Answer:
B) The rod is subject to tensile stress
Explanation:
Tensile stress is the force induced in the body per unit area in response to externally applied force, which stretches the body.
Tensile stress is attached with tensile forces. It is makes material elongate along the axis of the applied load.
It is given as magnitude (F) of the force that is applied along an elastic rod divided by the cross sectional area (A) of the rod in a direction that is perpendicular to the applied force.
materials such as Doctile have the ability to withstand the load while brittle materials who not have the ability they fail before reaching the ultimate material strength.
A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a momentum of magnitude 4.0 kg∙ m/s. What is the magnitude of the large object's momentum change?
Answer:
The magnitude of the large object's momentum change is 3 kilogram-meters per second.
Explanation:
Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:
[tex]p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2}[/tex] (1)
Where:
[tex]p_{S,1}[/tex], [tex]p_{S,2}[/tex] - Initial and final momemtums of the small object, measured in kilogram-meters per second.
[tex]p_{B,1}[/tex], [tex]p_{B,2}[/tex] - Initial and final momentums of the big object, measured in kilogram-meters per second.
If we know that [tex]p_{S,1} = 7\,\frac{kg\cdot m}{s}[/tex], [tex]p_{B,1} = 0\,\frac{kg\cdot m}{s}[/tex] and [tex]p_{S, 2} = 4\,\frac{kg\cdot m}{s}[/tex], then the final momentum of the big object is:
[tex]7\,\frac{kg\cdot m}{s} + 0\,\frac{kg\cdot m}{s} = 4\,\frac{kg\cdot m}{s}+p_{B,2}[/tex]
[tex]p_{B,2} = 3\,\frac{kg\cdot m}{s}[/tex]
The magnitude of the large object's momentum change is:
[tex]p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}-0\,\frac{kg\cdot m}{s}[/tex]
[tex]p_{B,2}-p_{B,1} = 3\,\frac{kg\cdot m}{s}[/tex]
The magnitude of the large object's momentum change is 3 kilogram-meters per second.
A 0.320 kg ball approaches a bat horizontally with a speed of 14.0 m/s and after getting hit by the bat, the ball moves in the opposite direction with a speed of 22 m/s. If the ball is in contact with the bat for 0.0600 s, determine the magnitude of the average force exerted on the bat.
Answer:
42.67NExplanation:
Step one:
Given
mass m= 0.32kg
intital velocity, u= 14m/s
final velocity v= 22m/s
time= 0.06s
Step two:
Required
Force F
the expression for the force is
F=mΔv/t
F=0.32*(22-14)/0.06
F=(0.32*8)/0.06
F=2.56/0.06
F=42.67N
The average force exerted on the bat 42.67N
The solid shaft with a 20 mm radius is used to transmit the torques applied to the gears. Determine the maximum torsional shear stress developed in in the shaft.
Answer:
τ = (7.96 x 10⁴ m⁻³)T
This is the expression for maximum allowable shear stress in terms of the maximum torque applied in Nm.
Explanation:
The maximum allowable shear stress on the solid shaft can be given by the torsional formula as follows:
τ = Tc/J
where,
τ = Maximum Allowable Shear Stress = ?
T = Maximum Torque Applied to the Shaft
c = maximum distance from center to edge = radius in this case = 20 mm = 0.02 m
J = Polar Moment of inertia = πr⁴/2 = π(0.02 m)⁴/2 = 2.51 x 10⁻⁷ m⁴
Therefore,
τ = T(0.02 m)/(2.51 x 10⁻⁷ m⁴)
τ = (7.96 x 10⁴ m⁻³)T
This is the expression for maximum allowable shear stress in terms of the maximum torque applied in Nm.
Which information can be determined using half-life?
In,radioactive decay,the half-life is the length of time after which there is a 50 percent chance that an atom will have undergone nuclear decay.It varies depending on the atom type and isotope,and is usually determined experimentally.
A stone is thrown horizontally. In
0.5s
after the stone began to move, the numerical value of its
velocity was 1.5 times its initial velocity. Find the initial velocity of the stone. Disregard the
resistance of the air.
Answer:
Approximately [tex]4.39 \; \rm m \cdot s^{-1}[/tex] (assuming that [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex].)
