Assuming the wave is being transmitted on the vacuum, we know that the velocity of an electromagnetic wave on vacuum is c. We can use the following relation:
[tex]v=\frac{\lambda}{T}\Rightarrow T=\frac{\lambda}{c}=\frac{469.06*10^3}{3*10^8}=1.5635ms=1563.5\mu s[/tex]Thus, our final answer is T=1563.5us
can somebody help me with this?
Answer: (D) thrust has overcome the drag force
Explanation: In order for the plane to move forward, the backward force will have to be overcome. As thrust is a force in the opposite direction of drag so it can cancel out the backward pull of drag and move forward.
Two litres of water, initially at 20 °C, is heated to 40 °C. Determine thevolume of water at 40 °C if the coefficient of volumetric expansion ofwater within this range is 30 x 10-9 °C.
Given data
*The given initial volume of water is V_1 = 2 L
*The given initial temperature of the water is T_i = 20 °C
*The given final temperature of the water is T_f = 40 °C
*The coefficient of volumetric expansion of water is
[tex]\alpha=30\times10^{-9}^{}\text{C}[/tex]The formula for the volumetric expansion is given as
[tex]\frac{\Delta V}{V}=\alpha(T_f-T_i)[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} \frac{\Delta V}{V_1}=30\times10^{-9}\times(40^0-20^0) \\ V_2-V_1=6\times10^{-7}V_1 \\ V_2=6\times10^{-7}V_1+V_1^{} \\ =V_1(6\times10^{-7}+1) \\ =2.00\text{ L} \end{gathered}[/tex]If centripetal force is towards the center and gravitational force is also towards the center how can we be lighter at the equator?
ANSWER and EXPLANATION
We want to determine why a person weighs lighter at the equator.
At the equator, an object will weigh slightly less as a result of the slightly greater centripetal force and the slight increase in distance from the center of gravity of the earth due to the equatorial bulge. This would imply a decrease in the gravitational force that an object experiences.
In other words, the earth is not a perfect sphere, and so, at the equator, an object will be slightly farther from the center.
That is the answer.
Kinetic theory told us that gas particles are moving constantly and bumping into anything in their path. The collisions of these particles in the gas result in ___________ .
a
Energy conversion
b
Pressure
c
Creation of new forms of matter
d
Potential energy
Answer:
a
Explanation:
because if gas particles move they move with speed and that is because they are free,so when they bump into something,they bump into it with force so it converts the energy
What is the acceleration of a car that goes from
rest to 25 m/s in 5.0 s?
Answer:
5 m/s^2
Explanation:
Acceleration is change in velocity / change in time
Accel = 25 m/s / 5 s = 5 m/s^2
What human activities increase the levels of greenhouse gases released?
Answer: There are many human activities that have negative effects on the environment such as the release of greenhouse gas. Some include; transporting, industry and factories, burning of fossil fuels, agriculture, deforestation, and the release of chemicals into the atomsphere.
Explanation:
Transporting and factories emit gasses such as carbon dioxide, which is toxic and can affect the environment. The burning of fossil fuels also releases carbon dioxide into the air, leading to the massive spread of global warming. Deforestation can lead to less oxygen and more carbon dioxide in the air, and can also cause global warming. Animals can also lose their habitats.
Answer:
humans burn fossil fuels to generate electricity, to keep buildings warm, power cars etc. As a result, waste gases are being produced such as carbon dioxide
Human Activities:
- driving cars, carbon dioxide is released
-burning things e.g wood
-small loads of laundry, wasting
-heating
a powerful motorcycle can produce an acceleration of while traveling at 90.0 km/h. at that speed, the forces resisting motion, including friction and air resistance, total 400.0 n. (air resistance is analogous to air friction. it always opposes the motion of an object.) what is the magnitude of the force that motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 kg? 34. a car with a mass o
1257.5 N is the magnitude of the force that motorcycle exerts backward on the ground to produce its acceleration.
What is a magnitude in physics?The definition of magnitude is simply "distance or quantity." It displays an object's size, direction, or speed, whether relative or absolute. It's used to indicate that something has a certain size or scope.
