Answers
Answer:(b)
Explanation:
Given
Electric current flowing through the horizontal wire towards the observer's face.
The direction of the magnetic field is given by the right-hand thumb rule, i.e. place the thumb in the direction of current and the wrapping of fingers will give the direction of the magnetic field
the direction of the magnetic field will be counterclockwise as observed by an observer.
Related Questions
A small mailbag is released from a helicopter that is descending steadily at 2.00 m/s. After 3.00 s.
Required:
a. What is the speed of the mailbag?
b. How far is it below the helicopter?
Answers
Answer:
Explanation:
Initial velocity of mailbag u = 2 m/s
acceleration downwards a = g = 9.8 m/s²
time t = 3 s
a ) final velocity v = ?
v = u + at
= 2 + 9.8 x 3
= 31.4 m /s
b )
s = ut + 1/2 g t²
s is relative displacement of mailbag
u = relative initial velocity of mailbag = 0
relative acceleration = g = 9.8 m /s²
time t = 3 s
s = 0 + 1/2 x 9.8 x 3²
= 44.1 m
relative displacement of mailbag = 44.1 m .
HELP ASAP WILL GIVE BRAINLIEST TO WHOEVER ANSWERS FIRST!!!!
The Left and right forces acting on this cart are
A. balanced
B. super-positioned
C. unbalanced
D. counter-balanced
Answers
Answer:
B. Super-Positioned
Explanation:
a car with a mass of 2050 kg is traveling at 4.5 m/s. what is the cars momentum
Answers
Answer:
9225 kg m/s
Explanation:
momentum = mass * velocity
momentum = 2050 * 4.5
momentum = 9225
If a force acts constantly on a stationary object, the object will
Answers
Answer:
accelerate in the direction of the force
Explanation:
An alarm clock is dropped off the edge of a tall building. You, standing directly under it, hear a tone of 1350 Hz coming from the clock at the instant it hits the ground. Since you know the building is 25 m tall, you can find out what the frequency of the alarm would be if you had just held it in your hands. What would that frequency be
Answers
Answer:
the frequency clock would be 1262.85 Hz
Explanation:
Given data;
height of building h = 25 m
from the third equation of motion;
v² = u² + 2as
Since the Alarm clock falls with an acceleration equal to the acceleration due to gravity; a = g = 9.81 m/s²
initial velocity u = 0
so we substitute our values into the kinematic equation
v² = (0)² + 2 × 9.81 × 25
v² = 490.5
v = √490.5
v = 22.1472 m/s
Now, since the alarm clock is moving both I am stationary;
my velocity will be zero.
so Frequency of the alarm clock will be;
f' = [ (v - [tex]v_{s}[/tex] ) / ( v + [tex]v_{0}[/tex] ) ] × f
we know that; speed of sound is 343 m/s, so v = 343 m/s, [tex]v_{s}[/tex] is 22.1472 m/s, f is 1350 Hz, [tex]v_{0}[/tex] is 0 m/s
so we substitute the values into the equation
f' = [ (343 - 22.142 ) / ( 343 + 0 ) ] × 1350
f' = [ 320.858 / 343 ] × 1350
f' = 0.935446 × 1350
f' = 1262.85 Hz
Therefore, the frequency clock would be 1262.85 Hz
The source frequency or frequency of the alarm is 1,262.25 Hz.
The given parameters:
Observed frequency, Fo = 1350 HzHeight of the building, h = 25 mSpeed of sound, V = 343 m/sThe source velocity is calculated as follows;
[tex]v^2 = u^2 + 2gh\\\\ v^2 = 0 + 2gh\\\\ v = \sqrt{2gh} \\\\ v = \sqrt{2 \times 9.8 \times 25} \\\\ v = 22.14 \ m/s[/tex]
The source frequency or frequency of the alarm is calculated by applying Doppler effect as follows;
[tex]f_s = f_o (\frac{v- v_s}{v+ v_0} )\\\\ f_s = 1350 (\frac{343-22.14}{343} )\\\\ f_s = 1,262.25 \ Hz[/tex]
Learn more about Doppler effect here: https://brainly.com/question/3841958
A 40.0-kg packing case is initially at rest on the floor of a 1500-kg pickup truck. The coefficient of static friction between the case and the truck floor is 0.30, and the coefficient of kinetic friction is 0.20. Before each acceleration given below, the truck is traveling due north at constant speed. Find the magnitude and direction of the friction force acting on the case
(a) when the truck accelerates at 2.20m/s22.20m/s 2 northward and
(b) when it accelerates at 3.40m/s23.40m/s 2 southward.
Answers
Answer:
Before providing an answer to the question, the values for acceleration given in questions A and B were written twice. So correction would go like this: For (a) when the truck accelerates at 2.20m/s2 northward, and for (b) when it accelerates at 3.40m/s2 southward.
The answer:
(a) 88N, northward.
(b) 78.4, southward.
Explanation:
(a) Maximum frictional force acting on the packing case= (coefficient of static friction) X (Normal force)
Normal force = mass X acceleration due to gravity
Maximum static frictional force acting on the packing case = (coefficient of static friction) X (mass of packing case X acceleration due to gravity)
Maximum static frictional force = (0.30) X (40.0-kg) X (9.8m/s 2) = 117.6N
While Reaction force acting on the packing case = (mass of packing case) x (acceleration generated by the pickup truck)
Reaction force acting on the case = (40.0-kg) X (2.20m/s2) = 88N
With these values, one can conclude that the packing case is at rest since the reaction force of the case acting in the opposite direction is lesser than the frictional force. Making the magnitude and direction of the friction force acting on the case still move northward, and the static frictional force acting on equals the reaction force.
The answer is 88N, northward.
(b) Here too, we need to still compare the reaction force with the value of the already determined Maximum static frictional force (117.6N) above. This is necessary to know the frictional force between the pickup truck"s floor and the packing case.
Reaction force acting on the case when acceleration is 3.40m/s2 = (40.0-kg) X (3.40m/s2) = 136 N
We can conclude that the reaction force (136 N) is greater than the maximum static frictional force (117.6N), suggesting that the packing case is in motion and the frictional force is no longer static.
This means a kinetic force is now acting on the pickup truck"s floor causing the packing case to also move. This kinetic force can be calculated as:
kinetic force = (coefficient of kinetic friction) X (mass of packing case X acceleration due to gravity)
= (0,20) X (40.0-kg) X (9.8m/s 2) = 78.4N
You and a friend each hold a lump of wet clay. Each lump has a mass of 30 grams. You each toss your lump of clay into the air, where the lumps collide and stick together. Just before the impact, the velocity of one lump was < 3, 3, -3 > m/s, and the velocity of the other lump was < -4, 0, -4 > m/s. What is the velocity of the stuck-together lump just after the collision
Answers
Answer:
[tex]<-0.5, 1.5, -3.5>\ \text{m/s}[/tex]
Explanation:
[tex]u_1[/tex] = Velocity of one lump = [tex]3x+3y-3z[/tex]
[tex]u_2[/tex] = Velocity of the other lump = [tex]-4x+0y-4z[/tex]
m = Mass of each lump = [tex]30\ \text{g}[/tex]
The collision is perfectly inelastic as the lumps stick to each other so we have the relation
[tex]mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=<-0.5, 1.5, -3.5>\ \text{m/s}[/tex]
The velocity of the stuck-together lump just after the collision is [tex]<-0.5, 1.5, -3.5>\ \text{m/s}[/tex].
