Answer: See explanation
Explanation:
The scalar quantity has magnitude and isn't dependent on the direction. A vector quantity is dependent on the direction as well as the magnitude.
Speed: This is a scalar quantity as it has to do with the magnitude of an objects change in position.
Velocity: This is a vector quantity. It is the speed at which a particular object moves in a certain direction.
Instantaneous velocity: This is a vector quantity.
Acceleration: This is a vector. This is the rate at which velocity changes.
A Shaolin monk of mass 60 kg is able to do a ‘finger stand’: he supports his whole weight on his two index fingers, giving him a total contact area of 4 cm 2 with the ground. Calculate the pressure he exerts on the ground (include units), and write your answer to two significant figures.
Answer:
1500000 Pa
Explanation:
The formula for pressure is force per unit area.
P=F/A where F is force and A is area
Given that ;
F= mass * acceleration due to gravity
F= 60 * 9.81 = 588.6 = 589 N
A= area = 4cm² = 0.0004 m²
P= F/A = 589 / 0.0004
P= 1471500
P=1500000 Pa
A Shaolin monk of mass 60 kg is able to do a ‘finger stand’: he supports his whole weight on his two index fingers, giving him a total contact area of 4 cm 2 with the ground. Calculate the pressure he exerts on the ground (include units), and write your answer to two significant figures.
Answer:
P = 1471500 [Pa]
Explanation:
We must remember that pressure is defined as the relationship between Force over the area.
[tex]P=F/A[/tex]
where:
P = pressure [Pa] (units of pascals)
F = force [N] (units of Newtons)
A = area of contact = 4 [cm²]
But first we must convert from cm² to m²
[tex]A = 4[cm^{2}]*\frac{1^{2} m^{2} }{100^{2} cm^{2} }[/tex]
A = 0.0004 [m²]
Also, the weight should be calculated as follows:
[tex]w = m*g[/tex]
where:
m = mass = 60 [kg]
g = gravity acceleration = 9.81 [m/s²]
Now replacing:
[tex]w = 60*9.81\\w = 588.6[N][/tex]
And the pressure:
[tex]P=588.6/0.0004\\P=1471500 [Pa][/tex]
Because 1 [Pa] = 1 [N/m²]
A car was moving at 14 m/s After 30 s, its speed increased to 20 m/s. What was the acceleration during this time ( need help fast!!!)
Answer:let initial velocity u=14m/s
Final velocity v=20m/s
Time taken t=30
Acceleration =a
V=u +at
a= (20-14)/30
a=0.2m/s^2
Explanation:
Acceleration is the change in velocity with respect to time.
Two charges q1 and q2 are separated by a distance r and apply a force F to each other. If both charges are doubled, and the distance between them is halved, the new force between them is
Answer:
The new force between them is increased by a factor of 16.
Explanation:
According to Coulombs law, the force of attraction between two (2) charges is given by the formula;
F = Kq1q2/r²
Given the following data;
q1 = 2q1
q2 = 2q2
r = r/2
Substituting into the equation, we have;
F = 2q1*2q2/(r/2)²
F = 4q1q2/r²/4
F = 4q1q2 * 4/r²
F = 16q1q2/r²
Therefore, the new force between them is increased by a factor of 16.
A toy car, mass of 0.025 kg, is traveling on a horizontal track with a velocity of 5 m/s. If
the track then starts to climb upwards, how high up the track can the car reach?
Answer:
1.25 m
Explanation:
This is the vertical height not the distance along the slope.
[tex]K=U\\\frac{1}{2}mv^{2} = mgh\\h = \frac{v^{2}}{2g}=\frac{25}{20}=1.25 m[/tex]
The height the car can reach if the the track starts to climb upwards is 1.2742 meters up.
What is kinetic and potential energy?Kinetic energy is energy possessed by a body by virtue of its movement. Potential energy is the energy possessed by a body by virtue of its position or its relation with its surrounding systems.
P.E. = mass × g × height
K.E. = 0.5 × mass × (velocity)²
Given that the toy car has a mass of 0.025 kg and is traveling on a horizontal track with a velocity of 5 m/s. Now, the car starts to climb up vertically, therefore, the kinetic energy will be converted to potential energy.
Kinetic Energy = Potential Energy
0.5 × mass × (velocity)² = mass × g × height
Cancel mass from both the sides,
0.5 × (velocity)² = g × height
0.5 × (5 m/s)² = 9.81 m/sec² × height
height = 1.2742 meters
Hence, the car will travel 1.2742 meters up.
