Answer:
speed = 7.9 m/s
Explanation:
speed = total distance / time taken
speed = 300 / 38
speed = 7.89473684 m/s
to the nearest tenth
speed = 7.9 m/s
It's just in the name! Accurate data is helpful, and correct, but reproducible data is all of that, and is able to be given to other people through different sources! At least, that's what my understanding of them are. Hope it helps!
Answer:
The strength of magnetic field is 0.2 Tesla.
Explanation:
Data from the question is
Length (L) of wire ; L=0.10 m
Current in wire ; I= 2.0 A
Force on wire ; F = 0.04 N
Angle = Right angle So, 
Now ,
We have to find the magnetic Field strength (B)
For this formula for Force on wire in magnetic field is
Further modified as
Now insert values in the formula
So, the strength of magnetic field is 0.2 Tesla.
Answer:
a) 17.8 m/s
b) 28.3 m
Explanation:
Given:
angle A = 53.0°
sinA = 0.8
cosA = 0.6
width of the river,d = 40.0 m,
the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,
The river itself was 100 m below the ramp H = 100 m,
(a) find speed v
vertical displacement
putting values h=15 m, v=0.8
............. (1)
horizontal displacement d = vcosA×t = 0.6×v ×t
so v×t = d/0.6 = 40/0.6
plug it into (1) and get
solving for t we get
t = 3.734 s
also, v = (40/0.6)/t = 40/(0.6×3.734) = 17.8 m/s
(b) If his speed was only half the value found in (a), where did he land?
v = 17.8/2 = 8.9 m/s
vertical displacement = 
⇒ 
t = 5.30 s
then
d =v×cosA×t = 8.9×0.6×5.30= 28.3 m
<h2>Isaac Newton's First Law of Motion states, "A body at rest will remain at rest, and a body in motion will remain in motion unless it is acted upon by an external force." What, then, happens to a body when an external force is applied to it? That situation is described by Newton's Second Law of Motion. </h2><h2>
equation as ∑F = ma
</h2><h2>
</h2><h2>The large Σ (the Greek letter sigma) represents the vector sum of all the forces, or the net force, acting on a body. </h2><h2>
</h2><h2>It is rather difficult to imagine applying a constant force to a body for an indefinite length of time. In most cases, forces can only be applied for a limited time, producing what is called impulse. For a massive body moving in an inertial reference frame without any other forces such as friction acting on it, a certain impulse will cause a certain change in its velocity. The body might speed up, slow down or change direction, after which, the body will continue moving at a new constant velocity (unless, of course, the impulse causes the body to stop).
</h2><h2>
</h2><h2>There is one situation, however, in which we do encounter a constant force — the force due to gravitational acceleration, which causes massive bodies to exert a downward force on the Earth. In this case, the constant acceleration due to gravity is written as g, and Newton's Second Law becomes F = mg. Notice that in this case, F and g are not conventionally written as vectors, because they are always pointing in the same direction, down.
</h2><h2>
</h2><h2>The product of mass times gravitational acceleration, mg, is known as weight, which is just another kind of force. Without gravity, a massive body has no weight, and without a massive body, gravity cannot produce a force. In order to overcome gravity and lift a massive body, you must produce an upward force ma that is greater than the downward gravitational force mg. </h2><h2>
</h2><h2>Newton's second law in action
</h2><h2>Rockets traveling through space encompass all three of Newton's laws of motion.
</h2><h2>
</h2><h2>If the rocket needs to slow down, speed up, or change direction, a force is used to give it a push, typically coming from the engine. The amount of the force and the location where it is providing the push can change either or both the speed (the magnitude part of acceleration) and direction.
</h2><h2>
</h2><h2>Now that we know how a massive body in an inertial reference frame behaves when it subjected to an outside force, such as how the engines creating the push maneuver the rocket, what happens to the body that is exerting that force? That situation is described by Newton’s Third Law of Motion.</h2><h2 />
