(this is a practice not a test) I'm struggling to find the answers to this question, could you guys help me?

Dos cargas puntuales están fijas en el eje x: q1 = 6.0µC está en el origen, O, con x1 = 0.0 cm, y q2 = –3.0 µC está situada en e

erik [133]

Answer:

E_total = 1.30 10¹⁰ C / m²

Explanation:

The intensity of the electric field is

E = k q / r²

on a positive charge proof

The total electric field at the midpoint is

as q₁= 6 10⁻⁶ C the field is outgoing to the right

for charge q₂ = -3 10⁻⁶ C, the field is directed to the right, therefore

E_total = E₁ + E₂

E_total = k q₁ / r₁² + k q₂ / r₂²

r₁ = r₂ = r = 4 10⁻² m

E_total = k/r² (q₁ + q₂)

we calculate

E_total = 9 10⁹ / (4 10⁻²)² (6.0 10⁻⁶ +3.0 10⁻⁶)

E_total = 1.30 10¹⁰ C / m²

8 0

3 years ago

An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be

lord [1]

Answer:

a)

the quarterback will be moving back at speed of 0.080625 m/s

b)

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

Given the data in the question;

a)

How fast will he be moving backward just after releasing the ball?

using conservation of momentum;

m₁v₁ = m₂v₂

v₂ = m₁v₁ / m₂

where m₁ is initial mass ( 0.43 kg )

m₂ is the final mass ( 80 kg )

v₁ is the initial velocity ( 15 m/s )

v₂ is the final velocity

so we substitute

v₂ = ( 0.43 × 15 ) / 80

v₂ = 6.45 / 80

v₂ = 0.080625 m/s

Therefore, the quarterback will be moving back at speed of 0.080625 m/s

b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

we make use of the relation between time, distance and speed;

s = d/t

d = st

where s is the speed ( 0.080625 m/s )

t is time ( 0.30 s )

so we substitute

d = 0.080625 × 0.30

d = 0.0241875 m or 2.41875 cm

Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

5 0

3 years ago

A fan cart with the fan set to high rolled across the floor. The cart's speeds with the fan on high are shown below. If the fan

Vadim26 [7]

Answer:

Speed of cart's might be less than the high speed after 5 seconds.

Explanation:

Given that,

A fan cart with the fan set to high rolled across the floor.

Let the speed of fan cart with set to high is $x\ cm$ per second.

The fan supplies a force to the cart. If a lower fan speed were used, less force would be applied. This would cause a slower change in the cart's speed. So, the cart would be rolling more slowly than $x\ cm$ per second after 5 seconds. The speed of cart's might be less than $x\ cm$ per second.

Force is needed

A. for a moving object to keep moving at the same speed and direction

B. for a moving object to change its speed

C. for a motionless object to remain still

D. to prevent a moving object from turning

Hence,

Speed of cart's might be less than the high speed after 5 seconds.

3 0

3 years ago

Which will a positively charged objects attract

Vitek1552 [10]

Me and you! enjoy your day girllllaa or boy

6 0

3 years ago

Read 2 more answers

The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5T, collide with molecules of t

ollegr [7]

Answer:

8.79*10^6 rad/s

Explanation:

To find the frequency of the circular orbit for an electron you use the following expression, for the radius of the trajectory of an electron, that travels trough a constant magnetic field:

$r=\frac{mv}{qB}$ (1)