Answer:
0.967 m/s
1855 N
[tex]46.375\ \text{MPa}[/tex]
Explanation:
v = Final velocity
u = Initial velocity = 0
s = Displacement = 4.77 cm
g = a = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
From the kinematic equations
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.0477+0}\\\Rightarrow v=0.967\ \text{m/s}[/tex]
The velocity of the object at the moment of impact is 0.967 m/s
Now
[tex]\Delta v[/tex] = Change in velocity = 1 m/s
t = Time taken = 20 ms
m = Half mass of the person = [tex]\dfrac{74.2}{2}=37.1\ \text{kg}[/tex]
[tex]F=\dfrac{m}{t}\\\Rightarrow F=\dfrac{37.1\times 1}{20\times 10^{-3}}\\\Rightarrow F=1855\ \text{N}[/tex]
The force associated with a single step of the person is 1855 N
A = Area = [tex]0.4\ \text{cm}^2[/tex]
Stress is given by
[tex]\sigma=\dfrac{F}{A}\\\Rightarrow \sigma=\dfrac{1855}{0.4\times 10^{-4}}\\\Rightarrow \sigma=46375000\ \text{Pa}=46.375\ \text{MPa}[/tex]
The stress on the tendon is [tex]46.375\ \text{MPa}[/tex]
The speed of object during falling is 0.967 m/s.
(A) The magnitude of force associated with this single step for a person is 1855 N.
(B) The required value of stress at tendons is [tex]4.70 \times 10^{7} \;\rm Pa[/tex].
Given data:
The height of fall is, h = 4.77 cm = 0.0477 m.
The magnitude of downward velocity is, v' = 1.0 m/s.
The duration of impact is, [tex]t = 20 \;\rm ms =20 \times 10^{-3} \;\rm s[/tex].
The mass of person is, m = 74.2 kg.
The magnitude of force is, F' = 1880 N.
The cross-sectional area is, [tex]A =0.4 \;\rm cm^{2} = 0.4 \times 10^{-4} \;\rm m^{2][/tex].
The problem has several parts using different concepts. First obtain the final speed of object to fall by using the second kinematic equations of motion as,
[tex]v^{2}=u^{2}+2gh[/tex]
Solving as,
[tex]v^{2}=0^{2}+(2 \times 9.8 \times 0.0477)\\\\v = \sqrt{(2 \times 9.8 \times 0.0477)} \\v = 0.967 \;\rm m/s[/tex]
Thus, the speed of object during falling is 0.967 m/s.
(A)
Now coming to next part, the half of mass means, m' = m/2 = 74.2/2 = 37.1 kg.
Apply the expression of average force as,
[tex]F =\dfrac{m'v'}{t}[/tex]
Solving as,
[tex]F =\dfrac{37.1 \times 1}{20 \times 10^{-3}}\\\\F = 1855 \;\rm N[/tex]
Thus, the magnitude of force associated with this single step for a person is 1855 N.
(B)
The expression for the stress is given as,
[tex]\sigma = \dfrac{F'}{A}[/tex]
Solving as,
[tex]\sigma = \dfrac{1880}{0.4 \times 10^{-4}}\\\\\sigma =4.70 \times 10^{7} \;\rm Pa[/tex]
Thus, the required value of stress at tendons is [tex]4.70 \times 10^{7} \;\rm Pa[/tex].
Learn more about the Stress force here:
https://brainly.com/question/18274389
A long wire carries a current 5 A from west to east. A magnetic compass pointing North is placed underneath the wire at a distance of 2 mm. What is the deflection of the compass when it is placed under the wire?
Answer:
no deflection
Explanation:
current is flowing from west to east. As the magnetic field of a long wire carrying current is circular, its direction will be north below the wire and south above the wire (according to the right hand rule). So, when the compass is placed underneath the wire, it will still point towards the north direction.
If the weight of a person is 500 newton what is his mass on the earth ?
