Answer:
22 N upward
Explanation:
From the question,
Applying newton's second law of motion
F = m(v-u)/t....................... Equation 1
Where F = Average force exerted by the ground on the ball, m = mass of the baseball, v = final velocity, u = initial velocity, t = time of contact
Note: Let upward be negative and downward be positive
Given: m = 0.14 kg, v = -1.00 m/s, u = 1.2 m/s, t = 0.014 s
Substitute into equation 1
F = 0.14(-1-1.2)/0.014
F = 0.14(-2.2)/0.014
F = 10(-2.2)
F = -22 N
Note the negative sign shows that the force act upward
A proton moves at a speed 1.4 × 10^7 m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius 0.85 m. What is the field strength?
0.17T
Explanation:
When a charged particle moves into a magnetic field perpendicularly, it experiences a magnetic force [tex]F_{M}[/tex] which is perpendicular to the magnetic field and direction of the velocity. This motion is circular and hence there is a balance between the centripetal force [tex]F_{C}[/tex] and the magnetic force. i.e
[tex]F_{C}[/tex] = [tex]F_{M}[/tex] --------------(i)
But;
[tex]F_{C}[/tex] = [tex]\frac{mv^2}{r}[/tex] [m = mass of the particle, r = radius of the path, v = velocity of the charge]
[tex]F_{M}[/tex] = qvB [q = charge on the particle, B = magnetic field strength, v = velocity of the charge ]
Substitute these into equation (i) as follows;
[tex]\frac{mv^2}{r}[/tex] = qvB
Make B subject of the formula;
B = [tex]\frac{mV}{qr}[/tex] ---------------(ii)
Known constants
m = 1.67 x 10⁻²⁷kg
q = 1.6 x 10⁻¹⁹C
From the question;
v = 1.4 x 10⁷m/s
r = 0.85m
Substitute these values into equation(ii) as follows;
B = [tex]\frac{1.67 * 10 ^{-27} * 1.4 * 10^{7}}{1.6 * 10^{-19} * 0.85}[/tex]
B = 0.17T
Therefore, the magnetic field strength is 0.17T
We observe that a small sample of material placed in a non-uniform magnetic field accelerates toward a region of stronger field. What can we say about the material?
Answer:
C) It is either ferromagnetic or paramagnetic
Explanation:
The complete question is given below
We observe that a small sample of material placed in a non-uniform magnetic field accelerates toward a region of stronger field. What can we say about the material?
A) It must be ferromagnetic.
B) It must be paramagnetic.
C) It is either ferromagnetic or paramagnetic.
D) It must be diamagnetic.
A ferromagnetic material will respond towards a magnetic field. They are those materials that are attracted to a magnet. Ferromagnetism is associated with our everyday magnets and is the strongest form of magnetism in nature. Iron and its alloys is very good example of a material that readily demonstrate ferromagnetism.
Paramagnetic materials are weakly attracted to an externally applied magnetic field. They usually accelerate towards an electric field, and form internal induced magnetic field in the direction of the external magnetic field.
The difference is that ferromagnetic materials can retain their magnetization when the externally applied magnetic field is removed, unlike paramagnetic materials that do not retain their magnetization.
In contrast, a diamagnetic material is repelled away from an externally applied magnetic field.
6a. A special lamp can produce UV radiation. Which two statements
describe the electromagnetic waves emitted by a UV lamp? *
They have a higher frequency than X-rays.
They have the same wave speed as visible light
They have a longer wavelength than microwaves.
They have a lower frequency than gamma rays.
They have a greater wave speed than radio waves.
Answer:
The correct options are:
B) They have the same wave speed as visible light
D) They have a lower frequency than gamma rays.
Explanation:
B) Ultraviolet rays, commonly known as UV rays, are a type of electromagnetic ways. As electromagnetic waves, in the layman's term, are all kinds of life that can be identified, all electromagnetic waves (UV rays, visible light, infrared, radio etc) all travel with the same velocity, that is the speed of light, given as v = 3 × 10⁸ m/s
D) The frequency of all electromagnetic rays can be found by electromagnetic spectrum (picture attached below).
We can clearly see in the picture that the frequencies of UV rays lie at about 10¹⁵ - 10¹⁶ Hz which is lower than the frequency of Gamma ray, which lie at about 10²⁰ Hz.
