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1 year ago

Choice are 1.3 1.0 17

Physics

1 answer:

viva [34]1 year ago

8 0

The current flowing in resistor 1 in the given series circuit is determined as 1.0 A.

<h3>Equivalent resistance of the circuit</h3>

The equivalent resistance of the circuit is determined as follows;

Rt = R1 + R2

Rt = 15 + 5

Rt = 20ohms

<h3>Current flowing in the circuit</h3>

The current flowing in the circuit is determined from Ohm's law.

V = IR

I = V/R

I = 20/20

I = 1.0 A

since it is series circuit, same current will flow in R1 and R2.

Thus, the current flowing in resistor 1 in the given series circuit is determined as 1.0 A.

Learn more about current here: brainly.com/question/24858512

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A 4.00-kg model rocket is launched, expelling 64.0 g of burned fuel from its exhaust at a speed of 675 m/s. What is the velocity

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Answer:

The velocity of the rocket is 7.8 m/s

Explanation:

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2 years ago

A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/

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First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation

$KE_i + PE_i = KE_f + PE_f$

$\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 + 0$

$\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2$

So by solving above equation we will have

$v_f = 84.3 m/s$

now in order to find the momentum we can use

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3 years ago

An event occurs in system K' at x' = 2 m, y' = 3.7 m, z' = 3.7 m, and t' = 0. System K' and K have their axes coincident at t =

fiasKO [112]

Answer:

Coordinates of event in system K are (x,y,z,t)=(5.103m , 3.7m , 3.7m , 1.57×10⁻⁸s)

Explanation:

To find the coordinates of event in system K ,we have to use inverse Lorentz transformation

$x=\frac{x^{|}+vt^{|} }{\sqrt{1-\frac{v^{2} }{c{2} } } } \\x=\frac{2m+0.92c(0) }{\sqrt{1-\frac{(0.92c)^{2} }{c{2} } } }\\x=5.103m\\y=y^{|}\\ y=3.7m\\z=z^{|}\\ z=3.7m$

for t

$r=\frac{1}{\sqrt{1-v^{2} } } \\r=\frac{1}{\sqrt{1-(0.92)^{2} } } \\r=2.551\\t=r(t^{|}+vx^{|}/c^{2} )\\t=2.551(0s+(0.92c)(2)/c^{2} )\\t=1.57*10^{-8}s$

Coordinates of event in system K are (x,y,z,t)=(5.103m , 3.7m , 3.7m , 1.57×10⁻⁸s)

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3 years ago

Block A has a mass of 0.5kg, and block B has a mass of 2kg. Block is is released at a height of 0.75 meters above B. The coeffic

VikaD [51]

Answer:

0.075 m

Explanation:

The picture of the problem is missing: find it in attachment.

At first, block A is released at a distance of

h = 0.75 m

above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

$m_Agh=\frac{1}{2}m_Av_A^2$

where

$g=9.8 m/s^2$ is the acceleration due to gravity

$m_A=0.5 kg$ is the mass of the block

$v_A$ is the speed of the block A just before touching block B

Solving for the speed,

$v_A=\sqrt{2gh}=\sqrt{2(9.8)(0.75)}=3.83 m/s$

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

$e=\frac{v'_B-v'_A}{v_A-v_B}$

where:

e = 0.7 is the coefficient of restitution in this case

$v_B'$ is the final velocity of block B

$v_A'$ is the final velocity of block A

$v_A=3.83 m/s$

$v_B=0$ is the initial velocity of block B

Solving,

$v_B'-v_A'=e(v_A-v_B)=0.7(3.83)=2.68 m/s$

Re-arranging it,

$v_A'=v_B'-2.68$ (1)

Also, the total momentum must be conserved, so we can write:

$m_A v_A + m_B v_B = m_A v'_A + m_B v'_B$

where

$m_B=2 kg$

And substituting (1) and all the other values,

$m_A v_A = m_A (v_B'-2.68) + m_B v_B'\\v_B' = \frac{m_A v_A +2.68 m_A}{m_A + m_B}=1.30 m/s$

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

$\frac{1}{2}m_B v_B'^2 = \frac{1}{2}kx^2$