Answer:
2.86 m/sExplanation:
The velocity of the car can be found by using the formula
[tex]v = \frac{p}{m} \\ [/tex]
p is the momentum
m is the mass
From the question we have
[tex]v = \frac{100}{35} = \frac{50}{7} \\ = 2.857142...[/tex]
We have the final answer as
2.86 m/sHope this helps you
a tiger moving with constant accleration covers the distance between two points 70 meter apart in 7 seconds . its speed as it pass the second point 15 meter per second . 1) what is the speed at the first point. 2) what us its accleration
Answer:
a = 1.428 [m/s²]
v₀ = 5 [m/s]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]x=x_{o}+v_{o}*t+\frac{1}{2}*a*t^{2}[/tex]
where:
x = final point [m]
x₀ = initial point [m]
v₀ = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]
But we need to use this additional equation.
[tex]v_{f}=v_{o}+a*t[/tex]
where:
vf = final velocity = 15 [m/s]
Now we can use this equation, replacing it, in the first one. We must bear in mind that the difference among x - x₀ is equal to 70 [m]
[tex]x-x_{o}=v_{o}*t+\frac{1}{2}*a*t^{2} \\x-x_{o}=(v_{f}-a*t)*t+\frac{1}{2} *a*t^{2}\\70=(15-a*t)*t+\frac{1}{2}*a*t^{2}\\70=15*t-a*t^{2} +\frac{1}{2}*a*t^{2} \\70=15*t-\frac{1}{2}*a*t^{2}\\70=15*(7)-\frac{1}{2} *a*(7)^{2}\\105-70=0.5*a*49\\35=24.5*a\\a=1.428[m/s^{2} ][/tex]
Now replacing this value in the second equation, we can find the initial velocity.
[tex]15=v_{o}+1.428*7\\v_{o}=5[m/s][/tex]
how high above the ground is a 3 kg object that has 147 J (joules) of potential energy
Answer:
4.9949
Explanation:
M = 3 kg
P.E = 147 joules
G = 9.81 ms^-2
H = ?
P.E = MGH
147 = 3 x 9.81 x H
147 = 29.43H
147/29.43 = 29.43H/29.43
4.9949 = H
:. H = 4.9949
An inner city revitalization zone is a rectangle that is twice as long as it is wide. The width of the region is growing at a rate of 24 m per year at a time when the region is 300 m wide. How fast is the area changing at that point in time
Answer:
[tex]\frac{dA}{dt} = 28800 \ m^2/year[/tex]
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightEquality Properties
Geometry
Area of a Rectangle: A = lwAlgebra I
Exponential Property: [tex]w^n \cdot w^m = w^{n + m}[/tex]Calculus
Derivatives
Differentiating with respect to time
Basic Power Rule:
f(x) = cxⁿ f’(x) = c·nxⁿ⁻¹Explanation:
Step 1: Define
Area is A = lw
2w = l
w = 300 m
[tex]\frac{dw}{dt} = 24 \ m/year[/tex]
Step 2: Rewrite Equation
Substitute in l: A = (2w)wMultiply: A = 2w²Step 3: Differentiate
Differentiate the new area formula with respect to time.
Differentiate [Basic Power Rule]: [tex]\frac{dA}{dt} = 2 \cdot 2w^{2-1}\frac{dw}{dt}[/tex]Simplify: [tex]\frac{dA}{dt} = 4w\frac{dw}{dt}[/tex]Step 4: Find Rate
Use defined variables
Substitute: [tex]\frac{dA}{dt} = 4(300 \ m)(24 \ m/year)[/tex]Multiply: [tex]\frac{dA}{dt} = (1200 \ m)(24 \ m/year)[/tex]Multiply: [tex]\frac{dA}{dt} = 28800 \ m^2/year[/tex]Answer:
28,800 m²/yr
Explanation:
This rectangle has dimensions such that:
width = wlength = 2wWe are given [tex]\displaystyle \frac{dw}{dt} = \frac{24 \ m}{yr}[/tex] and want to find [tex]\displaystyle \frac{dA}{dt} \Biggr | _{w \ = \ 300 \ m} = \ ?[/tex] when w = 300 m.
