Answer:
Ba đặc trưng sinh lí của âm là độ cao, độ to và âm sắc.
Explanation:
Một học sinh làm thí nghiệm sóng dừng trên dây cao su dài L với hai đầu A và B cố định . Xét điểm M trên dây sao cho khi sợi dây duỗi thẳng thì M cách B một khoảng a < L/2 . Khi tần số sóng là f = f1 = 60 Hz thì trên dây có sóng dừng và lúc này M là một điểm bụng . Tiếp tục tăng dần tần số thì lần tiếp theo có sóng dừng ứng với f = f2=72 Hz và lúc này M không phải là điểm bụng cũng không phải điểm nút . Thay đổi tần số trong phạm vi từ 73 Hz đến 180 Hz , người ta nhận thấy với f = fo thì trên dây có sóng dừng và lúc này M là điểm nút . Lúc đó , tính từ B ( không tính nút tại B ) thì M có thể là nút thứ ?
What is this sport ⚽⚾
Answer:
sports are all forms of physical activity that contribute to physical fitness, mental well-being and social interaction.
hope it is helpful to you
What is the current in the 30 resistor?
A. 0.0833 A
B. 12 A
C. 2 A
D. 10 A
Answer:
Explanation:
Step 1) Combine all resistors into an equivalent overall resistor. These are all in series so you just add them up. Req = 10Ω + 20Ω + 30Ω = 60Ω:
Step 2) Using Ohm's Law, I = V/R = 120/60 = 2 A
Now you know how much current is flowing, and that current flows through each resistor the same. So the current in the 30 Ω resistor is 2.00 amps.
A man pulls his dog (m=20kg) on a sled with a force of 100N at a 60° angle from the horizontal. What is the horizontal component of the force?
A) 100N
B) 196N
C) 50N
D) 86N
show your work please
Answer:
the horizontal component of the force is 50 N
Explanation:
Given;
force applied by the man, F = 100 N
angle of inclination of the force, θ = 60⁰
mass of the dog, m = 20 kg
The horizontal component of the force is calculated as;
[tex]F_x = F\times cos(\theta)\\\\F_x = 100 \ N \times cos(60^0)\\\\F_x = 100\ N \times 0.5\\\\F_x = 50 \ N[/tex]
Therefore, the horizontal component of the force is 50 N
Find the X and Y components of the following:
A. 35 m/s at 57q from the x-axis.
Explanation:
Given that,
35 m/s at 57° from the x-axis.
Speed, v = 35 m/s
Angle, θ = 57°
Horizontal component,
[tex]v_x=v\cos\theta\\\\=35\times \cos(57)\\\\=19.06 m/s[/tex]
Vertical component,
[tex]v_y=v\sin\theta\\\\v_y=35\times \sin(57)\\\\=29.35\ m/s[/tex]
Hence, this is the required solution.
A 5.0-kg solid cylinder of radius 0.25 mis free to rotate about an axle that runs along the cylinders length and passes through its center. A thread wrapped around the cylinder is weighed down by a mass of 2.0 kg so as to unwrap and make the cylinder rotate as this mass falls. Ignore any friction in the axle. If there is no slippage between the thread and the cylinder, and the cylinder starts from rest (a) Calculate the velocity of the block after it has fallen a distance of 2.0m. Give your answer in m.s (b) Calculate the total work done by the rope on the cylinder after the block has fallen a distance of 2.0 m. Give your answer in Joule.
Answer:
157n is the correct answer
Nick and Chloe left their campsite by canoe and paddle downstream at an average speed of 12 km/h. They turned around and paddled back upstream at an average speed of 4 km/h. The total trip took 1 hour. After how much time did the campers turn around downstream
The time spent by the campers when they turn around downstream is 15 minutes.
Total distance traveled by Nick and Chloe
The concept of total distance traveled by Nick and Chloe can be used to determine the time they turn around downstream.
Let time for downstream = t1
Let time for upstream = t2
distance covered in upstream = distance covered in downstream = d
12(t1) = d
4(t2) = d
12t1 = 4t2
t1 + t2 = 1
t2 = 1 - t1
12t1 = 4(1 - t1)
12t1 = 4 - 4t1
16t1 = 4
t1 = 4/16
t1 = 0.25 hours
t1 = 0.25(60 min) = 15 mins
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An exoplanet has three times the mass and one-fourth the radius of the Earth. Find the acceleration due to gravity on its surface, in terms of g, the acceleration of gravity at Earth's surface. A planet's gravitational acceleration is given by gp = G Mp/r^2p
a. 12.0 g.
b. 48.0 g.
c. 6.00 g.
d. 96.0 g.
e. 24.0 g.