Explanation:
Let the initial velocity of this stone be [tex]v\; \rm m \cdot s^{-1}[/tex] ([tex]v \ge 0[/tex] because speed should be non-negative.) That should be numerically equal to the initial horizontal speed of this stone. Because the stone was thrown horizontally, its vertical speed would initially be zero ([tex]0\; \rm m \cdot s^{-1}[/tex].)
The weight of this stone will accelerate the stone downwards. Because the weight of the stone is in the vertical direction, it will have no effect on the horizontal speed of the stone.
Therefore, if air resistance is indeed negligible, the horizontal speed of this stone will stay the same. The horizontal speed of the stone after [tex]0.5\; \rm s[/tex] should still be [tex]v\; \rm m \cdot s^{-1}[/tex]- same as its initial value.
On the other hand, if [tex]g = 9.81\; \rm m \cdot s^{-2}[/tex], the (downward) vertical speed of this stone will increase by [tex]9.81\; \rm m \cdot s^{-2}[/tex] every second. After [tex]0.5\; \rm s[/tex], the vertical speed of this stone would have become:
[tex]0.5\; \rm s \times 9.81\; \rm m \cdot s^{-2} = 4.905\; \rm m \cdot s^{-1}[/tex].
[tex]t = 0\; \rm s[/tex]:
Horizontal speed: [tex]v\; \rm m \cdot s^{-1}[/tex].Vertical speed: [tex]0\; \rm m \cdot s^{-1}[/tex].[tex]t = 0.5\; \rm s[/tex]:
Horizontal speed: [tex]v\; \rm m \cdot s^{-1}[/tex].Vertical speed: [tex]4.905\; \rm m \cdot s^{-1}[/tex].Refer to the diagram attached. Consider the vertical speed and the horizontal speed of this rock as lengths of the two legs of a right triangle. The numerical value of the velocity of this stone would correspond to the length of the hypotenuse of that right triangle. Apply the Pythagorean Theorem to find that numerical value:
[tex]\begin{aligned} &\text{numerical value of velocity} \\ &= \sqrt{{\left(\text{horizontal speed}\right)}^2 + {\left(\text{vertical speed}\right)}^2}\end{aligned}[/tex].
At [tex]t = 0\; \rm s[/tex], the magnitude of the velocity of this stone would be [tex]\sqrt{v^{2} + 0^2} = v\; \rm m \cdot s^{-1}[/tex].
At [tex]t = 0.5\; \rm s[/tex], the magnitude of the velocity of this stone would be [tex]\left(\sqrt{v^2 + 4.905^2}\right)\; \rm m \cdot s^{-1}[/tex].
The question requires that the magnitude of the velocity of the stone at [tex]t = 0.5\; \rm s[/tex] should be [tex]1.5[/tex] times the value at [tex]t = 0\; \rm s[/tex]. In other words:
[tex]\sqrt{v^2 + 4.905^2} = 1.5 \, v[/tex].
Square both sides of this equation and solve for [tex]v[/tex]:
[tex]v^2 + 4.905^2 = 2.25\, v^2[/tex].
[tex]v \approx 4.39[/tex] ([tex]v \ge 0[/tex] given that speed should be non-negative.)
Therefore, the initial velocity of this stone should be approximately [tex]4.39\; \rm m \cdot s^{-1}[/tex].
There are_
periods included in the periodic table.
Answer here
SUBMIT
s
Answer:
7
Explanation
There are currently seven complete periods in the periodic table, comprising the 118 known elements. Any new elements will be placed into an eighth period; see extended periodic table.
Which major regions had the Romans controlled?
Why is people to come together and combine their efforts?
Answer:
people who had a hard time to help other people because they feel that if they help
that the work of the person is much harder than his own work
Individuals who find it difficult to help others because they believe that if they do, the person's job will be much tougher than their own.
A group of people working together to achieve a shared purpose.
Collaboration, cooperation, and coordination are all words that come to mind when thinking of teaming.
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-Two pickup trucks each have a mass of 2,000 kg. The gravitational force between the
trucks is 3.00 x 10-5 N. One pickup truck is then loaded with 1,000 kg of bricks. Which
of the following would be the new gravitational force between the two trucks?
