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Scorpion4ik [409]
2 years ago
8

Help ASAP Just answer the first question for me please!

Physics
1 answer:
Usimov [2.4K]2 years ago
4 0

Answer:

Im pretty sure its b

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Light travels in a straight line at a constant speed of 3.0 x 10 8 m/s for 4.1
Over [174]

Answer:

Explanation:

As the velocity of light is constant so the acceleration of the light is equal to zero.

a=

dt

dv

​

3 0
3 years ago
A puck of mass 0.70 kg approaches a second, identical puck that is stationary on frictionless ice. The initial speed of the movi
natali 33 [55]

Answer:

  • v_1  =  \ 5.196 \frac{m}{s}
  • v_2 =  3 \frac{m}{s}

Explanation:

For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:

\vec{p}_i = \vec{p}_f

where the suffix i  means initial, and the suffix f means final.

The initial momentum will be:

\vec{p}_i = m_1 \ \vec{v}_{1_i} + m_2 \ \vec{v}_{2_i}

as the second puck is initially at rest:

\vec{v}_{2_i} = 0

Using the unit vector \vec{i} pointing in the original line of motion:

\vec{v}_{1_i} = 6.0 \frac{m}{s} \hat{i}

\vec{p}_i = 0.70 \ kg  \ 6.0 \frac{m}{s} \ \hat{i} + 0.70 \ kg \ 0

\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i}

So:

\vec{p}_i =  4.2 \ \frac{kg \ m}{s} \ \hat{i} = \vec{p}_f

\vec{p}_f =  4.2 \ \frac{kg \ m}{s} \ \hat{i}

Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

So, our velocity vectors will be:

\vec{v}_{1_f} = v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )

\vec{v}_{2_f} = v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )

We got

\vec{p}_f = 0.7 \ kg \ \vec{v}_{1_f} + 0.7 \ kg \ \vec{v}_{2_f}

4.2 \ \frac{kg \ m}{s} \ \hat{i} = 0.7 \ kg \   v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )  + 0.7 \ kg \ v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )

So, we got the equations:

4.2 \ \frac{kg \ m}{s}  = 0.7 \ kg \   v_1 \  cos(30 \°) + 0.7 \ kg \ v_2 \  cos(-60 \°)

and

0  = 0.7 \ kg \   v_1 \  sin(30 \°) + 0.7 \ kg \ v_2 \  sin(-60 \°).

From the last one, we get:

0  = 0.7 \ kg \  ( v_1 \  sin(30 \°) +  \ v_2 \  sin(-60 \°) )

0  =  v_1 \  sin(30 \°) +  \ v_2 \  sin(-60 \°)

v_1 \  sin(30 \°) = -  \ v_2 \  sin(-60 \°)

v_1  =  \ v_2 \  \frac{sin(60 \°)}{ sin(30 \°) }

and, for the first one:

4.2 \ \frac{kg \ m}{s}  = 0.7 \ kg  \ (  v_1 \  cos(30 \°) + v_2 \  cos(60 \°) )

\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} =    v_1 \  cos(30 \°) + v_2 \  cos(60 \°)

\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} =    v_1 \  cos(30 \°) + v_2 \  cos(60 \°)

6 \ \frac{m}{s} =    (\ v_2 \  \frac{sin(60 \°)}{ sin(30 \°) } ) \  cos(30 \°) + v_2 \  cos(60 \°)

6 \ \frac{m}{s} = v_2     (\   \frac{sin(60 \°)}{ sin(30 \°) } ) \  cos(30 \°) +   cos(60 \°)

6 \ \frac{m}{s} = v_2  * 2

so:

v_2 = 6 \ \frac{m}{s} / 2 = 3 \frac{m}{s}

and

v_1  =  \ 3 \frac{m}{s}  \  \frac{sin(60 \°)}{ sin(30 \°) }

v_1  =  \ 5.196 \frac{m}{s}

3 0
3 years ago
A 7.3 cm diameter loop of wire is initially oriented so that its plane is perpendicular to a magnetic field of 0.61 T pointing u
kenny6666 [7]

Answer:

induced emf =  28.65 mV

Explanation:

given data

diameter = 7.3 cm

magnetic field = 0.61

time period = 0.13 s

to find out

magnitude of the induced emf

solution

we know radius is diameter / 2

radius = 7.3 / 2

radius = 3.65 m

so induced emf is dπ/dt  = Adb/dt

induced emf =  A × ΔB / Δt

induced emf =  πr² × ΔB / Δt

induced emf =  π (0..65)² × ( 0.61 - (-0.28))  / 0.13

induced emf =  0.0286538 V

so induced emf =  28.65 mV

3 0
2 years ago
What is the oxidation state of a hydrogen atom bound to an iron atom.?
tresset_1 [31]
the answer is rust so the answer is rust
8 0
3 years ago
Question is down below
rosijanka [135]

The vertical components of velocity is 10.35 m/s and the horizontal component of velocity is 38.6 m/s

<h3>What are the components of velocity?</h3>

We know that velocity is a vector quantity, a vector often can be resolved into its components. The vertical components is V sinθ while the horizontal component is vcosθ.

Hence;

Vertical component = 40 m/s sin 15 degrees = 10.35 m/s

Horizontal component = 40 cos 15 degrees = 38.6 m/s

Learn more about components of velocity:brainly.com/question/14478315

#SPJ1

7 0
1 year ago
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