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zvonat [6]
3 years ago
11

What do we call the force used when someone is pushing a lawnmower across the yard?

Physics
2 answers:
raketka [301]3 years ago
7 0
C.) applied force is the answer

dimaraw [331]3 years ago
7 0
I think your answer is C. applied, that is the force we use when pushing a lawnmower across the yard. 
Hope this helps.
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A car speeds up from rest to +16 m/ s in 4s. calculate the acceleration
Pani-rosa [81]

The magnitude of acceleration is (change in speed) / (time for the change).

Change in speed = (speed at the end) - (speed at the beginning) =

                                   (16 m/s)  -  (0)  =  16 m/s .

Time for the change  =  4 s .

Magnitude of acceleration = (16 m/s) / (4 s) = 4 m/s per sec = 4 m/s² .


6 0
3 years ago
Read 2 more answers
A 12.0N force with a fixed orientation does work on a
kvasek [131]

Answer:

(a) \theta=62.31^{\circ}

(b) \theta=117.68^{\circ}

Explanation:

It is given that,

Force acting on the particle, F = 12 N

Displacement of the particle, d=(2.00i -4.00j+3.00k)\ m

Magnitude of displacement, d=\sqrt{2^2+4^2+3^2}= 5.38\ m

(a) If the change in the kinetic energy of the particle is +30 J. The work done by the particle is given by :

W=Fd\ cos\theta

\theta is the angle between force and the displacement

According to work energy theorem, the charge in kinetic energy of the particle is equal to the work done.

So,

cos\theta=\dfrac{W}{Fd}

cos\theta=\dfrac{+30\ J}{12\times 5.38}

\theta=62.31^{\circ}

(b) If the change in the kinetic energy of the particle is (-30) J. The work done by the particle is given by :

cos\theta=\dfrac{W}{Fd}

cos\theta=\dfrac{-30\ J}{12\times 5.38}

\theta=117.68^{\circ}

Hence, this is the required solution.

8 0
3 years ago
Coulomb's law states that the force F of attraction between two oppositely charged particles varies jointly as the magnitude of
Mars2501 [29]

Answer: Force F will be one-sixteenth of the new force when the charges are doubled and distance halved

Explanation:

Let the charges be q1 and q2 and the distance between the charges be 'd'

Mathematical representation of coulombs law will be;

F1=kq1q2/d²...(1)

Where k is the electrostatic constant.

If q1 and q2 is doubled and the distance halved, we will have;

F2 = k(2q1)(2q2)/(d/2)²

F2 = 4kq1q2/(d²/4)

F2 = 16kq1q2/d²...(2)

Dividing equation 1 by 2

F1/F2 = kq1q2/d² ÷ 16kq1q2/d²

F1/F2 = kq1q2/d² × d²/16kq1q2

F1/F2 = 1/16

F1 = 1/16F2

This shows that the force F will be one-sixteenth of the new force when the charges are doubled and distance halved

4 0
3 years ago
Which of the following statements best explains why elements in the same family of the periodic table have similar bonding prope
slava [35]

Answer: D. The elements have the same number of valence electrons

Explanation: The chemical reactivity of elements is governed by the valence electrons present in the element.

The elements present in the same group or vertical column have similar valence configurations and thus behave similarly in chemical reactions or have similar bonding properties.

For Example: Both fluorine and chlorine belong to same family or group and both have 7 electrons in their valence shell and thus accept single electron to attain noble gas configuration.

F:9:1s^22s^22p^5

F^-:10: 1s^22s^22p^6

Cl:17:1s^22s^22p^63s^23p^5

Cl^-:18:1s^22s^22p^63s^23p^6

thus both would bond with a cation bearing a single positive charge.

3 0
3 years ago
Which of the four fundamental forces is responsible for holding together the molecules of the pizza dough as it is spinning in t
defon

Answer:

None.

Explanation:

Molecules are formed by an element's need or excess of electrons. For example, in nature oxygen generally exists as 02. Other molecules are formed via chemical reaction. The example here is the burning of gasoline. Gasoline's two main byproducts are water and carbon dioxide.

Hydrogen as an atom has one electron making it unstable. Put a second hydrogen atom next to the first and the two atoms will share electrons to fill the first energy level the atom needs to be stable.

5 0
2 years ago
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