The area of a rectangle is denoted by Area = length * width.
Let's multiply the width and length (with respect to w) together to have an area equation in terms of w:
[tex]A=2w^2[/tex]Differentiate this equation with respect to time t.
[tex]\displaystyle \frac{dA}{dt} =4w \cdot \frac{dw}{dt}[/tex]Let's plug known values into the equation:
[tex]\displaystyle \frac{dA}{dt} =4(300) \cdot (24)[/tex]Simplify this equation.
[tex]\displaystyle \frac{dA}{dt} =1200 \cdot 24[/tex] [tex]\displaystyle \frac{dA}{dt} =28800[/tex]The area is changing at a rate of 28,800 m²/yr at this point in time.
Give the Concept of heat and temperature on the basis of molecular motion. Please give long answer
Explanation:
Molecular motion is defined as the movement of constituent particles or molecules in a certain direction. The molecular motions are affected by heat and temperature. This is because the temperature is the measurement of the average kinetic energy of the molecules and represents the motion of molecules.
Answer:
Heat and temperate are directly related As the temperature increases then the heat inside of the tube or container also increase. When a liquid is at a sub-cooled liquid temperature and increases then the molecules begin to move faster and fill up more space or volume. Hence as temperature increases then the molecules kinetic energy increases and activity in terms of molecules significantly increases.Un corp de masa m=5 kg se afla pe o suprafata orizontala pe care se poate deplasa cu frecare (µ = 0,02). Calculati forta de frecare.
Răspuns:
0,98 N
Explicaţie:
Dat fiind:
µ = Coeficientul de frecare = 0,02
Masa (m) = 5 kg
Folosind relația:
Forța de frecare (F) = µ * N
N = reacție normală = masă * accelerație datorată gravitației
Prin urmare,
F = u * m * g
g = 9,8 m / s²
F = 0,02 * 5 kg * 9,8 m / s²
F = 0,98 kgm / s²
F = 0,98N
A horizontal disc of radius 45 cm rotates about a vertical axis through its centre. The disc makes one full
revolution in 1.40 s. A particle of mass 0.054 kg is placed at a distance of 22 cm from the centre of the disc.
The particle does not move relative to the disc.
a On a copy of the diagram draw arrows to represent the velocity and acceleration of the particle. [2]
b Calculate the angular speed and the linear speed of the particle. [2]
c The coe cient of static friction between the disc and the particle is 0.82. Determine the largest distance
from the centre of the disc where the particle can be placed and still not move relative to the disc. [3]
d The particle is to remain at its original distance of 22cm from the centre of the disc.
i Determine the maximum angular speed of the disc so that the particle does not move relative to
the disc. [2]
ii The disc now begins to rotate at an angular speed that is greater than the answer in d i. Describe
qualitatively what happens to the particle. [2]
Answer:
a. Please find attached the diagram of the disc, having arrows that represent the velocity and the acceleration of the particle placed on it
b. The angular speed is approximately 4.488 rad/s
The linear speed is approximately 0.987 m/s
c. The largest distance from the center of the disc where the particle can be placed and still not move is approximately 0.399 m from the center of the disc
d. i The maximum angular speed of the disk so that the particle does not move relative to the disk is approximately 6.044 rad/sec
ii When the angular speed with which the disc rotates is more than the the answer of question d i above, the particle slips on the disc, and the disc begins to rotate faster than the particle, while the particle is swung in an outward radial direction off the disc due to the centrifugal forces
Explanation:
The given parameters are;
The radius of the horizontal disc, r = 45 cm = 0.45 m
The time the disc makes one full revolution, T = 1.40 s
The mass of the particle placed on the disc = 0.054 kg
The location on the disc the particle is placed = 22 cm from the disc's center