An ideal gas in a cubical box having sides of length L exerts a pressure p on the walls of the box. If all of this gas is put into a box having sides of length 0.5L without changing its temperature, the pressure it exerts on the walls of the larger box will be...
p.
2p.
4p.
8p.
12p.
Answer:
2P
Explanation:
See attached file
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression, x = 8.00 cos 5t + π 8 where x is in centimeters and t is in seconds. (a) At t = 0, find the position of the piston. cm (b) At t = 0, find velocity of the piston. cm/s (c) At t = 0, find acceleration of the piston. cm/s2 (d) Find the period and amplitude of the motion. period s amplitude cm
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression, x = 8.00 cos (5t + π / 8) where x is in centimeters and t is in seconds. (a) At t = 0, find the position of the piston. cm (b) At t = 0, find velocity of the piston. cm/s (c) At t = 0, find acceleration of the piston. cm/s2 (d) Find the period and amplitude of the motion. period s amplitude cm
Answer:(a) 7.392cm
(b) -15.32 cm/s
(c) -184cm/s²
(d) 0.4πs and 8.00cm
Explanation:The general equation of a simple harmonic motion (SHM) is given by;
x(t) = A cos (wt + Φ) --------------(i)
Where;
x(t) = position of the body at a given time t
A = amplitude or maximum displacement during oscillation
w = angular velocity
t = time
Φ = phase constant.
Given from question:
x(t) = 8.00 cos (5t + π / 8) ---------------(ii)
(a) At time t = 0;
The position, x(t), of the body (piston) is given by substituting the value of t = 0 into equation (ii) as follows;
x(0) = 8.00 cos (5(0) + π / 8)
x(0) = 8.00 cos (π /8)
x(0) = 8.00 x 0.924
x(0) = 7.392 cm
Therefore, the position of the piston at time t = 0 is 7.392cm
(b) To get the velocity, v(t), of the piston at t = 0, first differentiate equation (ii) with respect to t as follows;
v(t) = [tex]\frac{dx(t)}{dt}[/tex]
v(t) = [tex]\frac{d(8.00cos(5t + \pi / 8 ))}{dt}[/tex]
v(t) = 8 (-5 sin (5t + π / 8))
v(t) = -40sin(5t + π / 8) --------------------(iii)
Now, substitute t=0 into the equation as follows;
v(0) = -40 sin(5(0) + π / 8)
v(0) = -40 sin(π / 8)
v(0) = -40 x 0.383
v(0) = -15.32 cm/s
Therefore, the velocity of the piston at time t = 0 is -15.32 cm/s
(c) To find the acceleration a(t) of the piston at t = 0, first differentiate equation (iii), which is the velocity equation, with respect to t as follows;
a(t) = [tex]\frac{dv(t)}{dt}[/tex]
a(t) = [tex]\frac{d(-40sin (5t + \pi /8))}{dt}[/tex]
a(t) = -200 cos (5t + π / 8)
Now, substitute t = 0 into the equation as follows;
a(0) = -200 cos (5(0) + π / 8)
a(0) = -200 cos (π / 8)
a(0) = -200 x 0.924
a(0) = -184.8 cm/s²
Therefore, the acceleration of the piston at time t = 0 is -184cm/s²
(d) To find the period, T, first, let's compare equations (i) and (ii) as follows;
x(t) = A cos (wt + Φ) --------------(i)
x(t) = 8.00 cos (5t + π / 8) ---------------(ii)
From these equations it can be deduced that;
Amplitude, A = 8.00cm
Angular velocity, w = 5 rads/s
But;
w = [tex]\frac{2\pi }{T}[/tex] [Where T = period of oscillation]
=> T = [tex]\frac{2\pi }{w}[/tex]
=> T = [tex]\frac{2\pi }{5}[/tex]
=> T = 0.4π s
Therefore, the period and amplitude of the piston's motion are respectively 0.4πs and 8.00cm
A dielectric material such as paper is placed between the plates of a capacitor holding a fixed charge. What happens to the electric field between the plates
Answer:
Majorly the electric field is reduced among other effect listed in the explanation
Explanation:
In capacitors the presence of di-electric materials
1. decreases the electric fields
2. increases the capacitance of the capacitors.
