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andrew11 [14]
3 years ago
5

A uniform electric field is oriented in the −z direction. The magnitude of the electric field is 6500 N/C.

Physics
1 answer:
nignag [31]3 years ago
5 0

Answer:

In a plane parallel to the xy plane.

Explanation:

Direction of uniform electric field is oriented in the −z direction.

a) We know that the direction of equipotential surfaces are always normal to electric field direction. So, equipotential surface will be in a plane perpendicular to electric field that in a plane perpendicular to z-axis that is parallel to the xy plane.

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A projectile is launched horizontally from a height of 8.0 m. The projectile travels 6.5 m before hitting the ground.
spayn [35]

Answer:

5.09 m/s

Explanation:

Use the height to find the time it takes to land:

y = y₀ + v₀ᵧ t + ½ gt²

0 = 8.0 m + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 1.28 s

Now use the horizontal distance to find the initial velocity.

x = x₀ + v₀ₓ t + ½ at²

6.5 m = 0 m + v₀ (1.28 s) + ½ (0 m/s²) (1.28 s)²

v₀ = 5.09 m/s

7 0
3 years ago
Given vectors D (3.00 m, 315 degrees wrt x-axis) and E (4.50 m, 53.0 degrees wrt x-axis), find the resultant R= D + E. (a) Write
Eva8 [605]

Answer:

  • R = ( 4.831 m , 1.469 m )
  • Magnitude of R = 5.049 m
  • Direction of R relative to the x axis= 16°54'33'

Explanation:

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

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

where | \vec{A} | its the magnitude and θ.

So, for our vectors, we will have:

\vec{D}= 3.00 m \ ( \ cos(315) \ , \ sin (315) \ )

\vec{D}=  ( 2.121 m , -2.121 m )

and

\vec{E}= 4.50 m \ ( \ cos(53.0) \ , \ sin (53.0) \ )

\vec{E}= ( 2.71 m , 3.59 m )

Now, we can take the sum of the vectors

\vec{R} = \vec{D} + \vec{E}

\vec{R} = ( 2.121 \ m , -2.121 \ m ) + ( 2.71 \ m , 3.59 \ m )

\vec{R} = ( 2.121 \ m  + 2.71 \ m , -2.121 \ m + 3.59 \ m )

\vec{R} = ( 4.831 \ m , 1.469 \ m )

This is R in Cartesian representation, now, to find the magnitude we can use the Pythagorean theorem

|\vec{R}| = \sqrt{R_x^2 + R_y^2}

|\vec{R}| = \sqrt{(4.831 m)^2 + (1.469 m)^2}

|\vec{R}| = \sqrt{23.338 m^2 + 2.158 m^2}

|\vec{R}| = \sqrt{25.496 m^2}

|\vec{R}| = 5.049 m

To find the direction, we can use

\theta = arctan(\frac{R_y}{R_x})

\theta = arctan(\frac{1.469 \ m}{4.831 \ m})

\theta = arctan(0.304)

\theta = 16\°54'33''

As we are in the first quadrant, this is relative to the x axis.

3 0
3 years ago
When water vapor is cooled it three forms into water droplets why does this represent a conservation in mass
AVprozaik [17]
In the conservation of mass, mass is never created or destroyed in chemical reactions in the same way water is not created or destroyed it is only transferred from one form to another and its mass is always conserved.
8 0
3 years ago
Which objects in space formed from the huge disk cisce and debris beyond the outer planets? Select two option
Dennis_Churaev [7]

Answer:

Comets

Explanation:

The Kuiper Belt is a collection of trans-Neptunian objects that consist of comets and other dwarf planets, including Pluto.

4 0
3 years ago
Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30∘ above the
Pavlova-9 [17]

Answer:1.71 m/s

Explanation:

Given

mass of Susan m=12 kg

Inclination \theta =30^{\circ}

Tension T=29 N

coefficient of Friction \mu =0.18

Resolving Forces Along x axis

F_x=T\cos \theta -f_r

where f_r=friction\ Force  

F_y=mg-N-T\sin \theta

since there is no movement in Y direction therefore

N=mg-T\sin \theta

and f_r=\mu N

Thus F_x=T\cos \theta -\mu N

F_x=29\cos (30)-\0.18\times (12\times 9.8-29\sin (30))                

F_x=25.114-18.558

F_x=6.556 N

Work done by applied Force is equal to change to kinetic Energy

F_x\cdot x=\frac{1}{2}\cdot mv_f^2-\frac{1}{2}\cdot mv_i^2

6.556\times 2.7=\frac{1}{2}\cdot 12\times v_f^2

v_f^2=\frac{6.556\times 2.7\times 2}{12}

v_f^2=2.95

v_f=1.717 m/s        

8 0
3 years ago
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