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
Lerok [7]
3 years ago
10

What must the diver's minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, wh

ich is 1.50 m wide and 9.50 m below the top of the cliff?
Physics
1 answer:
Dafna1 [17]3 years ago
4 0

Answer:

1.08 m/s

Explanation:

This can be solved with two steps, first we need to find the time taken to fall 9.5 m, then we can divide the horizontal distance covered with time taken to calculate the velocity.

Time taken to fall 9.5 m

vertical acceleration = a = 9.8 m/s^2.

vertical velocity = 0, (since there is only horizontal component for velocity, )

distance traveled  s = 9.5 m.

Substituting these values in the equation

s= u \timest+0.5at^{2}

t= \sqrt{\frac{2s}{g} }

t=\sqrt{\frac{2\times9.5}{9.8} }

⇒ t= 1.392 sec

Velocity needed

We know the time taken (1.392 s) to travel 1.5 m,

So velocity = 1.5 m / 1.392 s = 1.08 m/s

hence velocity of the diver must be at least 1.08 m/s

You might be interested in
The main difference between a series and a parallel circuit is that
ale4655 [162]
C) is correct

series circuit - in the same path : current flow on one path so they are equal on each component and equal to the source's. voltage on each components may be different.

parallel circuit - between same nodes : voltage of the components are equal and equal to the source's. current on each components may be different.
3 0
3 years ago
A moving fan continues to move for a while even after switched off, why? ​
dem82 [27]

Answer:

due to the inertia of motion, the fan continues to move for some time even after switching it off.

4 0
3 years ago
Read 2 more answers
Your friend, who is in a field 100 meters away from you, kicks a ball towards you with an initial velocity of 16 m/s. Assuming t
LekaFEV [45]

Answer:

Time, t = 5.355 seconds

Explanation:

Given the following data;

Distance = 100 m

Initial velocity = 16 m/s

Deceleration = 1 m/s²

To find the time, we would use the second equation of motion;

But since the ball is decelerating, it's acceleration would be negative.

S = ut + ½at²

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

100 = 16t - 0.5t²

200 = 32t - t²

t² + 32t - 200 = 0

Solving the quadratic equation using the quadratic formula;

The quadratic equation formula is;

x = \frac {-b \; \pm \sqrt {b^{2} - 4ac}}{2a}

Substituting into the equation, we have;

x = \frac {-32 \; \pm \sqrt {32^{2} - 4*1*(-200)}}{2*1}

x = \frac {-32\pm \sqrt {1024 - (-800)}}{2}

x = \frac {-32 \pm \sqrt {1024 + 800}}{2}

x = \frac {-32 \pm \sqrt {1824}}{2}

x = \frac {-32 \pm 42.71}{2}

x_{1} = \frac {-32 + 42.71}{2}

x_{1} = \frac {10.71}{2}

x1 = 5.355

We do not need the negative value of x, so we proceed.

Therefore, time = 5.355 seconds

3 0
3 years ago
Current evidence suggests that many massive jovian planets orbit at very close orbital distances to their stars. How do we think
ivanzaharov [21]

Answer:

In the Solar system, the Jovian planets are farther from the Sun. Majority of the extrasolar Jovian planets are closer to their stars. These are known as "Hot Jupiters". From the studies, the reason for the existence of massive Jovian planets to be closer to their star is found to be the gravitational interaction of these planets with other massive planets which pushes them closer to their stars. These planets are formed beyond the frost line initially but later on migrate inwards.

4 0
3 years ago
How does being on the moon effect gravitational field strength
SVETLANKA909090 [29]

Your being on the moon has no effect on the moon's
gravitational field strength, or on the Earth's for that
matter.

However, YOU notice a change on YOU when YOU move
from one to the other, because of the effect of the gravitational
field strength on you and your internal organs.

If you could stand on the moon, you would experience an incredible
sense of lightness, since the forces of attraction between the moon
and anything else are only 16% as great as the same forces are on
Earth.

5 0
3 years ago
Other questions:
  • Which structure is found in all EUKARYOTIC cells? Large central vacuole, Golgi apparatus, flagella, cilia
    7·1 answer
  • What is your question?
    11·1 answer
  • Which one of the following is a balanced equation?
    5·1 answer
  • What wave have a frequency of less then 20hz
    10·1 answer
  • Which type of energy is commonly referred to as kinetic energy?
    7·2 answers
  • Myopia is a condition of the eye where the crystalline lens focuses the light rays to a position between the lens and the retina
    8·1 answer
  • How does a theory differ from a theory
    14·2 answers
  • An object moves with a positive acceleration. Could the object be moving with increasing speed, decreasing speed or constant spe
    13·1 answer
  • The guitarist shortens the oscillating length of the properly tuned D-string by 0.15 m by pressing on the string with a finger.
    11·1 answer
  • Student Exploration: Nuclear Decay. Has anyone done a Gizmos lab on this?
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!