The amount of time that the ball took in air is 1 second.
We must recall that the ball was thrown as a projectile and its time of flight must be obtained from the formula for the time of flight of a projectile as follows;
T = 2usinθ/g
u = initial velocity = 19.6 m/s
θ = angle of projection = 30°
g = acceleration due to gravity = 9.8 m/s2
Substituting values;
T = 2 × 19.6 m/s sin (30°)/9.8 m/s2
T = 1 second
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Can yall help me please
Answer:
physical
Explanation:
it is a real life model that was built
Three forces act on a flange as shown below. Determine the magnitude of the unknown force F (in lb) such that the net force acting on the flange is a minimum.
Answer:
The unknown force will be 18.116 lb.
Explanation:
Given that,
Three forces act on a flange as shown in figure.
The net force acting on the flange is a minimum.
[tex]\dfrac{dF_{net}}{df}=0[/tex]
We need to calculate the unknown force
Using formula of net force
[tex]\vec{F_{net}}=\vec{F_{x}}+\vec{F_{y}}[/tex]
Put the value into the formula
[tex]\vec{F_{net}}=(F\cos45+70\cos30-40)\hat{i}+(70\sin30-F\sin45)\hat{j}[/tex]
[tex]\vec{F_{net}}=(F\cos45+70\times\dfrac{\sqrt{3}}{2})\hat{i}+(70\times\dfrac{1}{2}-F\sin45)\hat{j}[/tex]
The magnitude of net force,
[tex]F_{net}=\sqrt{F_{x}^2+F_{y}^2}[/tex]
[tex]F_{net}=\sqrt{(F\times\dfrac{1}{\sqrt{2}}+60.62)^2+(35-F\times\dfrac{1}{\sqrt{2}})^2}[/tex]
[tex]F_{net}=\sqrt{F^2+(60.62)^2+121.24\times\dfrac{F}{\sqrt{2}}+(35)^2-70\times\dfrac{F}{\sqrt{2}}}[/tex]
[tex]F_{net}=\sqrt{F^2+4899.78+36.232F}[/tex]
On differentiating w.r.to F
[tex](\dfrac{dF_{net}}{dF})^2=2F+36.232[/tex]
[tex]0=2F+36.232[/tex]
[tex]F=-\dfrac{36.232}{2}[/tex]
[tex]F=-18.116\ lb[/tex]
Negative sign shows the direction of force which is downward.
Hence, The unknown force will be 18.116 lb.
Following are the solution to the given question:
Calculating the Net force on flange:
[tex]\overrightarrow{F_{net}} = ( F \cos 45^{\circ} + 70 \cos 30^{\circ} -40 ) \hat{i}+(70 \sin 30^{\circ} - F \sin 45^{\circ} )\hat{y} \\\\ \overrightarrow{F_{net}} = ( F \cos 45^{\circ} + 20.62 ) \hat{i}+(35 -F\sin 45^{\circ})\hat{y} \\\\ [/tex]
Calculating the magnitude:
[tex]= \sqrt{f_{X}^2 +f_{y}^{2}}\\\\ = \sqt{( F \cos 45^{\circ} + 20.62 )^2+(35 -F\sin 45^{\circ})^2} \\\\ = \sqt{( F^2+20.62^2+ 35^2+ 41.24 F \cos 45^{\circ} -70 F\sin 45^{\circ})}\\\\ = \sqt{( F^2+20.33F +1650.1844)} \\\\[/tex]
Differentiate the value
[tex]\to \frac{F_{net}}{df}=0 [/tex]
[tex]\to 2F-20.33=0\\\\ \to 2F=20.33\\\\ \to F=\frac{20.33}{2}\\\\ \to F= 10.165\ lb [/tex]
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What is the net force?
60N left
4ON, right
60N, right
0N
Answer:
60n left
Explanation:
Do
The x-axis of a position-time graph represents
Answer:
The y-axis represents position relative to the starting point, and the x-axis represents time.
Explanation:
What are the characteristics of supernovae?
release of matter and energy
provide new material for future solar systems
move on to form a black hole or neutron star
might be seen from earth
Answer:
c
Explanation:
is c20h12 organic or not organic
Answer:
.
Explanation:
Calculate the difference in blood pressure between the feet and top of the head for a person who is 1.70 m tall.
Answer:
[tex]P_2 - P_1 = 1.8 * 10^4\ Pa[/tex]
Explanation:
Given
[tex]Height (h) = 1.70m[/tex]
Required
Determine the difference in the blood pressure from feet to top
This is calculated using Pascal's second law.
