(a) The change in momentum of the trunk is 19800 kgm/s,
(b) The force exerted on the trunk is 3960 N
Change in momentum:The can be defined as the product of the mass of a body and its change in velocity. The S.I unit of change in momentum is kgm/s
(a) To calculate the change in momentum of the trunk, we use the formula below.
Formula:
ΔM = m(v-u)............... Equation 1Whee:
ΔM = change in momentum of the trunkm = mass of the trunkv = final velocity of the trunku = initial velocity of the trunk.From the question,
Given:
m = 5500 kgv = 7.8 m/su = 4.2 m/sSubstitute these values into equation 1
ΔM = 5500(7.8-4.2)ΔM = 5500(3.6)ΔM = 19800 kgm/s.(b) To calculate how large is the force was exerted on the trunk, we use the formula below.
Formula:
F = ΔM/t................. Equation 2Where:
F = Force exerted on the trunkt = time.From the question,
Given:
t = 15.0 s.Substitute into equation 2
F = 19800/5F = 3960 N.
Hence, (a) The change in momentum of the trunk is 19800 kgm/s, (b) The force exerted on the trunk is 3960 N
Learn more about change in momentum here: https://brainly.com/question/7538238
Which electron dot diagram shows the bonding between 2 chlorine atoms?2 dots then C l with 2 dots above and 1 dot below then 2 dots then 2 dots then C l with 2 dots above and 1 dot below then 2 dots.2 dots then C l with 2 dots above and 2 dots below then 2 dots then C l with 2 dots above and 2 dot below then 2 dots.2 dots then C l with 2 dots above and 2 dots below then 1 dot then C l with 2 dots above and 2 dots below then 2 dots.2 dots then C l with 2 dots above and 1 dot below then 3 dots then 3 dots then C l with 2 dots above and 1 dot below then 2 dots.
Answer:
It is B
Explanation:
Answer: 2nd answer
Explanation: took exam
I need help with science homework
Answer:
4. atmosphere and geosphere
5. atmosphere and hydrosphere
6. hydrosphere and geosphere (? not sure about this one sorry)
7. hydrosphere and geosphere
8. biosphere and geosphere
The slope of a position-time graph can be used to find the moving obiects
Answer:
Is this a true and false statement?
Explanation:
The circumference of the Earth is
approximately 40,075 kilometers.
How many significant figures are
there in the numerical value?
Answer:
5 sig figures
Explanation:
A man speeding at 40m/s decides to outrun the cops and starts to
accelerate at a rate of 2.5m/s2 for 12 seconds. What is the criminal's new
speed?
Answer:
70 m/s.
Explanation:
Given that,
Initial speed, u = 40 m/s
Acceleration = 2.5 m/s²
Time, t = 12 s
We need to find criminal's new speed. Let it is v. Using equation of motion to find it as follows :
v = u +at
Substitute all the values
v = 40 + 2.5(12)
v = 70 m/s
So, the new speed is 70 m/s.
what are the effects of moon rotation and revolution
Answer:
The effects of the Moon's rotation includes;
1) The Moon rotation and revolution gives the appearance of a perfectly still Moon to observers of the Moon from the Earth
2) The Moon has two sides, the near side that continuously faces the Earth and the "back" or far side, which is also known as the dark side of the Moon
The effect of the Moon's revolution
1) The tides in the ocean and water bodies, due to the gravitational forces from the Moon
2) The changes in the observed shape of the Moon due to the amount of Sunlight that is able to be reflected from the Moon as a result of the relative position of the Moon, the Earth and the Sun, at a given point in time
3) Lunar and Solar eclipse, when the Earth and the Moon blocks the light coming from the Sun respectively, due to their combined revolution
Explanation:
The duration of the Moon's orbit round the Earth = 27.322 days
The time it takes the Moon to rotate round its axis = 27 days
The Moon is the closest cosmic body to the Earth.