Briefing:A = 3.50 m/s² is the motorcycle's acceleration.
f = 400.0 N, the force of resistance to motion
M = 245 kg for the motorcycle's mass
F is the amount of force a motorcycle exerts when moving rearward.
The amount of force a motorcycle exerts when moving rearward is described as
F = f + ma
Inserting the values
F = 400.0 + (245) (3.50)
F = 1257.5 N
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The complete question is -
A powerful motorcycle can produce an acceleration of 3.50 m/s² while traveling at 90.0 km/h. At that speed, the forces resisting motion, including friction and air resistance, total 400.0 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What is the magnitude of the force that motorcycle exerts backward on the ground to produce its acceleration if the mass of the motorcycle with rider is 245 kg?
There is a ball of fresh fruit at the top of a small cliff that is 45 m high. The fresh fruit baskets mass is 10kg's. Identify the energy (PE or KE) and calculate the it.
Given,
The mass of the fruit basket, m=10 kg
The height of the cliff, h=45 m
Kinetic energy is the energy possessed by an object due to its motion. Kinetic energy is directly proportional to the square of the velocity of the object.
Potential energy is the energy possessed by an object due to its position. The gravitational potential energy is directly proportional to the height at which the object is situated.
As the basket is at rest, its velocity is zero and hence it does not have kinetic energy.
Thus the energy possessed by the fruit basket is the potential energy.
The potential energy is given by,
[tex]E_P=\text{mgh}[/tex]Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} E_P=10\times9.8\times45 \\ =4410\text{ J} \end{gathered}[/tex]The potential energy of the fruit basket is 4410 J
jumping up before the elevator hits. after the cable snaps and the safety system fails, an elevator cab free-falls from a height of 36 m. during the collision at the bottom of the elevator shaft, a 90 kg passenger is stopped in 5.0 ms. (assume that neither the passenger nor the cab rebounds.) what are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision? if the passenger were to jump upward with a speed of 7.0 m/s relative to the cab floor just before the cab hits the bottom of the shaft,
a) The magnitude of the impulse is 2.39 × [tex]10^{3}[/tex] N. b) Average force on the passenger during the collision is 4.78 × [tex]10^{5}[/tex] N. c) impulse is 1.6 × [tex]10^{3}[/tex] N.s d) The corresponding average force would be 3.52 × [tex]10^{5}[/tex] N.
a) By energy conservation, the speed of the passenger when the elevator hits the floor is
1/2 mv² = mgh
v = √2gh = [tex]\sqrt{2(9.8)(36)}[/tex]
= 26.6 m/s
The magnitude of the impulse is
J = |Δp| = m|Δv| = mv = (90kg) (26.6m/s) ≈ 2.39 × 10³ N.s
b) With duration of Δt = 5.0 × [tex]10^{-3}[/tex] s for the collision, the average force is
[tex]F_{avg}[/tex] = J/ Δt = 2.39×10³ N.s / 5.0 × [tex]10^{-3}[/tex] s
≈ 4.78 × [tex]10^{5}[/tex] N
c) If the passenger were to jump upward with a speed of [tex]v^{'}[/tex]= 7.0 m/s, then the resulting downward velocity would be
[tex]v^{n}[/tex] = v- [tex]v^{'}[/tex]
= 26.6m/s - 7.0 m/s
= 19.6 m/s
The magnitude of the impulse becomes
[tex]J^{n}[/tex] = |Δ[tex]p^{n}[/tex]| = m|Δ[tex]v^{n}[/tex]| = m[tex]v^{n}[/tex] = (90kg) (19.6 m/s)
≈ 1.76 × 10³³N.s
d) The corresponding average force which is
[tex]F_{avg } ^{n}[/tex] = [tex]J^{n}[/tex]/ Δt = 1.76 × 10³ N.s/ 5.0 × [tex]10^{-3}[/tex] s
≈ 3.52 × [tex]10^{5}[/tex]N
Therefore the magnitude of the impulse is 26.6 m/s, average force on the passenger during the collision is 4.78 × [tex]10^{5}[/tex]N, the magnitude of the impulse is 1.76×[tex]10^{3}[/tex]N.s, average force is 3.52 × [tex]10^{5}[/tex].
