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3 years ago

14

The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the locat

ion of the nucleus?

1 answer:

Nonamiya [84]3 years ago

7 0

Answer:

E=0

Explanation:

The electric field at the centre of the shell is zero because total enclosed charge in the nucleus is zero

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Asteroids do not travel in a straight line as they orbit the Sun. Per Newton's second law of motion, this implies

Morgarella [4.7K]

A force is acting on the planet

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2 years ago

How to do this, i'm completely lost

vaieri [72.5K]

There are two torques t1 and t2 on the beam due to the weights, one torque t3 due to the weight of the beam, and one torque t4 due to the string.

You need to figure out t4 to know the tension in the string.

Since the whole thing is not moving t1 + t2 + t3 = t4.

torque t = r * F * sinФ = distance from axis of rotation * force * sin (∡ between r and F)

t1 =3.2 * 44g
t2 = 7 * 49g
t3 = 3.5 * 24g

t4 = t1 + t2 + t3 = 5570,118

The t4 also is given by:

t4 = r * T * sin Ф

r = 7
Ф = 32°
T: tension in the string

T = t4 / (r * sinФ)

T = t4 / (7 * sin(32°))

T = 1501,6 N

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3 years ago

A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th

NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

$v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2$

$v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)$

$v_{pf}=-31.4\ m/s$

$v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}$

$v_{gf} = 0.0988\ m/s$

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3 years ago

What is the motion of the particles in this kind of wave? A hand holds the left end of a set of waves. The waves themselves make

Sonja [21]

Answer:

The transverse wave

Explanation:

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3 years ago

Two boats start together and race across a 48-km-wide lake and back. boat a goes across at 48 km/h and returns at 48 km/h. boat

jolli1 [7]

Answer:

Time required by boat 1 for the round trip is less than that of boat 2.

Hence, boat 1 wins.

Explanation:

Case 1: Boat 1

Speed of boat = $\frac{distance of river}{time}$

time = $\frac{distance of river}{speed of boat}$

While going to another end

time = $\frac{distance of river}{speed of boat}$

time = $\frac{48}{48}$

time = 1 hour

While going back,

time = $\frac{distance of river}{speed of boat}$

time = $\frac{48}{48}$

time = 1 hour

Total time taken by boat 1 is,

Total time by boat 1 = 1 hour + 1 hour = 2 hour

Total time by boat 1 = 2 hour

Total time taken by boat 1 for the round trip is 2 hour.

Case 2: Boat 2

Speed of boat = $\frac{distance of river}{time}$

time = $\frac{distance of river}{speed of boat}$

While going to another end

time = $\frac{distance of river}{speed of boat}$

time = $\frac{48}{24}$

time = 2 hour

While going back,

time = $\frac{distance of river}{speed of boat}$

time = $\frac{48}{72}$

time = 0.66 hour

Total time taken by boat 2 is,

Total time by boat 1 = 2 hour + 0.66 hour

Total time by boat 1 = 2.66 hour

Total time taken by boat 2 for the round trip is 2.66 hour.

Time required by boat 1 for the round trip is less than that of boat 2.

Hence, boat 1 wins.

5 0

3 years ago