a. Please find attached the diagram of the disc created with Microsoft Visio, with arrows representing the velocity and the acceleration of the particle placed on the disk
b. The angular speed, ω = 2·π/T = 2 × π/1.4 ≈ 4.488 rad/s
The linear speed, v = ω × r = 4.488 rad/s × 0.22 m ≈ 0.987 m/s
The linear speed, v ≈ 0.987 m/s
c. The given coefficient of static friction = 0.82
Therefore;
The frictional force that prevents motion = Weight of the particle × The coefficient of static friction
The frictional force that prevents motion is [tex]F_f[/tex] = 0.054 × 9.8 × 0.82 ≈ 0.434 N
[tex]F_f[/tex] ≈ 0.434 N
Therefore, for the largest distance from the center of the disc where the particle can be placed and still not move, r, is given by the formula for the centripetal force, [tex]F_c[/tex], acting on the particle as follows;
For static equilibrium, no movement of the particle relative to the disc, we have;
[tex]F_f[/tex] = [tex]F_c[/tex]
Where;
[tex]F_c = \dfrac{m \times v^2}{r} = m \times \omega ^2 \times r[/tex]
Which gives;
[tex]F_c = {0.054 \ kg \times (4.488 \ rad/s)^2} \times r = F_f = 0.434 \ N[/tex]
r = 0.434 N/(0.054 kg × (4.488 rad/s)²) ≈ 0.399 m
The largest distance from the center of the disc where the particle can be placed and still not move, r = 0.399 m from the center of the disc
d. i From the static equilibrium equation where r = 0.22 m, we have;
[tex]F_c = {0.054 \ kg \times \omega ^2} \times 0.22 \ m = F_f = 0.434 \ N[/tex]
ω = √(0.434 N/(0.054 kg × (0.22 m))) ≈ 6.044 rad/sec
The maximum angular speed of the disk so that the particle does not move relative to the disk, ω ≈ 6.044 rad/sec
ii When the angular speed with which the disc rotates is more than the the answer of question d i above, we have
The particle begins to slip on the disc such that the disc rotates faster than the particle and the particle tends to rotate slower than the speed pf the disc and is swung off the disc by centripetal force acting on the particle due to the rotational motion of the disc.
Hair sticking to a balloon after it
has been rubbed on a girl's head is
an example of
Answer:
Friction
Explanation:
Answer:
It causes electrostatic charging, which makes it stick to your hair,
Which two elements would most likely form an ionic compound?
a
From the Periodic Table of Elements - #6 carbon and #8 oxygen
b
From the Periodic Table of Elements - #3 lithium and #9 fluorine
c
From the Periodic Table of Elements - #1 hydrogen and #7 nitrogen
d
From the Periodic Table of Elements - #5 boron and #10 neon
Answer:
boron and neon.
Explanation:
two element compounds are usually ionic when one element is a metal and the other is a non-metal. boron is a metal, whereas neon is a non-metal. i might be wrong but hope it helps.
Which ball has larger Inertia before touching the ground when it falls freely from a
certain height (air friction is zero)
a. mass 50 kg b. mass 25 kg
c. mass 10 kg d. All of these
Answer:
[tex]Mass = 50kg[/tex]
Explanation:
Required
Which has the larger inertia
Inertia (I) is calculated as follows:
[tex]I = mr^2[/tex]
Where m represents the mass of the object and r represents the distance from the axis.
Take for instance m = 10 and r = 4
The inertia would be:
[tex]I = 10 * 4^2[/tex]
[tex]I = 160[/tex]
Assume r is constant and m is increased to 25
The inertia becomes:
[tex]I = 25 * 4^2[/tex]
[tex]I = 25 * 16[/tex]
[tex]I = 400[/tex]
Notice that there is an increment in inertia, when the mass increased.
Hence, mass of 50kg will have the larger inertia.
please help answer brainlest
each rock you need to tell where OK can you find them
like: your backyard or the road
Flakes of brick
Cracks in rocks
Lichen growing on rocks
Roots or plant growth making cracks in concrete
Worn away statues
Exposed rocks
Rust on rocks
Example You Found:
Answer:
Flakes of brick: Athens, Ohio
Cracks in rocks: Sandston, Mudcracks
Lichen growing on rocks: Antarctica living below the surface
Roots or plant growth making cracks in concrete: In the driveway, sidewalk, or road.
