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
lana [24]
3 years ago
9

World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude

15m/s^2 .
Part A
How much horizontal force F must a sprinter of mass 64 kg exert on the starting blocks to produce this acceleration?
Express your answer in newtons using two significant figures.

Part B
Which body exerts the force that propels the sprinter, the blocks or the sprinter?
Which body exerts the force that propels the sprinter, the blocks or the sprinter?
Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
8 0

Answer:

A. The force exerted by the sprinter must be  9.6 × 10² N.

B. The force that propels the sprinter is exerted by the blocks.

Explanation:

Hi there!

Let´s begin with part B:

The sprinter exerts a force on the blocks and, as a reaction, the blocks exert a force on the sprinter that is of equal magnitude but opposite direction (Newton´s third law). This reaction of the blocks causes the acceleration of the sprinter.

Part A

The force exerted by the blocks can be calculated using Newton´s second law:

F = m · a

Where:

F = exerted force.

m = mass of the object being accelerated.

a = acceleration of the object after applying the force on the object.

F = m · a

F = 64 kg · 15 m/s²

F = 9.6 × 10² N

You might be interested in
How much heat is needed to raise the temperature of 1 kg of iron from 25 degree to 30 degrees?
stiv31 [10]

Answer:300 joules

Explanation:The formula for specific heat is:

E

=

m

c

θ

, where:

E

is the energy needed in joules,

m

is the mass in kilograms,

c

is the specific heat,

θ

is the change in temperature, Celsius of Kelvin doesn't matter.

Here,

m

=

0.125

kg

c

=

480

J/kg

∘

C

θ

=

5

∘

C

, and inputting:

E

=

0.125

⋅

480

⋅

5

E

=

300

J

of energy is needed.

4 0
4 years ago
The coefficients of friction between the 20-kg crate and the inclined surface are µ,8 = 0.24 and J.lk = 0.22. If the crate start
Yanka [14]

Answer:5.60 m/s

Explanation:

Given

Coefficient of static friction \mu _s=0.24

Coefficient of kinetic friction \mu _k=0.22

mass of crate m=20\ kg

Force applied F=200\ N

maximum static Friction F_s=\mu _sN

N=mg

F_s=0.24\times 20\times 9.8

F_s=47.04\ N

thus applied force is greater than Static friction therefore kinetic friction will come into play

F_k=\mu _kN

F_k=0.22\times 20\times 9.8=43.12\ N

net Force on crate F-F_k=ma

a=\frac{200-43.12}{20}=7.84\ m/s^2

Magnitude of velocity can be obtained by using

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

here initial velocity is zero as crate start from rest

v^2-0=2\times 7.84\times 2

v=\sqrt{31.37}

v=5.60\ m/s                                          

7 0
4 years ago
Calculate the magnitude and the direction of the resultant forces​
snow_tiger [21]

answer:

resultant = 127.65 in the positive direction

explanation:

F1 = 50N , F2 = 40N, f3 = 55N , f4 = 60N

Fy = 50 sin 50 = 50 × -0.26 = -13

Fx = 40 cos 0 = 40×1 = 40

fx = 55 cos 25 = 55×0.99 = 54.45

Fy = 60 sin 70 = 60 × 0.77 = 46.2

resultant = -13+40+54.45+46.2 = 127.65 in the positive direction

6 0
3 years ago
The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de
spin [16.1K]

Answer:

the spring constant k = 5.409*10^4 \ N/m

the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

Explanation:

From Hooke's Law

F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m

Thus; the spring constant k = 5.409*10^4 \ N/m

The amplitude is decreasing 37% during one period of the motion

e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}

b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) }    }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s

Therefore; the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

5 0
4 years ago
At night, large bodies of water release heat to the atmosphere quickly. True False
Nikolay [14]

Answer:

false I hope. have a good day!

3 0
2 years ago
Other questions:
  • Hey guys I really really need help with this question for ASAP! Explain what chart junk is and how it differs from the kind of i
    13·2 answers
  • An objects speed is the distance it travels____the amount of times it takes?
    7·1 answer
  • What is the value of (g) at north pole and at equator?
    9·1 answer
  • During an ,
    7·1 answer
  • What is the effect that causes objects to move in a curved direction due to earths rotation
    9·2 answers
  • A certain elevator cab has a total run of 214 m and a maximum speed is 315 m/min, and it accelerates from rest and then back to
    7·1 answer
  • Two diamonds begin a free fall from rest from the same height 0.9 s apart. How long after the first object begins to fall will t
    6·1 answer
  • When a rocket lifts off what provides the momentum match of the upper lift
    6·1 answer
  • Find the energy turned into heat.
    9·1 answer
  • Equation: S=D]
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!