Answer:
The person is on the Moon having a weight of 500 N. The acceleration of gravity on the Moon is approximately 1.6 m/s2. What is your his, which includes his space suit?
f= Force (of gravity)=500N
g=acceleration of gravity=1.6m/s^2
m=mass=312kg
m=f/a= 500N/1.6 m/s^2 = 500 (kg-m/1.6m/s^2) = 500/1.6kg = 312kg
his mass is 312kg
9. What distance is a book from the floor if the book contains 196 Joules of potential energy and has a mass of 5 kg?
Answer:
the book is 4 meters from the floor
Explanation:
Use the formula for potential energy:
U = m g h
196 J = 5kg 9.8m/s^2 h
h = 196 / (5*9.8 ) m
h = 4 m
What was used as a basket at each end of the gym?
Group of answer choices
bucket
hula hoop
apple basket
peach basket
Answer: apple basket ✨
Explanation:
The statement "46 m, East" refers to the ______________ of an object.
velocity
distance
displacement
speed
acceleration
Answer:
Distance
Example:
Answer:
Displacement
Explanation:
displacement implies a vector quantity that expresses the distance (in a straight line) between the starting and finishing points. Hence, the statement "46 m, East" refers to a displacement of 46 m toward the East.
On a surface of a planet of radius R and mass M the acceleration due to gravity is 7m/s?. Consider another planet of radius 2R and mass 0.4M. What would the acceleration due to gravity be on this new planet? Show your calculations.
Answer:
0.7 m/[tex]s^{2}[/tex]
Explanation:
From Newton's law of universal gravitation,
F = [tex]\frac{GMm}{r^{2} }[/tex]
and from Newton's second law of motion,
F = mg
So that;
mg = [tex]\frac{GMm}{r^{2} }[/tex]
⇒ g = [tex]\frac{GM}{r^{2} }[/tex]
For the first planet,
7 = [tex]\frac{GM}{R^{2} }[/tex]
⇒ G = [tex]\frac{7R^{2} }{M}[/tex] .............. 1
For the second planet,
g = [tex]\frac{0.4GM}{(2R)^{2} }[/tex]
= [tex]\frac{0.4GM}{4R^{2} }[/tex]
⇒ G = [tex]\frac{4gR^{2} }{0.4M}[/tex] ............. 2
Equating 1 and 2, we have;
[tex]\frac{7R^{2} }{M}[/tex] = [tex]\frac{4gR^{2} }{0.4M}[/tex]
g = [tex]\frac{7R^{2} *0.4M}{4R^{2}M }[/tex]
= [tex]\frac{7*0.4}{4}[/tex]
= [tex]\frac{2.8}{4}[/tex]
g = 0.7
Therefore, the acceleration due to gravity on the new planet is 0.7 m/[tex]s^{2}[/tex].
How long does it take a vehicle to reach a velocity of 32 m/s if it accelerates from rest at a rate of 4.2 m/s^2? ANSWER ALL PLEASE!! Im dumb:(
What is the initial velocity of the vehicle?
What is the final velocity of the vehicle?
What is the acceleration of the vehicle?
Write the equation you will use to solve the problem.
How long does it take the vehicle to reach its final velocity?
0.13 seconds
18.1 seconds
7.62 seconds
134.4 seconds
Answer: Givens
a = 4.4 m/s^2 This is an acceleration and is positive.
vi = 10.2 m/s
t = 4.2 seconds
vf = ????? The cruising speed in this case is vf.
Formula
a = (vf - vi)/t Notice the 3 givens and what you seek determine the formula
Solve
4.4 m/s^2 = (vf - 10.2)/4.2 Multiply both sides by 4.2
4.4 * 4.2 = vf - 10.2
18.48 = vf - 10.2 Add 10.2 to both sides
18.48 + 10.2 = vf 8.28 is the second best answer.
28.68 = vf This is your cruising speed.
C <<<< Answer
7.62 (2dp)
Explanation:
U = 0m/s
V = 32m/s
A = 4.2m/s^2
T = ?
[tex]a = \frac{v - u}{t}[/tex]
[tex]t = \frac{v - u}{a} [/tex]
[tex]t = \frac{32 - 0}{4.2} [/tex]
[tex]t = \frac{32}{4.2} [/tex]
[tex]t = 7.62 \: (2dp)[/tex]
If you liked this answer, brainliest?
A bicycle mechanic is checking a road bike's chain. He applies force F = 48 N to a pedal at the angle shown in (Figure 1) while keeping the wheel from rotating. The pedal is 17 cm from the center of the crank; the gear has a diameter of 16 cm.