Define relative motion??
Answer:
The laws of physics which apply when you are at rest on the earth also apply when you are in any reference frame which is moving at a constant velocity with respect to the earth. For example, you can toss and catch a ball in a moving bus if the motion is in a straight line at constant speed.
Explanation:
El motor de una licuadora gira a 3600 rpm, disminuye su velocidad angular hasta 2000 rpm realizando 120 vueltas. Calcular: a) La aceleración angular b) El tiempo transcurrido
Answer:
a) α = -65,2 rad/s².
b) t = 2,57 s.
Explanation:
a) La aceleración angular se puede calcular usando la siguiente ecuación:
[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta [/tex]
En donde:
[tex]\omega_{f}[/tex]: es la velocidad angular final = 2000 rpm = 209,4 rad/s
[tex]\omega_{0}[/tex]: es la velocidad angular inicial = 3600 rpm = 377,0 rad/s
α: es la aceleración angular=?
θ: es el desplazamiento o número de vueltas = 120 rev = 754,0 rad
Las conversiones de unidades se hicieron sabiendo que 1 revolución = 2π radianes y que 1 minuto = 60 segundos.
Resolviendo la ecuación (1) para α, tenemos:
[tex]\alpha = \frac{\omega_{f}^{2} - \omega_{0}^{2}}{2\theta} = \frac{(209,4 rad/s)^{2} - (377,0 rad/s)^{2}}{2*754,0 rad} = -65,2 rad/s^{2}[/tex]
Entonces, la aceleración angular es -65,2 rad/s². El signo negativo se debe a que el motor está desacelerando.
b) El tiempo transcurrido se puede encontrar como sigue:
[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]
Resolviendo para t, tenemos:
[tex]t = \frac{\omega_{f} - \omega_{0}}{\alpha} = \frac{209,4 rad/s - 377,0 rad/s}{-65,3 rad/s^{2}} = 2,57 s[/tex]
Por lo tanto, el tiempo transcurrido fue 2,57 s.
Espero que te sea de utilidad!
Dylan has two cubes of iron. The larger cube has twice the mass of the smaller cube. He measures the smaller cube. Its mass is 20 grams, and its density is 7.87 g/cm3. What’s the larger cube’s volume? The larger cube’s volume is about cm3.
Answer:
5.08cm³
Explanation:
Given parameters
The larger cube has twice the mass of the smaller cube
Mass of smaller cube = 20g
Mass of the larger cube = 2 x 20 = 40g
Density of the smaller cube = 7.87g/cm³
Unknown:
Volume of the larger cube = ?
Solution:
Density is the mass per unit volume of substance. For all samples of a substance, the density value is the same.
So, the density of the small and large iron is the same
Density = [tex]\frac{mass}{volume}[/tex]
Volume = [tex]\frac{mass}{density}[/tex]
So;
Volume of larger cube = [tex]\frac{40}{7.87}[/tex] = 5.08cm³
Answer:
5.08cm^3
Explanation:
Give three examples of how a cell's shape makes possible the cell's function
Answer:
A) a nerve cells long extensions enable it to transmit electrical impulses from one body part to another
B) the sheet like organization of epithelial cells enables them to protect underlying cells
C) the alignment of contractile proteins within muscle cells enables them to contract, pulling closer together the structures to which they attach
Explanation:
Explain how to change the pitch of the sound produced by a violin string when it is plucked or rubbed with the bow.
Answer:
Sound is produced by moving the bow over the string or by plucking it with the right hand. Pitch alterations are achieved by pressing down the string with the fingertips of the left hand on the fingerboard. This shortens the vibrating portion of the string and raises the pitch
Explanation:
PLZ HELP I DON'T UNDERSTAND!! a boy is playing catch with his friend. He throws the ball straight up. When it leaves his hand, the ball (2kg) is traveling 10m/s. Whats the ball's kinetic energy just as it leaves the boy's hand? What is the potential energy of the ball when it reaches the highest point?