3. decreases the voltage hence limiting the flow of electric current.
The di-electric material serves as an insulator between the metal plates of the capacitors
At what minimum speed must a roller coaster be traveling when upside down at the top of a 7.4 m radius loop-the-loop circle so the passengers will not fall out?
Answer:
v = 8.5 m/s
Explanation:
In order for the passengers not to fall out of the loop circle, the centripetal force must be equal to the weight of the passenger. Therefore,
Weight = Centripetal Force
but,
Weight = mg
Centripetal Force = mv²/r
Therefore,
mg = mv²/r
g = v²/r
v² = gr
v = √gr
where,
v = minimum speed required = ?
g = 9.8 m/s²
r = radius = 7.4 m
Therefore,
v = √(9.8 m/s²)(7.4 m)
v = 8.5 m/s
Minimum speed for a roller coaster while travelling upside down so that the person will not fall out = 8.5 m/s
For a roller coaster be traveling when upside down the Force balance equation can be written for a person of mass m.
In the given condition the weight of the person must be balanced by the centrifugal force.
and for the person not to fall out centrifugal force must be greater than or equal to the weight of the person
According to the Newton's Second Law of motion we can write force balance
[tex]\rm mv^2/r -mg =0 \\\\mg = mv^2 /r (Same\; mass) \\\\\\g = v^2/r\\\\v = \sqrt {gr}......(1)[/tex]
Given Radius of loop = r = 7.4 m
Putting the value of r = 7.4 m in equation (1) we get
[tex]\sqrt{9.8\times 7.4 } = \sqrt{72.594} = 8.5\; m/s[/tex]
For more information please refer to the link below
https://brainly.com/question/13259103
Value of g in CGS system
Answer:
in CGS system G is denoted as gram
An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the speed remained the same, by what factor would its centripetal acceleration change?
Answer:
The centripetal acceleration changed by a factor of 0.5
Explanation:
Given;
first radius of the horizontal circle, r₁ = 500 m
speed of the airplane, v = 150 m/s
second radius of the airplane, r₂ = 1000 m
Centripetal acceleration is given as;
[tex]a = \frac{v^2}{r}[/tex]
At constant speed, we will have;
[tex]v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1[/tex]
a₂ = 0.5a₁
Therefore, the centripetal acceleration changed by a factor of 0.5
In which direction does a bag at rest move when a force of 20 newtons is applied from the right?
ОА.
in the direction of the applied force
OB.
in the direction opposite of the direction of the applied force
OC. perpendicular to the direction of the applied force
OD
in a circular motion
Answer:
in the direction of the applied force
Explanation:
An ultrasound machine uses 1.64 × 105 watts of power. If it draws 12.0 amps of current, what is the resistance?
Answer:
R = 1138.9 Ω
Explanation:
Hello,
In this case, for the given power (P) and current (I), we can compute the resistance (R) via:
R = P / I²
Thus, we obtain:
R = 1.64x10⁵ W / (12.0 A)²
R = 1138.9 Ω
Best regards.
A total electric charge of 2.00 nC is distributed uniformly over the surface of a metal sphere with a radius of 26.0 cm . The potential is zero at a point at infinity.
a) Find the value of the potential at 45.0 cm from the center of the sphere.
b) Find the value of the potential at 26.0 cm from the center of the sphere.
c) Find the value of the potential at 16.0 cm from the center of the sphere.
Answer:
a) 40 V
b) 69.23 V
c) 69.23 V
Explanation:
See attachment for solution
7. Which statement is true about teens that are in Marcia’s final state of identity formation?
Answer:
D. All of the above
Explanation:
The last stage in the Marcia's identity formation theory is Identity achievement. In this last stage, teens have made a thorough search or exploration about their identity and have made a commitment to that identity. This identity represents their values, beliefs, and desired goals. At this point, they know want they want in life, and can now make informed decisions based on their belief and ideology.
James Marcia is a psychologist known mainly for his research and theories in human identity. Identity according to him is the sum total of a person's beliefs, values, and ideologies that shape what a person actually becomes and is known for. Occupation and Ideologies primarily determine identity. The four stages of Identity status include, Identity diffusion, foreclosure, moratorium, and achievement.
