Answer:
A. 456 seconds
Explanation:
We are given that two students walk in the same direction along a straight path at a constant speed.
One student walks with a speed=0.90 m/s
second student walks with speed=1.9 m/s
Total distance covered by each students=780 meter
We have to find who is faster and how much time extra taken by slower student than the faster student.
Time taken by one student who travel with speed 0.90 m/s=
Time=
Time taken by one student who travel with speed 0.90 m/s
=
Time taken by one student who travel with speed 0.90 m/s
=866.6 seconds
Time taken by second student who travel with speed 1.9 m/s=
=410.5 seconds
The second student who travels with speed 1.9 m/s is faster than the student travels with speed 0.90 m/s .
Extra time taken by the student travels with speed 0.90 m/s=866.6-410.5=456.1 seconds
Extra time taken by the student travels with speed 0.90 m/s=456 seconds
Hence, option A is true.
The force on a charged particle in a magnetic field is given by
the speed of the charged particle = 10842 m/s.
Explanation:
F= q V B sinθ
F=force=3.5 x 10⁻²N
q= charge= 8.4 x 10⁻⁴ C
B= magnetic field= 6.7 x 10⁻³ T
θ=35⁰
Thus the velocity is given by V=
V=(3.5 x 10⁻²)/[(8.4 x 10⁻⁴)(6.7 x 10⁻³)(sin35)]
V=10842 m/s
Convex.
Concave curved inward (like how a cave foes in) and convex curves outward. Reflected and refracted do not apply to a lens.
I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
1. a. longitudinal waves.
There are two types of waves:
- Transverse waves: in transverse waves, the oscillations of the wave occur in a direction perpendicular to the direction of propagation of the wave
- Longitudinal waves: in longitudinal waves, the oscillations of the waves occur parallel to the direction in which the waves are travelling.
So, these types of waves are called longitudinal waves.
2. d. a medium
There are two types of waves:
- Electromagnetic waves: these waves are produced by the oscillations of electric and magnetic field, and they can travel both in a medium and also in a vacuum (they do not need a medium to propagate)
- Mechanical waves: these waves are produced by the oscillations of the particles in a medium, so they need a medium to propagate - therefore, the correct choice is d. a medium
3. a. AM/FM radio
Analogue signals consist of continuous signals, which vary in a continuous range of values. On the contrary, digital signals consist of discrete signals, which can assume only some discrete values. For AM and FM radios, signals are transmitted by using analogue signals.