Answers
Answer:
The beetle
Explanation:
The beetle is correct because of the the force of the Throckmortan was a lot heavier and thus making the beetle feel the force.
Momentum before = momentum after
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(70.0 kg) (3.0 m/s) + (2.0 kg) (3.0 m/s) = (70.0 kg) (3.1 m/s) + (2.0 kg) v
210 kg m/s + 6 kg m/s = 217 kg m/s + (2.0 kg) v
-1 kg m/s = (2.0 kg) v
v = -0.5 m/s
The skateboard's velocity is 0.5 m/s west. Hope I helped and have a good day!
________________________________________________________
前の勢い=後の勢い
m₁u₁+m₂u₂=m₁v₁+m₂v₂
（70.0 kg）（3.0 m / s）+（2.0 kg）（3.0 m / s）=（70.0 kg）（3.1 m / s）+（2.0 kg）v
210 kg m / s + 6 kg m / s = 217 kg m / s +（2.0 kg）v
-1 kg m / s =（2.0 kg）v
v = -0.5 m / s
スケートボードの速度は西に0.5m / sです。私が助けて、良い一日を過ごせることを願っています！
Related Questions
The study of heat is ____?
Answers
Explanation:
thermodynamics is the study of heat.
Answer The study of heat and its relationship to useful work is called thermodynamics and involves macroscopic quantities such as pressure, temperature, and volume without regard for the molecular basis of these quantitie
Explanation:
4. Sally applies a horizontal force of 462 N with a rope to drag a wooden crate across a floor with a constant speed. The rope tied to the crate is pulled at an angle of 56.00 . c.What work is done by the floor through force of friction between the floor and the crate
Answers
Answer:
-6,329.5Joules
Explanation:
Complete question:
Sally applies a horizontal force of 462N with a rope to drag a wooden crate across a floor with a constant speed the rope tied to the crate is pulled at an angle of 56.0degree and sally moves the crate 24.5m. What work is done by the floor through the force of friction between the floor and crate.
Work done = Fd cos theta
F is the horizontal force
d is the distance covered
theta angle of inclination
Substituting into the formula
Workdone = 462(24.5)cos 56
Workdone = 11,319(0.5592)
Workdone = 6,329.5Joules
Hence the workdone by sally is 6,329.5Joules
The work done by friction will be opposite the work done by sally, hence work done by the floor through force of friction between the floor and the crate is -6,329.5Joules
Which option tells the forces that influence the movement of earths plates
Answers
Answer:
Gravity
Explanation:
A wall clock has a minute hand with a length of 0.46 m and an hour hand with a length of 0.24 m. Take the center of the clock as the origin, and use a Cartesian coordinate system with the positive x axis pointing to 3 o'clock and the positive y axis pointing to 12 o'clock. Write the vector that describes the displacement of a fly if it quickly goes from the tip of the minute hand to the tip of the hour hand at 3:00 P.M. (Let vector D represents the displacement of the fly.)
Answers
Answer:
the vector that describes the displacement of a fly if it quickly goes from the tip of the minute hand to the tip of the hour hand at 3:00 P.M is ; { 0.24 m(i) - 0.46 m(j) }
Explanation:
Given the data in the question;
as illustrated in the image below,
3:00 pm means
the hour hand is on 3 i.e along x-axis
while the minute hand is on 12 i.e along y-axis
so Displacement will be;
D = ( 0.24 + 0i) - ( 0 + 0.46j )
D = { 0.24 m(i) - 0.46 m(j) }
Therefore, the vector that describes the displacement of a fly if it quickly goes from the tip of the minute hand to the tip of the hour hand at 3:00 P.M is ; { 0.24 m(i) - 0.46 m(j) }
Which elements have one valence electron?
A. Sodium
B. Carbon
C. Fluorine
D. Magnesium
The answer is A
Answers
Answer:
Well, you said the answer is A, so it’s A!
a. As you coast down a hill on your bicycle, you accelerate at 0.5 m/s2. If the total mass of your body and the bicycle is 80 kilograms what is the net force pulling you down the hill (gravity - friction)?
Answers
F=ma
=80*0.5
=40N
The net force pulling you down the hill will be = 40 N
What is Newton's Second Law of motion?The Second Law of motion states that the acceleration of an object depend upon the object and the mass of the object.
F = Mass * acceleration
Given
Total mass : 80 kg
Acceleration : 0.5 m / s^2
Net Force = mass * acceleration (Second Law of motion )
Net Force = 80 * 0.5 = 40 N
The net force pulling you down the hill will be = 40 N
Learn more about Newton's Second Law of motion:
https://brainly.com/question/11553356?referrer=searchResults
#SPJ2
Problem 2: Estimate the electric power requirement, in kW, of a 1,400 ft2floor area (three bedroomshome) with three occupants. Using your home power estimate, predict the power requirementsfor a city of 300,000 people. Use these results to estimate the area (inkm2)of silicon solar cells requiredto satisfy the community power requirements. Assume, thepower requirements for an average single family home of 3 are 108.4 x 106BTU per year and solar panels insolation
Answers
Answer:
a) 3170 kw
b) 377 km^2
Explanation:
Estimate of electric power
a) Given :
Average power consumption for a family of 3 = 108.4 * 106 BTU per year = 0.0317 kw = 31.7 watts
The power requirement for a city of 300000 people
= 31.7 watts * 100000 = 3170000 watts = 3170 kw
b) Given :
Average solar panel insulation = 8.4 W /m^2
Estimate the area of silicon solar cells required to satisfy community power requirement
= (1 * 3170) * (1000/8.4 )
= 377.380 * 10^3 m2 = 377 km^2
A basketball with a mass of 20 kg is accelerated with a force of 10 N. If resisting forces are ignored, what is the acceleration of the basketball?
Answers
Which cell line is pointing to the body?
Answers
Answer:
The answer is B .........number 2
Explanation:
A weightlifter holds a 1,500 N barbell 1 meter above the ground. One end of a 2-meter-long chain hangs from the center of the barbell. The chain has a total weight of 400 N. How much work (in J) is required to lift the barbell to a height of 2 m
Answers
Answer:
1800 J
Explanation:
We are given that
Weightlifter hold weight, w=1500N
Height of weightlifter from ground,h=1m
Length of chain=2m
Total weight of chain, w'=400 N
We have to find the work done required to lift the barbell to height of 2m.
Total work done=Work done to lift a barbell and half of the chain+ work done to lift the upper half of the chain
Total work done=[tex]1500+\frac{1}{2}(400)+\frac{1}{2}(400)(\frac{1}{2}(1))[/tex]
Total work done=[tex]1700+100[/tex]
Total work done=1800 J
Hence, work done is required to lift the barbell to a height of 2 m=1800 J
Please Help! Will mark brainliest.
Answers
Answer:W = m*g*h
19*9.8*32.4 = 6,032.9 rounded
honestly, I do not know if this is correct so please don't come back at me
hopefully this helps
Explanation: [do the following, if you think I am wrong]
just pick a formula,
plug in the number to the mass, gravity, and height
than multiply
get your answer, but don't forget to round to the nearest tenth
Why do birds not get shock when they
sit on high power live wire but we do?
Answers
Answer:
Their bodies don't conduct electricity like we do.
Explanation:
Answer:
birds dont get shocked because they sit on their talons and their talons are a different type of skin then the rest of their body
Explanation:
3. If the gravitational force between 2 objects is 50 N, what is the gravitational
force when the distance between the two is increased to four times the distance?
Answers
The gravitational force : 3.125 N
Further explanationGiven
F₁ = 50 N
Required
F₂
Solution
Newton's Gravity Law:
[tex]\rm F=G.\dfrac{m_1.m_2}{r^2}[/tex]
with F = gravitational force,
G = gravitational constant,
m1, m2 = mass of object,
r = distance between two objects.
The value of m and G are the same , so :
F₁ ≈ 1/r₁²
The distance between the two is increased to four times:
r₂ = 4r₁
F₂ = 1/(4r₁)²
F₂ = 1/16r₁²
F₂ = F₁ x 1/16
F₂ = 50 : 16
F₂ = 3.125 N
Each tire on a car has a radius of 0.330 m and is rotating with an angular speed of 15.6 revolutions/s. Find the linear speed v of the car, assuming that the tires are not slipping against the ground.
