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
DaniilM [7]
2 years ago
8

What is the gpe of a 200g ball that is 10m off the ground​

Physics
1 answer:
rodikova [14]2 years ago
8 0

Answer:

gpe gained = 20J

Explanation:

gpe = mass * gravity * height

gpe = (200/1000) kg * 10 * 10

gpe  = 20J

You might be interested in
The liquid pressure exerted in one direction only​
Wewaii [24]
The pressure exerted by a liquid on an object increases as we go more deep into the liquid and this pressure is called as hydro static pressure . if we consider a part of the static fluid then all the horizontal forces will cancel out while the vertical forces will add vectorilly and due to which a pressure difference is created . so as we go more deep the pressure increase .

Now pressure is a scalar so it does not depend on direction but when two objects are on the same level with respect to a reference level then the pressure exerted on them by fluid is always the same . hope this helps
8 0
3 years ago
A grindstone of radius 4.0 m is initially spinning with an angular speed of 8.0 rad/s. The angular speed is then increased to 12
PSYCHO15rus [73]

Answer: 6.36

Explanation:

Given

Radius of grindstone, r = 4 m

Initial angular speed of grindstone, w(i) = 8 rad/s

Final angular speed of the grindstone, w(f) = 12 rad/s

Time used, t = 4 s

Angular acceleration of the grinder,

α = Δw / t

α = w(f) - w(i) / t

α = (12 - 8) / 4

α = 4/4 = 1 rad/s²

Number of complete revolution in 4s =

Δθ = w(i).t + 1/2.α.t²

Δθ = 8 * 4 + 1/2 * 1 * 4²

Δθ = 32 + 1/2 * 16

Δθ = 32 + 8

Δθ = 40 rad/s

40 rad/s = 40/2π rpm = 6.36 rpm

Therefore, the grindstone does 6.36 revolutions during the 4 s interval

6 0
3 years ago
Which of the following scenarios would be optimal for obtaining a date from radioactive decay using these isotopes: 87Rb, 147Sm,
REY [17]

Answer:

a) 238U, 40K and 87Rb, b)   235U and to a lesser extent 40K , c)  he 235U,

d) possibility is 14C , e)this period would be ideal for 14C , f) 14C should be used since it is the one with the least average life time, even though the measurements must be very careful

Explanation:

One of the applications of radioactive decay is the dating of different systems.

To do this, the quantity of radioactive material in a meter is determined and with the average life time, the time of the sample is found.

Let's write the half-life times of the given materials

87Rb T ½ = 4.75 1010 years

147Sm T ½ = 1.06 1011 years

235U = 7,038 108 years

238U = 4.47 109 years

40K = 1,248 109 years

14C = 5,568 103 years

we already have the half-life of the different elements given

a) meteors. As these decomposed in the formation of the solar system, their life time is around 3 109 to 5 109 years, so it is necessary to look for elements that have a life time of this order, among the candidates we have 238U, 40K and 87Rb if these elements were at the moment of the formation of these meteors, there must still be rations in them, instead elements 14C already completely adequate

b) rock. The formation period is 4.20-108 years, therefore one of the most promising elements is 235U and to a lesser extent 40K since it is more abundant in rocks. The other elements with higher life times have not decayed and therefore will not give a true value and the 14C is completely decayed

c) volcanic ash. Formation time 6107 years, the only element that has the possibility of having a count is the 235U, the others have a life time so long that they have not decayed and the 14C is complete, unbent

d) scarp of an earthquake formation time 5 101 years, The only one that has any possibility is 14C even when it has declined very little, all the others, you have time to long that has not decayed

e) INCA excavation. The time of this civilization is about 10000 to 500 years (104 to 5 102 years), we see that this period would be ideal for 14C since it has some period of cementation, the others have not decayed

f) Tree in Blepharitis. 14C should be used since it is the one with the least average life time, even though the measurements must be very careful because of a period of disintegration. We have such a long time that they have not decayed

8 0
3 years ago
Alonzo sprints for 500 meters along a straight track during a race. After crossing the finish line, he to walks back along the t
ehidna [41]
450 meters is the displacement
8 0
3 years ago
Read 2 more answers
Willie, in a 100.0 m race, initially accelerates uniformly from rest at 2.00 m/s2 until reaching his top speed of 12.0 m/s. He m
Oduvanchick [21]

Answer:

The total time for the race is 11.6 seconds

Explanation:

The parameters given are;

Total distance ran by Willie = 100.0 m

Initial acceleration = 2.00m/s²

Top speed reached with initial acceleration = 12.0 m/s

Point where Willie start to fade and decelerate = 16.0 m from the finish line

Speed with which Willie crosses the finish line = 8.00 m/s

The time and distance covered with the initial acceleration are found using the following equations of motion;

v = u₀ + a·t

v² = u₀² + 2·a·s

Where:

v = Final velocity reached with the initial acceleration = 12.0 m/s

u₀ = Initial velocity at the start of the race = 0 m/s

t = Time during acceleration

a = Initial acceleration = 2.00 m/s²

s = Distance covered during the period of initial acceleration

From, v = u₀ + a·t, we have;

12 = 0 + 2×t

t = 12/2 = 6 seconds

From, v² = u₀² + 2·a·s, we have;

12² = 0² + 2×2×s

144 = 4×s

s = 144/4 =36 meters

Given that the Willie maintained the top speed of 12.0 m/s until he was 16.0 m from the finish line, we have;

Distance covered at top speed = 100 - 36 - 16 = 48 meters

Time, t_t of running at top speed = Distance/velocity = 48/12 = 4 seconds

The deceleration from top speed to crossing the line is found as follows;

v₁² = u₁² + 2·a₁·s₁

Where:

u₁ = v = 12 m/s

v₁ = The speed with which Willie crosses the line = 8.00 m/s

s₁ = Distance covered during decelerating = 16.0 m

a₁ = Deceleration

From which we have;

8² = 12² + 2 × a × 16

64 = 144 + 32·a

64 - 144 = 32·a

32·a = -80

a = -80/32 = -2.5 m/s²

From, v₁ = u₁ + a₁·t₁

Where:

t₁ = Time of deceleration

We have;

8 = 12 + (-2.5)·t₁

t₁ = (8 - 12)/(-2.5) = 1.6 seconds

The total time = t + t_t + t₁ =6 + 4 + 1.6 = 11.6 seconds.

6 0
3 years ago
Other questions:
  • True of false all matter requires it’s own space
    5·1 answer
  • Gravity is a force. <br> True <br> False
    14·2 answers
  • The batteries in the figure above are connected in​
    8·2 answers
  • When 1 kg of water and 1 kg of wood absorb the same amount of heat, the change in temperature of the wood is greater than the ch
    14·1 answer
  • What dose air taste like and wrong answer only
    14·2 answers
  • Suppose that you'd like to find out if a distant star is moving relative to the earth. The star is much too far away to detect a
    7·1 answer
  • Which of the following cart configurations could produce the X versus T graph shown
    5·1 answer
  • The angular quantities L, omega, and alpha are vectors. The vectors point along the axis of rotation with a direction determined
    11·1 answer
  • Need help with this question ​
    5·2 answers
  • A closed curve encircles several conductors. The line integral around this curve is (image attached below)
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!