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a car's bumper is designed to withstand a 5.40 km/h (1.5-m/s) collision with an immovable object without damage to the body of the car. the bumper cushions the shock by absorbing the force over a distance. calculate the magnitude of the average force on a bumper that collapses 0.225 m while bringing a 890 kg car to rest from an initial speed of 1.5 m/s.
The average force is F = 2722.5N
This law of interaction states that when one body exerts a force on a second body, the second body exerts an equal and opposite force on the first body. This is the law of action and reaction, which helps explain why you feel the impact when you collide with another bumper car.
When the car hits something at low speed, the foam and fenders absorb the energy, pushing the bumper back and using the crumple zone to cushion the impact. Shattering the bumpers, fenders, and foam limits the damage you can do to the cars and the people inside them. means equal to the sum.
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Johnny rides his bike South from home to school which is 5 km away. It takes him 30 minutes to arrive at school. What is Johnny's velocity in km/hr?
The velocity of Johnny is 10km/hr
The rate at which an object's position changes in relation to a frame of reference and time is the term meant by velocity. Although it may appear sophisticated, velocity is just the act of moving quickly in one direction. Since it is a vector quantity, the definition of velocity requires both magnitude (speed) and direction. Its SI equivalent is (ms-1). A body is considered to be accelerating if the magnitude or direction of its velocity changes.
We are given that,
The distance of the school = d = 5km
The time taken to reach school = t = 30min = 0.5hr
Velocity of Johnny = V = ?
V = d/t
V = (5km)/ (0.5hr)
V = 10km/hr
Therefore , the velocity would be 10km/hr
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Calculator A dog has a kinetic energy of 111J and is running at a speed of 10m/s. What is the mass of the dog? Give your answer to 2 decimal places.
Explanation:
MASS=2.22
VELOCITY=10m/s
KINETIC ENERGY=111
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the magnetic field of a long, straight wire has a value b0 at a distance of r0 when the current is i0. at a distance of 3.00 r0. what current is necessary to produce a field having the same strength b0 at this position?
The magnetic field of a long straight wire has a board b0 at a distance r0 when the cutting edge is i0. at a distance of 3.00 r0: B'=2B0.
A magnetic field is a vector domain that describes the magnetic effect on moving electricity costs, currents, and magnetic materials. Transfer fees in magnetic fields are under pressure perpendicular to personal speed and magnetic fields.
The magnetic field is the place around the magnet where the effect of the magnetism is felt. We use magnetic fields as tools to explain how magnetic forces are directed into circular space and are essentially magnetic. This force attracts or repels magnets to each other. Examples of magnetic forces include compasses, motors, magnets that keep things in refrigerators, learning tracks, and new roller coasters. All the carrying charges create a magnetic field, and the charges flowing through that area enjoy the force.
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Ethan slid a glass of OJ for peter across a counter with a speed of 2.1 m/s unfortunately peter missed it if the countertop was 1.1m above the floor how long did it take for the glass to hit the ground
Answer:
It takes 0.47 seconds for the OJ to reach the ground.
Explanation:
d = vi*t + (1/2)at^2
(-1.1) = (0)t + (1/2)(-9.8)t^2
-1.1 = -4.9t^2
0.22445 = t^2
0.47s = t
It takes 10 seconds to accelerate from rest to rate of 8 m/s. If the mass of the car is 1,200kg. Calculate the net force acting on the car .
