1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
REY [17]
3 years ago
5

Can someone please help me on this

Physics
1 answer:
Triss [41]3 years ago
4 0

Answer:

I am on the same question and I’m lost

Explanation:

You might be interested in
The courtroom role that possesses the most discretion is?
KonstantinChe [14]
The role of the Prosecutor's
5 0
4 years ago
An airplane flying at 57 m/s releases a crate of supplies that lands 821 m from where it was released. How high was the plane fl
Mashutka [201]

Answer:1020

Explanation:

5 0
3 years ago
What science projects should I do? Please just choose something basic!
Romashka-Z-Leto [24]
How does the different type of music such as classic, hiphop, or rock affect the human mind
4 0
3 years ago
A test charge of +4 µC is placed halfway between a charge of +6 µC and another of +2 µC separated by 20 cm. (a) What is the magn
S_A_V [24]

Answer:

(a) Magnitude: 14.4 N

(b) Away from the +6 µC charge

Explanation:

As the test charge has the same sign, the force that the other charges exert on it will be a repulsive force. The magnitude of each of the forces will be:

F_e = K\frac{qq_{test}}{r^2}

K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q and qtest is the charge of the particles, and r is the distance between the particles.

Let's say that a force that goes toward the +6 µC charge is positive, then:

F_e_1 = K\frac{q_1q_{test}}{r^2}=-9*10^9 \frac{Nm^2}{C^2} \frac{6*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =-21.6 N

F_e_2 = K\frac{q_2q_{test}}{r^2}=9*10^9 \frac{Nm^2}{C^2} \frac{2*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =7.2 N

The magnitude will be:

F_e = -21.6 + 7.2 = -14.4 N, away from the +6 µC charge

3 0
4 years ago
A diffusion couple composed of two silver– gold alloys is formed; these alloys have compositions of 98 wt% Ag–2 wt% Au and 95 wt
jeyben [28]

Answer:

for this problem, 2.5 = (5+2/2)-(5-2/2)erf (50×10-6m/2Dt)

It now becomes necessary to compute the diffusion coefficient at 750°C (1023 K) given that D0= 8.5 ×10-5m2/s and Qd= 202,100 J/mol.

we have D= D0exp( -Qd/RT)

=(8.5×105m2/s)exp(-202,100/8.31×1023)

= 4.03 ×10-15m2/s

4 0
4 years ago
Other questions:
  • If an electric circuit has 0 A of current running through it then .
    9·1 answer
  • A runner accelerated to 4 m/s^2 for 20 seconds before winning the race.How far did he/she run?
    13·1 answer
  • A stone thrown horizontally from a height of 6.07 m hits the ground at a distance of 12.30 m. Calculate the initial speed of the
    5·1 answer
  • If the wings have a surface area of 1600 m2 , how fast must the air flow over the upper surface of the wing if the plane is to s
    5·1 answer
  • Why is important the electromagnetism, diff usages, and how will be our World without it
    13·1 answer
  • The relative density ofbrass is 8.4. Which of
    11·1 answer
  • What are the two main processes carried out by the excretory system?​
    6·1 answer
  • Which examples are a COMPLETE description of motion? Select ALL that apply.
    13·1 answer
  • 2. If Peter is traveling .25 m/s on his bike, how long will it take him to reach
    13·2 answers
  • Please help! It has to deal with Newton's 3rd law! 30 points! and if right brainliest!
    8·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!