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A 5kg box is sliding down a ramp with a rough surface as seen below. The height of the ramp is 20m and the distance the box travels down the ramp (from A to B) is 15m. At point A the velocity of the box is 8 m/s. If the velocity at point B is 3m/s, what was the impulse caused by friction? If the force of friction is 5N, how long did it take the box to slide the 15 m?
Answer:
A Impulse = – 25 Ns
B. Time = 5 s
Explanation:
From the question given above, the following data were obtained:
Mass (M) = 5 Kg
Initial velocity (u) = 8 m/s
Final velocity (v) = 3 m/s
Impulse (I) =?
Time (t) =?
A. Determination of the Impulse.
Mass (M) = 5 Kg
Initial velocity (u) = 8 m/s
Final velocity (v) = 3 m/s
Impulse (I) =?
I = Ft = M(v – u)
I = M(v – u)
I = 5 (3 – 8)
I = 5 × – 5
I = – 25 Ns
NOTE: the negative sign indicates that the net force is acting in the negative direction.
B. Determination of the time.
Impulse (I) = 25 Ns
Force (F) = 5 N
Time (t) =?
I = Ft
25 = 5 × t
Divide both side by 5
t = 25 / 5
t = 5 s
Thus, it will take 5 s for the box to slide through the 15 m long ramp.
A steam catapult launches a jet aircraft from the aircraft carrier John C. Stennis, giving it a speed of 195 mi/h in 2.90 s. (a) Find the average acceleration of the plane.
Answer:
The average acceleration of the plane is 0.0186 [tex]\frac{mi}{s^{2} }[/tex]
Explanation:
The acceleration of an object is a magnitude that indicates how the speed of the object changes in a unit of time. That is, acceleration is a magnitude that relates changes in speed with the time it takes to occur.
The average acceleration of an object is calculated using the following expression:
[tex]acceleration=\frac{change of speed}{time}[/tex]
In this case:
change of speed=[tex]195 \frac{mi}{h} *\frac{1 h}{60 minutes} *\frac{1 minute}{60 s} =0.054 \frac{mi}{s}[/tex]time= 2.90 sReplacing:
[tex]acceleration=\frac{0.054 \frac{mi}{s} }{2.90 s}[/tex]
and solving you get:
acceleration=0.0186 [tex]\frac{mi}{s^{2} }[/tex]
The average acceleration of the plane is 0.0186 [tex]\frac{mi}{s^{2} }[/tex]
A 5kg box is sliding down a ramp with a rough surface as seen below. The height of the ramp is 20m and the distance the box travels down the ramp (from A to B) is 15m. At point A the velocity of the box is 8 m/s. If the velocity at point B is 3m/s, what was the impulse caused by friction? If the force of friction is 5N, how long did it take the box to slide the 15 m?
Answer:
the answer is b luv .
Explanation:
A car accelerates uniformly from rest and reaches a speed of 9.9 m/s in 11.4 s. The diameter of a tire is 86.9 cm. Find the number of revolutions the tire makes during this motion, assuming no slipping. Answer in units of rev.
Answer:
Number of revolutions = 20.71 rev.
Explanation:
Given the following data;
Initial speed, u = 0m/s
Final speed, v = 9.9m/s
Time, t = 11.4secs
Diameter = 86.9cm to meters = 86.9/100 = 0.869m
To find the acceleration;
Acceleration, a = (v - u)/t
Acceleration, a = (9.9 - 0)/11.4
Acceleration, a = 9.9/11.4
Acceleration, a = 0.87m/s²
Now we would find the distance covered by the tire using the second equation of motion.
S = ut + ½at²
S = 0(11.4) + ½*0.87*11.4²
S = 0 + 0.435*129.96
S = 56.53m
The circumference of the tire is calculated using the formula;
Circumference = 3.142 * diameter
Circumference = 3.142 * 0.869
Circumference = 2.73m
Number of revolutions = distance/circumference
Number of revolutions = 56.53/2.73
Number of revolutions = 20.71 rev.
Therefore, the number of revolutions the tire makes during this motion is 20.71 revolutions.