Answer:
T = 102.3 ≈ 100
Explanation:
r(pedal) = 17 cm = 0.017 m
r(gear) = 16 cm / 2 = 8 cm = 0.008 m
θ = 180°-110° = 70°
F = 48 N
T = ?
τ₁ = r(pedal)*F*cosθ = (0.017 m)(48 N)cos(70°) = 0.28 N*m (*negative because it is applied clockwise)
τ₂ = -τ₁ = -(-0.28 N*m) = 0.28 N*m
τ₂ = r(gear)*T*cosθ
T = τ₂÷(r(gear)*cosθ) = 0.28 N*m ÷ (0.008 m * cos(70°))
T = 102.3 ≈ 100 ( 2 sig figs)
A 806 kg automobile is sliding on an icy street. It collides with a parked car which has a mass of 682 kg. The two cars lock up and slide together with a speed of 23.1 km/h. What was the speed of the first car just before the collision?
Answer:
42.6km/h
Explanation:
Step one:
given data
mass of the first car M1= 806kg
the velocity of the first car V1=?
mass of the second car mass M2=682kg
velocity of the second car V2= 0km/h -----Note the car is parked
common velocity V=23.1km/h------Note: the two cars have common velocity
since the collision is inelastic:
Step two:
Required:
The velocity of the moving car
We know that the expression for the conservation of linear momentum for inelastic collision is given as
M1V1+M2V2=V(M1+M2)
substitute
806*V1+682*0=23.1(806+682)
806V1+0=23.1(1488)
806V1=34372.8
V1=34372.8/806
V1=42.6km/h
The speed of the first car just before the collision is 42.6km/h
A 67.6-kg boy is surfing and catches a wave which gives him an initial speed of 1.56 m/s. He then drops through a height of 1.65 m, and ends with a speed of 8.30 m/s. How much nonconservative work was done on the boy?
Answer:
The nonconservative work was done on the boy is 1154.87 J.
Explanation:
Given;
mass of the boy, m = 67.6 kg
initial speed of the boy, u = 1.56 m/s
height of fall, h = 1.65 m
final speed of the boy, v = 8.30 m/s
The initial energy of the boy is given by;
E₁ = K.E₁ + P.E₁
E₁ = ¹/₂mu² + mgh
E₁ = ¹/₂(67.6)(1.56)² + (67.6 x 9.8 x 1.65)
E₁ = 82.134 + 1091.475
E₁ = 1,173.61 J
The final energy of the boy is given by;
E₂ = K.E₂
E₂ = ¹/₂mv²
E₂ = ¹/₂(67.6)(8.3)²
E₂ = 2,328.482 J
The nonconservative work was done on the boy is given by;
W = E₂- E₁
W = 2,328.482 J - 1,173.61 J
W = 1154.87 J
Therefore, the nonconservative work was done on the boy is 1154.87 J.
Using a maximum allowable shear stress of 70 MPa, find the shaft diameter needed to transmit 40 kW when (a) The shaft speed is 2500 rev/min. (b) The shaft speed is 250 rev/min
Answer:
a
[tex]d = 0.0223 \ m[/tex]
b
[tex]d = 0.0481 \ m[/tex]
Explanation:
From the question we are told that
The maximum allowable shear stress is [tex]\sigma = 70 MPa = 70 *10^{6} \ Pa[/tex]
The power is [tex]P = 40 \ kW = 40 *10^{3} \ W[/tex]
considering question a
The shaft speed is given as [tex]v = 2500\ rev/min[/tex]
Generally the torque experienced by the shaft is mathematically represented as
[tex]\tau = \frac{ 9.55 * P}{v}[/tex]
=> [tex]\tau = \frac{ 9.55 * 40 *10^{3}}{ 2500}[/tex]
=> [tex]\tau = 152.8 \ N \cdot m[/tex]
Generally the maximum torque experienced by the shaft is mathematically represented as
[tex]\tau_m = \frac{2 \tau }{ \pi r^2 }[/tex]
Generally diameter = 2 * radius (r)
So
[tex]\tau_m = \frac{2 \tau }{ \pi 4 d^2 }[/tex]
Generally the maximum allowable shear stress is mathematically represented as
[tex]\sigma = \frac{2 \tau }{ \pi 4 d^2 } * \frac{32}{d}[/tex]
=> [tex]\sigma = \frac{16 \tau }{ \pi d^3}[/tex]
=> [tex]d = \sqrt[3]{\frac{16 * \tau }{ \pi \sigma } }[/tex]
