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
nikdorinn [45]
2 years ago
9

Please need in hurry

Physics
1 answer:
gulaghasi [49]2 years ago
3 0

Explanation:

i) center of gravity (or mass)

ii) m = W/g = (160 N)/(9.8 m/s^2)

= 16.3 kg

You might be interested in
Which of the following is the most accurate statement concerning the properties of matter?
DIA [1.3K]

Answer:

b. they can be observed and measured

Explanation:

Matter is anything that has weight and occupy space. There are three states of matter namely Solid, liquid and gas.

The properties of matter are both physical and chemical in nature. Both properties can be measured and observed. Phhysical properties are anything that can be measured without changing the state of the matter. Example of physical properties includes mass, volume, length, color etc.

Chemical properties is another properties of matter. This is the ability of the states of matters to combine with other substance to form a new product for example, rusting of iron, formation of salt etc.

All this as discussed are both measurable and can be observed.

8 0
3 years ago
in order to qualify for the finals in a racing event, a race car must achieve an average speed of 278 km/h on a track with a tot
Pachacha [2.7K]

The minimum average speed it must have in the second half of the event in order to qualify is 414.7 km/h.

<h3>What is average speed?</h3>

The average speed of an object is the ratio of total distance traveled by the object to the total time of motion of the object.

<h3>Total time taken by the car during the entire race</h3>

time = distance/average speed

time = (1.41 km) / (278 km/h)

time = 0.0051 hr

The car travels the first half of the race, d (¹/₂ x 1410 m) at 210 km/h;

d = 705 m = 0.705 km

t1 = 0.705/210

t1 = 0.0034 hr

<h3>time for the second half</h3>

t2 = 0.0051 - 0.0034 hr

t2 = 0.0017 hr

<h3>minimum average speed of the second half</h3>

v = d/t

v = 0.705 km / 0.0017 hr

v = 414.7 km/hr

Thus, the minimum average speed it must have in the second half of the event in order to qualify is 414.7 km/h.

Learn more about average speed here: brainly.com/question/4931057

#SPJ1

6 0
1 year ago
A football player runs at 8m/s and plows into a 80kg referee standing on the field causing the referee to fly forward at 5m/s. I
madreJ [45]

The topic here is momentum.

When a collision is said to be elastic, it means that the colliding objects now travel at their own new, indivual and distinct velocities, often in different directions.

So we write that as,

(mass of football player x velocity of football player) + (mass of referee x velocity of referee) = (mass of football player x velocity of football player) + (mass of referee x velocity of referee)

(M × 8) + (80 × 0) = (M× 0) + (80 × 5)

8M = 400

M = 50 kg

4 0
3 years ago
A sphere of mass m and radius r is released from rest at the top of a curved track of height H. The sphere travels down the curv
iren2701 [21]

Explanation:

<em>(a) On the dots below, which represent the sphere, draw and label the forces (not components) that are exerted on the sphere at point A and at point B, respectively.  Each force must be represented by a distinct arrow starting on and pointing away from the dot.</em>

At point A, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing left, and static friction force Fs pushing down.

At point B, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing down, and static friction force Fs pushing right.

<em>(b) i. Derive an expression for the speed of the sphere at point A.</em>

Energy is conserved:

PE = PE + KE + RE

mgH = mgR + ½mv² + ½Iω²

mgH = mgR + ½mv² + ½(⅖mr²)(v/r)²

mgH = mgR + ½mv² + ⅕mv²

gH = gR + ⁷/₁₀ v²

v² = 10g(H−R)/7

v = √(10g(H−R)/7)

<em>ii. Derive an expression for the normal force the track exerts on the sphere at point A.</em>

Sum of forces in the radial (-x) direction:

∑F = ma

N = mv²/R

N = m (10g(H−R)/7) / R

N = 10mg(H−R)/(7R)

<em>(c) Calculate the ratio of the rotational kinetic energy to the translational kinetic energy of the sphere at point A.</em>

RE / KE

= (½Iω²) / (½mv²)

= ½(⅖mr²)(v/r)² / (½mv²)

= (⅕mv²) / (½mv²)

= ⅕ / ½

= ⅖

<em>(d) The minimum release height necessary for the sphere to travel around the loop and not lose contact with the loop at point B is Hmin.  The sphere is replaced with a hoop of the same mass and radius.  Will the value of Hmin increase, decrease, or stay the same?  Justify your answer.</em>

When the sphere or hoop just begins to lose contact with the loop at point B, the normal force is 0.  Sum of forces in the radial (-y) direction:

∑F = ma

mg = mv²/R

gR = v²

Applying conservation of energy:

PE = PE + KE + RE

mgH = mg(2R) + ½mv² + ½Iω²

mgH = 2mgR + ½mv² + ½(kmr²)(v/r)²

mgH = 2mgR + ½mv² + ½kmv²

gH = 2gR + ½v² + ½kv²

gH = 2gR + ½v² (1 + k)

Substituting for v²:

gH = 2gR + ½(gR) (1 + k)

H = 2R + ½R (1 + k)

H = ½R (4 + 1 + k)

H = ½R (5 + k)

For a sphere, k = 2/5.  For a hoop, k = 1.  As k increases, H increases.

<em>(e) The sphere is again released from a known height H and eventually leaves the track at point C, which is a height R above the bottom of the loop, as shown in the figure above.  The track makes an angle of θ above the horizontal at point C.  Express your answer in part (e) in terms of m, r, H, R, θ, and physical constants, as appropriate.  Calculate the maximum height above the bottom of the loop that the sphere will reach.</em>

C is at the same height as A, so we can use our answer from part (b) to write an equation for the initial velocity at C.

v₀ = √(10g(H−R)/7)

The vertical component of this initial velocity is v₀ sin θ.  At the maximum height, the vertical velocity is 0.  During this time, the sphere is in free fall.  The maximum height reached is therefore:

v² = v₀² + 2aΔx

0² = (√(10g(H−R)/7) sin θ)² + 2(-g)(h − R)

0 = 10g(H−R)/7 sin²θ − 2g(h − R)

2g(h − R) = 10g(H−R)/7 sin²θ

h − R = 5(H−R)/7 sin²θ

h = R + ⁵/₇(H−R)sin²θ

4 0
2 years ago
Balances are used to measure what a. Mass of an object b. Volume c. Force acting upon it d. Level surfaces on the table
suter [353]
<span>a. Mass of an object </span>
4 0
2 years ago
Other questions:
  • A snowboarder travels 150 m down a mountain slope that is 65 degrees above horizontal. What is his vertical displacement?
    8·1 answer
  •   The diagram shows a tray of marbles being shaken from side to side.  As this happens some of the marbles jump out of the tray.
    11·1 answer
  • A clam of mass 0.12 kg dropped by a seagull takes 3.0 s to hit the ground. [Neglect friction.]
    11·1 answer
  • a student conducts an experiment to determine how the additional of salt to water affects the density of the water. the student
    9·1 answer
  • Hurryyyyyyyy
    13·2 answers
  • Which of the following would have the greatest force of gravitational attraction?
    6·1 answer
  • Match the units in the right hand column with the terms in the right hand column​
    14·2 answers
  • HELP ASAP !!
    11·2 answers
  • Which factor has the greatest effect on the strength of an electromagnet?
    15·2 answers
  • In an open circuit like the picture
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!