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KIM [24]
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?



Group of answer choices

inertia

mass

force

acceleration
Physics
2 answers:
JulsSmile [24]3 years ago
5 0

Answer:

inertia

first

Explanation

BigorU [14]3 years ago
4 0

Answer:

inertia

Explanation:

The property of matter that will keep the body in motion when the car comes to a halt is the inertia force.

Inertia is the ability of a body to remain in static position. It is the tendency to remain in a stable condition where there is no motion.

  • Newton's first law is the law of inertia and it states that a body remain in a state of rest or of uniform motion unless acted upon by an external force.
  • The ability to remain in state of rest by a body is predicated on the force of inertia.
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If 27 J of work are needed to stretch a spring from 15 cm to 21 cm and 45 J are needed to stretch it from 21 cm to 27 cm, what i
kupik [55]

Answer:

9 cm.

Explanation:

The energy used for stretch the spring from 15 cm to 21 cm will be , E_{1}=27J

The energy used for stretch the spring from 21 cm to 27 cm will be , E_{2}=45J

using the energy of spring formula ,we find that

27 = \frac{1}{2}K((21-L^{2})-(15-L^{2}))

45 = \frac{1}{2}K((27-L^{2})-(21-L^{2}))

Dividing both the equation will get,

\frac{3}{5}=\frac{(21-L)^{2}-(15-L)^{2}}{(27-L)^{2}-(21-L)^{2}}\\5((21-L)^{2}-(15-L)^{2})=3((27-L)^{2}-(21-L)^{2})\\3(729 - 54L + L^{2}- 441 + 42L - L^{2} ) = 5(441 - 42L + L^{2} - 225 + 30L - L^{2} )\\3(288 - 12L) = 5(216 - 12L)\\24L = 216\\L = 9 cm

Therefore, the natural length of the spring is, 9 cm.

4 0
3 years ago
What is the second law of thermodynamics
andriy [413]
It states that the total entropy of an isolated system can never decrease over time
5 0
3 years ago
What is the efficiency of a machine?​
Julli [10]

Answer:

Efficiency is the percent of work put into a machine by the user (input work) that becomes work done by the machine (output work).

Explanation:

It is a measure of how well a machine reduces friction.

7 0
3 years ago
PLEASE HELP ASAP Which one of the following is a step used for balancing chemical equations?
scZoUnD [109]

Answer:

C. Count the atoms in each substance in the reactants and products.

Explanation:

A chemical reaction can be defined as a chemical process which typically involves the transformation or rearrangement of the atomic, ionic or molecular structure of an element through the breakdown and formation of chemical bonds to produce a new compound or substance.

In order for a chemical equation to be balanced, the condition which must be met is that the number of atoms in the reactants equals the number of atoms in the products.

This ultimately implies that, the mass and charge of the chemical equation are both balanced properly.

In Chemistry, all chemical equation must follow or be in accordance with the Law of Conservation of Mass, which states that mass can neither be created nor destroyed by either a physical transformation or a chemical reaction but transformed from one form to another in an isolated (closed) system.

One of the step used for balancing chemical equations is to count the atoms in each substance in the reactants and products.

For example;

NH3 + O2 -----> NO + H2O

The number of atoms in each chemical element are;

For the reactant side:

Nitrogen, N = 1

Hydrogen, H = 3

Oxygen, O = 2

For the product side;

Nitrogen, N = 1

Hydrogen, H = 2

Oxygen, O = 2

When we balance the chemical equation, we would have;

NH3 + 3O2 -----> 4NO + 2H2O

3 0
3 years ago
A basketball leaves a player's hands at a height of 2.00 m above the floor. The basket is 3.05 m above the floor. The player lik
shutvik [7]

Answer:(10.69, 11.436)

Explanation:

Given

initial height of ball is 2 m

height of basket is 3.05 m

Launching angle=40^{\circ}

x =12\pm 0.27

y=1.05

equation of trajectory of ball is given by

y=xtan\theta -\frac{gx^2}{2u^2cos^2\theta }

for x=12.27

1.05=12.27\times tan40-\frac{g12.27^2}{2u^2cos^{2}40 }

u=10.69

for x=11.73

1.05=11.73\times tan40-\frac{g11.73^2}{2u^2cos^{2}40 }

u=11.436 m/s

Thus range of speed is (10.69, 11.436)

3 0
3 years ago
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