=> [tex]d = \sqrt[3]{\frac{16 * 152.8 }{ \pi * 70 *10^{6} } }[/tex]
=> [tex]d = 0.0223 \ m[/tex]
considering question b
The shaft speed is given as [tex]v = 250\ rev/min[/tex]
Generally the torque experienced by the shaft is mathematically represented as
[tex]\tau = \frac{ 9.55 * 40 *10^{3}}{250 }[/tex]
=> [tex]\tau = 1528 \ N \cdot m[/tex]
Generally the shaft diameter is mathematically represented as
[tex]d = \sqrt[3]{\frac{16 * \tau }{ \pi \sigma } }[/tex]
=>[tex]d = \sqrt[3]{\frac{16 * 1528 }{ 3.142 * 70 *10^{6} } }[/tex]
=>[tex]d = 0.0481 \ m[/tex]
a pot of water at 20.0ºC is warmed to 100º when 40000cal is added. what is the mass of the water?
Answer:
500 g
Explanation:
m = Mass of water
c = Specific heat of water = [tex]1\ \text{cal/g}^{\circ}\text{C}[/tex]
[tex]\Delta T[/tex] = Temperature difference = [tex]100-20=80^{\circ}\text{C}[/tex]
Q = Heat added = 40000 cal
Heat is given by
[tex]Q=mc\Delta T\\\Rightarrow m=\dfrac{Q}{c\Delta T}\\\Rightarrow m=\dfrac{40000}{1\times 80}\\\Rightarrow m=500\ \text{g}[/tex]
The mass of water is 500 g.
branches of sicence
Answer: Natural science can be divided into two main branches
Explanation:
life science and physical science. Life science is alternatively known as biology, and physical science is subdivided into branches: physics, chemistry, astronomy and Earth science.
A parallel-plate capacitor with circular plates of radius 40 mm is being discharged by a current of 6.0 A. At what radius (a) inside and (b) outside the capacitor gap is the magnitude of the induced magnetic field equal to 75% of its maximum value?(c) What is that maximum value?
Answer:
A) r = 0.03 m
B) r = 0.0533 m
C) B_max = 0.00003 T
Explanation:
Formula for magnetic field inside the capacitor when it is parallel to the length element is;
B_in = (μ_o•I•r/(2πR²)
Formula for maximum magnetic field is;
B_max = (μ_o•I/(2πR)
Formula for magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)
A) Magnetic field inside the capacitor is gotten from our first equation above;
B_in = (μ_o•I•r/R²)
Since we want to find the radius at which the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value.
Thus;
B_in = 0.75B_max
(μ_o•I•r/(2πR²) = 0.75((μ_o•I/(2πR))
μ_o•I and 2πR will cancel out to give;
r/R = 0.75
r = 0.75R
We are given R = 40 mm = 0.04 m
r = 0.75 × 0.04
r = 0.03 m
B) magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)
Thus for the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value:
B_out = 0.75B_max
(μ_o•I/(2πr) = 0.75((μ_o•I/(2πR))
μ_o•I and 2π will cancel out to give;
1/r = 0.75/R
r = R/0.75
r = 0.04/0.75
r = 0.0533 m
C) B_max = μ_o•I/(2πR)
μ_o is a constant known as vacuum of permeability with a value of 4π × 10^(-7) T.m/A
Thus;
B_max = (4π × 10^(-7) × 6)/(2π × 0.04)
B_max = 0.00003 T
A 500-N box is at rest on the floor. Dennis Elbo makes several
attempts to move the box, pushing against the box with varying
amounts of horizontal force. Yet the box never does move. In this
situation, the amount of static friction force experienced by the box
Select all that apply.
-
0 is 500 N
O is equal to the force with which Dennis exerts on the
box
has an upper limit and Dennis O has not yet exceeded the upper limit
Ois always the coefficient of friction multiplied by the normal force value
Answer:
Select the second and the third options you listed.