A spherical shell rolls without sliding along the floor. The ratio of its rotational kinetic energy (about an axis through its center of mass) to its translational kinetic energy is:
Answer:
The ratio is [tex]\frac{RE}{TE} = \frac{2}{3}[/tex]
Explanation:
Generally the Moment of inertia of a spherical object (shell) is mathematically represented as
[tex]I = \frac{2}{3} * m r^2[/tex]
Where m is the mass of the spherical object
and r is the radius
Now the the rotational kinetic energy can be mathematically represented as
[tex]RE = \frac{1}{2}* I * w^2[/tex]
Where [tex]w[/tex] is the angular velocity which is mathematically represented as
[tex]w = \frac{v}{r}[/tex]
=> [tex]w^2 = [\frac{v}{r}] ^2[/tex]
So
[tex]RE = \frac{1}{2}* [\frac{2}{3} *mr^2] * [\frac{v}{r} ]^2[/tex]
[tex]RE = \frac{1}{3} * mv^2[/tex]
Generally the transnational kinetic energy of this motion is mathematically represented as
[tex]TE = \frac{1}{2} mv^2[/tex]
So
[tex]\frac{RE}{TE} = \frac{\frac{1}{3} * mv^2}{\frac{1}{2} * m*v^2}[/tex]
[tex]\frac{RE}{TE} = \frac{2}{3}[/tex]
What happens when you increase the number of slits per millimeter (decrease the spacing between slits)?
Answer:
Increasing the number of slits not only makes the diffraction maximum sharper, but also much more intense. If a 1 mm diameter laser beam strikes a 600 line/mm grating, then it covers 600 slits and the resulting line intensity is 90,000 x that of a double slit. Such a multiple-slit is called a diffraction grating.
A 2kg block is sitting on a hinged ramp such that you can increase the angle of the incline. The coefficient of static friction between the block and the ramp is 0.67 and the coefficient of kinetic friction is 0.25.
a. What angle do you have to tilt the ramp to get the block to slide?
b. What acceleration does the block experience at this angle when kinetic friction takes over?
Answer:
θ = 33.8
a = 3.42 m/s²
Explanation:
given data
mass m = 2 kg
coefficient of static friction μs = 0.67
coefficient of kinetic friction μk = 0.25
solution
when block start slide
N = mg cosθ .............1
fs = mg sinθ ...............2
now we divide equation 2 by equation 1 we get
[tex]\frsc{fs}{N} = \frac{sin \theta }{cos \theta }[/tex]
[tex]\frac{\mu s N }{N}[/tex] = tanθ
put here value we get
tan θ = 0.67
θ = 33.8
and
when block will slide then we apply newton 2nd law
mg sinθ - fk = ma ...............3
here fk = μk N = μk mg cosθ
so from equation 3 we get
mg sinθ - μk mg cosθ = ma
so a will be
a = (sinθ - μk cosθ)g
put here value and we get
a = (sin33.8 - 0.25 cos33.8) 9.8
a = 3.42 m/s²
What fundamental frequency would you expect from blowing across the top of an empty soda bottle that is 24 cm deep, if you assumed it was a closed tube
Answer:
f = 357.29Hz
Explanation:
In order to calculate the fundamental frequency in the closed tube, you use the following formula:
[tex]f_n=\frac{nv}{4L}[/tex] (1)
n: order of the mode = 1
v: speed of sound = 343m/s
L: length of the tube = 24cm = 0.24m
You replace the values of the parameters in the equation (1):
[tex]f_1=\frac{(1)(343m/s)}{4(0.24m)}=357.29Hz[/tex]
The fundamental frequency of in the tube is 357.29Hz
key points that can be found in the realist philosophical position
Answer:
Key points that can be found in the realist philosophical position are as follows:
The view that we observe or identify is real, truly out there.The objects which are identified are independent of someone's perceptions, linguistic practices, conceptual scheme, and beliefs.Quantum mechanics is an example of philosophical realism that claims world is mind-independent.I WILL MARK YOU AS BRAINLIEST!!! An object is launched straight up into the air with an initial velocity of 40 meters per second, from a height 30 m above the ground. Assuming that gravity pulls it down, changing its position by about 4.9 /2, after how many seconds will the object hit the ground? Enter your answer as a number rounded to the nearest tenth, such as: 42.5
Answer:
8.9 seconds
Explanation:
The height of the object at time t is:
y = h + vt − 4.9t²
where h is the initial height, and v is the initial velocity.
