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3 years ago

True or False The greater the speed of an object, the less kinetic energy it possesses.

Physics

2 answers:

oksano4ka [1.4K]3 years ago

7 0

That Is A True Statement

Anettt [7]3 years ago

3 0

That is true because if the object is moving at Forceful speeds than it will lose more of its kinetic energy

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Hii please help i’ll give brainliest if you give a correct answer please please hurry it’s timed

Yanka [14]

Answer:

Sorry I'm wrong The person above is correct.

I tried.

4 0

3 years ago

Read 2 more answers

A ship travels with velocity given by 12, with current flowing in the direction given by 11 with respect to some co-ordinate axe

nataly862011 [7]

Answer:

$v_x = 11.78 m/s$

Explanation:

Velocity of the ship is given as

$v = 12 units$

the direction of the velocity of the ship is making an angle of 11 degree with the current

so we will have two components of the velocity

1) along the direction of the current

2) perpendicular to the direction of the current

so here we know that the component of the ship velocity along the direction of the current is given as

$v_x = v cos\theta$

$v_x = 12 cos11$

$v_x = 11.78 m/s$

7 0

3 years ago

Which statement is correct? Theories are accepted as true when a single experiment yields similar results to another one. When a

levacccp [35]

The third statement is correct.

3 0

3 years ago

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When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam

Len [333]

Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s$

b) the change in the combined kinetic energy of the two-car system during this collision

$\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))$

substitute the value in the equation above

$=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J$

Hence, the change in combine kinetic energy is -2534.78J

8 0

3 years ago

A closed container initially holds 50 monatomic Aparticles that have a combined energy of 480 units. After 100 monatomic B parti

Molodets [167]

Answer:

"8 units" is the appropriate answer.

Explanation:

According to the question,

Throughout equilibrium all particles are of equivalent intensity, and as such the integrated platform's total energy has been uniformly divided across all individuals.

Now,

The total energy will be:

= $480+720$