Two force vectors are oriented such that the angle between their directions is 46 degrees and they have the same magnitude. If their magnitudes are 2.81 newtons, then what is the magnitude of their sum
Answer:
F = 5.17 N
Explanation:
If we know the magnitudes of both vectors, and the angle between them, we can find the magnitude of their sum, applying the cosine theorem, as follows:[tex]F =\sqrt{F_{1} ^{2} +F_{2} ^{2} + 2*F_{1} * F_{2} * cos \theta} (1)[/tex]
Replacing by the givens, F₁ = F₂ = 2.81 N, θ = 46º, we get:[tex]F =\sqrt{2.81 N ^{2} +2.81 N ^{2} + 2*2.81N* 2.81N* cos 46} = 5.17 N (2)[/tex]
A car with mass 1141 kg, moving at 16 m/s, strikes a(n) 2916 kg car at rest. If the two cars stick together, with what speed do they move
Answer:
V = 4.49 m/s
Explanation:
Given that,
Mass of the car 1, m₁ = 1141 kg
Initial speed of car 1, u₁ = 16 m/s
Mass of car 2, m₂ = 2916 kg
Initial speed of car 2, u₂ = 0
We need to find the speed of the cars if they stick together. Let the speed be V. The momentum will remain conserved in the process. Using the conservation of momentum to find it.
m₁u₁ + m₂u₂ = (m₁+m₂)V
[tex]V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\=\dfrac{1141\times 16+2916 \times 0}{(1141 +2916 )}\\\\=4.49\ m/s[/tex]
So, the required speed of the two cars is 4.49 m/s.
Two charged particles attract each other with a force of magnitude F. If the distance between the charges is made 3.5 times as large and the charge on one of the particles is made 4 times as big, what is the ratio of the new F to the old one
Answer:
The ratio between the forces is:
[tex]\frac{F_{new}}{F_{old}}=0.33[/tex]
Explanation:
The electrostatic force equation is:
[tex]F_{old}=k\frac{q_{1}q_{2}}{d^{2}}[/tex]
Where:
q1 and q2 are the electric charges
d is the distance between them
k is the electrostatic constant
Now, the distance is 3.5 times as large and q1 is 4 times as big, then the new force will be:
[tex]F_{new}=k\frac{4q_{1}q_{2}}{(3.5d)^{2}}[/tex]
[tex]F_{new}=\frac{4}{3.5^{2}}k\frac{q_{1}q_{2}}{d^{2}}[/tex]
We can rewrite this equation in terms of F(old)
[tex]F_{new}=\frac{4}{3.5^{2}}F_{old}[/tex]
Therefore, the ratio between the forces is:
[tex]\frac{F_{new}}{F_{old}}=\frac{4}{3.5^{2}}[/tex]
[tex]\frac{F_{new}}{F_{old}}=0.33[/tex]
I hope it helps you!
which hand has a negatively charged?
Answer:
The dryer sheet is negatively charged and your hand is positively charged
Explanation:
A child is twirling a 0.0113-kg ball on a string in a horizontal circle whose radius is 0.147 m. The ball travels once around the circle in 0.388 s. (a) Determine the centripetal force acting on the ball. (b) If the speed is doubled, by what factor does the centripetal force increase
Answer:
0.435 N
The centripetal force increases by a factor of 4
Explanation:
Let the centripetal force be F
F =mv^2/r
m= mass of the object
v = linear velocity of the object
r = radius of the circular path
but v = 2πr/T = 2 * 3.142 * 0.147 / 0.388
v = 2.38 m/s
F = 0.0113 * (2.38)^2/0.147
F= 0.435 N
if v= 2v
Then;
F = m(2v)^2/r
F = m4v^2/r
F= 4mv^2/r
The centripetal force increases by a factor of 4
The Earth (geosphere) comprises several major layers, differing in chemical and mineral composition, material strength, and other physical properties. Which of the layers listed below constitutes the largest proportion of Earth's volume?
a) Lithosphere
b) Crust
c) Outer core ·
d) Inner core
e) Mantle
Answer:
Option E:
The mantle
Explanation:
The earth's mantle is the mushy, semi-solid portion of the earth that makes up most of the earth's volume. The mantle extends for a depth of about 2800km downwards into the earth, making it the largest internal portion of the earth. It makes up about 84 percent of the earth's structure, leaving the core and the crust with 15 percent and 1 percent respectively.
Due its nature, convection currents are set up predominantly in the mantle of the earth, which leads to movements in the upper layers of the earth (the crust).
