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
Readme [11.4K]
2 years ago
10

The maximum mass the can be hung vertically from a string without breaking the string is 10kg. A length of this string that is 2

m long is used to rotate a 0.5kg object in a circle on a frictionless table with the string horizontal. The maximum speed that the mass can attain under these conditions without the string breaking is most nearly:
Answer is 20 m/s
Please show me how I can get the answer. Thank you.
Physics
1 answer:
vichka [17]2 years ago
0 0

Answer:

Explanation:

Think about circular motion when answering this question. First determine the maximum force that can be applied on the string. F = mg so F = (10)(10) = 100 N. Then determine the centripetal acceleration of the .5 kg mass, a = F/m so a = 100/.5 = 200 m/s².  On the equation sheet, use equation a(centripetal acceleration) = v²/r so 200 = v²/2 therefore v = 20 m/s. Hope this helps!

You might be interested in
A historian claims that a cannonball fired at a castle wall would melt on impact with the wall. Let's examine this claim. Assume
marysya [2.9K]

Answer:

   v = 1,167 10³ m / s

Explanation:

The equation for heat is

             Q = m ce ΔT

Let's replace

              Q = m 450 (1811-298)

              Q = m 6.81 105

This is the energy of the ball, using energy conservation

              Q = Me = K = ½ m v²

              m 6.81 10⁵ = ½ m v²

               v = √ (2 6.81 105)

               v = 1,167 10³ m / s

This speed is greater than the speed of sound, so the process is unlikely to occur

4 0
1 year ago
Which of the following is true about a hypothesis?
Leviafan [203]

Answer:

i think you need to show a picture

Explanation:

7 0
1 year ago
T The force exerted by the wind on the sails of a sailboat is 390 N north. The water exerts a force of 180 N east. If the boat (
S_A_V [24]

Answer:

a= 1.59 m/s² : Magnitude  of the acceleration

β = 65.22°   (north of east) : Direction of the acceleration

Explanation:

Conceptual analysis

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Problem development

The acceleration is presented in the direction of the resultant force applied .

Calculation of the resultant forces (R)

R=\sqrt{(F_{N})^{2} +(F_{E})^{2} }

R=\sqrt{(390)^{2} +(180)^{2} }

R= 429.5 N

We apply the formula (1) to calculate the magnitude of the acceleration(a) :

∑F = m*a , m= 270 kg

R= m*a

429.5 =270*a

a=\frac{429.5}{270}  \frac{m}{s^{2} }

a= 1.59 m/s²

Calculation of the direction of the acceleration (β)

\beta = tan^{-1} (\frac{F_{N} }{F_{E}})

\beta = tan^{-1} (\frac{390 }{180})

β = 65.22°   (north of east)

5 0
1 year ago
13. A proton moves at 7.50×107 m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path o
diamong [38]

The magnetic force acting on a charged particle moving perpendicular to the field is:

F_{b} = qvB

F_{b} is the magnetic force, q is the particle charge, v is the particle velocity, and B is the magnetic field strength.

The centripetal force acting on a particle moving in a circular path is:

F_{c} = mv²/r

F_{c} is the centripetal force, m is the mass, v is the particle velocity, and r is the radius of the circular path.

If the magnetic force is acting as the centripetal force, set F_{b} equal to F_{c} and solve for B:

qvB = mv²/r

B = mv/(qr)

Given values:

m = 1.67×10⁻²⁷kg (proton mass)

v = 7.50×10⁷m/s

q = 1.60×10⁻¹⁹C (proton charge)

r = 0.800m

Plug these values in and solve for B:

B = (1.67×10⁻²⁷)(7.50×10⁷)/(1.60×10⁻¹⁹×0.800)

B = 0.979T

5 0
2 years ago
What must the charge (sign and magnitude) of a particle of mass 1.49 g be for it to remain stationary when placed in a downward-
Kipish [7]

Answer;

= -2.18 × 10^-5 C

Explanation;

m = 1.49 × 10^-3 kg  

Take downward direction as positive.  

Fg = m g  

E = 670 N/C  

Fe = q E  

Fe + Fg = 0  

q E + m g = 0  

q = -m g/E

   = -1.49 × 10^-3 × 9.81/670

    = -2.18 × 10^-5 C  

=  -2.18 × 10^-5 C

6 0
2 years ago
Other questions:
  • PLEASE help me with this science question and I attached a photo.
    12·1 answer
  • How do mass and speed affect kinetic energy?
    14·2 answers
  • A bobsled zips down an ice track, starting from rest at the top of a hill with a vertical height of 150m. Disregarding friction,
    6·2 answers
  • How to do this problem
    12·2 answers
  • Is it possible for a distance-versus-time graph to be a vertical line?explain
    8·1 answer
  • What is the kinetic energy of a 478 kg object that is moving with the speed of 15 m/s
    11·1 answer
  • Why is it colder as you get closer to the poles?
    8·2 answers
  • A trolley at rest is pushed to accelerate at a
    8·1 answer
  • An object 2cm high is placed 3cm in front of a concave lens of focal length 2cm, find the magnification?​
    6·1 answer
  • Please help meeeee!!! ​
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!