A car traveling with 500,000 J of kinetic energy is brought to a kinetic energy of
100,000 J in 12 seconds. What is the force acting on the car to accomplish this?
A:41,666.67
B:3x10^-5N
C:33,333.33N
D:8,333.33N
Answer:
33,333.33 N
Explanation:
Given that :
Initial kinetic energy = 500,000 J
Final kinetic energy = 100,000 J
Using the relation :
Force * time = change in momentum (Newton's law)
Force (F) * 0.12 = (500,000 - 100,000)
0.12F = 400,000 J
Force = (400,000 J) / 0.12s
Force = 33333.333
Force = 33,333.33 N
While delivering 125 kg blocks of ice to a local village, Kristoff and his family (and Sven too) come upon a cliff that is 5.7 m above them. To solve their problem, they build a catapult that will launch their blocks of ice with an initial velocity of 15 m/s and at an angle of 45 degrees above the ground.
Prove that the catapult will successfully launch the ice blocks up to the top of the cliff. Quantities you will need to solve for will be initial vertical velocity, horizontal velocity, the distance the catapult is from the cliff, and the amount of time it takes the ice to reach the cliff. Assume that they are experts and have arranged everything so that the ice blocks just barely reach the top of their parabolic path when they land at the top of the cliff and make a smooth landing. Draw a picture of the scene to help you visualize what is happening.
After landing on the flat land above, each block of ice travels 20 meters while slowing to a stop.
What is the rate of acceleration while the blocks slow to a stop?
How long do the blocks take to slow to a stop?
What is the amount of friction between the ice and the snowy ground?
How long do the blocks take to slow to a stop?
What is the amount of friction between the ice and the snowy ground?
Answer:
a. Since h = 5.74 m > 5.7 m, the height of the cliff, the block of ice will successfully launch to the top of the cliff.
b. -2.81 m/s²
c. 3.78 s
d. -351.25 N
Explanation:
a. After landing on the flat land above, each block of ice travels 20 meters while slowing to a stop.
For the block of ice to reach the top of the cliff, its maximum height, h should be greater than or equal to 5.7 m. That is, h ≥ 5.7 m.
The maximum height of a projection h, projected with an initial velocity v at an angle Ф is h = v²sin²Ф/2g where g = acceleration due to gravity = 9.8 m/s².
For the block of ice, v = 15 m/s and Ф = 45°. So,
h = v²sin²Ф/2g
= (15 m/s)²sin²45/(2 × 9.8 m/s²)
= 225 (m/s)²(1/√2)²/19.6 m/s²
= 225 (m/s)²(1/2)/19.6 m/s²
= 112.5 (m/s)²/19.6 m/s²
= 5.74 m
Since h = 5.74 m > 5.7 m, the height of the cliff, the block of ice will successfully launch to the top of the cliff.
The graph is in the attachment.
b. What is the rate of acceleration while the blocks slow to a stop?
Using v² = u² + 2as where u = initial horizontal velocity of block = 15m/scos45° = 10.61 m/s, v = final velocity of block = 0 m/s since it stops, a = acceleration and s = distance block moves = 20 m
So, a = (v² - u²)/2s
substituting the variables into the equation, we have
a = ((0 m/s)² - (10.61 m/s)²)/2(20 m)
= - 112.57 (m/s)²)/40 m
= -2.81 m/s²
c. How long do the blocks take to slow to a stop?
Using v = u + at where u = initial horizontal velocity of block = 10.61 m/s v = final velocity of block = 0 m/s since it stops, a = acceleration = -2.81 m/s² and t = time it takes block of ice to stop
So, making t subject of the formula,
t = (v - u)/a
substituting the values of the variables, we have
t = ( 0 m/s - 10.61 m/s)/-2.81 m/s²
= -10.61 m/s/-2.81 m/s²
= 3.78 s
d. What is the amount of friction between the ice and the snowy ground?
