Answer:
5m/s ez
Explanation:
plz make me brainliest
A skydiver that has a weight of 1050 N. jumps from a plane at an altitude of 5000 m. How much work is performed on the skydiver?
Answer:
5.25MJ
Explanation:
Considering the weight of the skydiver constant over the descent -ie, the acceleration of gravity not decreasing as you go up the earth surfice, the work is simply the product of the two:
[tex]w= \vec F\cdot \vec x = Fx = 1050\times 5000 = 5,250,000 = 5.25MJ[/tex]
During a collision, a 63.9-kilogram driver of a car moving at 24 m/s is brought to rest by an inflating airbag in 1.2 seconds. What is the magnitude of the force exerted on the driver by the airbag?
Hi there!
We can use impulse for this situation:
I = Δp = mΔv
Impulse = Force × time, so:
I = 63.9(24) = 1533.6 Ns
Find force by dividing by time:
I/t = 1533.6/1.2 = 1278 N
An "energy bar" contains 26 ggof carbohydrates.How much energy is this in joules?
An example of a conversation killer is?
making eye contact.
encouraging another person
talking about others
sharing your ideas and feelings
Answer: talking about others
An eraser is tied to a string swung in a horizontal circle. Identify the type of force which causes this object to travel along a circular path.
Answer:
A centripetal force
Explanation:
The type of force in the given scenario is tension. The correct option is c.
What is tension force?Tension is defined in physics as the pulling force conveyed axially by a string, cable, loop, or similar material, or by every end of a rod, truss member, or similar three-dimensional object.
Tension can also be defined as the action-reaction pair of forces acting at each end of said elements.
Newton's second law states that the tension in the rope must equal the weight of the backed mass.
Tension, the normal force, and friction are all examples of contact forces.
Since the weight is not moving, the acceleration is zero. Even if the acceleration is not zero, this equals zero.
Thus, the correct option is c.
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Your question seems incomplete, the missing options are:
a) gravityb) appliedc) tensiond) normalA crane lifts a 10000 N load through a distance of 30mins in 40s calculate work done
Given,
Force = 10,000 N
Distance = 30 m
Time = 40 s
Work done = force × distance
Work = 10,000 × 30
Work = 300,000 J
______
꧁✿ ᴿᴬᴵᴺᴮᴼᵂˢᴬᴸᵀ2222 ✬꧂
The work done on the object is equal to the power it experiences which is equal to 735 watts.
What is work done?Work done is defined as the product of force applied and the distance over which the force is applied on the object. Work is done on an object when a force is applied to an object and the object is moved through a particular distance.
The time rate of doing work is called as power, which is designated by the symbol P. It is measured in the units of watts, where one watt is equal to one joule energy applied per second time.
Work done on an object is equal to the force (F) multiplied by the distance d in the direction of the force. Where, time interval is Δt.
P = W/Δt = F × d/Δt = m × g × h/Δt = 100 × 9.8 × 30 / 40
Power = 735 watts
The power of the object is 735 watts.
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a block weighing (Fg) 500 N is resting on a steel table ( us=0.74) the minimum force start this block moving is?
The minimum force required to start this block moving is 370 N.
The given parameters;
weight of the block, W = 500 Ncoefficient of static friction, [tex]\mu_s[/tex] = 0.74The minimum force required to start this block moving is calculated as follows;
[tex]F= \mu_s F_n[/tex]
where;
[tex]F_n[/tex] is the normal force on the block which is equal to the weight of the block
[tex]F= \mu_s F_n \\\\F= \mu_s W\\\\F = 0.74 \times 500 \\\\F= 370 \ N[/tex]
Thus, the minimum force required to start this block moving is 370 N.
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Using your outline and the materials you’ve gathered, write a 500- to 750-word paper using word processing software. Be sure to proofread and revise your writing to catch any errors in grammar, spelling, logic, or organization. Add a works cited page at the end to give credit to your sources. Submit your completed paper along with this activity to your teacher for evaluation.
please do it based off of Werner Heisenberg the scientist.