The second law is represented as:
[tex]P_2 = P_1 + pgd[/tex]
Subtract P1 from both sides
[tex]P_2 - P_1 = pgd[/tex]
Where
[tex]p = blood\ density = 1.06 * 10^3kg/m^3[/tex]
[tex]g = acceleration\ of\ gravity = 9.8N/kg[/tex]
[tex]d =height = 1.70m[/tex]
P2 - P1 = Blood Pressure Difference
So, the expression becomes:
[tex]P_2 - P_1 = 1.06 * 10^3 * 9.8 * 1.70[/tex]
[tex]P_2 - P_1 = 17659.6Pa[/tex]
[tex]P_2 - P_1 = 1.8 * 10^4\ Pa[/tex]
Hence, the difference in blood pressure is approximately [tex]1.8 * 10^4\ Pa[/tex]
Which of the following best describes the charge of the nucleus of an atom?
A. The nucleus can have a positive, neutral, or negative charge.
B. The nucleus always has a positive charge.
c. Jahe nucleus always has a neutral charge.
D. The nucleus always has a negative charge.
SUBMIT
what is the force of gravitational attraction between an object with a mass of 0.5kg and another object that had a mass of 0.33kg and a distance between them of 0.002m
Answer:
1.1x10^-4 N
Explanation:
F = G[(m_1*m_2)/r^2]
G = Gravitational constant 6.67433x10^-11 (N*m^2)/(kg^2)
m_1 = mass 1
m_2 = mass 2
r = radius between objects
F = G[(0.5kg*0.33kg)/(0.001m)^2]
F = 1.1x10^-4 N
1. A sailor pulls a boat along a dock using a rope at an angle of 60.0º with the horizontal. How much work does the sailor do if he exerts a force of 255 N on the rope and pulls the boat 3.00 m?
W = FdcosO =255 x 3 x cos 60 =
2. An elephant pushes with 2000 N at an angle of 33o above the horizontal on a load of trees. It then pushes these trees for 150 m. How much work did the elephant do?
3. Hans Full is pulling on a rope to drag his backpack to school across the ice. He pulls upwards and rightwards with a force of 22.9 Newtons at an angle of 35o above the horizontal to drag his backpack a horizontal distance of 129 m. Determine the work done upon the backpack.
4. If 100 N force has 30o angle pulling on a 15 kg block for 5 m. What’s the work?
Answer:
(1)Work done by snail is 382.5 N
(2)Work done by ELEPHANT is 25160 N
(3)Work done by HANS is 2420 N
(4)Work done on block is 433 N
Explanation:
The work done due to force F applied at an angle θ from the horizontal to the body is give by
Work done = Fscosθ where, s is distance traveled by body
Case 1: F= 255N , s= 3.00m and θ = 60.[tex]0^0[/tex]
Work done = Fscosθ = 255 x 3.00 x cos60.[tex]0^0[/tex] = 382.5N
thus work done by snail is 382.5 N
Case 2: F= 200N , s= 150m and θ = 33.[tex]0^0[/tex]
Work done = Fscosθ = 200 x 150 x cos33.[tex]0^0[/tex] = 25160N
thus work done by elephant is 25160 N
Case 3: F= 22.9N , s= 129m and θ = 35.[tex]0^0[/tex]
Work done = Fscosθ = 22.9 x 129 x cos35.[tex]0^0[/tex] = 2420N
thus work done by Hans is 2420 N
Case 4: F= 100N , s= 5m and θ = 30.[tex]0^0[/tex]
Work done = Fscosθ = 100 x 5.00 x cos30.[tex]0^0[/tex] = 433N
thus work done on block is 433 N
(1) The work done by the sailor is 382.5 J.
(2) The work done by elephant is 25160 J.
(3) The work done by Hans upon the backpack is 2420 J.
(4) The required work done on block is 433 J.
Let us solve these questions in parts. All these questions are based on the work done. The work done due to force F applied at an angle θ from the horizontal to the body is give by,
[tex]W =F \times s \times cos \theta[/tex]
Here, s is the distance covered by the body.
(1)
Given data:
F= 255N , s= 3.00m and θ = 60.
The work done is calculated as,
W = Fscosθ
W= 255 x 3.00 x cos60 = 382.5 J.
Thus, the work done by the sailor is 382.5 J.
(2)
Given data:
F= 200N , s= 150m and θ = 33.
The work done by the elephant is calculated as,
W = Fscosθ
W= 200 x 150 x cos33. = 25160 J
Thus, the work done by elephant is 25160 J.
(3)
Given data:
F= 22.9N , s= 129m and θ = 35.
