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
mihalych1998 [28]
3 years ago
5

a 455n gymnast jumps upward a distance of 1.5 meters to reach the uneven parallel bars. how much work did she do before she even

began her routine
Physics
1 answer:
Taya2010 [7]3 years ago
3 0
Work = force x distance
Work = 455N (weight force) x 1.5m (distance)
Work done = 682.5J
Hope this helped!
You might be interested in
What is the differ of fermentation from respirstion
Andrej [43]
This may help u... :P

8 0
3 years ago
When a surfer rides an ocean wave on her surfboard, she is actually riding on. A. a crest that is toppling over. . B. a trough o
ruslelena [56]
The right answer to this question is A. a crest that is toppling over. When a surfer rides an ocean wave on her surfboard, she is actually riding on a crest. The crest is the point on a wave with the maximum value or upward displacement within a cycle.
8 0
3 years ago
Read 2 more answers
A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long
Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

W=-\mu mg\times l

Put the value into the formula

W=-0.30\times62\times9.8\times3.50

W=-637.98\ J

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}

\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}

Final potential energy is zero

So, \dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl

Put the value into the formula

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50

v_{2}=\sqrt{2\times55.915}

v_{2}=10.57\ m/s

The speed of skier is 10.57 m/s.

Hence,  (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0
3 years ago
Two vectors A and B are at right angles to each other. The magnitude of A is 3.00. What should be the length of B, so that the m
AfilCa [17]

Answer: length of B =4.00

Explanation:

for  the vectors A and B and the angle between them as  x.

Magnitude of the sum of A and B is  given as = √(A²+B²+2ABcosx

where

Magnitude of A  = 3.00

Magnitude of the sum of A and B is  5.00

5.00=√(A²+B²+2ABcos90°

5.00= √3² +b² +0

5²= 3² +b²

25=9+b²

b²= 25-9

b² = 16

b=  √16

b= 4

7 0
3 years ago
Please need help fast
iVinArrow [24]

(a) See graph in attachment

The appropriate graph to draw in this part is a graph of velocity vs time.

In this problem, we have a horse that accelerates from 0 m/s to 15 m/s in 10 s.

Assuming the acceleration of the horse is uniform, it means that the velocity (y-coordinate of the graph) must increase linearly with the time: therefore, the velocity-time graph will appear as a straight line, having the final point at

t = 10 s

v = 15 m/s

(b) 1.5 m/s^2

The average acceleration of the horse can be calculated as:

a=\frac{v-u}{t}

where

v is the final velocity

u is the initial velocity

t is the time interval

In this problem,

u = 0

v = 15 m/s

t = 10 s

Substituting,

a=\frac{15-0}{10}=1.5 m/s^2

(c) 75 m

For a uniformly accelerated motion, the distance travelled can be calculated by using the suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance travelled

u is the initial velocity

t is the time interval

a is the acceleration

In this problem,

u = 0

t = 10 s

a=1.5 m/s^2

Substituting,

s=0+\frac{1}{2}(1.5)(10)^2=75 m

(d) See attached graphs

In a uniformly accelerated motion:

- The distance travelled (x) follows the equation mentioned in part c,

x=ut+\frac{1}{2}at^2

So, we see that this has the form of a parabola: therefore, the graph x vs t will represents a parabola.

- The acceleration is constant during the motion, and its value is

a=1.5 m/s^2 (calculated in part b)

therefore, the graph acceleration vs time will show a flat line at a constant value of 1.5 m/s^2.

6 0
3 years ago
Other questions:
  • Identify the basic structure of a chemical equation
    5·1 answer
  • Which layer of ground contain organic matter and tiny rock particles
    8·1 answer
  • Barium chlorate (ba(clo3)2) breaks down to form barium chloride and oxygen. what is the balanced equation for this reaction? whe
    12·2 answers
  • What energy transformation takes place when an object is pushed up an inclined plane?
    10·2 answers
  • A calorimeter uses the temperature change of water to determine the _____ of another substance.
    10·1 answer
  • Number of vibrations (waves) in an amount of time
    13·1 answer
  • 11. Find the direction of the Force
    11·1 answer
  • 2. If you are 5'10" tall, that is, 5 feet 10 inches, what is your height in meters? (2 54 cm = 1.00 in)
    11·2 answers
  • Find the distance from a point charge q=100nC where the field intensity is equal to E=6kN/C. please include description of the r
    14·1 answer
  • if the electron affinity for an element is a negative number, does it mean that the anion of the element is more stable than the
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!