Answer:
im half new how does this work
Explanation:
Answer:Werner Heisenberg was born on 5th December, 1901, at Wurzburg. He was the son of Dr. August Heisenberg and his wife Annie Wecklein. His father later became Professor of the Middle and Modern Greek languages in the University of Munich. It was probably due to his influence that Heisenberg remarked, when the Japanese physicist Yukawa discovered the particle now known as the meson and the term “mesotron” was proposed for it, that the Greek word “mesos” has no “tr” in it, with the result that the name “mesotron” was changed to “meson”.Heisenberg went to the Maximilian school at Munich until 1920, when he went to the University of Munich to study physics under Sommerfeld, Wien, Pringsheim, and Rosenthal. During the winter of 1922-1923 he went to Göttingen to study physics under Max Born, Franck, and Hilbert. In 1923 he took his Ph.D. at the University of Munich and then became Assistant to Max Born at the University of Göttingen, and in 1924 he gained the venia legendi at that University.From 1924 until 1925 he worked, with a Rockefeller Grant, with Nlels Bohr, at the University of Copenhagen, returning for the summer of 1925 to Göttingen.In 1926 he was appointed Lecturer in Theoretical Physics at the University of Copenhagen under Nlels Bohr and in 1927, when he was only 26, he was appointed Professor of Theoretical Physics at the University of Leipzig.In 1929 he went on a lecture tour to the United States, Japan, and India.In 1941 he was appointed Professor of Physics at the University of Berlin and Director of the Kaiser Wilhelm Institute for Physics there.At the end of the Second World War he, and other German physicists, were taken prisoner by American troops and sent to England, but in 1946 he returned to Germany and reorganized, with his colleagues, the Institute for Physics at Göttingen. This Institute was, in 1948, renamed the Max Planck Institute for Physics.In 1948 Heisenberg stayed for some months in Cambridge, England, to give lectures, and in 1950 and 1954 he was invited to lecture in the United States. In the winter of 1955-1956 he gave the Gifford Lectures at the University of St. Andrews, Scotland, these lectures being subsequently published as a book.During 1955 Heisenberg was occupied with preparations for the removal of the Max Planck Institute for Physics to Munich. Still Director of this Institute, he went with it to Munich and in 1958 he was appointed Professor of Physics in the University of Munich. His Institute was then renamed the Max Planck Institute for Physics and Astrophysics.
Help meeeeeeeeeeeeee
A catamaran with a mass of 5.44×10^3 kg is moving at 12 knots. How much work is required to increase the speed to 16 knots? (One knot = 0.51 m/s.)
show all steps
The work that is required to increase the speed to 16 knots is 14,176.47 Joules
If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;
5.44×10^3 kg = 12 knots
For an increased speed to 16knots, we will have:
x = 16knots
Divide both expressions
[tex]\frac{5.44 \times 10^3}{x} = \frac{12}{16}\\12x = 16 \times 5.44 \times 10^3\\x = 7.23\times 10^3kg\\[/tex]
To get the required work done, we will divide the mass by the speed of one knot to have:
[tex]w=\frac{7230}{0.51}\\w= 14,176.47Joules[/tex]
Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules
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What is the acceleration as an object goes from +20 m/s to +26 m/s over 1.4 s?
Answer:
4.3 m/s^2
Explanation:
a = vf - vi / t
26 - 20 / 1.4
= 4.28 ( round your numbers)
Help me out with this physics please!!!
Answer: It was on between 7 and 7.30 and also between 8 and 8.30 (when the line on the graph goes up only). So it was on for an hour or 60 minutes
Help help fast please first answer gets brainiest also needs to be clearly stated
b. Ron bicycles forward with an acceleration of 2.1 m/s2. If he is applying a forward force of 195 N, what is his mass?