Answers
Answer:
v = 32.345 m/s
Explanation:
given data
radius r = 0.330 m
angular speed = 15.6 revolutions/s
solution
we will get here linear speed v, that is express as
v = r × w .....................1
here w = 17.5 × 2 × π radians
so now put value in eq1
v = 0.330 × 15.6× 2 × π
v = 32.345 m/s
What must the charge (sign and magnitude) of a particle of mass 1.48 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 640 N/C
Answers
Answer:
[tex]q=-2.26\times 10^{-5}\ C[/tex]
Explanation:
Given that,
The mass of a particle, m = 1.48 g = 0.00148 kg
The electric field, E = 640 N/C
We need to find the charge of the particle when placed in a downward-directed electric field.
The force of gravity is balanced by the electric force such that,
mg = qE
Where
q is the charge of the particle
[tex]q=\dfrac{mg}{E}\\\\q=\dfrac{0.00148\times 9.8}{640}\\\\q=2.26\times 10^{-5}\ C[/tex]
q must be negative, the force must be upward (opposite direction of the electric field).
two students sit on a see-saw. archie is a hulking football player with a mass of 120 kg. clementine is a dainty cheerleader with a mass of 40 kg. the see-saw is 3.5 m in total length with the fulcrum at the center. if clementine sits at the end on one side, where must archie sit relative to the center to keep the see-saw balanced
Answers
Answer:
Archie must sit 0.58 m relative to the center to keep the see-saw balanced
Explanation:
Given the data in the question;
Mass of Archie [tex]m_{a}[/tex] = 120 kg
Mass of clementine [tex]m_{c}[/tex] = 40 kg
total length of see-saw L = 3.5 m
as illustrated on the image below, Fulcrum is at the center,
suppose Archie sits at a distance x from center then for balancing, we will have;
[tex]m_{a}[/tex] × x = [tex]m_{c}[/tex] × ( one end = 3.5/2 = 1.75)
so we substitute
120kg × x = 40kg × 1.75m
x12okg = 70 kg.m
x = 70 kg.m / 120 kg
x = 0.58 m
Therefore, Archie must sit 0.58 m relative to the center to keep the see-saw balanced
A DC motor connected to a switch-mode dc-dc converter goes into regenerative braking mode. The average current being supplied by the dc motor is 20 A. In the equivalent circuit of the dc motor, Ea=103 V, Ra=0.19 Ohms, and La=4 mH. Calculate the average power flow into the converter. Round answer to the nearest whole number.
Answers
Answer:
The correct answer is "2524 W".
Explanation:
The given values are:
Ia = 20 A
Ea = 103 V
Ra = 0.19 Ω
La = 4 mH
According to KVL,
⇒ [tex]V_0=E_a-I_aR_a[/tex]
On substituting the given values in the above equation, we get
⇒ [tex]=103-(20)(0.19)[/tex]
⇒ [tex]=103-3.8[/tex]
⇒ [tex]=126.2 \ V[/tex]
Now,[tex]=126.2\times 20[/tex]
The average power flow into the converter will be:
⇒ [tex]P=V_0 I_a[/tex]
On substituting the given values, we get
⇒ [tex]=(126.2\times 20)[/tex]
⇒ [tex]=2524 \ W[/tex]
A car initially traveling 7 m/s speeds up uniformly at a rate of 3 m/s2 until it reaches a velocity of 22 m/s. How much time did it take the car to reach this final velocity?
Answers
Answer:
t = 5 s
Explanation:
Data:
Initial Velocity (Vo) = 7 m/sAcceleration (a) = 3 m/s²Final Velocity (Vf) = 22 m/sTime (t) = ?Use formula:
[tex]\boxed{t=\frac{Vf - Vo}{a}}[/tex]Replace:
[tex]\boxed{t=\frac{22\frac{m}{s} -7\frac{m}{s}}{3\frac{m}{s^{2}}}}[/tex]Solve the subtraction of the numerator:
[tex]\boxed{t=\frac{15\frac{m}{s}}{3\frac{m}{s^{2}}}}[/tex]It divides:
[tex]\boxed{t=5\ s}[/tex]How much time did it take the car to reach this final velocity?
It took a time of 5 seconds.
A spacecraft is flying away from the moon toward earth.
What will be true of the moon’s gravitational pull on the spacecraft?
It will decrease.
It will increase.
It will repel the spacecraft.
It will remain the same.