The frictional force, f = net force on block of ice
f = ma where m = mass of bock = 125 kg and a = acceleration of block = -2.81 m/s²
f = ma
= 125 kg(2.81 m/s²)
= -351.25 N
A wagon having a mass of 42 kg is accelerated across a level road at 1.50 m/s2. What net force acts on the wagon horizontally?
Answer:
The net force acting on the wagon is 63 N
Explanation:
Net Force
According to the second Newton's law, the net force exerted by an external agent on an object of mass (m) is:
F = m.a
Where (a) is the acceleration of the object.
The wagon has a mass of m=42 Kg is accelerated at [tex]a=1.5\ m/s^2[/tex]. The net force is:
F = 42*1.5
F = 63 N
The net force acting on the wagon is 63 N
A 250-kg moose stands in the middle of the railroad tracks in Sweden, frozen by the lights of an oncoming 10,000kg train traveling at 20m/s. Even though the engineer attempted in vain to slow the train down in time to avoid hitting the moose, the moose rides down the remaining track sitting on the train’s cowcatcher. What is the final velocity of the train and moose after the collision?
(Momentum & Impulse)
Answer:
The final velocity of the train and the moose after collision is approximately 19.51 m/s
Explanation:
The given mass of the moose, m₁ = 250 kg
The velocity of the moose, v₁ = 0
The mass of the oncoming train, m₂ = 10,000 kg
The velocity of the train, v₂ = 20 m/s
The velocity of the moose and the train after collision = v₃
By the principle of conservation of linear momentum, the total initial momentum before the collision = The total final momentum after collision
m₁·v₁ + m₂·v₂ = (m₁ + m₂)·v₃
Therefore, by substitution, we have;
250×0 + 10,000× 20 = (10,000 + 250) × v₃
200,000 = 10,250 × v₃
v₃ = 200,000/10,250 ≈ 19.51 m/s
The final velocity of the train and the moose after collision = v₃ ≈ 19.51 m/s
What travels by vibrating particles? Mechincal Waves or ElecrtoMagnetic Waves.
Answer:mechanical waves.
Explanation:
Mechanical waves require the particles of the medium to vibrate in order for energy to be transferred. For example, water waves, earthquake/seismic waves, sound waves, and the waves that travel down a rope or spring are also mechanical waves.
It takes serina 0.25 hours to drive to school. Her route is 16km long . What is serina’s average speed on her drive to school
Explanation:
distance=16km=16000m
time=0.25hours=1/4hr=15min=900sec
speed=dist./time
=16000/900=160/9=17.777m/s
speed=17.78m/s (rounded)
Answer:
64
Explanation:
Avg. speed = total distance/total time
Avg. speed= 16km/0.25h
Avg. speed= 64km/h
Which of the following is most likely to be a conductor?
OA. Aplastic
OB. A molecule containing only nonmetal atoms
OC. A covalently bonded molecule
D. Acompound containing metallic bonds
Answer:
D.
Explanation:
A compound containing metallic bonds
hope it helps!
The magnitude of vector vector A is 84.9 m and it points in the +y axis direction. The magnitude of vector vector B is 195.0 m and it points at an angle of 41.0° counterclockwise from +x axis. The magnitude of vector vector C is 126.2 m and it points in the +x axis direction.
Solution:
The magnitude of A vector is 84.9 m in the positive y-axis direction.
So the X component of A =0
the Y component of A = 84.9 m
Now the magnitude of B vector is 195 m and it makes an angle of 41° in the direction from the positive x-axis direction.
So the X component of B = B cos 41°
= 195 x cos 41°
= 195 x 0.75 = 146.25 m
the Y component of B = B sin 41°
= 195 x sin 41°
= 195 x 0.65 = 126.75 m
Now it is given that vector C has a magnitude of 126.2 m and it makes a direction towards the positive x-axis.