Answer:
b. 48.0 g.
Explanation:
Given;
mass of the exoplanet, [tex]M_p = 3M_e[/tex]
radius of the exoplanet, [tex]r_p = \frac{1}{4} r_e[/tex]
The acceleration due to gravity of the planet is calculated as;
[tex]g_p = \frac{GM_p}{r_p^2} \\\\for \ Earth's \ surface\\\\g = \frac{GM_e}{r_e^2} \\\\G = \frac{gr_e^2}{M_e} = \frac{g_pr_p^2}{M_p} \\\\\frac{gr_e^2}{M_e} = \frac{g_p(\frac{r_e}{4}) ^2}{3M_e} \\\\\frac{gr_e^2}{M_e} = \frac{g_pr_e ^2}{16\times 3M_e} \\\\g = \frac{g_p}{48} \\\\g_p = 48 \ g[/tex]
Therefore, the correct option is b. 48.0 g
Question 2:
Inclined Plane
A block (M) weighs 25-N, rests on an inclined plane when it is joined by a sting to a support
(S) as shown in the figure' below. Use g=10 N/Kg.
(S)
B
M
List and classify the forces acting on (M).
Représent, without scaling, the forces acting on (M).
Find the mass of (M).
74. If the string were cut, (M) does not slide. Explain this phenomenon.
15. Determine the mass and weight of (M) on moon.
06
Answer:
we need the block
Explanation:
1×2 =4 lest 74 =345
Three forces are pulling on the same object such that the system is in equilibrium. Their magnitudes are F1 = 2.83 N.F= 3.35 N. and F3 = 3.64 N, and they make angles of 0, = 45.0°, 02 = -63.43 and 03 =164.05° with respect to the x-axis, respectively.
Required:
a. What is the x-component of the force vector F1?
b. What is the y-component of the force vector F1?
(a) 2.001N
(b) 2.001N
Explanation:A sketch of the scenario has been attached to this response.
Since only the force vector F₁ is required, the only force shown in the sketch is F₁.
As shown in the sketch;
The x-component of the force vector F₁ = [tex]F_{x}[/tex]
The y-component of the force vector F₁ = [tex]F_{y}[/tex]
The magnitude of F₁ as given in the question = 2.83N
The angle that the force makes with respect to the x-axis = 45.0°
Using the trigonometric ratio, we see that;
(a) cos 45.0° = [tex]\frac{F_x}{F_1}[/tex]
=> [tex]F_{x}[/tex] = F₁ cos 45.0°
=> [tex]F_{x}[/tex] = 2.83 cos 45.0°
=> [tex]F_{x}[/tex] = 2.83 x 0.7071
=> [tex]F_{x}[/tex] = 2.001N
(b) Also;
sin 45.0° = [tex]\frac{F_y}{F_1}[/tex]
=> [tex]F_{y}[/tex] = F₁ sin 45.0°
=> [tex]F_{y}[/tex] = 2.83 sin 45.0°
=> [tex]F_{y}[/tex] = 2.83 x 0.7071
=> [tex]F_{y}[/tex] = 2.001N
Therefore, the x-component and y-component of the force vector F₁ is 2.001N
The x and y component of vector F1 is mathematically given as
F_x = 2.001N
F_y= 2.001N
What is the x and y component of vector F1?Question Parameters:
Generally, the equation for the x-component is mathematically given as
x=Fsin\theta
Therefore
F_x = F₁ cos 45.0°
F_x = 2.83 x 0.7071
F_x = 2.001N
For y component
x=Fcos\theta
F_y = F₁ sin 45.0
F_y = 2.83 x 0.7071
F_y= 2.001N
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You are a member of an alpine rescue team and must get a box of supplies, with mass 2.50 kg, up an incline of constant slope angle 30.0° so that it reaches a stranded skier who is a vertical distance 3.50 m above the bottom of the incline. There is some friction present; the kinetic coefficient of friction is 6.00x102. Since you can't walk up the incline, you give the box a push that gives it an initial velocity; then the box slides up the incline, slowing down under the forces of friction and gravity. Take acceleration due to gravity to be 9.81 m/s
Use the work-energy theorem to calculate the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier. Express your answer numerically, in meters per second.