Answer:
[tex]F'=4.5\times 10^{-5}\ N[/tex]
Explanation:
The gravitational force between two pickups is given by :
[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]
It means,
[tex]F\propto m_1m_2[/tex]
So,
[tex]\dfrac{F}{F'}=\dfrac{m_1m_2}{m_1'm_2'}[/tex]
We have,
[tex]m_1=m_2=2000\ kg\\\\F=3\times 10^{-5}\ N\\\\m_1'=2000+1000 =3000\ kg\\\\m_2'\ \text{remains the same i.e. 1000 kg}[/tex]
F' is the new force
So, putting all the values,
[tex]\dfrac{3\times 10^{-5}}{F'}=\dfrac{2000\times 2000}{3000\times 2000}\\\\\dfrac{3\times 10^{-5}}{F'}=\dfrac{2}{3}\\\\F'=\dfrac{9\times 10^{-5}}{2}\\\\F'=4.5\times 10^{-5}\ N[/tex]
So, the new force between the two trucks is [tex]4.5\times 10^{-5}\ N[/tex].
Water flows with constant speed through a garden hose that goes up to 27.5 cm high. if the water pressure is 132kpa at the bottom of the step, what is its pressure at the top of the step?
Answer:
The pressure at the top of the step is 129.303 kilopascals.
Explanation:
From Hydrostatics we find that the pressure difference between extremes of the water column is defined by the following formula, which is a particular case of the Bernoulli's Principle ([tex]v_{bottom}\approx v_{top}[/tex]):
[tex]p_{bottom}-p_{top} = \rho\cdot g\cdot \Delta h[/tex] (1)
[tex]p_{bottom}[/tex], [tex]p_{top}[/tex] - Total pressures at the bottom and at the top, measured in pascals.
[tex]\rho[/tex] - Density of the water, measured in kilograms per cubic meter.
[tex]\Delta h[/tex] - Height difference of the step, measured in meters.
If we know that [tex]p_{bottom} = 132000\,Pa[/tex], [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]\Delta h = 0.275\,m[/tex], then the pressure at the top of the step is:
[tex]p_{top} = p_{bottom}-\rho\cdot g\cdot \Delta h[/tex]
[tex]p_{top} = 132000\,Pa-\left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.275\,m)[/tex]
[tex]p_{top} = 129303.075\,Pa[/tex]
[tex]p_{top} = 129.303\,kPa[/tex]
The pressure at the top of the step is 129.303 kilopascals.
The pressure of the water at the top of the step is 1.293 x 10⁵ Pa.
The given parameters:
Height of the hose, h = 27.5 cm = 0.275 mPressure of the water at bottom, Pb = 132 kPaThe pressure of the water at the top of the step is calculated as follows;
[tex]P_b - P_t = \rho gh\\\\P_t = P_b - \rho gh[/tex]
where;
[tex]\rho[/tex] is the density of the water
[tex]P_t = (132,000 \ Pa) \ - \ (1000 \times 9.8 \times 0.275)\\\\P_t = 1.293 \times 10^5 \ Pa[/tex]
Thus, the pressure of the water at the top of the step is 1.293 x 10⁵ Pa.
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How much work is done when 425 N of force is applied for 60.0m
Answer:
Workdone = 25500Nm
Explanation:
Given the following data
Force = 425N
Distance = 60m
To find the workdone
Workdone = force *distance
Substituting into the equation, we have
Workdone = 425*60
Workdone = 25500Nm
The upper arm muscle is _______________ to the skin.
Answer:
The pectoralis major, latissimus dorsi, deltoid, and rotator cuff muscles connect to the humerus and move the arm. The muscles that move the forearm are located along the humerus, which include the triceps brachii, biceps brachii, brachialis, and brachioradialis.
A jet plane lands with a velocity of 100 m/s and can accelerate at a maximum of -9.0 m/s2 as it comes to rest. The minimum time needed before it can come to rest is seconds.
Explanation:
That must be the right ans.
1. Find the magnitude of the gravitational force (in N) between a planet with mass 8.00 ✕ 1024 kg and its moon, with mass 2.75 ✕ 1022 kg, if the average distance between their centers is 2.40 ✕ 108 m.2. What is the moon's acceleration (in m/s2) toward the planet? 3. What is the planet's acceleration (in m/s2) toward the moon?
Answer:
1.358 10e8 have good day please mark brainliest
If a 6 volt battery is connected in series to resistances of 2 ohms, 8 ohms, and 14 ohms, what is the amount of the current that is flowing?
