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timofeeve [1]
1 year ago
6

Emmy kicks a soccer ball up at an angle of 45° over a level field. She watches the ball's trajectory and notices that it lands,

two seconds after being kicked, about 20 m away to the north. Assume that air resistance is negligible, and plot the horizontal and vertical components of the ball's velocity as a function of time. Consider only the time that the ball is in the air, after being kicked but before landing. Take "north" and "up" as the positive x ‑ and y ‑directions, respectively, and use g≈10g≈10 m/s2 for the downward acceleration due to gravity

Physics
1 answer:
Elenna [48]1 year ago
7 0

Let u be the initial velocity of the soccer ball at an angle of inclination of \theta_0 with the positive x-axis.

Given that:

\theta_0=45^{\circ}

The horizontal distance covered by the projectile=20 m

Time of flight, t_f=2 seconds

Acceleration due to gravity, g= 10 m/s^2 downward.

As "north" and "up" as the positive x ‑ and y ‑directions, respectively.

So, g= -10 m/s^2

As the acceleration due to gravity is in the vertical direction, so the horizontal component of the initial velocity remains unchanged.

The x-component of the initial velocity, u_x=u\cos\theta_0.

The horizontal distance covered by the projectile = u_x\times t_f

\Rightarrow u_x\times t_f=20

\Rightarrow u_x\times 2=20

\Rightarrow u_x=10 m/s

So, the horizontal component of the velocity is 10 m/s which is constant and the graph has been shown in the figure (i).

Now,  u\cos(45^{\circ})=10 [as u_x=u\cos\theta_0]

\Rightarrow u=10\sqrt{2} m/s.

The vertical component of the initial velocity,

u_y= u\sin\theta_0

\Rightarrow u_y=10\sqrt{2}\sin(45^{\circ})

\Rightarrow u_y=10 m/s

Let v be the vertical component of the velocity at any time instant t.

From the equation of motion,

v=u+at

where u: initial velocity, v: final velocity, a: constant acceleration, and t: time taken to change the velocity from u to v.

In this case, we have u=u_y, a= -10 m/s^2.

So at any time instant, t.

v=u_y+(-10)t

\Rightarrow v=10-10t

The vertical component of the velocity, v, is the function of time and related as v=10-10t.

This is a linear equation.

At 2 second, the vertical component of the velocity

v=10-10x2=-10 m/s.

The graph has been shown in figure (ii).

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Answer:

Final speed of car = 12 m/s

Explanation:

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        u = 0 m/s

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         t = 5 s

         v = u + at = 0 + 4 x 5 = 20 m/s

b) Then maintains that velocity for 10 s

        v = ?

        u = 20 m/s

        a = 0 m/s²

         t = 10 s

         v = u + at = 20 + 0 x 10 = 20 m/s

c) Then decelerates at the rate of 2.0 m/s² for 4.0 s

        v = ?

        u = 20 m/s

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3 0
1 year ago
Calculate the distance the marble travels during the first 3.0 seconds. [Show all work, including the equation and substitution
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D = V0t + 0.5at^2

Where d is the distance

V0 is the initial velocity

A is the acceleration

T is time

From the graph a = 4/3 m/s2

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A 74.9 kg person sits at rest on an icy pond holding a 2.44 kg physics book. he throws the physics book west at 8.25 m/s. what i
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Answer:

The recoil velocity is 0.2687 m/s.

Explanation:

∵ The person is sitting on an icy surface , we can assume that the surface is frictionless.

∴ There is no force acting acting on the person and book as a system in horizontal direction.

Hence , momentum is conserved for this system in horizontal direction of motion.

If 'i' and 'f' be the initial and final states of this system , then by principle of conservation of momentum(p)  -

p_{i}=p_{f}

System initially is at rest

∴p_{i}=0

∴ From the above 2 equations

p_{f}=0

We know that ,

Momentum(p)=Mass of the body(m)×velocity of the body(v)

Let m_{1} and m_{2} be the mass of the person and the book respectively and v_{1} and v_{2} be the final velocities of the person and book respectively.

∴p_{f}=m_{1}v_{1}+m_{2}v_{2}=0

From the question ,

m_{1} = 74.9 kg

m_{2} = 2.44 kg

v_{2} = 8.25 m/s

Substituting these values in the above equation we get ,

(74.9 × v_{1} )+ (2.44×8.25) = 0

∴v_{1}  = - 0.2687 m/s (Negative sign suggests that the motion of  the person is opposite to that of the book)

∴ The recoil velocity is 0.2687 m/s.

4 0
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Fed [463]

To solve this problem it is necessary to apply the concepts related to the magnetic dipole moment in terms of the current and the surface area, as well as the current density, as a function of the current over the area.

Part A) By definition we know that magnetic dipole moment is

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Where,

I = Current

S = Area

m = IA \rightarrow m= I( \pi r^2)

Replacing with our values we have that,

8*10^{22} = I \pi(\frac{3000*10^3}{2})^2

Re-arrange to find I,

I = 1.1317*10^{10} A

Part B) To find the Current density we need to find the cross sectional area of the Wire:

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Finally the current density is simply J

J = \frac{I}{A}\\J = \frac{1.1317*10^{10}}{7.854*10^{11}}\\J = 0.0144A/m^2

PART C) Finally to make the comparison with the given values we have to cross-sectional area would be

A = \pi (10-3)^2 \\A = 49\pi

Therefore the current density would be

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Comparing the two values we can see that the 2mm wire has a higher current density.

4 0
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An owl has a mass of 4.00 kg. It dives to catch a mouse, losing 800.00 J of its GPE. What was the starting height of the owl, in
vesna_86 [32]

Answer:

height =20m

Explanation:

gpe=mgh

800=4×10×x

40x=800

x=20

3 0
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