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v = initial velocity + acceleration * time.

initial velocity = 15m/s.
acceleration is gravity, so you call it g.

v = 15m/s + gt.

It takes 20.0s to go up and down. It would take half the time to finish going up. When this occurs, it takes 10.0s. When the rock stops at the top, it has a velocity of 0.0m/s.

Therefore ,

0.0 = 15m/s + g(10.0s)
g = -15/10 = -1.5m/s^2

To get the magnitute, you take the absolute value of g. Therefore it is 1.5m/s^2.

To find the direction, you have to look at the sign of g. Since up is positive, and down is negative, gravity is going down. Just like on Earth.

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3) To find the final position, calculate the area under the velocity-time curve. The area of a triangle is 1/2 (base)(height)

= 24 m+ 1/2 (6 * 4 m/s)(6 * 4 s)

= 312 meters.

11) Again find the area under that shape until t = 6 seconds.

It looks like the area is 54 - 1/2(9)(2) - 1/2(4)(2) = 41

Hence, final position is 10 m + 41 m = 51 m

12) Take the 51 m and subtract the area of the triangle from 6 to 8 since the velocity is negative

51 m - 1/2(3)(4) = 45 m

Final Answers

3) 312 m
11) 51 m
12) 45 m

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r = 50 m
ω = 1 rpm = π / 30 rad/s
W = 882 N
m = 882 / 9.8 = 90 kg

a) v = r*ω = π * 50 / 30 = 5/3 * π ≈ 5.236 m/s

b)At the top of the ferris wheel, his weight is down. The centripetal acceleration is also pointing down (it always points to the center). However, the seat is holding him up, which is the apparent weight (the normal force)

remember, ac = v² / r

So

-W + N= -m * ac
-882 + N = -90 * (5.236)²/60
N = 840.876 N

At the bottom of the ride, his weight is still down, but the centripetal accelaration is now up! The normal force is still the apparent weight.

-W + N = +m ac
-883 + N = 90 * (5.236)²/60
N = 923.124 N

c) Use the first force equation, set N = 0, and solve for v.
-W = -m v² / r
882 = 90 * v² / 60
v = 24.249 m/s

d) Using the second force equation.....

-W + N = +m v² / 4

-882 + N = 90 * (24.249)²/60

N = 1764 N

Physics help needed ASAP!! I have broke it down but stuck?

The mysterious Planet X has a radius of 5.10e6 m and an acceleration due to gravity of 13.0 m/s2 at its surface. Calculate its mass and average density.
_____ kg
______ kg/m3

radius of planet R = 5.10 106 m
acceleration due to gravity on the surface of the planet g = 13.0 m/s2
we know g = GM / R 2
where G = gravitational conatant = 6.67e-11 N - m 2 / kg 2
from this we fnd mass M value
( b ) . average density ρ = M / V
where V = volume of the planet = ( 4/ 3) π R 3
substitute values we get answers i.e., ρ value


* But how do I isolate M in the first part*???

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Let
M = penguin's mass,
u = coefficient of friction between penguin and ice

Angle of incline theta = 6.9 deg

When the penguin is on the incline, then the forces on it are
1. Weight Mg downward
2. Normal force Fn perpendicular to incline
3. Friction force u*Fn up the incline

There is no acceleration perpendicular to incline. Therefore net force perpendicular to incline = 0
Therefore Fn - Mg*cos(theta) = 0
Or Fn = Mg*cos(theta)-----------------------(1)

Velocity is constant. Therefore net force parallel to incline = 0
Therefore Mg*sin(theta) - u*Fn = 0
Or u*Fn = Mg*sin(theta)-----------------(2)

Dividing equation (2) by equation (1),
u = tan(theta) = tan(6.9 deg) = 0.121

When the penguin is on the horizontal patch: -
Initial velocity vo = 1.4 m/s
Acceleration a = - u*g = -0.121 * 9.8 = -1.186 m/s^2
Final velocity v = 0
Time t = ?

v = vo + at
Or 0 = 1.4 - 1.186 * t
Or t = 1.4/1.186 = 1.2 s
Ans: 1.2 s

Mestering physics help needed asap !!!?

the position of a dragonfly that is flying parallel to the ground is given as a function of time by r⃗=[2.90m+(0.0900m/s2)t2]i^− (0.0150m/s3)t3j .

At what value of t does the velocity vector of the insect make an angle of 36.0 ∘ clockwise from the x-axis?

At the time calculated in part (a), what is the magnitude of the acceleration vector of the insect?

At the time calculated in part (a), what is the direction of the acceleration vector of the insect?

Physics help needed asap pleasssseee?

A rail gun uses electromagnetic forces to accelerate a projectile to very high velocities. The basic mechanism of acceleration is relatively simple and can be illustrated in the following example. A metal rod of mass 30.0 g and electrical resistance 0.100 Ω rests on parallel horizontal rails that have negligible electric resistance. The rails are a distance L = 7.00 cm apart. (Figure 1) The rails are also connected to a voltage source providing a voltage of V = 5.00 V .
The rod is placed in a vertical magnetic field. The rod begins to slide when the field reaches the value B = 4.20×10−2 T . Assume that the rod has a slightly flattened bottom so that it slides instead of rolling. Use 9.80 m/s2 for the magnitude of the acceleration due to gravity.


Find μs, the coefficient of static friction between the rod and the rails.