Explanation:
The mass of a ball, m = 2 kg
It is traveling with a speed of 10 m/s
The ball's kinetic energy just as it leaves the boy's hand is calculated as follows :
[tex]K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 2\times (10)^2\\\\=100\ J[/tex]
The ball's kinetic energy just as it leaves the boy's hand is 100 J. The potential energy of the ball when it reaches the highest point is same as the kinetic energy as it leaves the boy's hand.
Hence, the required kinetic and potential energy is 100 J.
Billy throws a peach vertically upward. It takes 3 seconds for the peach to land back in his hand. How fast did Billy throw the peach?
Answer:
48 ft/s
Explanation:
Acceleration due to gravity decreased the upward velocity to zero in 1.5 seconds (half the total time), so the initial velocity must have been ...
(32 ft/s^2)(1.5 s) = 48 ft/s
How long will it take a fired shell from a cliff at an initial velocity of 800m/s at an angle 30 degree below the horizontal to reach the ground 150m below
Answer:
0.373 seconds. First, calculate the initial vertical velocity of the shell. 800sin(30) = 800*0.5 = 400 m/s Now the formula for the distance traveled is d = 400 m/s * T + 0.5A T^2 Substituting known values gives.
Explanation:
PLZ MARK ME BRAINLIST GUYSPLZ FOLLOW ME GUYSA tank holding 70 liters of water is heated by a 18000W electric immersion heater. If the specific heat capacity of water is 4200 J/(kg .℃), Calculate the time to change the temperature from 20 Degree Celsius to 100 degree Celsius.
Answer:
1306.67 s
Explanation:
From the question given above, the following data were obtained:
Volume of water = 70 L
Power (W) = 18000 w
Specific heat capacity (C) = 4200 J/g℃
Initial temperature (T1) = 20 ℃
Final temperature (T2) = 100 ℃
Time (t) =?
Next, we shall determine the mass of the water. This can be obtained by converting 70 L to kg. This is illustrated below:
1 L = 1 kg
Therefore,
70 L = 70 Kg
Thus, the mass of the water is 70 kg.
Next, we shall determine the heat energy involved. This is can be obtained as follow:
Mass (M) of water = 70 Kg
Specific heat capacity (C) = 4200 J/g℃
Initial temperature (T1) = 20 ℃
Final temperature (T2) = 100 ℃
Heat Energy (Q) =?
Q = MC(T2 – T1)
Q = 70 × 4200 × (100 – 20)
Q = 294000 × 80
Q = 3520000 J
Thus, the heat energy involved is 3520000 J
Finally, we shall determine the time taken.
Power (W) = 18000 w
Heat Energy (Q) = 3520000 J
Time (t) =?
Power = energy / time
18000 = 3520000 / time
Cross multiply
18000 × time = 3520000
Divide both side by 18000
Time = 3520000 / 18000
Time = 1306.67 s
If you have a liquid with a mass of 6g and a density of 25g/ml, what volume of liquid do you have?
Hello!!
For calculate the volume of the liquid, lets applicate the formula:
[tex]\boxed{V=m/ p }[/tex]
[tex]\textbf{Being:}[/tex]
[tex]\sqrt{}[/tex] V = Volume = ?
[tex]\sqrt{}[/tex] m = Mass = 6 g
[tex]\sqrt{}[/tex] p = Density = 25 g/ml
[tex]\text{Let's \textbf{replace} and resolve: }[/tex]
[tex]V = 6 \ g / 25 \ g / m l[/tex]
[tex]V = 0,24 \ m l[/tex]
[tex]\textbf{Result:}[/tex]
The volume of the liquid is 0,24 ml.
A 100 kg skateboarder sits at the top of a 10 m halfpipe, ready to go down the ramp. What will be his velocity when
he is 4 m off of the ground?