The mantle is large enough for the lighter crust to float on its surface.
A jumbo egg (80 grams) is dropped from a height of 15 meters onto a 1 inch of foam. Using kinematics, determine the velocity of the egg the instant before impact.
Answer:
the velocity of the egg the instant before impact is 17.15 m/s.
Explanation:
Given;
mass of the egg, m = 80 g = 0.08 kg
height through which the egg was dropped, h = 15 m
The velocity of the egg before impact will be maximum, and the final velocity is given by the following kinematic equation;
v² = u² + 2gh
where;
u is the initial velocity of the egg = 0
v is the final velocity of the egg before impact
v² = 0 + 2 x 9.8 x 15
v² = 294
v = √294
v = 17.15 m/s
Therefore, the velocity of the egg the instant before impact is 17.15 m/s.
3.
A person drives north 6 blocks, then drives west 6 blocks.
The displacement is a straight line from the starting point to the finish in a
direction
O A northwesterly
O B southwesterly
O c northeasterly
OD southeasterly
A Discuss the possibility of fracture of two leg bones that have a length of about 70cm and an average area of
about 4cm
2 when a 80kg person jump from a height of 300cm.
Noting: The breaking stress of the bone ϬB =1.5×108 N/m2 , and
Young’s modulus for the bone is Y=1.5×1010 N/m2
Answer: The bones won't fracture.
Explanation: Stress, in Physics, is a quantity describing forces that can cause deformation. Strain is the measure of how muc an object can be stretched or deformed. The ratio between stress and strain is called Young's modulus or elastic modulus
Breaking Stress of Bone is the maximum stress a bone can take before a rupture occur.
To determine if a person will break his/her bones by jumping from a height, we determine the energy necessary for that jump and compare it with the energy necessary to break a bone.
The energy for breaking a bone is calculated as
[tex]E=\frac{Al_{0}\sigma_{B}^{2}}{2Y}[/tex]
A is the area in m²
l₀ is length in m
[tex]\sigma_{B}[/tex] is breaking stress in N/m²
Y is Young's modulus in N/m²
Calculating energy to break a bone:
[tex]E=\frac{4.10^{-4}.7.10^{-1}.(1.5.10^{8})^{2}}{2.(1.5.10^{10})}[/tex]
[tex]E=210[/tex] J
This is the energy necessary to break one leg bone, so as there are 2, energy will be 420 Joules.
Potential energy gained by jumping is calculated as
E = m.g.h
m is mass in kg
g is acceleration due to gravity in m/s²
h is height in m
Calculating
E = 80.(9.8)(0.3)
E = 235.2 J
Comparing the two energies, potential energy for jumping is less than maximum energy a bone can absorve without breaking, so the leg bones won't suffer a fracture.
While driving his sports car at 20.00 m/s down a four lane highway, a
man comes up behind a very slow school bus. The man is in a hurry so he
decides to pass the bus. If the man's car can accelerate at 6.00 m/s^2,
how long will it take for him to reach a speed of 35.00 m/s in order to pass
the bus? Round to the nearest hundredth. (2 decimal places) Type the
number only.
Answer:
2.50 s
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 20 m/s
Acceleration (a) = 6 m/s²
Final velocity (v) = 35 m/s
Time (t) =?
Thus, we can obtain the time taken for the man to reach a speed of 35 m/s as follow:
v = u + at
35 = 20 + 6t
Collect like terms
35 – 20 = 6t
15 = 6t
Divide both side by 6
t = 15/6
t = 2.50 s
Thus, it will take the man 2.50 s to reach a speed of 35 m/s.
100 POINTS. PLEASE EXPLAIN
Answer:
Explanation:
Note the charge balls on the top and bottom row are identical. So those charges cancel each other out. The only charges in the net electric field are the two in the middle row.
Electric field strength = k*Q/r^2
= (8.99 *10^9) * (3-(-3)) * 5*10^(-6) / (2*0.5)^2
= 269700 N/C
Answer:
Explanation:
(b) cuz the 1st n 3nd row cancel out, net electric field will go from +3Q to -3Q. the direction is right.
You want to build a snowman, so you accelerate a 2kg snowball across your yard at a rate of 0.5m/s2. Calculate the amount of force you applied to your friend.
Answer:
4
Units:
Newtons
Which statements correctly describe the formula or name of a compound? Select all that apply.
