Answer:
4500 N
Explanation:
When a body is moving in a circular motion it will feel an acceleration directed towards the center of the circle, this acceleration is:
a = v^2/r
where v is the velocity of the body and r is the radius of the circumference:
Therefore, a body with mass m, will feel a force f:
f = m v^2/r
Therefore we need another force to keep the body(car) from sliding, this will be given by friction, remember that friction force is given a the normal times a constant of friction mu, that is:
fs = μN = μmg
The car will not slide if f = fs, i.e.
fs = μmg = m v^2/r
That is, the magnitude of the friction force must be (at least) equal to the force due to the centripetal acceleration
fs = (1000 kg) * (30m/s)^2 / (200 m) = 4500 N
Answer:
4.4 cm
Explanation:
Given:
Distance of the screen from the slit, D = 1 m
Distance between two third order interference minimas, x = 22 cm
Let's say, minima occurs at:
We have:
Calculating further for the width of the central bright fringe, we have:
= 4.4 cm
Note: w in representswavelength
Gravity adds 9.8 m/s to the speed of a falling object every second.
An object dropped from 'rest' (v = 0) reaches the speed of 78.4 m/s after falling for (78.4 / 9.8) = <em>8.0 seconds</em> .
<u>Note:</u>
In order to test this, you'd have to drop the object from a really high cell- tower, building, or helicopter. After falling for 8 seconds and reaching a speed of 78.4 m/s, it has fallen 313.6 meters (1,029 feet) straight down.
The flat roof of the Aon Center . . . the 3rd highest building in Chicago, where I used to work when it was the Amoco Corporation Building . . . is 1,076 feet above the street.
Answer:
Interact with it
Explanation:
The body in a condensed form (often a dot or a box)
Straight arrows pointing in the direction in which forces act on the body are represented.
Moments are portrayed by curved arrows pointing in the direction in which they impact the body.
A coordinate system is a series of coordinates.
Answer:
a) t = 0.90 s, b) t = 0.815 s, c) t = 0.90 s, d) x = 3.6 m, e) t = 0.639 s
Explanation:
all these exercises are about kinematics
a) The body is released from rest,
y = y₀ + v₀ t - ½ g t²
in this case when reaching the ground y = 0 and its initial velocity is vo = 0
0 = y₀ + 0 - ½ g t²
t² = 2 y₀ / g
t² = 2 4 /9.81
t² = 0.815
t = √0.815
t = 0.90 s
b) It is thrown upwards at v₀ = 4 m / s
y = y₀ + v₀ t - ½ g t²
in this case the initial and final height is the same
y = y₀ = 0
0 = v₀ t -1/2 g t²
t = 2 v₀ / g
t = 2 4 /9.81
t = 0.815 s
c) the ball is at y₀ = 4 m and its initial velocity is horizontal v₀ = 4 m / s
y = y₀ + v_{oy} t - ½ g t²
0 = y₀ + 0 - ½ g t²
t² = 2 i / g
t² = 2 4 / 9.81
t² = 0.815
t = 0.90 s
d) the horizontal distance traveled is
x = v₀ₓ t
x = 4 0.90
x = 3.6 m
e) We can calculate the time to fall from I = 2 m
y = y₀ + v_{oy} t - ½ g t²
0 = y₀ + 0 - ½ g t²
t² = 2 y₀i / g
t² = 2 2 /9.81
t² = 0.4077
t = 0.639 s
Therefore, when making measurements, you should find readings around this value.
