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
Pie
2 years ago
15

In tae-kwon-do, a hand is slammed down onto a target at a speed of 10 m/s and comes to a stop during the 7.0 ms collision. Assum

e that during the impact the hand is independent of the arm and has a mass of 0.90 kg. What are the magnitudes of the (a) impulse and (b) average force on the hand from the target
Physics
1 answer:
wlad13 [49]2 years ago
5 0

Answer:

(A) Impulse = 9Ns

(B) F = 1286N

Explanation:

Impulse = change in momentum = m(v-u)

v = 0 (the hand comes to a stop)

u = -10m/s

Mass = 0.9kg

Impulse = 0.9 ×(0- (-10))

= 9Ns

(B) F×t = Impulse

F = Impulse/ t

t = 7ms = 7×10-³

F = 9/ (7×10-³)

F = 1286N.

You might be interested in
Organizing and interrupting information from the sense is called
NeTakaya

I think It is Perception


5 0
2 years ago
"What is the magnifying power of an astronomical telescope using a reflecting mirror whose radius of curvature is 5.9 m and an e
notka56 [123]

Answer:

The Magnifying power of a telescope is M = 109.26

Explanation:

Radius of curvature R = 5.9 m = 590 cm

focal length of objective f_{objective} = \frac{R}{2}

⇒ f_{objective} = \frac{590}{2}

⇒ f_{objective} = 295 cm

Focal length of eyepiece f_{eyepiece} = 2.7 cm

Magnifying power of a telescope is given by,

M = \frac{f_{objective} }{f_{eyepiece} }

M = \frac{295}{2.7}

M = 109.26

therefore the Magnifying power of a telescope is M = 109.26

4 0
3 years ago
The diagram shows two forces of equal magnitude acting on an object. If the common magnitude of the forces is 3.6 N and the angl
Nuetrik [128]
<h3>Answer</h3>

6.6 N pointing to the right

<h3>Explanation</h3>

Given that,

two forces acting of magnitude 3.6N

angle between them = 48°

To find,

the third force that will cause the object to be in equilibrium

<h3>1)</h3>

Find the vertical and horizontal components of the two forces

vertical force1 = sin(24)(3.6)

vertical force2= -sin(24)(3.6)

<em>(negative sign since it is acting on opposite direction)</em>

vertical force3 = sin(24)(3.6) - sin(24)(3.6)

                        = 0

<h3>2)</h3>

horizontal force1 = cos(24)(3.6)

horizontal force2= cos(24)(3.6)

horizontal force3 = cos(24)(3.6) + cos(24)(3.6)

                            = 2(cos(24)(3.6))

                            = 6.5775 N

                            ≈ 6.6 N

<em />

<em />

4 0
3 years ago
Which conditions are low air pressure systems usually associated with?
Inga [223]

cloudy, wet weather                                          

4 0
3 years ago
Read 2 more answers
A substance that produces H+ ion in a solution is an (n)
SpyIntel [72]

Answer:

acid

Explanation:

acid provides hydrigen ions

3 0
3 years ago
Other questions:
  • Answer for brainliest and mad points pllllssssssss
    7·1 answer
  • Swimming is a great way to improve cardiovascular/respiratory health.
    7·1 answer
  • Fred points a toy laser gun at a wall. Considering that the frequency of the gun's light is 4.91 × 1014 hertz and that Planck’s
    11·2 answers
  • Suppose an event is measured to be at a = (0,-2, 3, 5) in one reference frame. Find the components of this event in another refe
    5·1 answer
  • You are standing 1 meter from a squawking parrot. If you move to a distance three meters away, the sound
    12·2 answers
  • An ideal gas initially at 4.00atm and 350 K is permitted
    5·1 answer
  • Which object is accelerating downward at the slowest rate? (1 point)
    8·1 answer
  • When holes are drilled through the wall of a water tower, water will spurt out the greatest horozontal distance from the hole cl
    11·1 answer
  • How do i convert 0.25hr into minutes
    5·1 answer
  • An object's
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!