Then the work done upon the backpack is calculated as,
W= Fscosθ
W= 22.9 x 129 x cos35. = 2420 J
Thus, the work done by Hans upon the backpack is 2420 J.
(4)
Given data:
F= 100N , s= 5m and θ = 30.
The work done is calculated as,
W = Fscosθ
W= 100 x 5.00 x cos30. = 433 J
Thus, the required work done on block is 433 J.
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The density of a solid or liquid material divided by the density of water is called
Answer:
I believe the answer is specific gravity
Explanation:Hope this helps :)
What height would a 4 kg book need to be to have a potential energy of
235.2 J on earth?*
need help!!
Answer:
5.99 m = 6 m
Explanation:
PE = m*g*h
235.2 J = (4 kg)(9.81 m/s^2)(h)
h = (235.2 J)/(9.81*4)
h = 5.99 m
h = 6 m
What does a light-year measure? A. volume B. weight C. circumference D. distance
Answer:
Distance. A light-year is the distance a single photon of light can travel in empty space in a year.
Answer:
D-Distance
Explanation:
pls help, i have no idea where to start
Answer:
Use equation 1/2 * m * v²
Explanation:
where m = mass in kg
and v = velocity in m/s
plug accordingly.
1. What is the momentum of a 1550 kg car that is traveling leftward at a velocity of 15 m/s?
Answer:
Momentum, p = 23250 kg m/s
Explanation:
Given that
Mass of a car, m = 1550 kg
Speed pf car, v = 15 m/s
We need to find the momentum of the car. The formula for the momentum of an object is given by :
p = mv
Substituting all the values in the above formula
p = 1550 kg × 15 m/s
p = 23250 kg m/s
So, the momentum of the car is 23250 kg m/s.
Homeostasis refers to the ability of the body to maintain a stable internal environment despite changes in external conditions.
True
False
Answer:
True
Explanation:
Juans mother drives 7.25 miles southwest to her favorite shopping mall. What is the average velocity of her automobile if she arrives at the mark in 20min?
Problem 1. A tugboat exerts a constant force of 4000N toward the right on a ship, moving it
a distance of 15m. What work is done?
Answer:
75,000 JExplanation:
The work done by an object can be found by using the formula
workdone = force × distance
From the question we have
workdone = 5000 × 15
We have the final answer as
75,000 JHope this helps you
A north magnetic pole brought near a south magnetic pole will...
A)
cancel the south pole.
B)
attract the south pole.
C)
repel the south pole.
EASY!!
A 5000kg truck and a 50 kg person have the same momentum. How is this possible?
Explanation:
The mass of a truck = 5000 kg
Mass of a person = 50 kg
The momentum of a body is given by :
p = mv
m is mass
If momentum of the truck and the person is same. Let v₁ and v₂ are velocity of the truck and the person.
[tex]5000\times v_1=50\times v_2\\\\\dfrac{v_1}{v_2}=\dfrac{50}{5000}\\\\\dfrac{v_1}{v_2}=\dfrac{1}{100}\\\\v_2=100v_1[/tex]
So, it can only possible if the velocity of the person is 100 times of the velocity of the truck.
Example
A 1.8 m tallman stand in an elevator accelerating upward at 12 m/s?,
what is the blood pressure in the brain and foot.
Take the height difference between the heart and the brain to be
0.35 m?
13.3 x 103 Pa] & [Pblood = 1060 kg.m-3]
Note: [: P Heart
Answer:
Explanation:
From the given information; the diagram below shows a clearer understanding.
The blood pressure in the brain [tex]P_{brain} = P_{heart} - \delta ( a - g ) h[/tex]
= 13300 - 1060 (12-9.81) 0.35
= 13300 - 1060 (2.19) 0.35
= 13300 - 812.49
= 12487.51 Pa
The blood pressure in the feet [tex]P_{feet} = P_{heart} + \delta (a + g) h[/tex]
= 13300 + 1060 (12 + 9.81) 1.45
= 13300 + 1060( 21.81 ) 1.45
= 13300 + 33521.97
= 46821.97 Pa
Answer:
The blood pressure in the brain = [tex]12487.51 pa[/tex]The blood pressure in the feet = [tex]46821.97pa[/tex]Explanation:
[tex]P_brain = P_heart - y(a - g)h\\\\P_brain = 13300 - 1060 (12-9.81)0.35\\\\P_brain = 12487.51 pa\\\\[/tex]
[tex]P_feet = P_heart + y(a+g)*h_r\\\\P_feet = 13300 + 1060(12+9.81)*(1.8-0.35)\\\\P_feet = 46821.97pa[/tex]
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Explain why crinoid fossils can be found on the shore of Lake Michigan, which is a fresh water lake.