Answer:
I = 0.25 [amp]
Explanation:
To solve this problem we must use ohm's law which tells us that the voltage is equal to the product of the current by the resistance.
V = I*R
where:
V = voltage [Volt]
I = amperage or current [amp]
R = resistance [ohm]
Since all resistors are connected in series, the total resistance will be equal to the arithmetic sum of all resistors.
Rt = 2 + 8 + 14
Rt = 24 [ohm]
Now clearing I for amperage
I = V/Rt
I = 6/24
I = 0.25 [amp]
Answer: 0.25
Explanation:
lucia raced her car on a raceway
Answer:
Good question to ask in physics, sir maam
Answer:
That's good
Explanation:
A cart is pulled by a force of 250 N at an angle of 35° above the horizontal. The cart accelerates at 1.4 m/s2. The free-body diagram shows the forces acting on the cart.
A free body diagram with 3 force vectors. The first vector is pointing downward, labeled F Subscript g Baseline. The second vector is pointing up, labeled F Subscript N Baseline. The third vector is pointing up to the right at an angle of 35 degrees, labeled F Subscript p Baseline = 250 N. The up and down vectors are the same length.
The mass of the cart, to the nearest whole number, is
Answer:
m=146.277kg which is rounded to 146kg
Explanation:
Remember that F=ma
But F represents not 250N, but 250cos(35)N since the force is being pulled above the horizontal.
So 250cos(35)=204.7880111 approximately, and since a=1.4m/s^2, we have 204.7880111=m(1.4m/s^2). Then we divide both sides by the acceleration to get the mass. So m=146.2771508kg which the nearest number is 146kg
Mass is always in kg, unless stated otherwise.
Answer:
146kg
Explanation:
right on edge
A 7.5 Kg cannon ball leaves a canon with a speed of 185 meters per second. Find the average net force applied to the ball is the canon muzzle is 3.6 meters long
Answer:
F = 35651 [N]
Explanation:
To solve this problem we must use the following equation of kinematics
[tex]v_{f} ^{2} =v_{o} ^{2} +2*a*x[/tex]
where:
Vf = final velocity = 185 [m/s]
Vo = initial velocity = 0 (starts from the rest)
a = acceleration [m/s²]
x = distance = 3.6 [m]
Now we can find the acceleration.
185² = 0 + 2*a*3.6
34225 = 7.2*a
a = 4753.47 [m/s²]
Now we must use Newton's second law which tells us that the total force on a body is equal to the product of mass by acceleration.
F = m*a
where:
m = mass = 7.5 [kg]
F = force [N] (units of Newtons)
F = 7.5*4753.47
F = 35651 [N]
Another name for a pivot point is the:
a.
output
b.
torque
c.
fulcrum
Answer:
I think it's c .... fulcrum
In certain ranges of a piano keyboard, more than one string is tuned to the same note to provide extra loudness. For example, the note at 110 Hz has two strings at this frequency. If one string slips from its normal tension of 600 N to 540 N, what beat frequency is heard when the hammer strikes the two strings simultaneously?
Answer:
f = 5.65 Hz
Explanation:
The fundamental frequency of a string is given by following formula:
f = v/2L
where,
f = fundamental frequency
v = speed of wave = √(TL/m)
L = Length of String
m = Mass of String
T = Tension in String
Therefore,
f = √(TL/m)/2L
2f = √(T/Lm)
For initial condition:
T₁ = 600 N
f₁ = 110 Hz
2(110 Hz) = √(600 N/Lm)
√(600 N)/220 Hz = √Lm
Lm = 0.01239 N/s²
Now, for changed tension:
2f₂ = √(540 N/0.01239 N/s²)
f₂ = 208.7 Hz/2
f₂ = 104.35 Hz
So, the beat frequency will be:
f = f₁ - f₂
f = 110 Hz - 104.35 Hz
f = 5.65 Hz
A spring-loaded gun has a spring for which
k =180N /m
and is initially compressed by
14cm
. It fires
a
0.024kg
projectile vertically. Find the maximum height above the initial position?
