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jekas [21]
3 years ago
8

You should always wear your seatbelt just in case the car comes to an abrupt stop. The seatbelt will hold you in place so that y

our body does not continue moving when the car stops. What property of matter will keep your body in motion when the car comes to a halt? Which of Newton's three laws does this example most closely represent?
Physics
2 answers:
vazorg [7]3 years ago
6 0
Inertia
first law of motion
rodikova [14]3 years ago
4 0

Answer: The correct answers are inertia and first law of motion.

Explanation:

Newton's first law is also known as inertia.

Inertia resists the change in the state of the object.

If the object is in the uniform motion or in rest then it will continue in that state unless it is acted upon by the external force.

In the given problem, the seat belt in the car will hold you in place so that your body does not continue moving when the car stops.

It is due to the inertia. Therefore, Newton's first law represents the given example in the problem.

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What type of bond is formed if atoms donate electrons to other atoms when the elements are combined?
natita [175]
Ionic bond.

In the ionic bond one atom loses one or more electrons, leaving the atom with positive charge, and the other atom accepts those electrons standing with negative charge.
4 0
2 years ago
if an object is moving with a velocity of 24m/s and has an acceleration of -4m/s how long will it take to stop
ololo11 [35]
The negative sign on the acceleration is only a vector quantity that means the object is accelerating to the left. Hence, we can only focus on it magnitude which is 4 m/s^2. Acceleration is the change in velocity over time. The change in velocity must be 24 m/s - 0 m/s, if you want the object to stop. Therefore,

a = (v2 - v1)/t
4 = (24 - 0)t
t = 6 seconds

The object will stop after 6 seconds.
7 0
3 years ago
An electric generator transforms mechanical energy into electrical energy. This process could be done by which of these?
Dimas [21]

Answer:

its d

Explanation:

6 0
3 years ago
Read 2 more answers
Two point charges of equal magnitude are 8.0 cm apart. At the midpoint of the line connecting them, their combined electric fiel
bagirrra123 [75]

Answer:

r = 8/2 = 4cm = 0.04m

k = 9×10^9

Enet = 51 N/C

Enet = E1 + E2

since E1 = E2

E1 = Enet/2 = 51/2

E/2 = kq/r²

q = Er²/2k

q = (51 × 0.04²)/(2×9×10^9)

q = 4.5×10^-12 C

q1 = q2 = 4.5 pC

Explanation:

The electric field is a region around a

charge in which it exerts electrostatic force

on another charges. While the strength of

electric field at any point in space is called

electric field intensity. It is a vector

quantity. Its unit is NC¯¹.

According to coulomb’s law ,if a unit

positive charge q (call it a test charge) is

brought near a charge q (call a field

charge) placed in space,the charge q will

experience a force. The value of this force

depends upon the distance between the

two charges. If the charge q is moved

away from q ,this force would decrease till

at a certain distance the force would be

practically reduced to zero. The charge q

is then out of the influence of charge q.

The region of space surrounding the charge

q in which it exerts a force on the charge

q is known as E.F of the charge

q. Mathematically it is expressed as:

E =F/q

The direction of the vector E is the same

as the direction of F,because q is a

positive scalar. Dimensionally,the E.F is

force per unit charge,and its SI unit is the

newton/coulomb (N/C).

7 0
2 years ago
The car's initial speed was 15 m / s and the distance the car travels before it comes to a complete stop after the driver applie
pentagon [3]

Initial speed of the car (u) = 15 m/s

Final speed of the car (v) = 0 m/s (Car comes to a complete stop after driver applies the brake)

Distance travelled by the car before it comes to halt (s) = 63 m

By using equation of motion, we get:

\bf \longrightarrow  {v}^{2}  =  {u}^{2}  + 2as \\  \\ \rm \longrightarrow  {0}^{2}  =  {15}^{2}  + 2 \times a \times 63 \\  \\ \rm \longrightarrow 0 = 225 + 126a \\  \\ \rm \longrightarrow 126a =  - 225 \\  \\ \rm \longrightarrow a =  -  \dfrac{225}{126}  \\  \\ \rm \longrightarrow a =  - 1.78 \: m {s}^{ - 2}

\therefore Acceleration of the car (a) = -1.78 m/s²

Magnitude of the car's acceleration (|a|) = 1.78 m/s²

5 0
2 years ago
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