Given h = 30 and v = 40:
y = 30 + 40t − 4.9t²
When y = 0:
0 = 30 + 40t − 4.9t²
4.9t² − 40t − 30 = 0
Solving with quadratic formula:
t = [ -(-40) ± √((-40)² − 4(4.9)(-30)) ] / 2(4.9)
t = [ 40 ± √(1600 + 588) ] / 9.8
t = 8.9
It takes 8.9 seconds for the object to land.
What must the charge (sign and magnitude) of a 1.60 g particle be for it to remain balanced against gravity when placed in a downward-directed electric field of magnitude 680 N/C
Answer:
Explanation:
The charge must be negative so that force in a downward electric field will be upward so that its weight is balanced .
Let the charge be - q .
force on charge
= q x E where E is electric field
= q x 680
weight = 1.6 x 10⁻³ x 9.8
so
q x 680 = 1.6 x 10⁻³ x 9.8
q = 1.6 x 10⁻³ x 9.8 / 680
= 23 x 10⁻⁶ C
- 23 μ C .
Point A of the circular disk is at the angular position θ = 0 at time t = 0. The disk has angular velocity ω0 = 0.17 rad/s at t = 0 and subsequently experiences a constant angular acceleration α = 1.3 rad/s2. Determine the velocity and acceleration of point A in terms of fixed i and j unit vectors at time t = 1.7 s.
Given that,
Angular velocity = 0.17 rad/s
Angular acceleration = 1.3 rad/s²
Time = 1.7 s
We need to calculate the angular velocity
Using angular equation of motion
[tex]\omega=\omega_{0}+\alpha t[/tex]
Put the value in the equation
[tex]\omega=0.17+1.3\times1.7[/tex]
[tex]\omega=2.38(k)\ m/s[/tex]
We need to calculate the angular displacement
Using angular equation of motion
[tex]\theta=\theta_{0}+\omega_{0}t+\dfrac{\alpha t^2}{2}[/tex]
Put the value in the equation
[tex]\theta=0+0.17\times1.7+\dfrac{1.3\times1.7^2}{2}[/tex]
[tex]\theta=2.1675\times\dfrac{180}{\pi}[/tex]
[tex]\theta= 124.18^{\circ}[/tex]
We need to calculate the velocity at point A
Using equation of motion
[tex]v_{A}=v_{0}+\omega\times r[/tex]
Put the value into the formula
[tex]v_{A}=0+2.38(k) \times0.2(\cos(124.18)i+\sin(124.18)j))[/tex]
[tex]v_{A}=0.476\cos(124.18)j+0.476\sin(124.18)i[/tex]
[tex]v_{A}=(-0.267j-0.393i)\ m/s[/tex]
We need to calculate the acceleration at point A
Using equation of motion
[tex]a_{A}=a_{0}+\alpha\times r+\omega\times(\omega\times r)[/tex]
Put the value in the equation
[tex]a_{A}=0+1.3(k)\times0.2(\cos(124.18)i+\sin(124.18)j)+2.38\times2.38\times0.2(\cos(124.18)i+\sin(124.18)j)[/tex]
[tex]a_{A}=0.26\cos(124.18)i+0.26\sin(124.18)j+(2.38)^2\times0.2(\cos(124.18)i+\sin(124.18)j)[/tex]
[tex]a_{A}=-0.146j-0.215i−0.636i+0.937j[/tex]
[tex]a_{A}=0.791j-0.851i[/tex]
[tex]a_{A}=-0.851i+0.791j\ m/s^2[/tex]
Hence, (a). The velocity at point A is [tex](-0.267j-0.393i)\ m/s[/tex]
(b). The acceleration at point A is [tex](-0.851i+0.791j)\ m/s^2[/tex]
A 2.0-kg object moving 5.0 m/s collides with and sticks to an 8.0-kg object initially at rest. Determine the kinetic energy lost by the system as a result of this collision.
