Gravity is a force that pulls things

Determine the direction of the force that will act on the charge in each of the following situations. A negative charge moving t

wlad13 [49]

Answer:

a) DOWN direction, b) directed INTO THE SCREEN, c) F = 0

Explanation:

The direction of the force is

for electric force

F = q E

where we assume a positive test charge, for which the force has the direction of the electric field.

For a magnetic field

in this case the direction of the force is given by the right hand rule.

For a positive test charge, the thumb points in the direction of velocity, the other fingers extended in the direction of the magnetic field, and the palm gives the direction of force for a positive charge.

F = q v x B

Let us apply these considerations to our case.

a) negative charge moving to the left

in a magnetic field points away from the screen

In this case the thumb goes to the left, the fingers extended outwards and the palm points upwards, but since the charge is negative the force has a DOWN direction.

b) negative charge moves to the left

in electric field it points off the screen.

The outside is in the direction of the electric field and since the charge is negative, the force is directed INTO THE SCREEN

c) positive charge moves down

in magnetic field points up

in this case the velocity and the field have the same direction so the vector product of them is zero

F = q v B sin 0

F = 0

6 0

3 years ago

Daring Darless wishes to cross the Grand Canyon of the Snake River by being shot from a cannon. She wishes to be launched at 65

guajiro [1.7K]

Answer:

She must be launched with minimum speed of <u>57.67 m/s</u> to clear the 520 m gap.

Step-by-step explanation:

Given:

The angle of projection of the projectile is, $\theta =65$ °

Range of the projectile is, $R=520$ m.

Acceleration due to gravity, $g=9.8\ m/s^2$

The minimum speed to cross the gap is the initial speed of the projectile and can be determined using the formula for range of projectile.

The range of projectile is given as:

$R=\frac{v_{0}^2\sin2\theta}{g}$

Plug in all the given values and solve for minimum speed, $v_0$ .

$520=\frac{v_{0}^2\sin(2(65))}{9.8}\\520\times 9.8=v_{0}^2\sin(130)\\5096=1.532v_{0}^2\\v_0^2=\frac{5096}{1.532}\\v_0^2=3326.371\\v_0=\sqrt{3326.371}=57.67\textrm{ m/s}$

Therefore, she must be launched with minimum speed of 57.67 m/s to clear the 520 m gap.

3 0

3 years ago

If f(x) = x2 - 3x + 1 and f(2 a) = 2f(a) then a is equal to

loris [4]

For these question, it has two separate equations: 2f(a) and f(2a) .

For f(2a) equations its x=2a, so you must substitute 2a into the f(x) equation

For 2f(a), it means the two time of f(a) equation with x=a, so you substitute a inti f(x) equation first, then you multiply it by 2.

5 0

3 years ago

A bicyclist travels 30 km in 2 hours, what is their average speed?

Mamont248 [21]

Explanation:

average speed = total distance ÷ total time taken

=30÷2 = 15 km/h

3 0

3 years ago

A person stands on top of a tall building holding two rocks at a height of 50 meters. At the same moment, one rock is dropped fr

labwork [276]

Answer

The second rock will land 2.4s after the first rock

Explanation:

Given that

Height of the building s=50m

We assume that the first rock is acting with gravity so that a=9.81m/s

And initial velocity u=0

Applying the equation of motion

S=ut+1/2at²

50=0*t+1/2(9.81)t²

50=4.905t²

t²=50/4.905

t²=10.19

t=√10.19

t=3.19sec

For the second rock initial velocity u=8m/s and v=0 and a=9.81

Applying the equation of motion

v=u+at

0=8+9.81t

t=-8/9.81

t=0.81sec

Hence the second rock will land 2.4s after the first rock

I.e

3.19-0.81

3 0

3 years ago