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
Irina-Kira [14]
3 years ago
5

A box is sitting stationary on a ramp that is 42° to the horizontal. The box has a gravitational force of 112.1 N. What is the m

inimum amount of force that can be applied to the box to get the box to move?
Physics
2 answers:
ale4655 [162]3 years ago
5 0

Answer:

Answer is 80

Explanation:

I just took the tets

nordsb [41]3 years ago
4 0

Force of gravity along the inclined plane

F_x = F_g sin\theta

here we have

F_g = 112.1 N

\theta = 42 degree

now we have

F_x = (112.1 N)sin42

F_x = 75 N

so here we require 75 N force along the incline to move the box upwards

You might be interested in
A 4.2 kg parachutist is moving straight downward with a speed of 3.85 m/s
statuscvo [17]
Kinetic energy= .5 x m x v^2
KE=.5 x 4.2 x 3.85^2
KE=31.13

4 0
3 years ago
How much force is needed to accelerate a 1.5 kg physics book to an acceleration of 6<br> m/s^2?
aleksley [76]

Answer:

Force = 8.0 k g m / s

Explanation:

Force = mass x acceleration

Mass = 4.0 k g Acceleration = 2.0 m / s 2

Hence,force = ( 4.0 x 2.0 ) k g m / s 2 = 8.0 k g m / s 2

3 0
3 years ago
2.(Ramp section) Suppose the height of the ramp is h1= 0.40m, and the foot of the ramp is horizontal, and is h2= 1.5m above the
frozen [14]

Answer:

a) the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b) the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

c) the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

 

Explanation:

Given that;

height of the ramp h1 = 0.40 m

foot of the ramp above the floor h2 = 1.50 m

assuming R = 15 mm = 0.015 m

density of steel = 7.8 g/cm³

density of aluminum =  2.7 g/cm³

a) distance that the solid steel sphere sliding down the ramp without friction;

we know that

distance = speed × time

d = vt --------let this be equ 1

according to the law of conservation of energy

mgh₁ = \frac{1}{2} mv²

v² = 2gh₁  

v = √(2gh₁)

from the second equation; s = ut +  \frac{1}{2} at²

that is; t = √(2h₂/g)

so we substitute for equations into equation 1

d = √(2gh₁) × √(2h₂/g)

d = √(2gh₁) × √(2h₂/g)

d = 2√( h₁h₂ )    

we plug in our values

d = 2√( 0.40 × 1.5 )

d = 1.55 m

Therefore, the distance that the solid steel sphere sliding down the ramp without friction is 1.55 m

b)

distance that a solid steel sphere rolling down the ramp without slipping;

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{5}mR²) ω²

v = √( \frac{10}{7}gh₁  )

so we substitute √( \frac{10}{7}gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( \frac{10}{7}gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid steel sphere rolling down the ramp without slipping is 1.31 m

 

c)

distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping;

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{3}mR²) ω²

v = √( \frac{6}{5}gh₁ )

so we substitute √( \frac{6}{5}gh₁ ) for v and t = √(2h₂/g) in equation 1 again

d = vt

d = √( \frac{6}{5}gh₁ ) × √(2h₂/g)

d = 1.549√( h₁h₂ )

d = 1.549√( 0.4 × 1.5 )

d = 1.2 m

Therefore, the distance that a spherical steel shell with shell thickness 1.0 mm rolling down the ramp without slipping is 1.2 m

d) distance that a solid aluminum sphere rolling down the ramp without slipping.

we know that;

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} I_{}ω²

mgh₁ = \frac{1}{2} mv² + \frac{1}{2} (\frac{2}{5}mR²) ω²

v = √( \frac{10}{7}gh₁  )

so we substitute √( \frac{10}{7}gh₁  ) for v and  t = √(2h₂/g) in equation 1;

d = vt

d = √( \frac{10}{7}gh₁  ) × √(2h₂/g)  

d = 1.69√( h₁h₂ )

we substitute our values

d = 1.69√( 0.4 × 1.5 )  

d = 1.31 m

Therefore, the distance that a solid aluminum sphere rolling down the ramp without slipping is 1.31 m

8 0
3 years ago
Vectors A and B lie in the xy-plane. Vector A has a magnitude of 19.1 and is at an angle of 125.5º counterclockwise from the +x-
Nady [450]

Answer:

a!aaaaaaaaaaaa!aaaaaa

Explanation:

l

8 0
3 years ago
24. While measuring the length of a book, the reading of the scale at one end is 5.0 cm and at the other end is 20.5
il63 [147K]

Answer:

...do

Explanation:

24. While measuring the length of a book, the reading of the scale at one end is 5.0 cm and at the other end is 20.5

cm. What is the length of the book in mm?

25. Explain the modifications

3 0
3 years ago
Other questions:
  • The purpose of a fuse or circuit breaker is _____. to speed up the current to slow down the current to stop current that is flow
    14·2 answers
  • If a pigeon flew 1000 meters north in 48 seconds, what is its average velocity?
    5·2 answers
  • A plastic rod that has been charged to -15.0nC touches a metal sphere. Afterward, the rod's charge is -10.0nC. How many charged
    10·1 answer
  • A cyclist moves at 10 m / s for 5 minutes and at 18 m / s for 6 minutes in the same direction and direction. Get the average spe
    13·1 answer
  • One strategy in a snowball fight is to throw
    6·1 answer
  • What happens when all the external forces on a system are balanced ?
    8·1 answer
  • A spring has an unstretched length of
    6·1 answer
  • 25 points please help PLEASE
    13·1 answer
  • How do I calculate the amount of work done using the information on the graphs given?​
    13·1 answer
  • How high would a skater need to start on a previous incline to make it up and around a loop that is 6.1meters high?
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!