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3 years ago

Kinetic energy....

Physics

1 answer:

Solnce55 [7]3 years ago

7 0

Explanation:

Given parameters:

mass of bullet 1 = 0.0014kg

velocity of bullet 1 = 396m/s

mass of bullet 2 = 7.1 x 10⁷kg

velocity of bullet 2 = 13m/s

Unknown;

Comparing the kinetic energy

Solution:

Kinetic energy is the energy due to the motion of a body.

K.E = $\frac{1}{2}$ m v $^{2}$

bullet 1:

K.E = $\frac{1}{2}$ x 0.0014 x 396 $^{2}$ = 109.8J

bullet 2:

K.E = $\frac{1}{2}$ 7.1 x 10⁷ x 13 $^{2}$ = 6 x 10⁹J

The second bullet has more kinetic energy than the first one.

Learn more:

Kinetic energy brainly.com/question/6536722

#learnwithBrainly

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A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object mo

Zinaida [17]

Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

$E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2$

$E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2$

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

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3 years ago

If an object has a mass of 10 kilograms, how much does it weigh in newtons?

schepotkina [342]

10 kilograms of mass weighs 98.1 newtons on Earth,
16.2 newtons on the Moon, 37.1 newtons on Mars,
and other weights in other places.

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3 years ago

How much mass should be attached to a vertical ideal spring having a spring constant (force constant) of 39.5 n/m so that it wil

mrs_skeptik [129]

The frequency of a simple harmonic oscillator such as a spring-mass system is given by
$f= \frac{1}{2 \pi} \sqrt{ \frac{k}{m} }$
where
k is the spring constant
m is the mass attached to the spring.

Re-arranging the formula, we get:
$m= \frac{k}{4 \pi^2 f^2}$
and since we know the constant of the spring:
$k=39.5 N/m$
and the frequency of oscillation:
f=1.00 Hz
we can find the value of the mass attached to it:
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3 years ago

Which statement about the volume and pressure of a gas is the most accurate? Volume of a gas is equal to pressure of that gas in

Julli [10]

Answer: Volume of a gas is inversely proportional to pressure of that gas in any container.

Explanation:

Hi, according to Boyle's Gas law, the volume of a gas is inversely proportional to the pressure of that gas, at a constant temperature.

The expression is:

P1.V1= P2.V2

V= 1/P

PV = k

Where:

P = pressure of a gas

V = volume of a gas

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Feel free to ask for more if needed or if you did not understand something.

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2 years ago

Read 2 more answers

The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5

ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E = K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.89e5)^{2} + (7.89e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) net magnitude = 1.115e6 $\frac{N}{C}$ @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.88e5)^{2} + (7.88e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) and (b) the net magnitude = 1.242e6 $\frac{N}{C}$ @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0

2 years ago