Answer:
Explanation:
Crinoid fossils can be found on the shore of Lake Michigan because during the Carboniferous period, all of what is now the United States, except for a small part of the upper Midwest, and all of the states along the East Coast, except for Florida, was covered by a warm, shallow inland sea(seawater), and since crinoid fossils live in sea water, they washed up on many places like Lake Michigan.
Answer:
Explanation:
Crinoid fossils can be found on the shore of Lake Michigan because during the Carboniferous period, all of what is now the United States, except for a small part of the upper Midwest, and all of the states along the East Coast, except for Florida, was covered by a warm, shallow inland sea(seawater), and since crinoid fossils live in sea water, they washed up on many places like Lake Michigan.
A stationary 12.5 kg object is located on a table near the surface of the earth. The coefficient of static friction between the surfaces is 0.50 and of kinetic friction is 0.30. Show all work including units.
A horizontal force of 15 N is applied to the object.
a. Draw a free body diagram with the forces to scale.
b. Determine the force of friction.
c. Determine the acceleration of the object.
When the object is at rest, there is a zero net force due the cancellation of the object's weight w with the normal force n of the table pushing up on the object, so that by Newton's second law,
∑ F = n - w = 0 → n = w = mg = 112.5 N ≈ 113 N
where m = 12.5 kg and g = 9.80 m/s².
The minimum force F needed to overcome maximum static friction f and get the object moving is
F > f = 0.50 n = 61.25 N ≈ 61.3 N
which means a push of F = 15 N is not enough the get object moving and so it stays at rest in equilibrium. While the push is being done, the net force on the object is still zero, but now the horizontal push and static friction cancel each other.
So:
(a) Your free body diagram should show the object with 4 forces acting on it as described above. You have to draw it to scale, so whatever length you use for the normal force and weight vectors, the length of the push and static friction vectors should be about 61.3/112.5 ≈ 0.545 ≈ 54.5% as long.
(b) Friction has a magnitude of 15 N because it balances the pushing force.
(c) The object is in equilibrium and not moving, so the acceleration is zero.
1. When you talk into your paper cup telephone, the person on the other end can feel the bottom
of their telephone vibrate. Why do you think this happens?
Answer:
The correct answer is - the sound waves make vibration that travels through the string.
Explanation:
When an individual person talks into your paper cup telephone the person on the other end can feel the bottom of their cup vibrate. The sound waves create vibration go through the string that travels through the string to the end of the cup where vibrations can feel.
The sound waves are longitudinal waves that move or travel through different mediums like air, solid, or gas. The waves create vibration in the particles.
You could create a paper cup telephone but instead of using string, test out different materials and see if those materials will allow sound vibrations to travel through them
Numerical Problems
A bus is moving with the initial velocity 10 m/s. After 2 seconds, the velocity
becomes 20 m/s. Find the acceleration and distance moved by the bus.
Answer:
a = 5 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f}=v_{o}+a*t[/tex]
where:
Vf = final velocity = 20 [m/s]
Vo = initial velocity = 10 [m/s]
t = time = 2 [s]
a = acceleration [m/s²]
Now replacing:
[tex]20 =10 +a*2\\10=2*a\\a=5[m/s^{2} ][/tex]
I NEED HELP WITH 8 I WILL GIVE U BRAINLIEST
Answer:
u will need 1/2 cells
Explanation:
A steel cable of diameter 3.0 cm supports a load of 2.0kN. What is the fractional length increase of the cable compared with the length when there is no load if Y=2.0 x 10^11 Pa?
Answer:
1.429*10^-5 m
Explanation:
From the question, we are given that
Diameter of the cable, d = 3 cm = 0.03 m
Force on the cable, F = 2 kN
Young Modulus, Y = 2*10^11 Pa
Area of the cable = πd²/4 = (3.142 * 0.03²) / 4 = 0.0028 / 4 = 0.0007 m²
The fractional length = Δl/l
Δl/l = F/AY
Δl/l = 2000 / 0.0007 * 2*10^11
Δl/l = 2000 / 1.4*10^8
Δl/l = 1.429*10^-5 m
Therefore, the fractional length is 1.429*10^-5 m long
When the torques on the wheel and axle equal zero, the machine has
_____
Answer:
8
Explanation:
Answer:
rotational equilibrium
Explanation:
Which waves are electromagnetic and can travel through a vacuum?
Light and heat waves
Longitudinal and transverse waves
Sound waves
Surface waves
The correct answer is 40 or - 40??
Answer:
40 cm
hope it's help you but you can choose if your ans. is -40 cm