_{1}=KE_{2}\\\\m*g*h=1450J\\\\59.3kg*9.8\frac{m}{s^{2}}*h=1450J\\ \\h=\frac{1450J}{59.3kg*9.8\frac{m}{s^{2}}}=2.50m

Hence, we have that the board is 2.50m high.

Have a nice day!

You might be interested in
Which group contains elements that have the following characteristics:
Harrizon [31]

actually the answer is B because Chlorine, sulfur, and silicon. Chlorine is a halogen and gas. Sulfur forms an ion with a -2 charge in ionic bonds. Silicon is a well-known metalloid.

6 0
3 years ago
P2O5 is a covalent compound used to purify sugar. What is the name of this compound?
AfilCa [17]

Answer:

B) Diphosphorus pentoxide

Explanation:

8 0
3 years ago
Read 2 more answers
Organic macromolecules called _______ are insoluble in water
NikAS [45]

Answer:

lipids are insoluble in water which is why lipids are often found in biological membranes and other waterproof coverings.

3 0
3 years ago
A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit
inn [45]

Answer:twice of initial value

Explanation:

Given

spring compresses x_1 distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

kinetic energy converted into spring Elastic potential energy

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1

When speed doubles

\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2

divide 1 and 2

\dfrac{1}{4}=\dfrac{x_1^2}{x_2^2}

x_2=2x_1

Therefore spring compresses twice the initial value

   

7 0
3 years ago
1. A 2.5 kg led projector is launched as a projectile off a tall building. At one point, as it
spin [16.1K]

Answer:

Explanation:

I got everything but i. Don't know why but it's eluding me. So let's do everything but that.

a. PE = mgh so

   PE = (2.5)(98)(14) and

   PE = 340 J

b. KE=\frac{1}{2}mv^2 so

   KE=\frac{1}{2}(2.5)(14)^2 and

   KE = 250 J

c. TE = KE + PE so

   TE = 340 + 250 and

   TE = 590 J

d. PE at 8.7 m:

   PE = (2.5)(9.8)(8.7) and

   PE = 210 J

e. The KE at the same height:

   TE = KE + PE and

   590 = KE + 210 so

   KE = 380 J

f. The velocity at that height:

   380=\frac{1}{2}(2.5)v^2 and

   v=\sqrt{\frac{2(380)}{2.5} } so

   v = 17 m/s

g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:

   590 = KE + PE and

   PE = (2.5)(9.8)(11.6) so

   PE = 280 then

   590 = KE + 280 so

   KE = 310 then

   310=\frac{1}{2}(2.5)v^2 and

   v=\sqrt{\frac{2(310)}{2.5} } so

   v = 16 m/s

h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:

   v=v_0+at and

   26 = 0 + 9.8t and

   26 = 9.8t so the time at 26 m/s is

   t = 2.7 seconds. Now we use that in the equation for displacement:

   Δx = v_0t+\frac{1}{2}at^2 and filling in the time the object was at 26 m/s:

   Δx = 0t + \frac{1}{2}(-9.8)2.7)^2 so

   Δx = 36 m

i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.

4 0
2 years ago
Other questions:
  • A block slides from rest with negligible friction down the track above, descending a vertical height of 5.0 m to point P at the
    5·1 answer
  • Physic help????????????????
    11·2 answers
  • The weight of a person is 500N and his foot imprint area is 0.5m^2.Calculate the total pressure exerted by person when he stands
    10·1 answer
  • A pendulum has 573 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o
    13·1 answer
  • Philip drives his car at a velocity of 28 m/s he applies the break which slows the vehicle down at a rate of six. 4 m/s2 and it
    6·1 answer
  • how long would it take for a radio wave sent from a space satellite circling mars to reach Earth? Assume that radio waves (a for
    8·1 answer
  • Put the balloon near (BUT NOT TOUCHING) the wall. Leave about as much space as the width of your pinky finger between the balloo
    11·1 answer
  • Help<br>pls give me an honest answer ​
    12·1 answer
  • Some scientists believe that:
    7·2 answers
  • What factors determine climate?
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!