1. How to approach the problem
2. Find the total work done on the box
3. Initial kinetic energy
4. What is the final kinetic energy?
Answer:
v₀ = 2.67 m / s
Explanation:
This problem can be solved using the Kinetic Enemy Work Theorem
W = ΔK
Work is defined by the relation
W = fr. d
The bold letters indicate vectors, in this case the blow is in the direction of the slope of the ramp and the displacement is also in the direction of the ramp, therefore the angle between the force and the displacement is zero.
the friction force opposes the displacement therefore its angle is 180º
W = - fr d
Let's use Newton's second law, we define a reference frame with the horizontal axis parallel to the plane
Y axis
N- Wy = 0
N - W cos tea = 0
the friction force has the expression
fr = μ N
fr = μ W cos θ
we substitute
W = - μ W cos θ d
let's look for kinetic energy
the minimum velocity at the highest point is zero
K_f = 0
the initial kinetic energy is
K₀ = ½ m v₀²
we substitute energy in the work relationship
- μ W cos θ d = 0 - ½ m v₀²
v₀² = - μ W cos θ 2d / m
Let's use trigonometry to find distance d
sin θ= y / d
d = y /sin θ
d = 3.50 / sin 30
d = 7 m
let's calculate
v₀² = (6 10⁻² 2.50 9.8 cos 30) 2 7 / 2.50
v₀ = √7.129
v₀ = 2.67 m / s
Which of these hazmat products are allowed in your FC?
Please choose all that apply.
A GPS unit (lithium batteries)
A subwoofer (magnetized materials)
A can of hairspray (flammable/aerosols)
Fireworks (explosives)
Answer: Hazmat products are allowed in your FC are:
A GPS unit (lithium batteries) A subwoofer (magnetized materials)Explanation:
Hazmat products consist of flammable, corrosive and harmful substances which are actually very hazardous to human health and environment.
Hazardous material allowed in FC are as follows.
Magnetized material products like as speakers.Non-spillable battery products like toy cars.Lithium-ion battery containing products like laptops, mobile phones etc.Non-flammable aerosol.So, hazmat allowed products are GPS unit (lithium batteries) and subwoofer (magnetized materials).
Thus, we can conclude that hazmat products are allowed in your FC are:
A GPS unit (lithium batteries) A subwoofer (magnetized materials)What is the change in internal energy if 70 J of heat is added to a system and
the system does 30 J of work on the surroundings. Uze al-Q-W.
O A. 40 J
O B. -40.3
O C. 100.
D. -1003
Answer:
A. 40 J
Explanation:
Given;
heat added to the system, Q = 70 J
work done by the system, W = 30 J
The change in the internal energy of the system is calculated using the first law of thermodynamic as shown below;
ΔU = Q - W
ΔU = 70 J - 30 J
ΔU = 40 J
Therefore, the change in the internal energy of the system is 40J
Mary applies a force of 73 N to push a box with an acceleration of 0.48 m/s^2. When she increases the pushing force to 84 N, the box's acceleration changes to 0.64 m/s^2. There is a constant friction force present between the floor and the box.
Required:
a. What is the mass of the box?
b. What is the coefficient of kinetic friction between the floor and the box?
Answer: [tex]68.75\ kg, 0.06[/tex]
Explanation:
Mary applies a force of 73 N to create an acceleration of [tex]0.48\ m/s^2[/tex]
When She increases force to 84 N, it creates an acceleration of [tex]0.64\ m/s^2[/tex]
Friction opposes the motion of box
[tex]\Rightarrow 73-f=m\times 0.48\quad \ldots(i)\\\Rightarrow 84-f=m\times 0.64\quad \ldots(ii)[/tex]
Subtract (i) from (ii)
[tex]\Rightarrow 11=m(0.64-0.48)\\\Rightarrow m=68.75\ kg[/tex]
Therefore friction is
[tex]\Rightarrow f=73-68.75\times 0.48\\\Rightarrow f=73-33\\\Rightarrow f=40\ N[/tex]
Here, friction is kinetic friction which is given by
[tex]\Rightarrow f=\mu_kmg\\\Rightarrow 40=\mu_k 68.75\times 9.8\\\Rightarrow \mu_k=0.061[/tex]
A permanent magnet is pushed into a wire, left there for a while, and then pulled out. During which time does a current run though the wire? A from the time that the magnet is pushed into the coil to the time it is pulled out B while the magnet remains within the coil C while the magnet is moving D only while the magnet is being pulled out of the coil
Answer:
C. while the magnet is moving
Explanation:
Electromagnetic induction implies the production of electric current by mere movement of a magnet with respect to a coil or wire.