Answer:
v = 10.85 m/s
Explanation:
We will apply the law of conservation of energy to the skateboarder. Neglecting the frictional effects, the law of conservation of energy can be written as:
[tex]Loss\ in\ Potential\ Energy\ of\ Skateboarder = Gain\ in\ Kinetic\ Energy\ of\ Skateboarder[/tex][tex]mg\Delta h = \frac{1}{2}mv^2\\\\v^2 = 2g\Delta h\\v = \sqrt{2g\Delta h} \\[/tex]
where,
v = velocity of skateboarder = ?
g = acceleration due to gravity = 9.81 m/s²
Δh = change in height = 10 m - 4 m = 6 m
Therefore,
[tex]v = \sqrt{{(2)(9.81\ m/s^2})({6\ m})}}[/tex]
v = 10.85 m/s
Anyone....................
Answer:
h₁ = 18 [cm]
Explanation:
This problem can be solved by applying the principles of the static pressure of a liquid column. The key to the solution is to know that the pressure on the horizontal line should be equal for both liquids, this must be met regardless of their density.
[tex]P_{2}=P_{1}\\[/tex]
where:
P₂ = Pressure exerted by the liquid 2 in the given point L₂
P₁ = Pressure exerted by the liquid 2 in the given point L₁
Now we can calculate the pressure in each point using the following expression:
[tex]P_{1}=Ro_{1}*g*h_{1}\\P_{2}=Ro_{2}*g*h_{2}[/tex]
Where:
Ro₁ = density of the liquid 1 = 0.8 [g/cm³]
Ro₂ = density of the liquid 2 = 1.8 [g/cm³]
g = gravity acceleration = 9.81 [m/s²]
h₁ = column of the liquid 1 [m]
h₂ = column of the liquid 2, = 8 [cm] = 0.08 [m]
But first we must convert the units from grams per cubic centimeter to kilograms per cubic meter.
[tex]0.8[\frac{g}{cm^{3}}]*[\frac{1kg}{1000g} ]*[\frac{100^{3}cm^{3} }{1m^{3} } ]=800[kg/m^{3} ]\\1.8[\frac{g}{cm^{3}}]*[\frac{1kg}{1000g} ]*[\frac{100^{3}cm^{3} }{1m^{3} } ]=1800[kg/m^{3} ][/tex]
Now replacing in the first equation:
[tex]800*9.81*h_{1}=1800*9.81*0.08\\h_{1}=\frac{1800}{800} *0.08\\h_{1}=0.18 [m] = 18 [cm][/tex]
Answer:
[tex]h_1 = 18 \: cm[/tex]
Explanation:
See image for explanation
I hope my working is correct.
A student designs an experiment to demonstrate the conservation of gravitational potential and kinetic energy. She drops a wooden block from rest through a photogate 2 m below. If energy is conserved, how fast should the block be traveling when it passes through the photogate?
A.
2.0 m/s
B.
6.3 m/s
C.
9.8 m/s
D.
19.6 m/s
Answer:
6.3 m/s
Explanation:
A student drops a wooden block from rest through a photogate 2 m below. Since there is a conservation of the gravitational potential energy and the kinetic energy, the final speed of the block is B. 6.3 m/s.
Gravitational potential energy and kinetic energyGravitational potential energy: is the potential energy a massive object has in relation to another massive object due to gravity.Kinetic energy: is a form of energy that an object or a particle has by reason of its motion.A wooden block is dropped from rest. If energy is conserved, the initial gravitational potential energy (U₁) will be equal to the final kinetic energy (K₂).
U₁ = K₂
m × g × h₁ = 1/2 × m × v₂²
where,
m is the mass of the object.g is the gravity.h₁ is the initial height.v₂ is the final speed.m × g × h₁ = 1/2 × m × v₂²
g × h₁ = 1/2 × v₂²
v₂ = √(2 × g × h₁)
v₂ = √(2 × (9.8 m/s²) × 2 m) = 6.3 m/s
A student drops a wooden block from rest through a photogate 2 m below. Since there is a conservation of the gravitational potential energy and the kinetic energy, the final speed of the block is B. 6.3 m/s.