Answer:
960 N
Explanation:
Find acceleration
Change in velocity / change in time
8 /10 = .8 m/s^2
Then F = ma
= 1200 kg * .8 m/s^2 = 960 N
two objects with equal masses are in motion. which object will have more kinetic energy? (1 point) responses the object with the greater speed the object with the greater speed the object with the greater density the object with the greater density the object with the greater acceleration the object with the greater acceleration the object with the greater volume the object with the greater volume
Answer:
I rephrased your question:
Two objects with equal masses are in motion. which object will have more kinetic energy?
A) the object with the greater speed
B) the object with the greater density
C) the object with the greater acceleration
D) the object with the greater acceleration
E) the object with the greater volume
F) the object with the greater volume
Explanation:
Kinetic Energy is equal to (1/2)mv^2, where m is the mass and v is the velocity.
-----
A) the object with the greater speed YES
B) the object with the greater density NO
C) the object with the greater density NO
D) the object with the greater acceleration MAYBE. Acceleration means energy is being added, so it depends on how much and the masses of each object.
E) the object with the greater volume NO
Answer:
Explanation: Two objects with equal masses are in motion. Which object will have more kinetic energy?
Responses
the object with the greater density
the object with the greater volume
the object with the greater acceleration
the object with the greater speed
Answer: The object with the greater speed
an electron accelerated from rest through a voltage of 790 v enters a region of constant magnetic field. part a if the electron follows a circular path with a radius of 21 cm , what is the magnitude of the magnetic field? express your answer using two significant figures. b
The electron is accelerated through a potential difference of , so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
1/2mv²=eΔV
where
m is the electron mass
v is the final speed of the electron
e is the electron charge
is the potential difference
Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
v=√2eΔv/m=√2(1.6-10-19c)(790v)/9.1-10-31=1.66.10 m/s
Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
evb=mv²/r
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
b=m/er =3.8.10 t
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You want to lift a heavy box with a mass L = 64.0 kg using the two-ideal pulley system as shown. With what minimum force do you have to pull down on the rope in order to lift the box at a constant velocity? One pulley is attached to the ceiling and one to the box.
The given problem can be solved using the following free-body diagram:
The diagram is the free-body diagram for the pulley that is holding the weight. Where:
[tex]\begin{gathered} T=\text{ tension} \\ m=\text{ mass} \\ g=\text{ acceleration of gravity} \end{gathered}[/tex]Now we add the forces in the vertical direction:
[tex]\Sigma F_v=T+T-mg[/tex]Adding like terms:
[tex]\Sigma F_v=2T-mg[/tex]Now, since the velocity is constant this means that the acceleration is zero and therefore the sum of forces is zero:
[tex]2T-mg=0[/tex]Now we solve for "T" by adding "mg" from both sides:
[tex]2T=mg[/tex]Now we divide both sides by 2:
[tex]T=\frac{mg}{2}[/tex]Now we substitute the values and we get:
[tex]T=\frac{(64\operatorname{kg})(9.8\frac{m}{s^2})}{2}[/tex]Solving the operations:
[tex]T=313.6N[/tex]Now we use the free body diagram for the second pulley:
Now we add the forces in the vertical direction:
[tex]\Sigma F_v=T-F[/tex]The forces add up to zero because the velocity is constant and the acceleration is zero:
[tex]T-F=0[/tex]Solving for the force:
[tex]T=F[/tex]Therefore, the pulling force is equal to the tension we determined previously and therefore is:
[tex]F=313.6N[/tex]
A child boards a bus for summer camp. His mom is waiting next to the bus. The bus begins
moving 3 m/s east with respect to the ground. The child is walking 0.5 m/s west towards the back
of the bus with respect to the bus. The mother begins running 1 m/s east with respect to the
ground, chasing after the bus and thoroughly humiliating her child. What is the child's velocity
a) relative to the ground?
b) relative to the mother?