In the given question, current would be induced in the wire only when the magnet moves. That is either when the magnet is pushed into a wire, or when pulled out. But no current would flow through the wire when the magnet is left there for a while.
The current is induced because of the motion involved. Thus, the appropriate option is C.
A car has a mass of 900 kg is accelerated from rest at a rate of 1.2 m/s calculate the time taken to reach 30/s
Answer:
12+2=24+30+2=66
Explanation:
For the following questions, assume the wavelengths of visible light range from 380 nm to 760 nm in a vacuum. (a) What is the smallest separation (in nm) between two slits that will produce a ninth-order maximum for any visible light
Answer:
Explanation:
This is an interference exercise, which the case of constructive interference is described by the expression
d sin θ = m λ
in this case they indicate that we are in the ninth order (m = 9).
To be able to observe the pattern, the dispersion angle must be less than 90º
we substitute
sin 90 = 1
d = m lang
let's calculate
d = 9 λ
d = 9 380 10⁻⁰
d = 3.42 10⁻⁶
d2 = 9 760 10⁻⁹
d2 = 6.84 10₋⁶
A boy pushes his little brother on a sled. The sled accelerates from rest to (4 m/s). If the combined mass of his brother and the sled is (40.0 kg) and (20 W) of power is developéd, how long time does boy push the sled?
16s
300s
15s
23s
The boy pushed the sled for 16 seconds.
We have a boy who pushes his little brother on a sled.
We have to determine for how long time does boy push the sled.
State Work - Energy Theorem.The Work - Energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.
According to the question -
The sled is initially at rest → initial velocity (u) = 0.
Final velocity (v) = 4 m/s
Mass of boy and sled (M) = 40 kg
Power developed (P) = 20 W = 20 Joules/sec
According to work - energy theorem -
Work done (W) = Δ E(K) = E(f) - E(i)
Therefore -
W = ([tex]\frac{1}{2} \times 40 \times 4 \times 4 - \frac{1}{2}[/tex] x 40 x 0) = 320 Joule
Now, Power is defined as the rate of doing work -
P = [tex]\frac{dW}{dt}[/tex] = [tex]\frac{W}{t}[/tex]
20 = [tex]\frac{320}{t}[/tex]
t = 16 seconds
Hence, the boy pushed the sled for 16 seconds.
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A bucket of mass m in a well is held up by a rope. the rope is wound around a drum of radius r there is also a handle of length R attached to the drum. the tension in the rope is equal to T. If the buket is allowed to fall into the well, which point will have the greatest angular acceleration, a point on the rim of the drum (at radius r) or a point on the end of the handle (at radius R)?
a. The point on the rim of the drum.
b. The point at the end of the handle.
c. They will both have the same angular acceleration.
Answer:
the correct answer is C
Explanation:
This is a system with circular motion, there is a relationship between the linear and angular variables
a = α r
with the cube going down the well, the tension of the leather is maintained therefore the acceleration of the cube is
W = m a
-mg = ma
a = -g
this acceleration a is the same as that at the edge of the drum.
α = a / r
where we can see that the angular acceleration is constant
consequently the correct answer is C
What is the electric field 3.9 m from the center of the terminal of a Van de Graaff with a 6.60 mC charge, noting that the field is equivalent to that of a point charge at the center of the terminal
Answer:
the electric field is 3.91 x 10⁶ N/C
Explanation:
Given the data in the question;
Electric field at a point due to point charge is;
E = kq/r²
where k is the constant, r is the distance from centre of terminal to point where electric field is, q is the excess charge placed on the centre of terminal of Van de Graff,a generator
Now, given that r = 3.9 m, k = 9.0 x 10⁹ Nm²/C², q = 6.60 mC = 6.60 x 10⁻³ C
so we substitute into the formula
E = [(9.0 x 10⁹ Nm²/C²)( 6.60 x 10⁻³ C)] / ( 3.9 )²
E = 59400000 / 15.21
E = 3.91 x 10⁶ N/C
Therefore, the electric field is 3.91 x 10⁶ N/C
The equation of damped oscillations is given in the form x=0.05e^-0.25sin½πt (m). Find the velocity of an oscillating point at the moments of time: 0, T, 2T, 3T and 4T.