Answer:
7.5 m
Explanation:
k = Spring constant = 180 N/m
x = Displacement of spring = 14 cm
m = Mass of projectile = 0.024 kg
a = g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
s = Displacement of projectile
v = Final velocity = 0
u = Initial velocity
The potential energy of the spring will be equal to the kinetic energy of the object
[tex]\dfrac{1}{2}kx^2=\dfrac{1}{2}mu^2\\\Rightarrow u=\sqrt{\dfrac{kx^2}{m}}\\\Rightarrow u=\sqrt{\dfrac{180\times 0.14^2}{0.024}}\\\Rightarrow u=12.12\ \text{m/s}[/tex]
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0-12.12^2}{2\times -9.81}\\\Rightarrow s=7.5\ \text{m}[/tex]
The maximum height reached above the initial position is 7.5 m
Calculate the drag force acting on a 105 kg (including the bicycle) cyclist moving at 10 m/s, 20 m/s, an at 30 m/s. Use a drag coefficient of 0.8, a frontal surface area of 1.0 m2 , and a density of 1.0 kg/m3.
Answer:
40N
160N
360N
Explanation:
The drag force can be calculated using the formula below
FD = 0.5CPAV²
The drag coefficient = c = 0.8
The frontal density = p = 1.0kg
The frontal surface area = A = 1.0
The velocity v = 10m/s, 20m/s, 30m/s
When V = 10m/s
Fd = 0.5(0.8)(1)(1)(10)²
Fd = 40N
When V = 20m/s
FD = 0.5(0.8)(1)(1)(20)²
FD = 160N
When V = 30m/s
FD = 0.5(0.8)(1)(1)(30)²
FD = 360N
Thank you
. State any three applications of conduction
A car is traveling at 15m/s on a horizontal road. the brakes are applied and the car skids to a stop in 4.0s . the coefficient of Kinetic friction between the tires and road is:_________.
A) .38
B) .69
C) .76
D) .92
E) 1.11
Answer:
the coefficient of Kinetic friction between the tires and road is 0.38
Option A) .38 is the correct answer
Explanation:
Given that;
final velocity v = 0
initial velocity u = 15m/s
time taken t = 4 s
acceleration a = ?
from the equation of motion
v = u + at
we substitute
0 = 15 + a × 4
acceleration a = -15/4 = - 3.75 m/s²
the negative sign tells us that its a deacceleration so the sign can be ignored.
Deacceleration due to friction a = μ × g
we substitute
3.75 = μ × 9.8
μ = 3.75 / 9.8 = 0.3826 ≈ 0.38
Therefore the coefficient of Kinetic friction between the tires and road is 0.38
Option A) .38 is the correct answer
A ball is attached to one end of a string such that the ball travels in a vertical circular path near Earth's surface. The force diagram
of the ball at its lowest point in the circular path is shown above. What is the net centripetal force exerted on the ball?
A) 10N
B) 15 N
C
25 N
D
35 N
Complete Question
The complete question is shown on the first uploaded image
Answer:
The value is [tex]F_{centripetal} = 15 \ N[/tex]
Explanation:
From the diagram we see that
The tension force on the ball is [tex]F_{Tension} = 25 \ N[/tex]
The gravitational force on the ball is [tex]F_{Gravity } = 10 \ N[/tex]
Generally the net centripetal force exerted on the ball is mathematically represented as
[tex]F_{centripetal} = F_{Tension} - F_{Gravity}[/tex]
=> [tex]F_{centripetal} = 25 - 10[/tex]
=> [tex]F_{centripetal} = 15 \ N[/tex]
The net centripetal force exerted on the ball is 15 Newton.
Hence, Option B) 15N is the correct answer.
Given the data in the question;
Tension force on the ball; [tex]F_{Tension} = 25N[/tex]Gravitational force on the ball; [tex]F_{Gravity} = 10N[/tex]Net centripetal force exerted on the ball; [tex]F_{centripetal} = \ ?[/tex]
From the diagram below, net force acting on the ball gives rise to centripetal force. Hence
[tex]F{net} = F_1 + f_2 = F_{centripetal}[/tex]
Now, from the diagram, force acting towards the center of the circular track is is positive while force directly pointing away from center is negative.
Hence
[tex]F{net} = F_1 + f_2 = F_{centripetal} = F_{Tension} - F_{Gravity}[/tex]
We substitute in our given values
[tex]F_{centripetal} = 25N - 10N\\\\ F_{centripetal} = 15N[/tex]
The net centripetal force exerted on the ball is 15 Newton.
Hence, Option B) 15N is the correct answer.