So the X component of C =126.2 m
the Y component of C = 0
Comparing all these, we get
1. B vector has the largest X component
2. B vector has the largest Y component
A Chevy Camaro drives straight off the top level off a parking garage at 13 m/s. If the car landed 90 meters away from the base of the parking garage, how high (height) was the top level?
a200.64
b234.64
c34.64
d134.64
Answer:
234.64 m
Explanation:
Using the formula for calculating range;
Range R = u√2H/g
u is the speed = 13m/s
H is the maximum height
g is the acceleration due to gravity = 9,8m/s²
R = 90m
Get the maximum height;
90 = 13√2H/9.8
90/13 = √2H/9.8
6.923 = √2H/9.8
square both sides
6.923² = 2H/9.8
469.698 = 2H
H = 469.698/2
H = 234.64
Hence the top level was 234.64 m high.
A hose on the ground projects a water current upwards at an angle 40 to the horizontal at velocity 20 m/s find height at which water hits a wall at 8 m away from the hose (consider that acceleration due to gravity =9.8 m/s2)
Answer:
The water hits the wall at a height of 5.38 m
Explanation:
Projectile Motion
It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.
The object describes a parabolic path given by the equation:
[tex]{\displaystyle y=\tan(\theta )\cdot x-{\frac {g}{2v_{0}^{2}\cos ^{2}\theta }}\cdot x^{2}}[/tex]
Where:
y = vertical displacement
x = horizontal displacement
θ = Elevation angle
vo = Initial speed
The hose projects a water current upwards at an angle of θ=40° at a speed vo=20 m/s.
The height at which the water hits a wall located at x=8 m from the hose is:
[tex]{\displaystyle y=\tan40^\circ\cdot 8-{\frac {9.8}{2*20^{2}\cos ^{2}40^\circ }}\cdot 8^{2}}[/tex]
Calculating:
y = 5.38 m
The water hits the wall at a height of 5.38 m
Lina holownicza wytrzymuje działanie siły o wartości co najwyżej 1400 N (większa siła powoduje rozerwanie liny). Oblicz maksymalne przyspieszenie, jakie może osiągnąć samochód o masie 2000 kg holowany na takiej linie po poziomej jezdni, jeżeli wiadomo, że łączna wartość sił oporu ruchu tego samochodu w czasie holowania wynosi 400 N.
Odpowiedź:
0,5 m / s²
Wyjaśnienie:
Jeśli się uwzględni:
Masa samochodu (m) = 2000 kg
Wartość siły oporu podczas holowania = 400N
Lina o maksymalnej sile może wytrzymać = 1400N
Maksymalne przyspieszenie = siła / masa samochodu
Siła = maksymalna siła - siła oporu
Siła = 1400 N - 400 N = 1000 N.
Stąd maksymalne przyspieszenie;
1000 N / 2000 kg
= 0,5 m / s²
Jerry is pushing a 50-kg box across a moth floor with an acceleration of 0.6 m/s2. What force is he applying to the box? *
83.3 N
0.012 N
0
30 N
Answer:
30 NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 50 × 0.6
We have the final answer as
30 NHope this helps you
an object has an mass of 15 kg and is falling at a rate of 2.0 m/s what is the momentum?
Answer:
30 kg.m/sExplanation:
The momentum of an object can be found by using the formula
momentum = mass × velocity
From the question we have
momentum = 15 × 2
We have the final answer as
30 kg.m/sHope this helps you
A ball is thrown up into the air. Ignore air resistance. When it is rising and reaches half of its maximum height,the net force acting on it is
Answer:
The net and only force acting on the ball thrown up into the air at half of its maximum height is the weight of the ball
Explanation:
When the ball is rising in the air, the force, F, acting on it is given by the product of the mass, m × acceleration, a
The acceleration of a body thrown in the air = Gravitational acceleration = g = Constant
Therefore;
The force acting on the body thrown in the air F = Constant = m × g (downwards)
The force acting on the ball thrown up into the air at half of its maximum height = The mass of the ball × The acceleration due to gravity = The weight of the ball.