Learn more about conservation of energy here: https://brainly.com/question/14525402
A rope hangs between two cliffs, edge A and edge B. There is a 250 kg hiker sliding across and is closer to
edge A than edge B. The angles that the ropes make below the horizontal are 35° below edge A and 65° below
edge B. What is the Tension in both ropes?
Answer:
1655.5 N
Explanation:
We resolve the forces on the rope into horizontal and vertical components. Since the only vertical force exerted on the rope is the weight of the man, this weight equals the vertical components of the tensions on both sides of the rope .
The only horizontal forces acting on the rope are horizontal components of both tensions on both sides. Since the hiker doesn't move sideways, they cancel out and he is balanced. So, the net horizontal force is zero.
Let T be the tension in the rope. So, each vertical component on side A is Tsin35° and that on side B is Tsin65°. Let the weight of the man be W = mg where m = mass of hiker = 250 kg and g = acceleration due to gravity = 9.8 m/s².
So, since the hiker balances, Tsin35° + Tsin65° = mg
0.5736T + 0.9063T = 250 kg × 9.8 m/s²
1.4799T = 2450 N
dividing both sides by 1.4799, we have
T = 2450 N/1.4799
T = 1655.52 N
T ≅ 1655.5 N
So, the tension in the rope equals T = 1655.5 N
what is the correct answer?
Answer:
4
Explanation:
I added all of them and them minus the 5.
11. A ball is thrown up and attains maximum height of 20 m. Calculate its initial speed. (g = 10 ms')
a) 19.7 m/s
b) 20 m/s
c) 25 m/s
d) 35m/s
final velocity = 0
acceleration = - 10 m/ s 2
distance. = 20 m
u = ?
v^2 - u ^2 = 2 a s
0^2 - u^ 2 = 2 * -10 * 20
-u^2 = -400
u = √ 400
u = 20m / s
A person who normally weighs 700 N is riding in an elevator that is moving upward but slowing down at a steady rate. If this person is standing on a bathroom scale inside the elevator, what would the scale read?
Answer:
The scale will read less than 700 N
Explanation:
Given;
normal weight of the person, W = 700 N
The upward acceleration of the elevator is given by Newton's second law of motion;
F = ma
Also, the weight of the person, W = mg
Net force on the person when the elevator accelerates upward is given as;
R = ma + mg
When the elevator slows down at a steady rate, then net force on the person is given as;
R = mg - ma
R = m(g - a), this net force on the person is also the reading of the scale.
Thus, the scale will read less than 700 N.
Answer: The answer is less than 700 N
In the middle of a turn, you should _____________________. A. brake hard to maintain complete control over your vehicle B. accelerate abruptly and straighten the wheel C. avoid using your brake or clutch, unless it is necessary to do so in order to avoid a collision D. downshift to allow you to turn more sharply
Answer:
D. Downshift to allow you to turn more sharply
D. downshift to allow you to turn more sharply
What is the purpose of downshifting?The purpose of "downshifting", or shifting the transmission down from a higher gear to a lower gear, is so that a driver can accelerate their car as fast as possible when exiting a corner that they had to slow down for.
The reason for ensuring your downshift is completed before turning into the corner is because letting the clutch out and engaging a lower gear while turning can cause the car to spin.
Two special conditions where you should downshift
Before Starting Down a Hill.
Slow down and shift down to a speed that you can control without using the brakes hard. Otherwise the brakes can overheat and lose their braking power.
Before Entering a Curve.
Slow down to a safe speed, and downshift to the right gear before entering the curve. This lets you use some power through the curve to help the vehicle be more stable while turning. It also allows you speed up as soon as you are out of the curve.
Therefore,
In the middle of a turn, you should downshift to allow you to turn more sharply .
Learn more about Downshift here:
https://brainly.com/question/10600958
#SPJ2
An object of mass m is traveling in a circle with centripetal force f c. If the velocity of the object is v, what is the radius of the circle?
Answer:
r = mv^2/ Fc
Explanation:
Got it right
Why does the Sun appear white at noon?