Explanation:
Finding the (maximum) respective prime powers would yield the answer. Also we need not ... Is perfectly divisible by 720^n? ... So we can say that for any positive value of n it not divisible.
A 92-kg man climbs into a car with worn out shock absorbers, and this causes the car to drop down 4.5 cm. As he drives along he hits a bump, which starts the car oscillating at an angular frequency of 4.52 rad/s. What is the mass of the car ?A) 890 kg
B) 1900 kg
C) 920 kg
D) 990 kg
E) 760 kg
Answer:
the mass of the car is 890 kg
Explanation:
Given;
mass of the man, m = 92 kg
displacement of the car's spring, x = 4.5 cm = 0.045 m
acceleration due to gravity, g = 9.8 m/s²
The spring constant of the car,
f = kx
where;
f is the weight of the man on the car = mg
mg = kx
k = mg/x
k = (92 x 9.8) / 0.045
k = 20,035.56 N/m
The angular speed of car, ω, when the is inside is given as 4.52 rad/s
The total mass of the car and the man is calculated as;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega^2 = \frac{k}{m} \\\\m = \frac{k}{\omega^2} = \frac{20,035.56}{(4.52)^2} = 980.7 \ kg[/tex]
The mass of the car alone = 980.7 kg - 92 kg
= 888.7 kg
≅ 890 kg
Therefore, the mass of the car is 890 kg
If you pull with your lower leg such that you exert a 90 N force on the cord attached to your ankle, determine the magnitude of the tension force of your hamstring on your leg and the compression force at the knee joint.
This question is incomplete, the missing diagram is uploaded along this answer below.
Answer:
- the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N
- the magnitude of compression force at the knee joint is 900 N
Explanation:
Given the data in the question and diagram below;
Net torque = 0
Torque = force × lever arm
so
F[tex]_{ConF[/tex] × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in
given that F[tex]_{ConF[/tex] = 90 N
90 × ( 15.0 in + 1.5 in ) = T[tex]_{HonL[/tex] × 1.5 in
90 N × 16.5 in = T[tex]_{HonL[/tex] × 1.5 in
T[tex]_{HonL[/tex] = ( 90 N × 16.5 in ) / 1.5 in
T[tex]_{HonL[/tex] = 990 N
Therefore, the magnitude of the tension force exerted by the hamstring muscles on the leg is 990 N
b) magnitude of compression force at the knee joint;
In equilibrium, net force = 0
along horizontal
F[tex]_{FonB[/tex] - T[tex]_{HonL[/tex] + F[tex]_{ConF[/tex] = 0
we substitute
F[tex]_{FonB[/tex] - 990 + 90 = 0
F[tex]_{FonB[/tex] - 900 = 0
F[tex]_{FonB[/tex] = 900 N
Therefore, the magnitude of compression force at the knee joint is 900 N
They create a heat engine where the hot reservoir is filled with water and steam at equilibrium, and the cold reservoir is filled with ice and water at equilibrium. What is the Carnot efficiency for their heat engine if the pressure is constant at 1.0 atmospheres?
Answer:
The efficiency of Carnot's heat engine is 26.8 %.
Explanation:
Temperature of hot reservoir, TH = 100 degree C = 373 K
temperature of cold reservoir, Tc = 0 degree C = 273 K
The efficiency of Carnot's heat engine is
[tex]\eta = 1-\frac{Tc}{T_H}\\\\\eta = 1 -\frac{273}{373}\\\\\eta = 0.268 =26.8 %[/tex]
The efficiency of Carnot's heat engine is 26.8 %.