A(n) ____________ stretch is one done where antagonist muscles are used to stretch the muscles. But the _____________ stretch is one that is done that the muscle needs help from something or someone to stretch the muscle.
Answer: Dynamic - Static Flexibility
Explanation:
A force of 30 N stretches a very light ideal spring 0.73 m from equilibrium. What is the force constant (spring constant) of the spring
The forces constant (spring constant) of the spring will be 41.09 N/m.
What is spring force?The force required to extend or compress a spring by some distance scales linearly concerning that distance is known as the spring force. Its formula is;
F = kx
The given data in the problem is;
F is the spring force = 30 N
K is the spring constant= ?
x is the displacement of spring = 0.73 m
The spring constant is;
K =F/x
K=30/0.73
K=41.09 N/m
Hence the force constant (spring constant) of the spring will be 41.09 N/m.
To learn more about the spring force refer to the link;
https://brainly.com/question/4291098
#SPJ1
please answer asignment due today
Answer:
Give me some time okkkk
It takes a crane 59s to lift a flagstone using 342 W of power. How much work is done on the flagstone?
Answer: The work done on the flagstone is 20178 J
Explanation:
Power is the rate at which work is done . It is equal to the amount of work done divided by the time it takes to do the work.
[tex]Work=Power\times time[/tex]
Given : work = ?
Power = 342 W = 342J/s
Time = 59 s
[tex]Work=342J/s\times 59s=20178J[/tex]
Thus the work done on the flagstone is 20178 J
In raising an object vertically at a constant speed of 2 m / s 10 watts of power is developed the weight of the object is
Answer:
5NExplanation:
Power = Workdone/Time
Since workdone = Force * distance = mgh
Power = mgh/t
Given
Power = 10watts
speed = 2m/s
Power = Weight * velocity
Substitute
10 = Weight * 2
Weight = 10/2
Weight = 5N
Hence the weight of the object is 5N
a metallic cube whose each side is 10 cm is subjected to a shearing force of 100 kg. The top force is displaced through 0.25 cm with respect to the bottom. calculate the shearing stress strain and modulus
Answer:
9.8×104Nm−2,0.025,3.92×106Nm−2
Solution :
Here, L=10cm=10×10−2m
F=100kgf=100×9.8N
ΔL=0.25cm=0.25×10−2m,
Shearing stress =FL2=100×9.8(10×10−2) Sheraing strain =ΔLL=0.25×10−210×10−2 = 0.025 Shear Modulus of elasticity, G=Shearing stressShearing strain=9.8×1040.025
=3.92×106Nm−2
Explanation:
(Blank) is caused by plate motion.
It’s please help.
Answer:
heat
Explanation:
a squirrel runs at a speed of 9.9 m/s with 25 J of kinetic energy
What is the squirrels mass
Answer:
yeet yeet yeet yeet
Explanation:
Kinetic energy (K.E):-
So, the Mass of the Squirrel is 0.51 Kg (or) 510 grams.
A squirrel runs at a speed of 9.9 m/s with 25 J of kinetic energy.
What is the squirrel’s mass?
Answer: 0.51 kg
Cannon ball of mass 3.5kg is fired at a speed of 22m/s2
Answer:
77N
Explanation:
Since we are not told what to find, we can as well look for the net force acting on the ball.
According Newton's second law;
F = mass × acceleration
Given
Mass = 3.5kg
Acceleration = 22m/s²
Net force F = 3.5×22
F = 35/10×22
F = 7/2×22
F = 7×11
F =77N
Hence the magnitude of force required is 77N
A friend is coming to Tim’s house to study after school. What directions would Tim give for reaching his house from the entrance of the school?
(I WILL GIVE BRAINLIEST)
Answer:
go up the street as you exit the house and make a right and keep going up for 3 blocks and you should see the school