Answer:
Explanation:
The only thing I can figure you need here is the accleration of the sled. The equation we need to find this is Newton's Second Law that says that sum of the forces acting on an object is equal to the object's mass times its acceleration. For us, that looks like this because of the friction working against the sled:
F - f = ma but of course it's much more involved than that simple equation! We have the F value as 230 N, and we have the mass as 105, but we do not have the frictional force, f, and we need it to solve for a in the above equation. We know that
f = μ
where μ is the coefficient of friction, and is the normal force, aka weight of the object. We will use the coefficient of friction and find the weight in order to fill in for f:
so
so the weight of the sled is
1.0 × 10³ with the correct number of sig dig there. Now to find f:
f = (.025)(1.0 × 10³) so
f = 25 to the correct number of sig fig. Now on to our "real" equation:
F - f = ma and
230 - 25 = 105a. We have to do the subtraction first, round, and then divide since the rules for addition and subtraction are different from the rules for dividing and multiplying.
230 - 25 will round to the tens place giving us 210. Then
210 = 105a. 210 has 2 sig figs in it while 105 has 3, so we will divide and round to 2 sig fig:
a = 2.0 m/sec²
Muscles function only by contracting. This makes it necessary for one end of the muscle to be fixed and the other mobile.
Take the bicep for example.
Its origin is at the shoulder and its two heads connect to the bones of the forearm, the radius and ulna.
Now, had the muscle not been fixed at one end, and contracted, it would pull both our shoulder and forearm together resulting in an ineffective movement. The desired motion is to lift the forearm (proximal and distal movement) which can only be achieved if the bicep is fixed at the shoulder and allowed to move at the forearm.
Complete question:
while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8.32 meters per second take the speed of sound as 340 meters per second. calculate the frequency reflected off the wall to the bat?
Answer:
The frequency reflected by the stationary wall to the bat is 41 kHz
Explanation:
Given;
frequency emitted by the bat, = 39 kHz
velocity of the bat, 
= 8.32 m/s
speed of sound in air, v = 340 m/s
The apparent frequency of sound striking the wall is calculated as;
The frequency reflected by the stationary wall to the bat is calculated as;
Explanation:
The Coulomb's law states that the magnitude of each of the electric forces between two point-at-rest charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:
In this case we have an electron (-e) and a proton (e), so:
In this case, the electric force is negative, therefore, the force is repulsive and its magnitude is:
Answer:
Explanation:
The electric field produced by a single point charge is given by:
where
k is the Coulomb's constant
q is the charge
r is the distance from the charge
In this problem, we have
E = 1.0 N/C (magnitude of the electric field)
r = 1.0 m (distance from the charge)
Solving the equation for q, we find the charge:
