As a roller coaster car crosses the top of a 40-m-diameter loop-the-loop, its apparent weight (the normal force) is the same magnitude as the car's weight. What is the car's speed at the top?

Answers

Answer 1

Answer:

40 because if it is the same weight then there is no weight to make the ride slower so it 40

Explanation:


Related Questions

Which person ha the most freedom to make his or her own lifestyle decisions?
A. Frieda is 10 years old and lives with her grandmother
B. Vladimir is 3 years old and attends preschool
C. Quincy is 16 years old and lives in a dormitory
D. Lucinda is 32 years old and has two children

Answers

Answer:

C Quincy

Explanation:

Both Frieda and Vladimir are too young to be making their own decisions. Lucinda has limited freedom due to having two children, while Quincy is just now becoming an adult and has his whole life still ahead of him. Therefore, Quincy has the most freedom to make their own lifestyle decisions.

I hope this helps!

Answer:

The correct answer is C. Quincy is 16 years old and lives in a dormitory

Explanation:

Quincy being 16 may be able to get a job and hold a membership at a gym even if not at a gym he still has the most freedom period with Lucinda having children and Frieda and Vladimir being to young to make the choices of exercising on their own.

Please tell me if I'm wrong so I may give you the correct answer!

Happy Holidays!!  

HEELLPPPPPpppppppppppppppp

Answers

Explanation:

Given:

[tex]A_1[/tex] = 4.5 cm[tex]^2[/tex]

[tex]v_1[/tex] = 40 cm/s

[tex]v_2[/tex] = 90 cm/s

[tex]A_2[/tex] = ?

a) The continuity equation is given by

[tex]A_1v_1 = A_2v_2[/tex]

Solving for [tex]A_2[/tex],

[tex]A_2 = \dfrac{v_1}{v_2}A1 = \left(\dfrac{40\:\text{cm/s}}{90\:\text{cm/s}}\right)(4.5\:\text{cm}^2)[/tex]

[tex]= 2\:\text{cm}^2[/tex]

b) If the cross-sectional area is reduced by 50%, its new area [tex]A_2'[/tex] now is only 1 cm^2, which gives us a radius of

[tex]r = \sqrt{\dfrac{A_2'}{\pi}} = 0.564\:\text{cm}[/tex]

What is the total kinetic energy of a 0.15 kg hockey puck sliding at 0.5 m/s and rotating about its center at 8.4 rad/s

Answers

Answer:

[tex]K=0.023J[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=0.15[/tex]

Velocity [tex]v=0.5m/s[/tex]

Angular Velocity [tex]\omega=8.4rad/s[/tex]

Generally the equation for Kinetic Energy is mathematically given by

[tex]K=\frac{1}{2}M(v^2+\frac{1}{2}R^2\omega^2)[/tex]

[tex]K=\frac{1}{2}0.15(0.5^2+\frac{1}{2}(0.038)^2.(8.4rad/s^2))[/tex]

[tex]K=0.023J[/tex]

Set the initial bead height to 3.00 m. Click Play. Notice that the ball makes an entire loop. What is the minimum height required for the ball to make an entire loop

Answers

Answer:

h> 2R

Explanation:

For this exercise let's use the conservation of energy relations

starting point. Before releasing the ball

       Em₀ = U = m g h

Final point. In the highest part of the loop

       Em_f = K + U = ½ m v² + ½ I w² + m g (2R)

where R is the radius of the curl, we are considering the ball as a point body.

      I = m R²

      v = w R

we substitute

       Em_f = ½ m v² + ½ m R² (v/R) ² + 2 m g R

       em_f = m v² + 2 m g R

Energy is conserved

       Emo = Em_f

       mgh = m v² + 2m g R

       h = v² / g + 2R

 

The lowest velocity that the ball can have at the top of the loop is v> 0

      h> 2R

Una bala de 10 g se dispara contra un bloque de madera de 102 g inicialmente en reposo sobre una superficie horizontal. Después del impacto el bloque se desliza 8 m antes de detenerse. Si el coeficiente de fricción entre el bloque y la superficie es 0,5, ¿Cuál es la velocidad de la bala inmediatamente antes del impacto?

Answers

Answer:

una ess abola cola sola answer

A small object with mass 0.20 kg swings as a pendulum on the end of a long light rope. For small amplitude of swing, the period of the motion is 3.0 s. If the object is replaced by one with mass 0.400 kg, what is the period for small amplitude of swing? (a) 1.5 s (b) 3.05 (c) 6.0 s (d) 12.0 s (e) none of the above answers

Answers

Answer:

The correct option is (e) "none of the above".

Explanation:

Given that,

A small object with mass 0.20 kg swings as a pendulum on the end of a long light rope. For small amplitude of swing, the period of the motion is 3.0 s.

If the object is replaced by one with mass 0.400 kg, then we have to find the period for small amplitude of the swing.

We know that the time period can be calculated as :

[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]

Where

l is the length

g is acceleration due to gravity

It means the time period is independent of the mass. So, if the mass is replaced by one with mass 0.400 kg, there is no effect on the time period.

Two 0.20-kg balls, moving at 4 m/s east, strike a wall. Ball A bounces backwards at the same speed. Ball B stops. Which statement correctly describes the change in momentum of the two balls?

a. |ΔpBl<|ΔPA|
b. |ΔpBl=|ΔPA|
c. |ΔpB|>|ΔPA|
d. ΔpB > ΔPA

Answers

Answer:

Option A

Explanation:

From the question we are told that:

Mass [tex]m=0.20kg[/tex]

Velocity [tex]v=4m/s[/tex]

Generally the equation for momentum for Ball A is mathematically given by

Initial Momentum

 [tex]M_{a1}=mV[/tex]

 [tex]M_{a1}=0.2*4[/tex]

 [tex]M_{a1}=0.8[/tex]

Final Momentum

 [tex]M_{a2}=-0.8kgm/s[/tex]

Therefore

 [tex]\triangle M_a=-1.6kgm/s[/tex]

Generally the equation for momentum for Ball B is mathematically given by

Initial Momentum

 [tex]M_{b1}=mV[/tex]

 [tex]M_{b1}=0.2*4[/tex]

 [tex]M_{b1}=0.8[/tex]

Final Momentum

 [tex]M_{b2}=-0 kgm/s[/tex]

Therefore

 [tex]|\triangle M_a|>|\triangle Mb|[/tex]

Option A

A force of 15 N toward the WEST is applied to a 4.0 kg box. Another force of 42 N toward the EAST is also applied to the 4.0 kg box. The net force on the 4.0 kg box
is

Answers

Answer :

[tex]\implies F_1 < F_2[/tex]

[tex] \implies F_{net} = F_2 - F1[/tex]

[tex]\implies F_{net} = 42 -15[/tex]

[tex]\implies \underline{ \boxed{ F_{net} = 27 \: N}}[/tex]

The net force on the 4.0 kg box is 27 N towards EAST.

A weight suspended from a spring is seen to bob up and down over a distance of 20 cm triply each second. What is the period? What is the amplitude?

Answers

The period and the amplitude of the weight suspended from spring are 0.33 seconds and 10 centimeters, respectively.

1) The period is given by:

[tex] T = \frac{1}{f} [/tex]

Where:

f: is the frequency = 3 bob up and down each second = 3 s⁻¹ = 3 Hz

[tex] T = \frac{1}{f} = \frac{1}{3 Hz} = 0.33 s [/tex]

Hence, the period is 0.33 s.

2) The amplitude is the distance between the equilibrium position and the maximum position traveled by the spring. Since the spring is moving up and down over a distance of 20 cm, then the amplitude is:          

[tex] A = \frac{20 cm}{2} = 10 cm [/tex]  

Therefore, the amplitude is 10 cm.          

You can learn more about the period and amplitude here: https://brainly.com/question/15169209?referrer=searchResults

I hope it helps you!                    

A 650-kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruis- ing speed of 1.75 m/s. (a) What is the average power of the elevator motor during this period

Answers

Answer: [tex]P=5573.43\ W[/tex]

Explanation:

Given

Mass of the elevator is [tex]M=650\ kg\\\[/tex]

Time period of ascension [tex]t=3\ s[/tex]

cruising speed [tex]v=1.75\ m/s[/tex]

Distance moved by elevator during this time

Suppose Elevator starts from rest

[tex]\Rightarrow v=u+at\\\Rightarrow 1.75=0+a(3)\\\Rightarrow a=0.583\ s[/tex]

Distance moved

[tex]\Rightarrow h=ut+0.5at^2\\\Rightarrow h=0+0.5\times 0.5833\times (3)^2\\\Rightarrow h=2.62\ m[/tex]

Gain in Potential Energy is

[tex]\Rightarrow E=mgh\\\Rightarrow E=650\times 9.8\times 2.62\\\Rightarrow E=16,720.3\ N[/tex]

Average power during this period is

[tex]\Rightarrow P=\dfrac{E}{t}\\\\\Rightarrow P=\dfrac{16,720.3}{3}\\\\\Rightarrow P=5573.43\ W[/tex]

Answer:

The power is 331.7 W.

Explanation:

mass, m = 650 kg

time, t= 3 s

initial  velocity,  u = 0 m/s

final velocity, v = 1.75 m/s

(a) The power is defined as the rate of doing work.

Work is given by the change in kinetic energy.

W = 0.5 m (V^2 - u^2)

W = 0.5 x 650 x 1.75 x 1.75 =  995.3 J

The power is given by

P = W/t = 995.3/3 = 331.7 W

Object A has a mass m and a speed v, object B has a mass m/2 and a speed 4v, and object C has a mass 3m and a speed v/3. Rank the objects according to the magnitude of their momentum.

Required:
Rank from smallest to largest.

Answers

Answer:

Momentum of object A = Momentum of object C < momentum of B.

Explanation:

The momentum of an object is equal to the product of mass and velocity.

Object A has a mass m and a speed v. Its momentum is :

p = mv

Object B has a mass m/2 and a speed 4v. Its momentum is :

p = (m/2)×4v = 2mv

Object C has a mass 3m and a speed v/3. Its momentum is :

p = (3m)×(v/3) = mv

So,

Momentum of object A = Momentum of object C < momentum of B.

A container is filled with water and the pressure at the container bottom is P. If the container is instead filled with a liquid having specific gravity 1.05, what new bottom pressure will be measured

Answers

eat ahah is when s susuehs

Which wave has the smallest amplitude?

I will give brainliest to the right answer.

Answers

Wave-D (the red one) has the smallest amplitude of the 4 waves on this graph.

12. A car travels in a straight line with an average velocity of 80 km/h for 2.5h and then an average velocity of 40 km/h from 1.5 h (a) what is the total displacement for the 4 h trip? (b) What is the average velocity for the total trip?

Answers

Answer:

a. Total displacement = 140 km/h

b. Average velocity = 35 km/h

Explanation:

Given the following data;

Average velocity A = 80 km/h

Time A = 2.5 hours

Average velocity B = 40 km/h

Time B = 1.5 hours

a. To find the total displacement for the 4 h trip;

Total time = Time A + Time B

Total time = 2.5 + 1.5

Total time = 4 hours

Next, we would determine the displacement at each velocity.

Mathematically, displacement is given by the formula;

Displacement = velocity * time

Substituting into the formula, we have;

Displacement A = 80 * 2.5

Displacement A = 200 km/h

Displacement B = 40 * 1.5

Displacement B = 60 km/h

Total displacement = Displacement A - Displacement B

Total displacement = 200 - 60

Total displacement = 140 km/h

b. To find the average velocity for the total trip;

Mathematically, the average velocity of an object is given by the formula;

[tex] Average \; velocity = \frac {total \; displacement}{total \; time} [/tex]

Substituting into the formula, we have;

[tex] Average \; velocity = \frac {140}{4} [/tex]

Average velocity = 35 km/h

what is the prefix notation of 0.0000738?​

Answers

Answer:

The scientific notation of 738 is 7.38 x 1002.

I agree with the answer above me…

How many joules of energy are required to accelerate one kilogram of mass from rest to a velocity of 0.866c?

Answers

Answer:

the amount of energy needed is 1.8 x 10¹⁷ J.

Explanation:

Given;

mass of the object, m₀ = 1 kg

velocity of the object, v = 0.866 c

By physics convection, c is the speed of light = 3 x 10⁸ m/s

The energy needed is calculated as follows;

E = Mc²

As the object approaches the speed of light, the change in the mass of the object is given by Einstein's relativity formula;

[tex]M = \frac{M_0}{\sqrt{1- \frac{v^2}{c^2} } } \\\\ M = \frac{1}{\sqrt{1- \frac{(0.866c)^2}{c^2} } }\\\\ M = \frac{1}{\sqrt{1- \frac{0.74996c^2}{c^2} } }\\\\ M = \frac{1}{\sqrt{0.25} } \\\\ M = 2 \ kg[/tex]

The energy required is calculated as;

E = 2 x (3 x 10⁸)²

E = 1.8 x 10¹⁷ J

Therefore, the amount of energy needed is 1.8 x 10¹⁷ J.

A compact disk with a 12 cm diameter is rotating at 5.24 rad/s.

a. What is the linear speed _______m/s
b. What is the centripetal acceleration of a point on its outer rim _______
c. Consider a point on the CD that is halfway between its center and its outer rim. Without repeating all of the calculations required for parts (a) and (b), determine the linear speed of this point. _______m/s
d. Determine the centripetal acceleration of this point. _______

Answers

Answer:

(a) 31.44 m/s (b) 164.74 m/s²

Explanation:

Given that,

The diameter of a disk, d = 12 cm

Radius, r = 6 cm

Angular speed = 5.24 rad/s

(a) Linear speed,

[tex]v=r\omega\\\\v=6\times 5.24\\\\v=31.44\ m/s[/tex]

(b) Centripetal acceleration,

[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{31.44^2}{6}\\\\a=164.74\ m/s^2[/tex]

The same constant force is used to accelerate two carts of the same mass, initially at rest, on horizontal frictionless tracks. The force is applied to cart A for twice as long a time as it is applied to cart B. The work the force does on A is WA; that on B is WB. Which statement is correct?

a. WA = WB
b. WA = 2WB.
c. WA=4WB
d. WB= 2WA

Answers

Answer:

Option (c).

Explanation:

Let the mass of each cart is m and the force is F.

Time for cart A is 2t and for cart B is t.

Work done is given by the

W= force x displacement

As the distance is given by

S= u t +0.5 at^2

So, when the time is doubled the distance is four times.

So, WA = F x 4 S

WB = F x S

WA= 4 WB

TRUE or FALSE: The acceleration of projectile is 0 m/s/s at the peak of the trajectory. Identify the evidence which supports your answer.

Answers

FALSE The vertical velocity of a projectile is 0 m/s at the peak of its trajectory; but the horizontal component of the velocity at the peak is whatever the value was when first launched

The vertical acceleration of the projectile is at 0 m/s while the horizontal acceleration of the projectile at the peak of the trajectory = initial acceleration  ( i.e. statement in the question is False )

Projectile motion follows a parabolic path with x and y components of its velocity and acceleration. also the acceleration of a projectile is subject only to the acceleration due to gravity unlike other kinds of motions.

In a parabolic motion an object ( projectile ) is thrown into the air and left to move through a parabolic path under the effect of acceleration due to gravity.

Hence we can conclude that the statement is false, because horizontal acceleration of the projectile at the peak of the trajectory = initial acceleration

Learn more : https://brainly.com/question/24658194

The _______ principle encourages us to resolve a set of stimuli, such as trees across a ridgeline, into smoothly flowing patterns
A.) depth perception.
B.) perception.
C.) similarity.
D.) continuity.

Answers

Answer:

The answer is continuity ( D )

Explanation:

PLZ MARK AS BRAINLIEST

please help me .finish this paper​

Answers

Solution-1:-

[tex]\boxed{\sf \dfrac{10\times 1000}{60\times 60}}[/tex]

Solution:-2

[tex]\boxed{\sf Sodium\:and\:Potassium}[/tex]

Solution:-3

[tex]\boxed{\sf 320m}[/tex]

Solution:-4

[tex]\boxed{\sf Rough\:tiles\:are\:used\:in\:bathroom}[/tex]

Solution:-5

[tex]\boxed{\sf Mg_3N_2}[/tex]

Solution:-6

[tex]\boxed{\sf Grapes\:and\:Rambutan}[/tex]

Solution:-7

[tex]\boxed{\sf {}^{}_{}N}[/tex]

Solution:-8

[tex]\boxed{\sf Galactuse}[/tex]

Solution:-9

[tex]\boxed{\sf Y-X}[/tex]

Solution:-10

[tex]\boxed{\sf Cell\:wall}[/tex]

I need the help
Please. I’m terrible at physics

Answers

Answer:

Explanation:

so opposite and equal , right?    forces are. soooo..

528+52= 580 N    is the force that is being exerted up on the scale

A spring of stiffness 11000 N/m is connected to a wall. The other end is attached to a block of mass 2 kg. A second spring of stiffness 12200 N/m extends between the first block and a second block of mass 2 kg. You are holding the second block with 300 N of force and everything is motionless.

Required:
How stretched is the spring connecting the first block to the wall?

Answers

Answer:

Explanation:

Let the stretch produced in first spring and second spring be x₁ and x₂ respectively. 300 N is the restoring force created in them .

k₁x₁ = k₂x₂ = 300 N .

k₁ = 11000 N/m

11000 x₁ = 300N

x₁ = 300 N / 11000

= 0.027 m

= 2.7 cm .

A barge is hauled along a straight-line section of canal by two horses harnessed to tow ropes and walking along the tow paths on
either side of the canal. Each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal. Find the sum of these
two forces on the barge.
answer in ___kN

Answers

Answer:

1.621 kN

Explanation:

Since each horse pulls with a force of 839 N at an angle of 15° with the centerline of the canal, the horizontal component of the force due to the first horse along the canal is F= 839cos15° N and its vertical component is F' = 839sin15° N(it is positive since it is perpendicular to the centerline of the canal and points upwards).

The horizontal component of the force due to the second horse along the canal is f = 839cos15° N and its vertical component is f' = -839sin15° N (it is negative since it is perpendicular to the centerline of the canal and points downwards).

So, the resultant horizontal component of force R = F + f = 839cos15° N + 839cos15° N = 2(839cos15°) N = 2(839 × 0.9659) = 2 × 810.412 = 1620.82 N

So, the resultant vertical component of force R' = F' + f' = 839sin15° N + (-839sin15° N) = 839sin15° N - 839sin15° N = 0 N

The magnitude of the resultant force which is the sum of the two forces is R" = √(R² + R'²)

= √(R² + 0²)  (since R' = 0)

= √R²

= R  

= 1620.82 N

= 1.62082 kN

≅ 1.621 kN

So, the sum of these  two forces on the barge is 1.621 kN

A 59.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 0.250 mins, what is the spring constant (in N/m) of the bungee cord, assuming it has negligible mass compared to that of the jumper

Answers

Answer:

The spring constant of the spring is 10.3 N/m.

Explanation:

Given that,

Mass of a bungee jumper, m = 59 kg

The period of oscillation, T = 0.25 min = 15 sec

We need to find the spring constant of the bungee cord. We know that the period of oscillation is given by :

[tex]T=2\pi\sqrt{\dfrac{m}{k}}[/tex]

Where

k is the spring constant

[tex]T^2=4\pi^2\times \dfrac{m}{k}\\\\k=4\pi^2\times \dfrac{m}{T^2}\\\\k=4\pi^2\times \dfrac{59}{(15)^2}\\\\k=10.3\ N/m[/tex]

So, the spring constant of the spring is 10.3 N/m.

A covalent bond is formet by of electrons..?​

Answers

Answer:

The covalent bond is formed by pairs of electrons that are shared between two atom

Explanation:

The covalent bond is formed by pairs of electrons that are shared between two atoms, in general the electrons must have opposite spins to have a lower energy state.

In this bond, the electrons are between the two atoms and are shared between them in such a way that there is a configuration of eight electrons in the orbit.

Physics question plz help

Answers

Answer:

B. [tex]30\,s\,< t \le 40\,s[/tex].

Explanation:

The vehicle is travelling eastwards when its velocity is positive and westwards when velocity is negative. According to the graph, [tex]v > 0[/tex] for [tex]30\,s\,< t \le 40\,s[/tex]. Hence, correct answer is B.

A baseball pitcher throws a fastball by spinning his arm at 27.7m/s. The ball has a mass of 0.700kg and experiences a net centripetal force of 625N. How long is the pitchers arm (the radius of the curve)?

Answers

In the historical sense, postmodern society is simply a society that occurs after the modern society. ... Many of the elements of a society like this are reactions to what the modern society stood for: industrialism, rapid urban expansion, and rejection of many past principles.

A person pushes horizontally on a heavy box and slides it across the level floor at constant velocity. The person pushes with a 60.0 N force for the first 16.4 m at which time he begins to tire. The force he exerts then starts to decrease linearly from 60.0 N to 0.00 N across the remaining 6.88 m. How much total work did the person do on the box

Answers

Over the first 16.4 m, the person performs

W = (60.0 N) (16.4 m) = 984 J

of work.

Over the remaining 6.88 m, they perform a varying amount of work according to

F(x) ≈ 60.0 N + (-8.72 N/m) x

where x is in meters. (-8.72 is the slope of the line segment connecting the points (0, 60.0) and (6.88, 0).) The work done over this interval can be obtained by integrating F(x) over the interval [0, 6.88 m] :

W = ∫₀⁶˙⁸⁸ F(x) dx ≈ 206.4 J

(Alternatively, you can plot F(x) and see that it's a triangle with base 6.88 m and height 60.0 N, so the work done is the same, 1/2 (6.88 m) (60.0 N) = 206.4 J.)

So the total work performed by the person on the box is

984 J + 206.4 J = 1190.4 J ≈ 1190 J

A hungry monkey is sitting at the top of a tree 69 m above ground level. A person standing on the ground wants to feed the monkey. He uses a tee-shirt cannon to launch bananas at the monkey. If the person knows that the monkey is going to drop from the tree at the same instant that the person launches the bananas, how should the person aim the banana cannon

Answers

Answer:

Well if you want to be sure you should just throw it to the ground so then when he lands he can catch it.

If the cannon throws the banana with the same force the monkey falls

(m.g=Fz <=> m.9,81N/kg=...N).

Then the throw will slow down because of the gravitational pull.

Because the banana cannon is selfmade you can choose what mass the bananas in question have, so let that be the same as the monkeys.

The monkey falls with the speed of 9,81m.s => so it takes the monkey 7,1s to land.

If the cannon can shoot the banana at the same speed the monkey falls then they would cross in the middle.

So to do so you need to throw the bananas with a speed of at least 9,81m.s

Soo ... throw them with a force of that is greater then the gravitational pull and things will work out.

I'm sorry I don't know why I wrote all of this irrelevant information it's 2:21 right now and I'm tired.

kind regards

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