Answer:
20J
Explanation:
In a collision, whether elastic or inelastic, momentum is always conserved. Therefore, using the principle of conservation of momentum we can first get the final velocity of the two bodies after collision. This is given by;
m₁u₁ + m₂u₂ = (m₁ + m₂)v ---------------(i)
Where;
m₁ and m₂ are the masses of first and second objects respectively
u₁ and u₂ are the initial velocities of the first and second objects respectively
v is the final velocity of the two objects after collision;
From the question;
m₁ = 2.0kg
m₂ = 8.0kg
u₁ = 5.0m/s
u₂ = 0 (since the object is initially at rest)
Substitute these values into equation (i) as follows;
(2.0 x 5.0) + (8.0 x 0) = (2.0 + 8.0)v
(10.0) + (0) = (10.0)v
10.0 = 10.0v
v = 1m/s
The two bodies stick together and move off with a velocity of 1m/s after collision.
The kinetic energy(KE₁) of the objects before collision is given by
KE₁ = [tex]\frac{1}{2}[/tex]m₁u₁² + [tex]\frac{1}{2}[/tex]m₂u₂² ---------------(ii)
Substitute the appropriate values into equation (ii)
KE₁ = ([tex]\frac{1}{2}[/tex] x 2.0 x 5.0²) + ([tex]\frac{1}{2}[/tex] x 8.0 x 0²)
KE₁ = 25.0J
Also, the kinetic energy(KE₂) of the objects after collision is given by
KE₂ = [tex]\frac{1}{2}[/tex](m₁ + m₂)v² ---------------(iii)
Substitute the appropriate values into equation (iii)
KE₂ = [tex]\frac{1}{2}[/tex] ( 2.0 + 8.0) x 1²
KE₂ = 5J
The kinetic energy lost (K) by the system is therefore the difference between the kinetic energy before collision and kinetic energy after collision
K = KE₂ - KE₁
K = 5 - 25
K = -20J
The negative sign shows that energy was lost. The kinetic energy lost by the system is 20J
Find the average power Pavg created by the force F in terms of the average speed vavg of the sled.
Complete Question
The complete question is shown on the first and second uploaded image
Answer:
The power created is [tex]P_{avg} = F * v_{avg}[/tex]
Explanation:
From the question we are told that
The that the average power is mathematically represented as
[tex]P_{avg} = \frac{W }{\Delta t }[/tex]
Where W is is the Workdone which is mathematically represented as
[tex]W = F * s[/tex]
Where F is the applies force and s is the displacement due to the force
So
[tex]P_{avg} = \frac{F *s }{\Delta t }[/tex]
Now this displacement can be represented mathematically as
[tex]s = v_{avg} * \Delta t[/tex]
Where [tex]v_{avg }[/tex] is the average velocity and [tex]\Delta t[/tex] is the time taken
So
[tex]P_{avg} = \frac{F *v_{avg} * \Delta t }{\Delta t }[/tex]
=> [tex]P_{avg} = F * v_{avg}[/tex]
Answer:
Pavg = Fvavg
Explanation:
Since the P (power) done by the F (force) is:
P = Fs/t
and we are looking for the velocity, so then it would be:
P = Fv
with the average velocity the answer is:
Pavg = Favg
If an object is moving at a constant speed, and the force F is also constant, this formula can be used to find the average power. If v is changing, the formula can be used to find the instantaneous power at any given moment (with the quantity v in this case meaning the instantaneous velocity, of course).
A trough is filled with a liquid of density 810 kg/m3. The ends of the trough are equilateral triangles with sides 8 m long and vertex at the bottom. Find the hydrostatic force on one end of the trough. (Use 9.8 m/s2 for the acceleration due to gravity.)
Answer:
The hydrostatic force on one end of the trough is 54994.464 N
Explanation:
Given;
liquid density, ρ = 810 kg/m³
side of the equilateral triangle, L = 8m
acceleration due to gravity, g = 9.8 m/s²
Hydrostatic force is given as;
H = ρgh
where;
h is the vertical height of the equilateral triangle
Draw a line to bisect upper end of the trough, to the vertex at the bottom, this line is the height of the equilateral triangle.
let the half side of the triangle = x
x = ⁸/₂ = 4m
The half section of the triangle forms a right angled triangle
h² = 8² - 4²
h² = 48
h = √48
h = 6.928m
F = ρgh
F = 810 x 9.8 x 6.928
F = 54994.464 N
Therefore, the hydrostatic force on one end of the trough is 54994.464 N
Which of the following technologies is based on the work of Ibn al-Haytham?