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If the population of penguins increased, then this would have a direct effect on the populations of?
Answer:
Globel warming
Explanation:
hope this helpex
Which statement best explains why running on a track with constant speed at 3m/s is not work but climbing a mountain at 1m/s is work?
1. At constant speed, change in the kinetic energy is zero but climbing a mountain produces change in the potential energy
2. At constant speed, change in the potential energy is zero, but climbing a mountain produces change in the kinetic energy.
3. At constant speed, change in the kinetic energy is finite, but climbing a mountain produces no change in the potential energy.
4. At constant speed, change in the potential energy is finite, but climbing a mountain produces no change in the kinetic energy.
Answer:
1. At constant speed, change in the kinetic energy is zero but climbing a mountain produces change in the potential energy
Explanation:
According to work-energy theorem, work done in moving an object from one point to another is equal to change in mechanical energy ( kinetic energy or potential energy ) of the object.
Kinetic energy is given by;
K.E = ¹/₂m(Δv)²
where;
Δv is change in speed
at constant speed, Δv = 0
Potential energy is given by;
P.E = mgΔh
where;
Δh is change in height,
there is change in height in climbing a mountain
Therefore, the best explanation in the given options is "1".
"At constant speed, change in the kinetic energy is zero but climbing a mountain produces change in the potential energy"
A merry-go-round rotates from rest with an angular acceleration of 1.45 rad/s2. How long does it take to rotate through (a) the first 3.70 rev and (b) the next 3.70 rev?
Answer:
(a) t = 5.66 s
(b) t = 8 s
Explanation:
(a)
Here we will use 2nd equation of motion for angular motion:
θ = ωi t + (1/2)∝t²
where,
θ = Angular Displacement = (3.7 rev)(2π rad/1 rev) = 23.25 rad
ωi = initial angular speed = 0 rad/s
t = time = ?
∝ = angular acceleration = 1.45 rad/s²
Therefore,
23.25 rad = (0 rad/s)(t) + (1/2)(1.45 rad/s²)t²
t² = (23.25 rad)(2)/(1.45 rad/s²)
t = √(32.06 s²)
t = 5.66 s
(b)For next 3.7 rev
θ = ωi t + (1/2)∝t²
where,
θ = Angular Displacement = (3.7 rev + 3.7 rev)(2π rad/1 rev) = 46.5 rad
ωi = initial angular speed = 0 rad/s
t = time = ?
∝ = angular acceleration = 1.45 rad/s²
Therefore,
46.5 rad = (0 rad/s)(t) + (1/2)(1.45 rad/s²)t²
t² = (46.5 rad)(2)/(1.45 rad/s²)
t = √(64.13 s²)
t = 8 s
The time taken in each case is 2.55 s
Let us recall that the equations of circular motion are almost like those of linear motion;
α = ω2 - ω1/t
α = angular acceleration
ω2 = final angular velocity
ω1 = initial angular velocity
t = time taken
a) 1.45 rad/s2 =3.70 rev/s - 0 rev/s/t
t = 3.70 rev/s/1.45 rad/s2
t = 2.55 s
b) For, ) the next 3.70 rev:
1.45 rad/s2 =3.70 rev/s - 0 rev/s/t
t = 3.70 rev/s/1.45 rad/s2
t = 2.55 s
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A freely falling object near the Earth's surface has a constant?
acceleration of 1.0 m/s
acceleration of 9.8 m/s
velocity of 1.0 m/s
velocity of 9.8 m/s
Answer:
answer is B (CROWN ME)
Explanation:
One should revise his/ her research questions very well in the week before the exam. They should contribute at least 15-20 hours for going through these questions in that particular week. They must put some extra efforts on weekends.
A freely falling object near the Earth's surface has a constant: B. acceleration of 9.8 [tex]m/s^2[/tex].
Gravity can be defined as a force that controls the movement of the planets such as Earth around the Sun, hold stars grouped in galaxies together, and galaxies grouped in clusters. This ultimately implies that, gravity is a universal force of attraction that acts between all objects that are having both mass and energy, and can occupy space.
Furthermore, the gravity near the Earth's surface makes it possible for all physical objects to possess weight and experience a free fall.
Generally, the acceleration due to gravity for an object experiencing a free fall near the Earth's surface is 9.8 [tex]m/s^2[/tex].
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Can someone let me know if I got these questions right please
