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2 years ago

WILL MARK BRAINLIEST!!!!

Physics

1 answer:

Ksju [112]2 years ago

8 0

Explanation:

kinetic energy = 1 /2 x mass x velocity²

substitute the values given into the equation

250 = 1/2 x 20 x velocity ² ( remember it is velocity squared )

5 m/s = velocity

hope this helps bro

please mark it brainliest

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A vector has a magnitude of 30m at an angle of 225 degrees with respect to the positive x-axis and has a magnitude of 13m.what a

Aneli [31]

Answer:

I'm pretty sure this is not a complete question. My guess is that you are trying to add/subtract vectors. Vectors have both magnitude and direction, so vector A is pretty clear, but a magnitude of 13 (i'm guessing a resultant) without a direction is weird.

IF 13 is the magnitude of the resultant, vector B added to vector A could have any magnitude 17 ≤ B ≤ 43

It could have any direction of

θ = (225 - 180) ± arcsin(13/30)

θ = 45 ± 25.679...

70.679 ≤ θ ≤ 19.321

components of vector B would be

Bx = |B|cosθ

By = |B|sinθ

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2 years ago

Hi so this is 15 points :D

mario62 [17]

Answer:

A. The core makes up the majority of Earth's volume.

Explanation:

Although the core and mantle are about equal in thickness, the core actually forms only 15% of earth's volume whereas the mantle takes up 84%

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2 years ago

Two high-current transmission lines carry currents of 29.0 A and 78.0 A in the same direction and are suspended parallel to each

jarptica [38.1K]

Answer with Explanation:

We are given that

$I_1=29 A$

$I_2=78 A$

$d=38 cm=\frac{38}{100}=0.38 m$

$1 m=100 cm$

a.Length of segment,l=20 m

Magnetic force ,F= $\frac{2\mu_0I_1I_2 l}{4\pi d}$

$\frac{\mu_0}{4\pi}=10^{-7}$

Substitute the values

$F=\frac{10^{-7}\times 29\times 78\times 20}{0.38}=0.0119 N$

Hence, the magnetic force exert by each segment on the other=0.0119 N

b.We know that when current carrying in the wires are in same direction then the force will attract to each other.

Hence, the force will be attractive.

4 0

2 years ago

Read 2 more answers

Who was the decline in the anchovy of peru blamed on

____ [38]

<span> The failure of the anchovy harvest in 1972 was blamed on El Niño.</span>

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3 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

2 years ago