Answer:
a) relative to the ground
Explanation:
A piece of rock weighs 5 Newtons. A force F is applied to it and it produces an acceleration of a. Now, if there is a second piece of rock, what force needs to be applied to the second piece of rock to produce an acceleration of 8a?
Given,
weight of a piece of rock, W = mg = 5N
So, mass of a piece of rock is m = W ÷ g = 5N ÷ 9.8 m/[tex]s^{2}[/tex] = 0.51 kg.
When a force F is applied, it produces an acceleration 'a'. From Newton's second law of motion, F = ma = 0.51 a.
⇒a = F / 0.51.
Acceleration produced by 2nd piece of the rock is a' = 8a.
Assuming the mass of 2nd piece of rock to be same as the 1st piece, force to be applied on 2nd piece is F' = m'a'.
⇒ F' = m' × 8a = 8m'a
If the mass of both pieces of rock are equal, then m' = m = 0.51 kg.
⇒ F' = 0.51kg × 8a
⇒ F' = 0.51kg × 8 (F / 0.51)
⇒ F' = 8F.
Thus, a force equal to 8 times the force applied on 1st piece should be applied on the 2nd piece of the rock to produce an acceleration of 8a (if their masses are equal.
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A 14 kg boulder is pushed off a cliff with velocity v = ( 14.0 m/s ) + ( 2.0m/s ) y . Will the object experience a larger vertical or horizontal acceleration?
The object will eventually experience a large vertical acceleration since the vertical velocity increases as the object moves downwards.
What is horizontal motion of a projectile?The horizontal motion of a projectile is the motion of the projectile along a horizontal path.
The horizontal distance of a projectile is not affected by gravity and hence the horizontal speed of a projectile remains constant. That is the initial horizontal velocity is equal to the final horizontal velocity of the projectile.
However, the during the vertical motion an object, the vertical velocity decreases as the object moves upwards and eventual becomes zero when the object reaches the maximum height.
As the object begins to move downwards the vertical velocity increase and eventually become maximum before the object hits the ground.
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a 7.50-nc point charge is located 1.80 m from a 4.20-nc point charge. (a) find the magnitude of the electric force that one particle exerts on the other. (b) is the force attractive or repulsive?
The electric force is 8.370 x 10^9N.
We need to know about the electric force to solve this problem. The electric force produced by charges interacting each other, whose strength is given by the equation
F = k.Q1 . Q2/r²
where F is the electric force, Q1 and Q2 is the charge and r is the radius.
From the question above, we know that
k = 9 x 10⁹ Nm²/C²
Q1 =e = 7.50 x 10^-9 C
Q2 = e = 4.02 x 10^-9 C
r = 1.80 m
The electric force will attract each other when the charges are different and will repel each other when the charges are same.
F = k.Q1 . Q2/r²
F = 9 x 10⁹ . 7.50 x 10^-9 . 4.02 x 10^-9/(1.80)²
F = 8.37 x 10^9 N
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You are driving at the speed of 29.5 m/s
(66.0037 mph) when suddenly the car in
front of you (previously traveling at the same
speed) brakes and begins to slow down with
the largest deceleration possible without skidding. Considering an average human reaction,
you press your brakes 0.558 s later. You also
brake and decelerate as rapidly as possible
without skidding. Assume that the coefficient
of static friction is 0.708 between both cars’
wheels and the road.
The acceleration of gravity is 9.8 m/s
2
.
Calculate the acceleration of the car in front
of you when it brakes.
Answer in units of m/s
2
.
The acceleration of the car in front before stopping is 6.94 m/s².
What is the acceleration of the car?The acceleration of the car is the rate of change of velocity of the car with time.
The acceleration of the car in front is determined by applying the Newton's second law of motion assuming the car decelerated at a constant rate before stopping.
