A 2.0 kg block is pulled across a horizontal surface by a 15 N force at a constant velocity. What is the force of friction actin
1 answer:
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Answer:t=0.3253 s
Explanation:
Given
speed of balloon is
speed of camera
Initial separation between camera and balloon is
Suppose after t sec of throw camera reach balloon then,
distance travel by balloon is
and distance travel by camera to reach balloon is
Now
There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .
(b)When passenger catches the camera time is
velocity is given by
and position of camera is same as of balloon so
Position is
Answer:
The position of the car at t = 1.5 s is at -8.1625 meters
Explanation:
The initial position of the car is 3.2 meters
The initial velocity is -8.4 m/s
The constant acceleration is 1.1 m/s²
We need to find the final position of the car at the time t = 1.5 seconds
The displacement <em>s</em> = final position - initial position
, where <em>u</em> is the initial velocity, <em>a</em> is the
constant acceleration and <em>t</em> is the time
So we can find the final velocity by using the rule:
final position - initial position =
initial position = 3.2 meters , u = -8.4 m/s , a = 1.1 ²m/s , t = 1.5 s
Substitute these values in the rule
final position - 3.2 =
final position - 3.2 = -12.6 + 1.2375
final position - 3.2 = -11.3625
add 3.2 for both sides
final position = -8.1625
<em>That means the car is at 8.1625 meters in opposite direction</em>
<em>The position of the car at t = 1.5 s is at -8.1625 meters </em>
Answer:
(a) Electric field at 0.250 m is zero.
(b) Electric field at 2.90 m is zero.
(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.
(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.
Explanation:
Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.
Electric field inside and outside the metal sphere is :
E = 0 for r ≤ R ( inside )
= for r > R ( outside )
Here K is electric constant and r is the distance from the center of the metal sphere.
(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.
(b) Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.
(c) Electric field at 3.10 m is given by the relation as r > R :
E =
Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.
E = -
E = - 5.15 x 10³ V/m
(d) Electric field at 8.00 m is given by the relation as r > R :
E =
Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.
E = -
E = - 0.77 x 10³ V/m