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
Ket [755]
3 years ago
6

A rope horizontally pulls a massive object lying on a surface with friction with a constant

Physics
1 answer:
SVEN [57.7K]3 years ago
5 0

Answer:

Equal to the frictional force

Explanation:

<u>Question</u>; The options given with regards to a similar question posted online are;

A. Equal (equivalent) to the frictional force

B. Larger than the frictional force

C. Equal to the object's weight

D. More than the object's weight

Explanation

According to Newton's first law of motion, every object shall remain at rest or continue moving with uniform (constant speed) motion unless there is a net force acting on the object

Given that the velocity of the massive block, lying on the surface that has friction, being pulled by the rope = Constant

Therefore;

The net force acting on the moving block while being pulled by the rope = 0

From which we have;

The pulling force = The resistive force

Where;

The pulling force = The (pulling) force (applied) on the rope

The resistive force = The frictional force of the surface which tends to prevent the motion of the block

Therefore, given that the net force acting on the block = 0

The force on the rope = The frictional force (of the surface)

The correct option is option A. Equal to the frictional force.

You might be interested in
El valor más preciso de la masa de un electron es 9.11*10^-11kg ¿ Cuánto aumentaría la masa de un cuerpo que se carga con -1 c d
chubhunter [2.5K]

Answer:

m = 569.375\times 10^{6}\,kg

Explanation:

Un electrón tiene una carga negativa de 1.6\times 10^{-19}\,C y la masa total es igual al producto del número de electrones y esa carga unitaria. El número de electrones se obtiene al dividir la carga total por la carga unitaria. (An electron has a negative charge of 1.6\times 10^{-19}\,C and the total mass is equal to the product of the qunatity of electrons and such unit charge. The quantity of electrons is found by diving the total charge by the unit charge):

m = \frac{Q}{q} \cdot m_{e}

m = \left(\frac{1\,C}{1.6\times 10^{-19}\,C} \right)\cdot (9.11\times 10^{-11}\,kg)

m = 569.375\times 10^{6}\,kg

7 0
3 years ago
A 97.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.63 r
sammy [17]

Answer:

the final angular velocity of the platform with its load is 1.0356 rad/s

Explanation:

Given that;

mass of circular platform m = 97.1 kg

Initial angular velocity of platform ω₀ = 1.63 rad/s

mass of banana m_{b} = 8.97 kg

at distance r = 4/5  { radius of platform }

mass of monkey m_{m} = 22.1 kg

at edge = R

R = 1.73 m

now since there is No external Torque

Angular momentum will be conserved, so;

mR²/2 × ω₀ = [ mR²/2 + m_{b} (\frac{4}{5} R)² + m_{m}R² ]w

m/2 × ω₀ = [ m/2 + m_{b} (\frac{4}{5} )² + m_{m} ]w

we substitute

w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1

w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )

w = 48.55 × [ 1.63 / ( 76.3908 ) ]

w = 48.55 × 0.02133

w = 1.0356 rad/s

Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s

8 0
3 years ago
A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a
tigry1 [53]

Explanation:

Given that,

Mass if the rock, m = 1 kg

It is  suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.

We need to find the mass of the meter stick. The force acting by the stone is

F = 1 × 9.8 = 9.8 N

Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}

W = 3.266 N

The mass of the meters stick is :

m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg

So, the mass of the meter stick is 0.333 kg.

5 0
3 years ago
Why are unit of length mass and time independent with each other?​
mariarad [96]
Масса (килограмм), длина (метр) и время (секунда) - это 3 из 7 основных единиц в системе СИ.

Базовые единицы произвольно определены как то, что они есть, поэтому они не могут быть получены.

Все другие единицы в системе СИ являются производными от 7 основных единиц, таких как скорость (метр / секунда).

Остальные 4 - это температура (Кельвин), количество вещества (моль), электрический ток (ампер) и светимость (кандела).
4 0
3 years ago
Ну car has 4 wheels is a<br> observation. *<br> 1 point<br> O Quantitative<br> Qualitative
sergiy2304 [10]
Quantitative because it tells you how many wheels
7 0
3 years ago
Other questions:
  • Which of the following describes an organ system?
    10·2 answers
  • A non-conducting sphere of radius R = 3.0 cm carries a charge Q = 2.0 mC distributed uniformly throughout its volume. At what di
    8·1 answer
  • Evaporation is a process that requires energy to occur
    8·1 answer
  • Which of these is closest to the age of our solar system?
    9·1 answer
  • Frost on the grass in autumn is an example of
    13·1 answer
  • The force that makes objects keep traveling is ____.
    12·2 answers
  • IDENTIFY WHAT IS BEING DESCRIBED IN EACH SENTENCE AND WRITE YOUR ANSWER ON THE BLANKS
    11·1 answer
  • The physical (natural) laws which establish how things work can be investigated by the scientific method. True or False
    12·2 answers
  • The melting of a glacier is an example of the interactions among which of Earth’s spheres?
    15·1 answer
  • Which combinations of forces would best move a large, heavy box across a room? Choose the three statements that apply.
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!