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3 years ago

Please help ASAP I will mark brainliest

Physics

2 answers:

likoan [24]3 years ago

5 0

Answer:

arrow c tell me if im incorect

Explanation:

chubhunter [2.5K]3 years ago

3 0

The answer is arrow c

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In a velocity versus time graph, what does the slope of a line at any point indicate?

VladimirAG [237]

The acceleration and speed

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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:

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

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

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

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

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

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

What is the acceleration experienced by a car that takes 10s to reach 27m/s from rest

Lerok [7]

|acceleration| = (change in speed) / (time for the change)

Change in the car's speed = (27 - 0) = 27 m/s
Time for the change = 10 sec

|acceleration| = (27 m/s) / (10 s) = 2.7 m/s² .

That's the magnitude of the car's acceleration.
We don't know anything about its direction.

5 0

3 years ago

A motorcycle is stopped at a traffic light. When the light turns green, the motorcycle accelerates to a speed of 91 km/h over a

zimovet [89]

Given :

Initial speed , u = 0 m/s .

Final speed , v = 91 km/h = 25.28 m/s .

To Find :

a) Average acceleration .

b ) Assuming the motorcycle maintained a constant acceleration, how far is it from the traffic light after 3.3 s .

Solution :

a )

We know ,by equation of motion :

$v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{25.28^2-0^2}{2\times 47}\ m/s^2\\\\a=6.8\ m/s^2$

Also , by equation of motion :

$s=ut+\dfrac{at^2}{2}\\\\s=0+\dfrac{6.8\times (3.3)^2}{2}\ m\\\\s=37.02\ m$

Hence , this is the required solution .

6 0

3 years ago

As a system expands, it absorbs 52.5 J of energy in the form of heat from the surroundings. The piston is working against a pres

Kaylis [27]

Answer:

Vi = 0.055 m³ = 55 L

Explanation:

From first Law of Thermodynamics, we know that:

ΔQ = ΔU + W

where,

ΔQ = Heat absorbed by the system = 52.5 J

ΔU = Change in Internal Energy = -102.5 J (negative sign shows decrease in internal energy of the system)

W = Work Done in Expansion by the system = ?

Therefore,

52.5 J = - 102.5 J + W

W = 52.5 J + 102.5 J

W = 155 J

Now, the work done in a constant pressure condition is given by:

W = PΔV

W = P(Vf - Vi)

where,

P = Constant Pressure = (0.5 atm)(101325 Pa/1 atm) = 50662.5 Pa

Vf = Final Volume of System = (58 L)(0.001 m³/1 L) = 0.058 m³

Vi = Initial Volume of System = ?

Therefore,

155 J = (50662.5 Pa)(0.058 m³ - Vi)

Vi = 0.058 m³ - 155 J/50662.5 Pa

Vi = 0.058 m³ - 0.003 m³

7 0

3 years ago