Answer:
v₀ = [tex]\sqrt{2gH}[/tex]
to determine this speed experimentally, the student must measure the height of the body as a function of time and with equation (1) he can find the speed for each point of interest
Explanation:
In this internal exercise the student must use the conservation of mechanical energy,
Starting point. Highest point of the trajectory
Em₀ = U = m g H
Point of interest. Point at height Y
[tex]Em_{f}[/tex] = K + U = ½ m v² + m g Y
energy is conserved
Em₀ = Em_{f}
m g H = ½ m v² + m g Y
v² = 2 g (H -Y) (1)
in this case they indicate that Y is the equilibrium position whereby Y = 0 and the velocity is v = v₀
v₀ = [tex]\sqrt{2gH}[/tex]
Therefore, to determine this speed experimentally, the student must measure the height of the body as a function of time and with equation (1) he can find the speed for each point of interest
A baby elephant trots in a straight line along a river. The horizontal position of the elephant in meters over
time is shown below.
Position
(meters)
15
12+
9+
6-
3+
+
4
+
16
8
12
20
24
-3+
-6+
Time
(seconds)
.92
Answer:
Displacement -6
Distance 24
Explanation:
Answer:
Explanation:
What is the displacement of the elephant between
0s
and
16s
displacement =9
distance traveled by the elephant between =9
Tarzan (who has mass 80.0 kg) is running across the jungle floor with speed 7.00 m/s as
shown in Figure 1. Tarzan grabs a large bunch of bananas (15.0 kg) and grabs a vine in an attempt to swing up to his monkey who is 3.00 m above him as in Figure 2.
i. What is Tarzan’s momentum before he grabs the bananas? [2 points]
ii. What is Tarzan’s speed just after he grabs the bananas? [4 points]
iii. Can Tarzan swing high enough to reach his monkey? Justify your answer. [2 points]
We are given:
Mass of Tarzan before swinging = 80 kg
Mass of Tarzan when swinging = 80 + 15 = 95 kg
Velocity of Tarzan = 7 m/s
The height of the rock Tarzan's monkey is sitting on = 3 m
__________________________________________________________
Momentum of Tarzan before swinging:
We know that:
Momentum = Mass*Velocity
Momentum = 80 * 7
Momentum = 560 kg m/s
__________________________________________________________
Speed of Tarzan after grabbing the bananas:
The momentum of Tarzan will remain the same but his mass will increase
So, Since Momentum = New Mass* velocity
560 = 95 * v [where v is the velocity of Tarzan]
v = 5.9 m/s
__________________________________________________________
Finding the Initial and Final KE and PE:Here, KE = Kinetic Energy and PE = Potential Energy
Initial and Final KE:
We know that KE = 1/2*(mv²)
Initial KE:
Initial KE = 1/2*(mv²) [where v is the velocity after picking the bananas]
Initial KE = 1/2*(95*5.9²)
Initial KE = 1653.5 Joules
Final KE:
Final KE = 1/2*(mv²)
[where v is the velocity at the maximum point of the swing]
Since Tarzan will be at rest at the maximum point of the swing, v = 0 m/s
Final KE = 1/2*(95*0²)
Final KE = 0 Joules
Initial and Final PE:
We know that:
PE = mgh
[where g is the acceleration due to gravity and h is the height]
Initial PE:
Since the height of Tarzan from the ground was 0 m at the beginning of the swing, h = 0
Initial PE = 95*10*0
Initial PE = 0 Joules
Final PE:
Let the maximum height of Tarzan be h m
Final PE = 95*10*h
Final PE = 950(h)
__________________________________________________________
Finding the maximum height Tarzan will reach:Here, KE = Kinetic Energy and PE = Potential Energy
From the law of conservation of momentum, we know that:
Initial KE + Initial PE = Final KE + Final PE
Replacing the variables:
1653.5 + 0 = 0 + 950h
1653.5 = 950h
h = 1653.5/950 [dividing both sides by 950]
h = 1.74 m
Therefore, the maximum height reached by Tarzan is 1.74 m
but since his monkey is sitting 3 m high, he will NOT be able to reach his monkey
Looking at the bank statement, what three items total that same amount?