OA. The formula of nitrogen trifluoride is NF 3
B. The formula of ammonia is NH3.
C. The name of AlF, is trialuminum fluoride.
D. The formula of calcium chloride is CaCl2
E. The name of Li, Se is lithium selenate.
OF. The formula of dinitrogen monoxide is NO
2
Results
G. The formula of sulfur trioxide is 30.
Cho
Аа
H. The formula of magnesium hydroxide is Mg(OH)2
G.
Answer:
A, B, D, and H
Explanation:
Statements A, B, D and H are all correct except the following:
Statement C is incorrect. The name of [tex] AlF_3 [/tex] is aluminium fluoride NOT "trialuminium fluoride".
Statement E is incorrect. The name of [tex] Li_2Se [/tex] is lithium selenide NOT "lithium selenate".
Statement F is incorrect. Dinitrigen monoxide, also known as nitrous oxide has a formula of [tex] N_2O [/tex] NOT [tex] NO_2 [/tex].
Statement G is incorrect. Sulfur trioxide formula is [tex] SO_3 [/tex].
g A box of mass 10 kg attached to a spring is pulled to a maximum of 44 cm. The box is released. What is the speed when the box reaches a point 10 cm from the equilibrium position
Answer:
The speed when the box reaches a point 10 cm from the equilibrium position is 2.02 m/s.
Explanation:
Mass attached to the spring, m = 10 kg
maximum displacement of the spring, A = 0.44 m
The spring constant is calculated from Hook's law;
F = kx
mg = kx
k = (mg) / x
k = (10 x 9.8) / 0.44
k = 222.73 N/m
The angular speed of the spring is calculated as;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega =\sqrt{\frac{222.73}{10} } \\\\\omega = 4.72 \ rad/s[/tex]
The speed when the box reaches a point 10 cm from the equilibrium position is calculated as;
[tex]v = \omega \sqrt{A^2-x^2} \\\\v = 4.72\sqrt{0.44^2-0.1^2}\\\\v = 2.02 \ m/s[/tex]
Therefore, the speed when the box reaches a point 10 cm from the equilibrium position is 2.02 m/s.
PLEASE ANSWER-Why are loose electrons needed for heat conduction?
Answer:
Why do metals conduct heat so well? The electrons in metal are delocalised electrons and are free moving electrons so when they gain energy (heat) they vibrate more quickly and can move around, this means that they can pass on the energy more quickly.
1. Amy uses 30N of force to push a lawn mower 10 meters. How much work does she do?
2. A girl of weight 400 N has feet of area 100 cm2. What
pressure does she put on the ground ?
Answer:
Explanation:
Given:
Weight = 400N
Area = 100cm^2
If 1cm^2 = 1×10^-4m^2
Hence 100 cm^2 = 0.01m^2
Pressure = force/area .........1
Force = ma = mg which is also the weight of the girl. Hence substitute the values into 1
P = 400/0.01
P = 40000 N/m^2
The pressure she puts on the ground is 40000 N/m^2
For which medical procedure would Doppler ultrasound be most useful?
A.
Finding a lung tumor
B.
Fixing a pulled muscle
C.
Locating a broken bone in a finger
D.
Detecting a blockage in a heart artery
Doppler ultrasound would be most useful in detecting a blockage in a heart artery.
What are the clinical uses of Doppler ultrasound?By monitoring the rate of change in pitch, a Doppler ultrasound may calculate how quickly blood flows (frequency). A sonographer with training in ultrasound imaging applies pressure to your skin with a tiny, hand-held instrument (transducer) roughly the size of a bar of soap across the area of your body being scanned, moving from one place to another as required.
As an alternative to more invasive treatments like angiography, which involves injecting dye into the blood arteries to make them visible on X-ray images, this test may be performed.
Your doctor may use a Doppler ultrasound to assess for artery damage or to keep track of specific vein and artery therapies.
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The velocity of a Froghopper flea, which is represented by a vector, has a direction and a magnitude. If a coordinate system is oriented where the x-axis is horizontal, and the y-axis is vertical, is it possible to write expressions for the x and y components of the velocity vector in terms of the vector's magnitude and direction.
Answer:
Explanation:
The solution of the question cab e found in attachment below:
Help can’t find the answer no where
i'm stuck on that question also
An athlete runs on track at a constant speed of 75 meters/min for 15 s. What is the total distance he covered during that time?
Answer:
45
Explanation: