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
Brut [27]
3 years ago
14

Give two examples of contact and non-contact forces and explain why they are contact and non-contact forces respectively.​

Physics
1 answer:
Sever21 [200]3 years ago
3 0

The Two examples of contact forces are:

  • frictional force
  • Contact force.

The two examples of non contact forces are:

  • Gravitational force
  • magnetic force.

Contact forces happens due to the contact between two objects

Non Contact forces happens because there is no contact between two objects. There is no attraction.

You might be interested in
Friction always opposes an objects?<br><br> A) Power<br> B) Weight<br> C) Motion<br> D) Net force
marin [14]
Friction always opposes motion
8 0
4 years ago
Read 2 more answers
Helllllllppppp
disa [49]

Answer: c and a

Explanation:

3 0
3 years ago
3. An electric motor is used to lift a 6.0 kg of mass through a height of 1 metre. The energy it uses is measure on an energy me
olga55 [171]

Answer:

Climbing stairs and lifting objects is work in both the scientific and everyday sense—it is work done against the gravitational force. When there is work, there is a transformation of energy. The work done against the gravitational force goes into an important form of stored energy that we will explore in this section.



Figure 1. (a) The work done to lift the weight is stored in the mass-Earth system as gravitational potential energy. (b) As the weight moves downward, this gravitational potential energy is transferred to the cuckoo clock.

Let us calculate the work done in lifting an object of mass m through a height h, such as in Figure 1. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight mg. The work done on the mass is then W = Fd = mgh. We define this to be the gravitational potential energy (PEg) put into (or gained by) the object-Earth system. This energy is associated with the state of separation between two objects that attract each other by the gravitational force. For convenience, we refer to this as the PEg gained by the object, recognizing that this is energy stored in the gravitational field of Earth. Why do we use the word “system”? Potential energy is a property of a system rather than of a single object—due to its physical position. An object’s gravitational potential is due to its position relative to the surroundings within the Earth-object system. The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the potential energy equal to 0. We usually choose this point to be Earth’s surface, but this point is arbitrary; what is important is the difference in gravitational potential energy, because this difference is what relates to the work done. The difference in gravitational potential energy of an object (in the Earth-object system) between two rungs of a ladder will be the same for the first two rungs as for the last two rungs.

Converting Between Potential Energy and Kinetic Energy

Gravitational potential energy may be converted to other forms of energy, such as kinetic energy. If we release the mass, gravitational force will do an amount of work equal to mgh on it, thereby increasing its kinetic energy by that same amount (by the work-energy theorem). We will find it more useful to consider just the conversion of PEg to KE without explicitly considering the intermediate step of work. (See Example 2.) This shortcut makes it is easier to solve problems using energy (if possible) rather than explicitly using forces.

More precisely, we define the change in gravitational potential energy ΔPEg to be ΔPEg = mgh, where, for simplicity, we denote the change in height by h rather than the usual Δh. Note that h is positive when the final height is greater than the initial height, and vice versa. For example, if a 0.500-kg mass hung from a cuckoo clock is raised 1.00 m, then its change in gravitational potential energy is

mgh=(0.500 kg)(9.80 m/s2)(1.00 m) =4.90 kg⋅m2/s2=4.90 Jmgh=(0.500 kg)(9.80 m/s2)(1.00 m) =4.90 kg⋅m2/s2=4.90 J

Note that the units of gravitational potential energy turn out to be joules, the same as for work and other forms of energy. As the clock runs, the mass is lowered. We can think of the mass as gradually giving up its 4.90 J of gravitational potential energy, without directly considering the force of gravity that does the

5 0
3 years ago
Basketball= What is the half of the court that is farthest from the offensive basket?
klasskru [66]
The answer would be half court I think
6 0
3 years ago
A body with mass of 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time tequals0 the bod
11Alexandr11 [23.1K]

Answer:

X(t) = 13/13 cos(12t+α)

C =13/13

π/6 s

Explanation:

(A) A body with mass 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time t = 0 the body is pulled 1 m in to the right, stretching the spring, 3 set in motion with an initial velocity of 5 m/s to the left.  

(a) Find X(t) in the form c • cos(w_o*t— α)  

(b) Find the amplitude 3 Period of motion of the body 1  

mass: m = 200g =  0.200 kg  

displacement: ΔX = 20 cm =  0.20 m

Spring Constant: K =  9/0.20 = 45 N/m

IV:   X(0) = 1m V(0) = -5 m/s

Simple Harmonic Motion: c•cos(cosw_t— α) = X(t)  

Circular Frequency: w_o = √k/m= √36/(0.20) = 13 rad/s

X(0) = 1m =c_1

X'(0) = V(0) = c_2*w_o/w_o

        = -5/12 =   c_2

"radians Technically Unitless"  

Amplitude: c = √ci^2 + c^2 ==> √1^2 + (-5/12)^2 = 1 m =13/13 = c

X(t) = 13/13 cos(12t+α)

since, C>0 : damped forced vibration c_1>0, c_2>0

phase angle 2π+tan^-1(c_2/c_1)

                        =2π+tan^-1(-5/12/1)= 5.884

period: T =2π/w_o

                =π/6 s

6 0
4 years ago
Other questions:
  • For the following elementary reaction 2br• -&gt; br2-. The rate of consumption of the reaction and the rate of formation of prod
    14·1 answer
  • Which of the following describes the way heat is transferred in a geyser?
    6·1 answer
  • What do you think would be the most important thing to know about understanding the life of Sioux (Lakota) children?
    7·1 answer
  • At summer camp, the swimming course runs the length (L) of a small lake. To determine the length of the course, the camp counsel
    6·1 answer
  • The amount of diffraction that a sound wave undergoes depends on
    9·1 answer
  • Plz answer this very soon
    9·2 answers
  • What unit must be used when measuring mass
    5·2 answers
  • A quantity that has both direction and magnitude is known as what
    15·1 answer
  • Difference Between Newton's first and third law​
    7·2 answers
  • Which statement best describes energy and matter in a closed system? (2 po
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!