Explanation:
Select the answer options:
"is equal to the force with which Dennis exerts on the box."
and
"has an upper limit and Dennis has not yet exceeded the upper limit."
In fact, this upper limit of the static friction force is the product of the coefficient of static friction ([tex]\mu[/tex]) times the weight of the box.
How do bones develop and grow?
Answer:
Explanation:
Bones grow in length at the epiphyseal plate by a process that is similar to endochondral ossification. The cartilage in the region of the epiphyseal plate next to the epiphysis continues to grow by mitosis. The chondrocytes, in the region next to the diaphysis, age and degenerate.
an object falls from a hovering helicopter and hits the ground at a speed of 30m per seconds. how long does it take the object to reach the ground and how far does it fall? sketch a velocity-time graph for the object ( ignore air resistance
Answer:
45.9m
Explanation:
Given parameters:
Final velocity = 30m/s
Initial velocity = 0m/s
Unknown:
Time it takes for the object of fall = ?
Height of fall = ?
Solution:
For the first problem, we use the equation below to solve for t;
V = U + gt
V is the final velocity
U is the initial velocity
g is the acceleration due to gravity
t is the time taken
30 = 0 + 9.8 x t
30 = 9.8t
t = [tex]\frac{30}{9.8}[/tex] = 3.1s
Now, height of fall;
V² = U² + 2gH
30² = 0² + 2 x 9.8 x H
900 = 19.6H
H = 45.9m
If a ball rolls down an incline with a starting velocity of 0m/s and a final velocity of 6m/s
and it takes a total of 1.4 seconds, calculate its acceleration.
Answer:
If a ball rolls down an incline with a starting velocity of 0m/s and a final velocity of 6m/s
and it takes a total of 1.4 seconds, calculate its acceleration.
Answer:
Acceleration is 4.28 m/s²
Explanation:
Acceleration is change of speed in time. To solve this, we will assume that the acceleration is constant, meaning that every second the velocity increases for the same constant value.
a = ∆v/t
∆v is the difference between two measured velocities:
a = (v2 - v1) / t
v1 = 0m/s
v2 = 6m/s
t = 1.4 s
Now, we only plug in the given values:
a = (6 - 0) / 1.4
a = 6 m/s / 1.4 s
a = 4.28 m/s²
A student wants to determine the coefficient of static friction μ between a block of wood and an adjustable inclined plane. Of the following, the minimum additional equipment the student needs to determine a value for μ is
Answer:
A protractor to measure the angle of the inclined plane with the horizontal
Explanation:
The student needs to lift the free end of the adjustable inclined plane until the object barely starts sliding, and measure the angle at which such happens. At that point, the force of friction equals the component of the weight in the direction of the incline. That is:
[tex]f=\mu\,* N = \mu * m g\, cos(\theta)[/tex]
and [tex]w_{//}= m\,g\,sin(\theta)[/tex]
Then
[tex]f = w_{//}\\\mu *\,m \,g\,cos(\theta) = m\,g\,sin(\theta)\\\mu = tan(\theta)[/tex]
and therefore, the coefficient of static friction is fully determined just by calculating the tangent of the angle that the incline forms with the horizontal.
Then the only extra instrument needed is a protractor to measure the angle.
In 1993, Wayne Brian threw a spear at a record distance of 201.24 m. (This is not an official sports record because a special device was used to “elongate” Brian’s hand.) Suppose Brian threw the spear at a 35.0° angle with respect to the horizontal. What was the initial speed of the spear? 2. Find the maximum height and time of flight of the spear in problem #1.
I really don't know how to do any of this please help me :(
Answer:
V₀ = 45.81 m/s
H = 70.45 m
T = 5.36 s
Explanation:
The motion of the spare is projectile motion. Therefore, we will first use the formula of range of projectile:
R = V₀² Sin 2θ/g
where,
R = Range of Projectile = 201.24 m
V₀ = Initial Speed = ?
