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3 years ago

____________occurs when an object travels in a curved path.

Physics

1 answer:

labwork [276]3 years ago

4 0

Answer:

Acceleration

Explanation:

can you mark me brainlies

So, if an object travels in a curved path, it changes velocity, and, thus, accelerates. This acceleration must be tied to a force. ... Therefore, whenever an object travels in a curved path, there must be an unbalanced force acting upon it. It is important to understand that all this may occur without a change in speed.t

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Five difference between elastic collision and inelastic collision?

olga_2 [115]

Answer:

Elastic Collision

Inelastic Collision

The total kinetic energy is conserved. The total kinetic energy of the bodies at the beginning and the end of the collision is different.

Momentum does not change. Momentum changes.

No conversion of energy takes place. Kinetic energy is changed into other energy such as sound or heat energy.

Highly unlikely in the real world as there is almost always a change in energy. This is the normal form of collision in the real world.

An example of this can be swinging balls or a spacecraft flying near a planet but not getting affected by its gravity in the end.

8 0

3 years ago

Give an example of something that is cyclic and non - cyclic

LiRa [457]

a clock .. and i guess a non functioning clock ?

7 0

3 years ago

Technician A says that low pressure smoke installed in the fuel system can be used to check for leaks. Technicians B says that n

pickupchik [31]

Answer:

A. Technician A only.B.

Explanation: The fuel system of a vehicle is made up of the fuel pump,the fuel filter,the injector or carburettor and the fuel tank. The main function of the fuel system is supply fuel to the engine of a vehicle. In order to check for leakage in the fuel system a small pressure smoke is usually installed in the fuel system.

Nitrogen at low pressure has also been used to check for leakage in the fuel system of a vehicle.

4 0

4 years ago

In the following diagrams the larger vector has a magnitude of 10, and the smaller

Anton [14]

the sum of vectors with the Pythagorean theorem allows us to find that the maximum magnitude occurs for the case:

d) the two vectors are parallel

Vectors are physical quantities that have modulus and direction, for example: force, velocity, acceleration, etc.

Vector algebra has defined the sum, the product by a scalar and by a vector. The modulus and the direction of the resulting vector must be encoded.

The sum of two quantities is done using the Pythagorean theorem

c² = a² + b²

where c is the resultant called hypotenuse, a and b are the summing vectors called legs; trigonometry is used for the direction.

Let's apply this expression to the present case

a, b) perpendicular vectors

c² = a² + b²

c = $\sqrt{6^2+10^2}$

c = 11.7

the magnitude is the same in both cases, changing the direction of the vector

c) Antiparallel vectors

For this case the vectors are collinear, so the sum reduces to the algebraic addition

c = a-b

c = 6 -10

c = -4

d) parallel vectors

c = a + b

c = 4 + 10

c = 14

We can see that the vectors addition gives their maximum and minimum values when the vectors are collinear.

In conclusion using the vectors addition we find that the correct answer is

d) the two vectors are parallel

learn more about vector addition here:

brainly.com/question/15074838

6 0

3 years ago

A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s

Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8 $^{-3}$

Explanation:

$m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg$

$v_{bullet}=341\frac{m}{s}$

Momentum of the motion the first part of the motion have a momentum that is:

$P_{1}=m_{bullet}*v_{bullet}$

$P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529$

The final momentum is the motion before the action so:

a).

$P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}$

$P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}$

$P_{1}=P_{2}$

$2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}$

b).

kinetic energy

$K=\frac{1}{2}*m*(v)^{2}$

Kinetic energy after

$Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J$

Kinetic energy before

$Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J$

Ratio = $\frac{Ka}{Kb}$

$R=\frac{1.14}{406.4}\\R=2.8x10^{-3}$

3 0

4 years ago