Answer:
92.86 kgExplanation:
lThe mass of an object given only it's force and acceleration can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force
a is the acceleration
From the question
f = 195 N
a = 2.1 m/s
We have
[tex]m = \frac{195}{2.1 } = 92.857142... \\ [/tex]
We have the final answer as
92.86 kgHope this helps you
What would be the acceleration of a block sliding down an inclined plane that makes an angle of 45° with the horizontal if the coefficient of sliding friction between two surfaces in 0.3? 141
Ans: 4.95 m/sec?
Answer:
unfortunately we have not received a reply to the park yet but
Explanation:
mohamedz and he had killed him
how do humans affect the lithosphere
Two towers of height 40 m and 30 m respectively support a transmission line conductor at water crossing. The horizontal distance between the towers is 300 m. If the tension in the conductor is 1590 kg, find the clearance of the conductor at a point mid-way between the supports. Weight of conductor is 0·8 kg/m. Bases of the towers can be considered to be at the water level.
The sag in the tensioned cable depends on the weight of the cable and
the applied tension.
The clearance midway between the support is approximately 29.34 metersReasons:
The given parameters are;
Height of the towers are; h₁ = 40 m and h₂ = 30 m
Horizontal distance between the towers, l = 300 m
The tension in the conductor, T = 1,590 kg
Weight of the conductor, w = 0.8 kg/m
Required:
The clearance space of the conductor mid-way between the support
Solution:
The distance, x₁, to the lowest point on the conductor, from the tower, h₁, is given by the formula;
[tex]\displaystyle x_1 = \mathbf{ \frac{l}{2} -\frac{T \cdot h}{w \cdot l}}[/tex]
Where;
h = h₂ - h₁ = 40 m - 30 m = 10 m
Which gives;
[tex]\displaystyle x_1 = \frac{300}{2} -\frac{1590 \times 10}{0.8 \times 300} = \mathbf{83.75}[/tex]
The sag at the lowest point is give by the formula;
[tex]\displaystyle d_1 = \mathbf{ \frac{W \cdot x_1^2}{2 \cdot T}}[/tex]
Therefore;
[tex]\displaystyle d_1 = \frac{0.8 \times 83.75^2}{2 \times 1590} = \frac{4489}{2544} \approx 1.765[/tex]
The lowest point sag, s₁ = 30 m - 1.765 m = 28.235 m
The distance of the midpoint from the lowest point, x, is therefore;
[tex]\displaystyle x = \frac{l}{2} - x_1 = \frac{300}{2} - 83.75 = \mathbf{ 66.25}[/tex]
Which gives;
[tex]\displaystyle d_x = \frac{0.8 \times 66.25^2}{2 \times 1590} = \frac{53}{48} \approx 1.104[/tex]
The clearance midway between the support ≈ 28.235 + 1.104 ≈ 29.34
The clearance midway between the support ≈ 29.34 m
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a 1200-kg pick-up truck traveling south at 15 m/s suddenly collides with a 750-kg car that is traveling east. the two vehicles stick together and slide along the road after colliding. a highway patrol officer investigating the accident determines that the final position of the wreckage after the collision is 25 m, at an angle of 50° south of east, from the point of impact. she also determines that the coefficient of kinetic friction between the tires and the road at that location was 0.40. what was the speed of the car just before the collision?
The conservation of momentum, Newton's second law and kinematics allows to find the result for the initial speed of the car is:
The speed of the car is 40 m/s at the East direction.
Given parameters
Mass of the truck M = 1200 kg. Truck speed v₀₁ = 15 m / s towards the South. Mass of the car m = 750 kg. Travel east The two vehicles unite. The braking distance d = 25m at 50º SE The friction coefficient μ = 0.40To find
The initial speed of the car.
The momentum is defined by the product of the mass and the velocity of the body, so it is a vector quantity. In the case of an isolated system the momentum is conserved.