Ff = F
μmg = ma
μg = a
where;
Ff is the frictional force on the car when it stopsF is the force of the carm is the mass of the cara is the acceleration of the carg is acceleration due to gravityμ is coefficient of static frictionSubstitute the given parameters and solve for the acceleration of the car.
a = (0.708) x (9.8 m/s²)
a = 6.94 m/s²
Thus, the acceleration of the car in front before stopping is 6.94 m/s².
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a wheel has a constant angular acceleration of 4 rad/s2. starting from rest, it turns through 300 rad. what is its final angular velocity? how much time elapses while it turns through the 300 radians?
If a wheel has a constant angular acceleration of 4 rad/s2. starting from rest, it turns through 300 rad , then the final angular velocity would be 48.96 rad /s and 12.24 seconds elapse while it turns through the 300 radians.
What are the three equations of motion?There are three equations of motion given by Newton,
v = u + at
S = ut + 1/2×a×t²
v² - u² = 2×a×s
By using the second equation of motion given by Newton,
S = ut + 1/2at²
300 = 0 + 0.5 × 4 × t²
t² =300 /2
t = 12.24 seconds
The final angular velocity = 0+ 12.24 × 4
= 48.96 rad /s
Thus, 12.24 seconds elapse while it turns through the 300 radians.
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after driving a portion of the route, the taptap is fully loaded with a total of 23 people including the driver, with an average mass of 65 kg per person. in addition, there are three 15- kg goats, five 3- kg chickens, and a total of 25 kg of bananas on their way to the market. assume that the springs have somehow not yet compressed to their maximum amount. how much are the springs compressed?
The compression of the spring at the given weight is 61 cm.
Given,
The spring constant is k =2.4 × 10⁴ N/m
mass of each of the 23 persons, = 65 kg
mass of each of the 3 goats, = 15 kg
mass of each of the 5 chickens, = 3 kg
mass of the bananas, = 25 kg
The total mass of all the passengers contained in the laptop is calculated as;
total mass of the persons = 23 x 65 = 1495 kg
total mass of the goats = 3 x 15 = 45 kg
total mass of the chickens = 5 x 3 = 15 kg
the total mass of the bananas = 25 kg
The total mass = (1495 + 45 + 15 + 25)kg
1The total mass = 580 kg
The calculation for the lengthy of spring:-
F = kx
x = F/k = mg/k
x = (1495*9.8)/2.4*10^4
x = 0.610 m
x = 61 cm
Hence, the compression of the spring at the given weight is 61 cm.
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An apple is resting on a table. Claire says that there are no forces acting on the apple because it is not moving. Is she right?
Answer:
no
Explanation:
there are forces , even when the apple isnt moving , for example gravity which is keeping the apple on the table
the barrel of a shotgun: group of answer choices is generally shorter than that of a rifle. is wider at the end to concentrate shot. is smooth without the grooves and lands found in rifles. is indistinguishable from that of a rifle.
Measures the diameter of a shotgun's barrel in terms of the quantity of lead balls needed to make the bore weigh one pound (12 gauge is the diameter of a lead ball weighing one-twelfth of a pound).
What differentiates a shotgun from a rifle?A shotgun has many projectile properties while a rifle only comes with one. Shotguns are helpful at close ranges when rifles are useful at a distance. Shotguns only have a front sight, whereas rifles have both front and rear sights.
Rifles are typically a little noisier than shotguns. Analyzing test results reveals that shotguns and rifles have decibel levels that are comparable. The typical decibel range
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A cart is set up as shown below, with three fans directed to the left and two of the fans running. The motion of the cart is represented by the v vs t graph shown. If the experiment were repeated with all three fans running, what might the resulting v vs t graph look like?
A. Graph A
B. Graph B
C. Graph C
D. Graph D
in february 1955, a paratrooper fell 370 m from an air- plane without being able to open his chute but happened to land in snow, suffering only minor injuries. assume that his speed at im- pact was 56 m/s (terminal speed), that his mass (including gear) was 85 kg, and that the magnitude of the force on him from the www.engineeringbookspdf problems 249 12.0 m/s and angle u1 35.0. just after, it is traveling directly upward with velocity of magnitude 10.0 m/s. the duration of the collision is 2.00 ms. what are the (a) magni- v : 2 snow was at the survivable limit of 1.2 105 n. what are (a) the minimum depth of snow that would have stopped him safely and (b) the magnitude of the impulse on him from the snow?
