Answers
Final answer:
The maximum frictional force acting on the object is 28 newtons.
Explanation:
The maximum frictional force acting on an object can be calculated using the formula:
Frictional force = coefficient of static friction x normal force
Given that the coefficient of static friction is 0.35 and the normal force is 80 newtons, we can calculate the maximum frictional force as follows:
Maximum frictional force = 0.35 x 80 = 28 newtons
Learn more about Calculating maximum frictional force here:
Related Questions
A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s.
15.Restore the battery setting to 10 V. Now change the number of loops from 4 to 3. Explain what happens to the magnitude and direction of the magnetic field. Now change to 2 loops, then to 1 loop. What do you observe the relationship to be between the magnitude of the magnetic field and the number of loops for the same current
A spring has a spring constant of 1350 N/m. You place the spring vertically with one end on the floor. You then drop a 1.3 kg book onto it from a height of 0.8 m above the top of the spring. Find the maximum distance the spring will be compressed. Express your answer with the appropriate mks units.
Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringeFor what wavelength of light will thefirst-order dark fringe be observed at this same point on thescreen?Express your answer in micrometers(not in nanometers).
Answers
Answer:
a) 8.99*10³ V b) 4.5*10⁻² J c) 0 d) 0
Explanation:
a)
- The electrostatic potential V, is the work done per unit charge, by the electrostatic force, producing a displacement d from infinity (assumed to be the reference zero level).
- For a point charge, it can be expressed as follows:
- As the electrostatic force is linear with the charge (it is raised to first power), we can apply superposition principle.
- This means that the total potential at a given point, is just the sum of the individual potentials due to the different charges, as if the others were not there.
- In our case, due to symmetry, the potential, at any corner of the triangle, is just the double of the potential due to the charge located at any other corner, as follows:
- The potential at point C is 8.99*10³ V
b)
- The work required to bring a positive charge of 5μC from infinity to the point C, is just the product of the potential at this point times the charge, as follows:
- The work needed is 0.045 J.
c)
- If we replace one of the charges creating the potential at the point C, by one of the same magnitude, but opposite sign, we will have the following equation:
- This means that the potential due to both charges is 0, at point C.
d)
- If the potential at point C is 0, assuming that at infinity V=0 also, we conclude that there is no work required to bring the charge of 5μC from infinity to the point C, as no potential difference exists between both points.
The three given moon phases are:
Waxing Crescent Moon
Waning Crescent Moon
Full Moon
The things we need to match to the above three topicsare:
*visible near western horizon an hour after sunset
*rises about the same time the sun sets
*visible near eastern horizon just before sun rises
*occurs about 3 days before new moon (i know this is waningcrescent)
*visible due south at midnite
*occurs 14 days after new moon (i know this is full moon)
Answers
Answer:
Explanation:
*visible near western horizon an hour aftersunset WAXING CRESCENT
*rises about the same time the sun sets FULLMOON
*visible near eastern horizon just before sun rises WANING CRESCENT
*occurs about 3 days before new moon WANING CRESCENT
*visible due south at midnite FULL MOON
*occurs 14 days after new moon FULL MOON
Answers
Answer:
V1 = -3.260 m/s, V2 = 1.303 m/s
Explanation:
Let mass of the left glider m1 = 0.157 kg and velocity v1 = 0.850 m/s
mass of the right glider m2 = 0.306 Kg and v2 = -2.26 m/s (-ve sign mean it is opposite to direction of left glider)
To Find: Final Velocity of Left Glider is V1=? m/s and Velocity of right Glider is V2 =? m/s (After Collision)
from law of conservation of momentum and energy we deduce a formula:
V1 = (m1-m2) v1 /(m1+m2) + 2 m2 v2/(m1+m2)
V1 = (0.157 kg - 0.306 Kg) × 0.850 m/s / (0.157 kg + 0.306 Kg) + 2 ×0.306 kg × -2.26 m/s / (0.157 kg + 0.306 Kg)
V1 = -0.273 -2.987
V1 = -3.260 m/s
and V2 Formula
V2 = (m2-m1) v2/(m1+m2) + 2 m1 v1/(m1+m2)
V2 = (0.157 kg - 0.306 Kg) × -2.26 m/s / (0.157 kg + 0.306 Kg) + 2 ×0.157 kg × 0.850 m/s / (0.157 kg + 0.306 Kg)
V2 = 0.727 + 0.576
V2 = 1.303 m/s
-0.149, 0.463
A. Consult with a friend and get their feeback
B. Dispute the beliefs by asking if these are true and examining the evidence
C. Seek mental health counseling
D. It is just too hard so let's just forget it.
Answers
Answer:
i believe the answer is B
Explanation:
Seeking the right answer is the best thing to do
Answers
Answer:
a)a=5.01m/s^2
b)t=11.26s
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)
{Vf^{2}-Vo^2}/{2.a} =X(2)
X=Xo+ VoT+0.5at^{2} (3)
X=(Vf+Vo)T/2 (4)
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
to solve the question a, we can use the ecuation number 2
Vo=0
Vf=10.5 m/s
x=11m
{Vf^{2}-Vo^2}/{2.a} =X
{Vf^{2}-Vo^2}/{2.x} =a
{10.5^{2}-0^2}/{2x11} =a
a=5.01m/s^2
to find the time we can use the ecuation number 1
Vf=Vo+a.t
t=(Vf-Vo)/a
t=(10.5-0)/5.01=2.09s
part b
in this case the spees is constant, so the movement is defined by the following ecuation
X=VT
t=x/v
t=96.3/10.5=9.17s
to find the total time we sum the times when the speed is constant and when the acceleration is constan
t=9.17+2.09
t=11.26s
Answers
Work = Force times Distance
Work = 200 x 30
Work = 6000
The work done by a force of 200N on a body that moved 30m is 6000J or 6000 Joules.