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3 years ago

80PTS

1 answer:

6 0

A) The magnitude of the electric field generated by a single point charge is given by
$E=k \frac{q}{r^2}$
where k is the Coulomb's constant, q is the charge and r is the distance at which we calculate the field.
Since we want to know the intensity of the field at distance of r=4.52 m from the charge of q=8.15 C, the intensity of the electric field is
$E=k \frac{q}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{8.15 C}{4.52 m}=3.6 \cdot 10^9 N/C$

b) Direction of the field
The charge that produces the field is negative, and we know that the electric field generated by a negative charge points toward the charge. Therefore, the direction of the electric field is towards the charge.

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A skier leaves the horizontal end of a ramp with a velocity of 25.0 m/s and lands 70.0 m from the base of the ramp. How high is

Valentin [98]

Answer:

Explanation:

<u>Horizontal Motion </u>

When an object is thrown horizontally with an initial speed v and from a height h, it follows a curved path ruled by gravity.

The maximum horizontal distance traveled by the object can be calculated as follows:

$\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}$

If the maximum horizontal distance is known, we can solve the above equation for h:

$\displaystyle h=\frac {d^2g}{2v^2}$

The skier initiates the horizontal motion at v=25 m/s and lands at a distance d=70 m from the base of the ramp. The height is now calculated:

$\displaystyle h=\frac {70^2\cdot 9.8}{2\cdot 25^2}$

$\displaystyle h=\frac {4900\cdot 9.8}{2\cdot 625}$

h= 38.416 m

The end of the ramp is 38.416 m high

8 0

2 years ago

A juggler is practicing with one ball. It takes 2.4 seconds for the ball to leave her hand and return to her hand. How fast did

nikklg [1K]

The kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s

Given parameters

Time t = 2.4 s

To find

Initial velocity

Kinematics is the part of physics that establishes the relationships between the position, velocity, and acceleration of bodies.

In this case we have a vertical launch

y = y₀ + v₀ t - ½ g t²

Where y and y₀ are the final and initial positions, respectively, v₀ the initial velocity, g the acceleration of gravity (g = 9.8 m / s²) and t the time

With the ball in hand, its position is zero

0 = 0 + v₀ t - ½ g t²

v₀ t - ½ g t² = 0

v₀ = ½ g t

Let's calculate

v₀ = ½ 9.8 2.4

v₀ = 11.76 m / s

In conclusion using kinematics for the vertical launch we can enter the initial velocity is 11.76 m / s

Learn more about vertical launch kinematics here:

brainly.com/question/15068914

5 0

2 years ago

IF YOUR GOOD AT SCIENCE THEN PLEASE ANSWER THIS ASAP I WILL MARK YOU THE BRAINLIEST

Digiron [165]

Answer:

This is an open circuit

Explanation:

An open circuit I believe

It needs to be closed for the bulb to be turned on

8 0

2 years ago

Use the diagram to answer each question.

valentina_108 [34]

Figure A shows cross section of a land form or rock. In Figure B, compression stress is applied on it. When compression stresses are applied on a rock, it squeezes the rock cause fold or fracture. The fault formed by compression stress is called thrust fault. If the compression stresses/ force continue to act on a rock it will converge and form thrust fault. In Figure C, tension stresses is applied on the rock. When a tension stress applied on a rock it deforms/ lengthen. There are three type of deformations occur due to tension stresses. One is elastic deformation, in which, rock retains it original shape when force/stresses are removed. Second is plastic deformation, in which rock lengthen and change occur permanently. Third type of deformation is result into fracture or breaking of rock. In Figure C, shear stresses are applied on rock. Shear stresses are applied with equal magnitude but in opposite direction. It cause breaking of rock.

3 0

3 years ago

Read 2 more answers

A man starts walking from home and walks 2 miles at 20° north of west, then 4 miles at 10° west of south, then 3 miles at 15° no

Rzqust [24]

Answer:

a) R = 2.5 mi b) To return to your case you must walk in the opposite direction or θ = 98º

This is 8º north west

Explanation:

This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately

To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.

First vector A = 2 to 20º north west

Measured from the positive x axis is θ = 180 -20 = 160º

We use trigonometry to find the components

Cos 20 = Aₓ / A

sin 20 = $A_{y}$ / A

Aₓ = A cos 160 = 2 cos 160

$A_{y}$ = A sin160 = 2 sin160

Aₓ = -1,879 mi

$A_{y}$ = 0.684 mi

Second vector B = 4 mi 10º west of the south

Angle θ = 270 - 10 = 260º

cos 2600 = Bₓ / B

sin 260 = $B_{y}$ / B

Bₓ = B cos 260

$B_{y}$ = B sin 260

Bₓ = 4 cos 260

$B_{y}$ = 4 sin 260

Bₓ = -0.6946mi

$B_{y}$ = - 3,939 mi

Third vector C = 3 mi to 15 north east

cos 15 = Cₓ / C

sin15 = $C_{y}$ / C

Cₓ = C cos 15

$C_{y}$ = C sin15

Cₓ = 3 cos 15

$C_{y}$ = 3 sin 15

Cₓ = 2,898 mi

$C_{y}$ = 0.7765 mi

Now we can find the final position of the person

X = Aₓ + Bₓ + Cₓ

X = -1.879 -0.6949 + 2.898

X = 0.3241 mi

Y = $A_{y}$ + $B_{y}$ + $C_{y}$

Y = 0.684 - 3.939 +0.7765

Y = -2.4785 mi

a) We use Pythagoras' theorem

R = √ (x2 + y2)

R = √ (0.3241 2 + (-2.4785) 2)

R = 2.4996 mi

R = 2.5 mi

b) let's use trigonometry

Tan θ = y / x

Tanθ = -2.4785 / 0.3241

θ = tan⁻¹ (-7,647)

θ = -82

Measured from the positive side of the x axis is Te = 360 - 82 = 278º

(90-82) south east

To return to your case you must walk in the opposite direction or Te = 98º

This is 8º north west

3 0

3 years ago