(a)1.1 m is the minimum depth of snow that would have stopped him safely. (b)The magnitude of the impulse on him from the snow[tex]|\Delta \overrightarrow{\mathrm{p}}|[/tex] -4.8 * 10³ kg. m/s.
What is magnitude ?A key concept in science is magnitude, which is employed in physics. Magnitude describes a general amount or a distance. We can relate magnitude to size and speed of the object is motion while considering the aspects of movement. The magnitude of the an object is defined by it's own size or quantity.
Briefing:We choose +y upward, which implies a>0 (the acceleration is upward since it represents a deceleration of his downward motion through the snow).
a) The maximum deceleration a[tex]_m_a_x[/tex] of the paratrooper (of mass m and initial speed ν=56m/s ) is found from Newton’s second law
F[tex]_s_n_o_w[/tex]−mg=ma[tex]_m_a_x[/tex]
where we require F[tex]_s_n_o_w[/tex] =1.2×10⁵N . Using Eq. 2 −15ν² =2a max d , we find the minimum depth of snow for the man to survive :
[tex]\mathrm{d}=\frac{v^2}{2 \mathrm{a}_{\text {max }}}=\frac{\mathrm{m} v^2}{2\left(\mathrm{~F}_{\text {snow }}-\mathrm{mg}\right)} \approx \frac{(85 \mathrm{~kg})(56 \mathrm{~m} / \mathrm{s})^2}{2\left(1.2 \times 10^5 \mathrm{~N}\right)}=[/tex]1.1 m.
(a) He experiences a change in momentum during his brief journey through the snow.
[tex]\Delta \overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{p}}_{\mathrm{f}}-\overrightarrow{\mathrm{p}}_{\mathrm{i}}[/tex] = 0 - (85 kg)(-56m/s)= -4.8 * 10³ kg. m/s.
or[tex]|\Delta \overrightarrow{\mathrm{p}}|[/tex] = -4.8 * 10³ kg. m/s.
Because downward is the negative direction, the starting velocity has a negative value. This equals the impulse caused by the net force F[tex]_s_n_o_w[/tex]−mg according to the impulse-momentum theorem, but since F[tex]_s_n_o_w[/tex] ≫mg we can approximate this as the impulse on him just from the snow.
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The complete question is -
In February 1955, a paratrooper fell 370m from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was 56m/s (terminal speed), that his mass (including gear) was 85kg, and that the magnitude of the force on him from the snow was at the survivable limit of 1.2×10⁵ N. What are (a) the minimum depth of snow that would have stopped him safely and (b) the magnitude of the impulse on him from the snow?
a 0.500 kg basketball is dropped out of a window that is 6.90 m above the ground. the ball is caught by a person whose hands are 1.62m above the ground. how much work is done on the ball by its weight
Work done on the ball by its weight is +25.872 J .
What is work done by gravity ?
Gravity is that force which pulls objects to the ground. There is a force that acts, so basically gravity acts. When you apply a power to an item, that power works for you. For example, if you throw a ball, the force on the ball will cause it to travel far and complete the task. Work is proportional to the applied force and the distance it covers or covers. For example, if you throw a ball with less force, the distance it travels will decrease proportionally to the force applied. Throwing the ball with more force increases the flight distance. Falling particles are forced to face the direction of gravity. The mass, the gravitational constant, and the height at which the falling object falls determine how large the falling object will be.
Work done by gravity is :
W= mg(h2-h1)
= (0.500)(9.80)(6.90-1.62)
= +25.872 J .
To learn more about the work done by gravity from the given link ;
https://brainly.in/question/33999289
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