Answer:
4.8967m
Explanation:
Given the following data;
M = 0.2kg
∆p = 0.58kgm/s
S(i) = 2.25m
Ratio h/w = 12/75
Firstly, we use conservation of momentum to find the velocity
Therefore, ∆p = MV
0.58kgm/s = 0.2V
V = 0.58/2
V = 2.9m/s
Then, we can use the conservation of energy to solve for maximum height the car can go
E(i) = E(f)
1/2mV² = mgh
Mass cancels out
1/2V² = gh
h = 1/2V²/g = V²/2g
h = (2.9)²/2(9.8)
h = 8.41/19.6 = 0.429m
Since we have gotten the heigh, the next thing is to solve for actual slant of the ramp and initial displacement using similar triangles.
h/w = 0.429/x
X = 0.429×75/12
X = 2.6815
Therefore, by Pythagoreans rule
S(ramp) = √2.68125²+0.429²
S(ramp) = 2.64671
Finally, S(t) = S(ramp) + S(i)
= 2.64671+2.25
= 4.8967m
As an object accelerates i.e., change it's velocity(either direction or speed), the position of the object depends on two factor; If the acceleration was direction based then it might have a zero displacement for eg: if it travels in circle. or it might have a net displacement if it travels in a straight line, quantitatively

where,
s = displacement
u = initial velocity
v = final velocity
a = acceleration
t = time
Now, for the hypothesis;
There is no direct relationship between fan speed and acceleration but anyways generally speaking if we do have a relationship that with more fan speed we have a larger displacement of air i.e., a more force i.e., greater acceleration
Thus, it can be said, well not exactly scientific, that with a greater fan speed there will be greater acceleration. if fan speed is increased then acceleration will be more.
:)
The question above can be answered through using the concept of Conservation of Momentum which may be expressed as,
m1v1 + m2v2 = mTvT
where m1 and v1 are mass and initial velocity of Tex, 2s are that of the bull, and the Ts are the total. Then substituting,
(85 kg)(3 m/s) + (520 kg)(13 m/s) = (520 + 85)(vT)
The value of vT obtained from above equation is 11.6 m/s
Answer:

Explanation:
Newton's law of universal gravitation states that the force experimented by a satellite of mass m orbiting Mars, which has mass
at a distance r will be:

where
is the gravitational constant.
This force is the centripetal force the satellite experiments, so we can write:

Putting all together:

which means:
![r=\sqrt[3]{\frac{GM}{4\pi^2}T^2}](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BGM%7D%7B4%5Cpi%5E2%7DT%5E2%7D)
Which for our values is:
![r=\sqrt[3]{\frac{(6.67\times10^{-11}Nm^2/kg^2)(6.39\times10^{23} kg)}{4\pi^2}(1.026\times24\times60\times60s)^2}=20395282m=20395.3km](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%286.67%5Ctimes10%5E%7B-11%7DNm%5E2%2Fkg%5E2%29%286.39%5Ctimes10%5E%7B23%7D%20kg%29%7D%7B4%5Cpi%5E2%7D%281.026%5Ctimes24%5Ctimes60%5Ctimes60s%29%5E2%7D%3D20395282m%3D20395.3km)
Since this distance is measured from the center of Mars, to have the height above the Martian surface we need to substract the radius of Mars R=3389.5 km
, which leaves us with:

Answer:
B_o = 1.013μT
Explanation:
To find B_o you take into account the formula for the emf:

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.
By applying the derivative you obtain:

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

hence, B_o = 1.013μT