<h2>
<em><u>⇒</u></em>Answer:</h2>
In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)
Step-by-Step Solution:
Solution 35PE
This question discusses about the increased range. So, we shall assume that the angle of jumping will be as the horizontal range is maximum at this angle.
Step 1 of 3<
/p>
The legs have an extension of 0.600 m in the crouch position.
So, m
The person is at rest initially, so the initial velocity will be zero.
The acceleration is m/s2
Acceleration m/s2
Let the final velocity be .
Step 2 of 3<
/p>
Substitute the above given values in the kinematic equation ,
m/s
Therefore, the final velocity or jumping speed is m/s
Explanation:
<span>Normal
fault. Normal (extensional ) fault is a
displacement of a rock as a result of rock-mass movement and occurs when the
crust is stretching. Because of the stretching the thickness of the crust is
reduced and the crust or horizontally extended. </span>
Answer:
Acceleration of gravity=
Explanation:
Newton's Second Law-acceleration is proportional to the net force acting on an object.
All objects usually free fall at the same acceleration of -this regardless of their mass. This acceleration is known as acceleration of gravity.
Answer:
2.69 m/s
Explanation:
Hi!
First lets find the position of the train as a function of time as seen by the passenger when he arrives to the train station. For this state, the train is at a position x0 given by:
x0 = (1/2)(0.42m/s^2)*(6.4s)^2 = 8.6016 m
So, the position as a function of time is:
xT(t)=(1/2)(0.42m/s^2)t^2 + x0 = (1/2)(0.42m/s^2)t^2 + 8.6016 m
Now, if the passanger is moving at a constant velocity of V, his position as a fucntion of time is given by:
xP(t)=V*t
In order for the passenger to catch the train
xP(t)=xT(t)
(1/2)(0.42m/s^2)t^2 + 8.6016 m = V*t
To solve this equation for t we make use of the quadratic formula, which has real solutions whenever its determinat is grater than zero:
0≤ b^2-4*a*c = V^2 - 4 * ((1/2)(0.42m/s^2)) * 8.6016 m =V^2 - 7.22534(m/s)^2
This equation give us the minimum velocity the passenger must have in order to catch the train:
V^2 - 7.22534(m/s)^2 = 0
V^2 = 7.22534(m/s)^2
V = 2.6879 m/s
Answer:
The strength of the magnetic field is 3.5 x 10⁻³ T
Explanation:
Given;
magnitude of the magnetic flux , Φ = 5.90 x 10⁻⁵ T·m²
angle of inclination of the field, θ = 42.0°
radius of the circular plate, r = 8.50 cm = 0.085 m
Generally magnetic flux in a uniform magnetic field is given as;
Φ = BACosθ
where;
B is the strength of the magnetic field
A is the area of the circular plate
Area of the circular plate:
A = πr²
A = π (0.085)² = 0.0227 m²
The strength of the magnetic field:
B = Φ / ACosθ
B = ( 5.90 x 10⁻⁵) / ( 0.0227 x Cos42)
B = 3.5 x 10⁻³ T
Therefore, the strength of the magnetic field is 3.5 x 10⁻³ T