A. Telescopes to observe the visible light of distant stars
B. Radiation treatments for breast cancer
C. Radar to detect the movement of storms
O D. An orbiting observatory to detect X-rays from space objects
Answer:
The answer is A
Explanation:
Its A because he created a telescope to be able to observe stars.
f the mass of the block is 2 kg, the radius of the circle is 0.8 m, and the speed of the block is 3 m/s, what is the tension in the string at the top of the circle
Answer:
the size are components relative to the whole.
Explanation:
they are particularly good at showing percentage or proportional data
A slender rod of length L has a varying mass-per-unit-length from the left end (x=0) according to dm/dx=Cx where C has units kg/m2. Find the total mass in terms of C and L, and then calculate the moment of inertia of the rod for an axis at the left end note: you need the total mass in order to get the answer in terms of ML^2
Answer:
ML²/6
Explanation:
Pls see attached file
The total mass is M = CL²/2, and the moment of inertia is I = ML²/2,
Moment of inertia:The length of the rod is L. It has a non-uniform distribution of mass given by:
dm/dx = Cx
where C has units kg/m²
dm = Cxdx
the total mass M of the rod can be calculated by integrating the above relation over the length:
[tex]M =\int\limits^L_0 {} \, dm\\\\M=\int\limits^L_0 {Cx} \, dx\\\\M=C[x^2/2]^L_0\\\\M=C[L^2/2]\\\\[/tex]
Thus,
C = 2M/L²
Now, the moment of inertia of the small element dx of the rod is given by:
dI = dm.x²
dI = Cx.x²dx
[tex]dI = \frac{2M}{L^2}x^3dx\\\\I= \frac{2M}{L^2}\int\limits^L_0 {x^3} \, dx \\\\I= \frac{2M}{L^2}[\frac{L^4}{4}][/tex]
I = ML²/2
Learn more about moment of inertia:
https://brainly.com/question/6953943?referrer=searchResults
in a certain region of space, the gravitational field is given by -k/r,where r=distance,k=const.if gravitational potential at r=r0 be v0,then what is the expression for the gravitational potential v?
options
1)k log(r/ro)
2)k log(ro/r)
3)vo+k log(r/ro)
4)vo+k log(ro/r)
plz help me out
I will mark u as brainliest if u answer correct
Answer:
The correct answer is option 3 .
Please check the answer once :)
⦁ A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a) If the coefficient of static friction is 0.5, what minimum force magnitude is required from the rope to start the crate moving? (b) If µk= 0.35, what is the magnitude of the initial acceleration of the crate?
Answer:
303.29N and 1.44m/s^2
Explanation:
Make sure to label each vector with none, mg, fk, a, FN or T
Given
Mass m = 68.0 kg
Angle θ = 15.0°
g = 9.8m/s^2
Coefficient of static friction μs = 0.50
Coefficient of kinetic friction μk =0.35
Solution
Vertically
N = mg - Fsinθ
Horizontally
Fs = F cos θ
μsN = Fcos θ
μs( mg- Fsinθ) = Fcos θ
μsmg - μsFsinθ = Fcos θ
μsmg = Fcos θ + μsFsinθ
F = μsmg/ cos θ + μs sinθ
F = 0.5×68×9.8/cos 15×0.5×sin15
F = 332.2/0.9659+0.5×0.2588
F =332.2/1.0953
F = 303.29N
Fnet = F - Fk
ma = F - μkN
a = F - μk( mg - Fsinθ)
a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0
303.29-0.35( 666.4 - 303.29*0.2588)/68.0
303.29-0.35(666.4-78.491)/68.0
303.29-0.35(587.90)/68.0
(303.29-205.45)/68.0
97.83/68.0
a = 1.438m/s^2
a = 1.44m/s^2
4. Chloe has a vertical velocity of 3 m/s when she leaves the 1 m diving board. At this instant, her center of gravity is 2.5 m above the water. How high above the water will Chloe go
Answer:
2.95m
Explanation:
Using h= 2.5+ v²/2g
Where v= 3m/s
g= 9.8m/s²
h= 2.95m