Two astronauts, each having a mass of 88.0 kg, are connected by a 10.0-m rope of negligible mass. They are isolated in space, moving in circles around the point halfway between them at a speed of 5.40 m/s. Treating the astronauts as particles, calculate each of the following.
a. the magnitude of the angular momentum of the system
b. the rotational energy of the system
c. What is the new angular momentum of the system?
d. What are their new speeds?
e. What is the new rotational energy of the system
Answer:
a) L = 4.75 103 kg m² / s, b) K_total = 2.57 10³ J,
c) L₀ = L_f =4.75 103 kg m² / s, d) K = 1.03 10⁴ J, K = 1.03 10⁴ J
Explanation:
a) the angular momentum is the sum of the angular momentum of each astronaut
the distance is measured from the center of the circle r = 10/2 = 5.0 m
L = 2m v r
L = 2 88.0 5.40 5.0
L = 4.75 103 kg m² / s
b) rotational kinetic energy
K = ½ I w²
As there are two astronauts, the total energy is the sum of the energy of each no.
The moment of inertia of a point mass
I = m r²
I = 88 5²
I = 2.2 10³ kg m²
the angular velocity is given by
v = w r
w = v / r
w = 5.40 / 5
w = 1.08 rad / s
the kinetic energy of the system
K_total = 2 K
K_total = 2 (½ I w²)
K_total = 2.2 10³ 1.08²
K_total = 2.57 10³ J
c, d) as astronauts are isolated in space, these speeds do not change unless there is an interaction between them, for example they approach each other, suppose they reduce their distance by half
r = 2.5 m
I = 88 2.5²
I = 5.5 10² kg m²
for the change in angular velocity let us use the conservation of moment
L₀ = L_f
2Io wo = 2 I w
w = Io / I wo
w = 2.2 10³ / 5.5 10² 1.08
w = 4.32 rad / s
linear velocity is
v = w r
v = 4.32 2.5
K = 1.03 10⁴ J
the kinetic energy of the system is
K = 5.5 10² 4.32²
K = 1.03 10⁴ J
which unit would be most suitable for its scale?
A mm
B
с
crn?
D
cm
[0625_504_9p_1].
8
A piece of cotton is measured between two points on a ruler.
1
coton
BAS
2
4
5
6
7
8
9
10
11
12
13
14
15 16
when the lenge of coton is wound closely around a pen, goes round six times.
pen
six turns of coton
दे-
What is the distance onde round the pen?
4 2.2 m
B 26 cm
с
13.2 cm
D 15.6 cm
Answer:
Mm, thats the answer trust me men
In a collision that is not perfectly elastic, what happens to the mechanical energy of the system?
a. All of the mechanical energy is converted into other forms
b. Some of the mechanical energy is converted into other forms
c. No mechanical energy is converted into other forms
In a collision that is not perfectly elastic, some of the mechanical energy is converted into other forms.
In a perfect elastic collision, both momentum and kinetic energy of the particles are conserved.
[tex]m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2[/tex]
[tex]\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2 ^2= \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2[/tex]
When the collision is not perfectly elastic, only momentum is conserved but the kinetic energy is not conserved.
Thus, we can conclude that in a collision that is not perfectly elastic, some of the mechanical energy is converted into other forms.
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How is fitness walking beneficial?
It can relieve stress and improve mood.
It can decrease energy levels.
It can decrease perspiration.
It can relieve allergy symptoms.
Answer:
It can relieve stress and improve mood.
A 75.0 kg diver falls from rest into a swimming pool from a height of 5.10 m. It takes 1.34 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.
Answer:
559.5 N
Explanation:
Applying,
v² = u²+2gs............. Equation 1
Where v = final velocity,
From the question,
Given: s = 5.10 m, u = 0 m/s ( from rest)
Constant: 9.8 m/s²
Therefore,
v² = 0²+2×9.8×5.1
v² = 99.96
v = √(99.96)
v = 9.99 m/s
As the diver eneters the water,
u = 9.99 m/s, v = 0 m/s
Given: t = 1.34 s
Apply
a = (v-u)/t
a = 9.99/1.34
a = -7.46 m/s²
F = ma.............. Equation 2
Where F = force, m = mass
Given: m = 75 kg, a = -7.46 m/s²,
F = 75(-7.46)
F = -559.5 N
Hence the average force exerted on the diver is 559.5 N
A meter stick has a mass of 0.30 kg and balances at its center. When a small chain is suspended from one end, the balance point moves 28.0 cm toward the end with the chain. Determine the mass of the chain.
Answer:
M L1 = m L2 torques must be zero around the fulcrum
M = m L2 / L1 = .3 kg * 28 cm / 22 cm = .382 kg
