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.
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
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.)
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
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