θ = Launch Angle = 35°
g = 9.8 m/s²
Therefore,
201.24 m = V₀²[Sin 2(35°)]/9.8 m/s²
V₀ = √[(201.24 m)/(0.095 m/s²)
V₀ = 45.81 m/s
Now, for maximum height:
H = V₀² Sin² θ/g
H = (45.81 m/s)² Sin² 35°/9.8 m/s²
H = 70.45 m
For the total time of flight:
T = 2 V₀ Sin θ/g
T = 2(45.81 m/s) Sin 35°/9.8 m/s²
T = 5.36 s
If a bicyclist has a mass of 70 kg and a velocity of 25 m/s, what is the momentum of the bicyclist? p=mv *
Answer:
1750Nm/s
Explanation:
70*25=1750Nm/s
Answer:
1750 kgm/s
Explanation:
The equation for momentum is p = mv = 70 * 25 = 1750
carrier concentration for n type
Answer:
Consider an n-type silicon semiconductor at T = 300°K in which Nd = 1016 cm-3 and Na = 0. The intrinsic carrier concentration is assumed to be ni = 1.5 x 1010 cm-3. - Comment Nd >> ni, so that the thermal-equilibrium majority carrier electron concentration is essentially equal to the donor impurity concentration.
Explanation:
Please help, I'm really struggling here, I can't do science :(
The mass of Jupiter is about 320 times the mass of Earth. However, Jupiter’s gravity affects Earth very little because_____________. a Earth is so far from Jupiter. b Earth is so small. c Jupiter is made of gas. d Jupiter is nearer to the sun than Earth is.
Answer:no sure sorry
Explanation:
What is potential energy? Use in your own words.
Answer:
Potential energy is the latent energy in an object at rest.
Explanation:
Hope this helps- this is how I would answer :)
A plane flying at a speed of 59.1 m/s is dropping a package 521 m above the intended target. How long does it take for the package to hit the ground?
A.) 10.3 seconds
B.) 0.1 seconds
C.) 106.3 seconds
D.) 8.8 seconds
Answer:
t = 10.31 seconds which agrees with answer option A)
Explanation:
Notice that the horizontal velocity of the plane imparted to the package, doesn't affect the vertical motion for which the original (vertical velocity) is zero.
Then the equation for the distance travelled by the package is:
d = (1/2) a t^2
in our case, d = 521 m, and a = g (acceleration of gravity 9.8 m/s^2)
then we can solve for t in the equation:
521 = 9.8/2 t^2
t^2 = 521/4.9
t^2 = 106.32 s^2
t = 10.31 seconds
if the motor m rotates in the direction shown by the arrow what is going on
The question is incomplete,I will complete the question and provide the answer.
Due to the nature of the question,I will sketch an answer/solution to the question and submit it as an attachment.
So you will be having two attachments,
1) The question
2) The solution
From the options given in the first attachment with is the question,the correct answer is option C.
1 AND 2 ARE GOING UP.
Proper explanation using sketch and arrows is given in the second attachment which shows the solution to the question.
in the woman's mouth if
she is drinking through a straw extending 0.085 m above the surface of the drink? Note:
Assume the drink has a density of 1015 kg/m^3. ????
Answer:
hi mate,
interesting question, first of all the pressure is determined by using the following formula:
Pg = p * G * h
where p is the density of the liquid, G is the gravity and h is the height difference, in you case you have:
p = 1015 kg/m3
G = 9.8m/s2
h = 0.085 m
insert these values into the equation above:
Pg = 1015 kg/m3 * 9.8m/s2 * 0.085 m = 849.81 kg·m-1·s-2 or 849.81 pascal
hope it helps, :-)
please mark me as brainliest
a force of 35N is exerted over a cylinder with an area of 5m^2. What pressure,in pascals, will be transmitted in the hydraulic system?
Answer:
The answer is 7 PaExplanation:
The pressure transmitted in the hydraulic system can be found by using the formula
[tex]p = \frac{f}{a} \\ [/tex]
f is the force
a is the area
From the question we have
[tex]p = \frac{35}{5} \\ [/tex]
We have the final answer as
7 PaHope this helps you
Friction that occurs in gases and liquids is called
5. Peter had an inflated balloon that he released and flies across the room. The balloon slows down and then stops on top of the dinning table, As the balloon slows downs the force becomes
A.balanced
B.frictional
C.restricted
D.unbalanced