In the attachment we see a diagram of the vehicle crash, let's write the moment for each axis.
y-axis
Initial instant. Before the shock.
[tex]p_{oy}[/tex] = M v₀₁
Final moment. After the crash
[tex]p_{fy}[/tex] = (M + m) [tex]v_{fy}[/tex]
The system is formed by the two vehicles for which during the crash it is isolated and the momentum is conserved.
[tex]p_{oy} = p_{fy} \\\\M v_{o1} = (M+m) v_{fy} \\v_{fy} = \frac{M}{M+m} \ v_{o1}[/tex]
Calculate us
[tex]v_{fy} = \frac{1200}{1200+750} \ 15[/tex]
vfx = 9.23 m / s
x-axis
Initial instant. Before the crash.
p₀ₓ = m v₀₂
Final moment. After the crash.
[tex]p_{fx}[/tex] = (M + m) [tex]v_{fx}[/tex]
The momentum is preserved.
[tex]p_{ox}= p_{fx} \\m v_{o2) = (M+m) \ v_{fx}[/tex]
Let's find the velocity just after the collision, let's use Newton's second law,
y-axis
N-W = 0
N = W = (M + m) g
x-axis
fr = (M + m) a
The friction force is the macroscopic manifestation of the interactions between the two bodies.
fr = μ N
We substitute.
μ g = a
Now we can use kinematics.
v² = v₀² - 2a d
When the vehicles stop their speed is zero.
v₀² = 2 a x
v₀ = [tex]\sqrt{2\ \mu \ g \ d}[/tex]
Let's calculate.
v₀ = [tex]\sqrt{2 \ 0.40 \ 9.8 25 }[/tex]
v₀ = 14 m / s
This is the speed of the two vehicles just after the collision, that is, let's use trigonometry to find their components.
cos 25 = [tex]\frac{v_{ox}}{v_o}[/tex]
v₀ₓ = v₀ cos 25
v₀ₓ = 14 cos 25
v₀ₓ = 12.69 m / s
We substitute in the expression of the conservation of the momentum in the x-axis.
m v₀₂ = (M + m) [tex]v_{fx}[/tex]
[tex]v_{o2} = \frac{M+m}{m} v_{fx}[/tex]
Let's calculate.
[tex]v_{o2} = \frac{1200+750}{750} \ 12.69[/tex]
vfy = 32.99 m / s = 40 m / s
In conclusion using the conservation of momentum, Newton's second law and kinematics we can find the result for the initial speed of the car is:
The speed of the car is 40 m/s at the East direction.
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(50 POINTS!!) 3. Make a model that shows the forces acting on two blocks on a flat, frictionless surface:
a. A 1 N block at rest
b. A 1 N block with 2 N of force applied in one horizontal direction
Include arrows to represent the forces and labels to indicate the magnitude of each force. Use your model to compare and explain the motion of each block.
The change in position of an object is known as motion. Force is external that causes a body to change its position
For a body is to be at rest, the sum of all the forces acting on the object must be zeroFor a 1N block to be at rest, the normal force acting opposite to the weight must also be 1N.For 1 N block with 2 N of force applied in one horizontal direction, the image of the body is as shown in the attachmentLearn more about force here: https://brainly.com/question/20955770
Answer:
In short terms. the 1N block at rest will stay still and the 1N block with 2N applied will move forward.
an object initially moving with 12m/s accelerates uniformly by3m/s^2 for 10 seconds how far does it travel in first 7 seconds
Answer:
157.5
Explanation:
X=vi *t + 0.5*a*t^2
vi=12, t=7, a=3
x= 12*7 + 0.5*3*7^2
= 84+ 73.5= 157.5 m
An object is at rest. There are several forces acting on the object, but the net force is zero. If all the
forces acting on the object were to suddenly double in size, would the object start moving? Why or why not?
Answer:
no
Explanation:
because the heaver an object the harder it is to move s the force will not be able to move it
the definition of convetion