The balance sheet, income statement, and cash flow statement each offer unique details with information that is all interconnected. Together the three statements give a comprehensive portrayal of the company's operating activities.
You push a manual lawn mower across the lawn at constant speed. What is the value of the coefficient of friction between the mower and the grass
Answer:
0.27Explanation:
The question is incomplete. Here is the complete question:
You are pushing a 13.3 kg lawn mower across the lawn with a force of 200 N. What is the value of the coefficient of friction between the mower and the grass if the mower moves with a constant velocity? The force is applied downward at an angle of 65° with the horizontal.
According to Newton's second law of motion:
[tex]\sum F_x= ma_x\\F_{app} - F_f = ma_x\\[/tex]
[tex]F_f = \mu R\\[/tex]
[tex]F_{app} - \mu Rcos \theta = ma_x[/tex]
Fapp is the applied force = 200N
Ff is the frictional force
[tex]\mu[/tex] is the coefficient of friction between the mower and the grass
R is the reaction
m is the mass of the object
ax is the acceleration
Given
R = mg = 13.3*9.8
R = 130.34N
m = 13.3kg
ax = 0m/s² (constant velocity)
Fapp = 200N
[tex]\theta = 65^0[/tex]
Substitute the given parameters into the formula and get the coefficient of friction as shown;
Recall that: [tex]F_f = \mu R\\[/tex]
[tex]\mu = \frac{F_f}{R}\\\mu = \frac{F_{x}cos65}{F_y+W} \\\mu =\frac{ 200cos65}{200sin65+13.3(9.8)}\\\mu = \frac{84.52}{181.26+130.34}\\\mu = \frac{84.52}{311.6}\\\mu = 0.27[/tex]
Hence the coefficient of friction between the mower and the grass is 0.27
g A 240-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s
Answer:
339.3 N
Explanation:
First, we start by converting the units.
1 rev/s = 2π rad/s, so
0.6 rev/s = 2π * 0.6 rad/s
0.6 rev/s = 1.2π rad/s
0.6 rev/s = 3.77 rad/s
Now we apply the equation of motion,
W(f) = w(o) + αt
3.77 = 0 + α * 2
3.77 = 2α
α = 3.77/2
α = 1.885 rad/s²
Torque = I * α
Torque = F * r
This means that
I * α = F * r, where I = 1/2mr²
Substituting for I, we have
1/2mr²α = F * r, making F the subject of formula, we have
F = 1/2mrα, then we substitute for the values
F = 1/2 * 240 * 1.5 * 1.885
F = 678.6 / 2
F = 339.3 N
A teacher does a demonstration in front of a class. For the demonstration, a motion sensor is placed at the end of a long, horizontal track of negligible friction. A force sensor is used to pull a cart of known mass along the track, and the motion sensor tracks the motion of the cart. For several positions of the cart, a computer records the reading on the force sensor and the reading on the motion sensor for both the speed of the cart and the distance the cart moves away from the motion sensor. Based on only these data, which of the following scientific hypotheses could a student test?
a. The force exerted on the cart is directly proportional to the time the cart has moved.
b. The gravitational acceleration for an object on Earth is constant near Earth's surface.
с. The net work done on the cart is equal to the change in kinetic energy of the cart.
d. As the force the sensor exerts on the cart increases, so does the acceleration of the cart.
e. Earth's gravitational force is a conservative force.
Answer:
с. The net work done on the cart is equal to the change in kinetic energy of the cart.
Explanation:
In the context, a motion sensor detects the motion of the cart where a cart is used to move using a force sensor that is attached to the cart of certain mass. As the cart moves with a speed at a particular time, the computer records both the readings on the force sensor as well as the motion sensor. From these readings the is can be concluded that the work done on the cart is same as the change in the kinetic energy in the cart.
Here's a quick story if a boy walks to the woods at night and slender man comes along what is most likely to happen put your answers in the comments.
Answer:
the boy would go missing
Answer: Slender man converts the boy to christianity
Two balls, one twice as massive as the other, are dropped from the roof of a building (freefall). Just before hitting the ground, the more massive ball has ______ the kinetic energy of the less massive ball. (Neglect air friction.)
If the Earths atmosphere is air, then why doesn't the land fall? How is all the lava and water IN earth?
Answer:
Its stored in there
Explanation:
And air has little mass