Answers
Answer:
it will decrease
Explanation:
According to the law of universal gravitation, the gravitational force exerted by the moon on the spacecraft is equal to the product of their masses and inversely proportional to the square of the distance that separates them. Therefore, as the spacecraft moves away, its distance increases and the force of attraction exerted by the moon decreases.
Answer:
A. It will increase
Explanation:
I took the quiz on K12 and this was the correct option.
Hope I helped
hmu if u brave shawtys
Answers
Answer:
BET, & done ✌
Answer:
boop
Explanation:
You are concerned about the moon roof on your new sports car. It seems to flex when driving at high speeds. Calculate how much net force the moon roof must withstand and in what direction. Assume the moon roof is flat with an area of 0.5 m^2 and the pressure and velocity over the moon roof is constant. Your driving speed is 20 m/s, and the velocity over the moon roof is 30 m/s. The pressure inside the car is 90,500 N/m^2 and the freestream pressure and density in front of the car are 90,000 N/m^2 and 1.1 kg/m^3.
Answers
Answer:
Explanation:
We shall apply Bernoulli's formula to solve the problem . It is as follows .
P + ρ gh + 1/2 ρ v² = constant .
P₁ + ρ gh + 1/2 ρ v₁² = P₂ + ρ gh + 1/2 ρ v₂²
P₁ + 1/2 ρ v₁² = P₂ + 1/2 ρ v₂²
P₁ - P₂ = 1/2 ρ (v₂² - v₁² )
= .5 x 1,1 ( 30² - 20² )
= 275 N / m²
velocity over moon roof is high , pressure will be lower there by 275 N / m²
Given pressure difference already existing = 90500 - 90000 = 500 N / m²
Additional pressure difference due to velocity difference = 275 N / m²
Total pressure difference = 275 + 500 = 775 N / m²
Area of roof = .5 m²
Total force acting upwards on the roof
= .5 x 775 N
= 387.5 N .
Two ropes are connected to a 200 kg dinghy. Two cousins each take one rope and pull. When the cousins pull in the same direction, the dinghy accelerates at a rate of 1.31 m/s2 to the east. If they pull in opposite directions, the dinghy has an acceleration of 0.526 m/s2 to the west. Assume the ropes are horizontal, and ignore any other horizontal forces acting on the dinghy. What is the magnitude of the force each cousin exerts on the dinghy
Answers
Answer:
The magnitude of the force each cousin exerts on the dinghy is 183.6 N and 78.4 N.
Explanation:
When the cousins pull in the same direction we have:
[tex] F_{1} + F_{2} = ma_{e} [/tex] (1)
Where:
F₁ and F₂ are the forces exerted by the two boys.
m: is the mass of the dinghy = 200 kg
[tex]a_{e}[/tex]: is the acceleration in the east direction
When the cousins pull in opposite directions we have:
[tex] F_{1} - F_{2} = ma_{w} [/tex] (2)
By adding equation (1) and (2):
[tex] 2F_{1} = m(a_{e} + a_{w}) [/tex]
[tex] F_{1} = \frac{200 kg(1.31 m/s^{2} + 0.526 m/s^{2})}{2} = 183.6 N [/tex]
Now, by entering F₁ into equation (1) we can find F₂:
[tex] F_{2} = ma_{e} - F_{1} = 200kg*1.31 m/s^{2} - 183.6 N = 78.4 N [/tex]
Therefore, the magnitude of the force each cousin exerts on the dinghy is 183.6 N and 78.4 N.
I hope it helps you!
you describe a friend’s position by including distance, direction, and what other term?
Answer choices:
A. Acceleration
B.displacement
C.Average speed
D. Reference point
PLEASE HELP I NEED THIS IN AN HOUR
Answers
Answer:
Acceleration
Explanation:
Answer: Acceleration
Explanation:
A student claims an object in motion must experience a force to stay in motion. Do you agree or disagree?
Answers
Answer:
agree because there is always a force that causes motion..
At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.70 m/s2. At the same instant a truck, traveling with a constant speed of 9.50 m/s, overtakes and passes the automobile. (a) How far beyond the traffic signal will the automobile overtake the truck
Answers
Answer:
66.85 m
Explanation:
We are given that
Acceleration ,a=[tex]2.7m/s^2[/tex]
Speed of truck, v=9.5 m/s
We have to find the distance beyond which the traffic signal will the automobile overtake the truck.
Initial speed of automobile, u=0
We know that
[tex]s=ut+\frac{1}{2}at^2[/tex]
Using the formula
[tex]s=0+\frac{1}{2}(27)t^2=\frac{27}{2}t^2[/tex]
For constant speed
Acceleration, a=0
Again
[tex]s=vt+0=9.5t[/tex]
[tex]9.5t=\frac{27}{2}t^2[/tex]
[tex]t=\frac{9.5\times 2}{2.7}=7.037s[/tex]
Substitute the value of t
[tex]x=9.5(7.037)=66.85m[/tex]
Hence, the distance beyond which the traffic signal will the automobile overtake the truck=66.85 m
what kind of lens curve inward toward its center
Answers
Answer:
A concave lens is exactly the opposite with the outer surfaces curving inward, so it makes parallel light rays curve outward or diverge. That's why concave lenses are sometimes called diverging lenses.
Answer: A concave lens
A concave lens is exactly the opposite with the outer surfaces curving inward, so it makes parallel light rays curve outward or diverge. That's why concave lenses are sometimes called diverging lenses.
Hopes this helps :)
What is the frequency of a wave of a light is with a wavelength of 4 x 10-7 m?
Answers
Answer:
7.5 × 10^14 Hz
Velocity of light = 3×10^8m/s
Frequency = (3×10^8)/(4 x 10^-7)
= 7.5 × 10^14 Hz
A box of bananas weighing 51.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.46 and the coefficient of kinetic friction is 0.23. Part A If no horizontal force is applied to the box and the box is at rest, how large is the frictional force exerted on the box by the surface
Answers
static friction force = The coefficient of static friction * normal force
static friction force = 0.46 * 51 = 23.46 N
kinetic friction force = The coefficient of kinetic friction * normal force
kinetic friction force = 0.23 * 51 = 11.73 N
the applied force acting on the object must be more than 23.5 N if the object was stationary to move it and must be more than 11.7 N during the movement to keep the object moving
A 97.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.63 rad/s . A monkey drops a 8.97 kg bunch of bananas vertically onto the platform. They hit the platform at 45 of its radius from the center, adhere to it there, and continue to rotate with it. Then the monkey, with a mass of 22.1 kg , drops vertically to the edge of the platform, grasps it, and continues to rotate with the platform. Find the angular velocity of the platform with its load. Model the platform as a disk of radius 1.73 m .
Answers
Answer:
the final angular velocity of the platform with its load is 1.0356 rad/s
Explanation:
Given that;
mass of circular platform m = 97.1 kg
Initial angular velocity of platform ω₀ = 1.63 rad/s
mass of banana [tex]m_{b}[/tex] = 8.97 kg
at distance r = 4/5 { radius of platform }
mass of monkey [tex]m_{m}[/tex] = 22.1 kg
at edge = R
R = 1.73 m
now since there is No external Torque
Angular momentum will be conserved, so;
mR²/2 × ω₀ = [ mR²/2 + [tex]m_{b}[/tex] ([tex]\frac{4}{5}[/tex] R)² + [tex]m_{m}[/tex]R² ]w
m/2 × ω₀ = [ m/2 + [tex]m_{b}[/tex] ([tex]\frac{4}{5}[/tex] )² + [tex]m_{m}[/tex] ]w
we substitute
w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1
w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )
w = 48.55 × [ 1.63 / ( 76.3908 ) ]
w = 48.55 × 0.02133
w = 1.0356 rad/s
Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s
An electric vehicle starts from rest and accelerates at a rate of 2.3 m/s2 in a straight line until it reaches a speed of 29 m/s. The vehicle then slows at a constant rate of 1.5 m/s2 until it stops. (a) How much time elapses from start to stop
Answers
Answer:
t = 12.6 seconds
Explanation:
Given that,
Initial velocity, u = 0
Acceleration of an electric vehicle, a = 2.3 m/s²
Final velocity, v = 29 m/s
We need to find the time elapses from start to stop. The acceleration of an object is given by the relation as follows :
[tex]a=\dfrac{v-u}{t}\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{29-0}{2.3}\\\\t=12.6\ s[/tex]
So, 12.6 seconds is elapsed from start to stop.