How much force is needed to accelerate a 65 kg rider AND her 215 kg motor scooter 8 m/s?? (treat
the masses as like terms)

Answer:

This idea helps students explain why more rain forms over West Ferris than East Ferris. ... Therefore, when students explain that water vapor condenses higher in the atmosphere, they are actually explaining that water vapor condenses high in the troposphere, which is relatively low in the atmosphere.

Explanation:

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An arrow pointing from the bottom of the ramp to the top, I assume it would be friction.

Static friction (magnitude Fs) keeps the car on the road, and is the only force acting on it parallel to the road. By Newton's second law,

Fs = m a = W a / g

(a = centripetal acceleration, m = mass, g = acceleration due to gravity)

We have

a = v ² / R

(v = tangential speed, R = radius of the curve)

so that

Fs = W v ² / (g R)

Solving for v gives

v = √(Fs g R / W)

Perpendicular to the road, the car is in equilibrium, so Newton's second law gives

N - W = 0

(N = normal force, W = weight)

so that

N = W

We're given a coefficient of static friction µ = 0.4, so

Fs = µ N = 0.4 W

Substitute this into the equation for v. The factors of W cancel, so we get

v = √((0.4 W) g R / W) = √(0.4 g R) = √(0.4 (9.80 m/s²) (105 m)) ≈ 20.3 m/s

Answer:

form 1 question??????????

Non-Malleable and Ductile: Non-metals are very brittle, and cannot be rolled into wires or pounded into sheets. Conduction: They are poor conductors of heat and electricity. Luster: These have no metallic luster and do not reflect light.

Group 15, the nitrogen family, contains two nonmetals: nitrogen and phosphorus. These non-metals usually gain or share three electrons when reacting with atoms of other elements. Group 16, the oxygen family, contains three nonmetals: oxygen, sulfur, and selenium.

Elements: Nitrogen; Oxygen; Phosphorus; Selenium...

Answer:

A. 72000 C

B. 1100 W

C. 26.4 cents.

Explanation:

From the question given above, the following data were obtained:

Current (I) = 10 A

Voltage (V) = 110 V

Time (t) = 2 h

A. Determination of the charge.

We'll begin by converting 2 h to seconds. This can be obtained as follow:

1 h = 3600 s

Therefore,

2 h = 2 h × 3600 s / 1 h

2 h = 7200 s

Thus, 2 h is equivalent to 7200 s.

Finally, we shall determine the charge. This can be obtained as follow:

Current (I) = 10 A

Time (t) = 7200 s

Charge (Q) =?

Q = It

Q = 10 × 7200

Q = 72000 C

B. Determination of the power.

Current (I) = 10 A

Voltage (V) = 110 V

Power (P) =?

P = IV

P = 10 × 110

P = 1100 W

C. Determination of the cost of operation.

We'll begin by converting 1100 W to KW. This can be obtained as follow:

1000 W = 1 KW

Therefore,

1100 W = 1100 W × 1 KW / 1000 W

1100 W = 1.1 KW

Thus, 1100 W is equivalent to 1.1 KW

Next, we shall determine the energy consumption of the range. This can be obtained as follow:

Power (P) = 1.1 KW

Time (t) = 2 h

Energy (E) =?

E = Pt

E = 1.1 × 2

E = 2.2 KWh

Finally, we shall determine the cost of operation. This can be obtained as follow:

1 KWh cost 12 cents.

Therefore, 2.2 KWh will cost = 2.2 × 12

= 26.4 cents.

Thus, the cost of operating the range for 2 h is 26.4 cents.

Answer:

1/9

Explanation:

Let A denote the bigger piece and let B denote the smaller piece.

We are told that one with three times the mass of the other.

Therefore, we have;

M_a = 3M_b

Firecracker is placed in the block and it explodes and thus, momentum is conserved.

Thus;

V_ai = V_bi = 0

Where V_ai is initial velocity of piece A and V_bi is initial velocity of piece B.

Since initial momentum equals final momentum, we have;

P_i = P_f

Thus;

0 = (M_a × V_af) + (M_b × V_bf)

Since M_a = 3M_b, we have;

(3M_b × V_af) + (M_b × Vbf) = 0

Making V_af the subject, we have;

V_af = -⅓V_bf

The kinetic energy gained by each block during the explosion will later be lost due to the negative work done by friction. Thus;

W_f = -½M_b•(v_bf)²

Now, let's express the work is in terms of the force and the distance.

Thus;

W_f = F_f × Δx × cos 180°

Frictional force is also expressed as μmg

Thus;

W_f = -μM_b × g × Δx

Earlier, we saw that;

W_f = -½M_b•(v_bf)²

Thus;

-½M_b•(v_bf)²= -μM_b × g × Δx

Δx = (v_bf)²/2μg

Let the distance travelled by block A be Δx_a and that travelled by B be Δx_b

Thus;

Δx_a/Δx_b = ((v_ba)²/2μg)/((v_bf)²/2μg)

Δx_a/Δx_b = ((v_af)²/((v_bf)²)

Δx_a/Δx_b = (-⅓V_bf)²/(V_bf)²

Δx_a/Δx_b = 1/9

Answer:

ummm hehe this is my time to shine

Explanation:

  MERICIA!!!!!!!!!!!!!!!!!!!!!!!

Answer:

Explanation:

equation of wave is given by the following equation

y = (2.6 cm) sin[1.8 - (5.8 s-1)t].

Comparing it with standard form of wave

y = A sin ( ωt - kx )

we get

ω = 5.8

2πn = 5.8

n = .92 per second

kx = 1.8

k x 6 = 1.8

k = 0.3

[tex]\frac{2\pi}{\lambda}[/tex] = 0.3

λ = 20.9 cm

Answer:

15km/h

Explanation:

→ Speed = Distance ÷ Time

30 ÷ 2 = 15km/h

Answer:

9.3m/s

Explanation:

Based on the law of conservation of momentum

Sum of momentum before collision = sum of momentum after collision

m1u1 +m2u2 = m1v1+m2v2

m1 = 8kg

u1 = 15.4m/s

m2 = 10kg

u2 = 0m/s(at rest)

v1 = 3.9m/s

Required

v2.

Substitute

8(15.4)+10(0) = 8(3.9)+10v2

123.2=31.2+10v2

123.2-31.2 = 10v2

92 = 10v2

v2 = 92/10

v2 = 9.2m/s

Hence the velocity of the 10.0 kg object after the collision is 9.2m/s

Answer:

force F = 1.66 × [tex]10^{-13}[/tex] N

Explanation:

given data

proton and an electron = 865 nm

solution

we get here force that is express as

force F = k q1 q2 ÷ r²      ......................1

put here value and we get

force F = 9 × [tex]10^{9}[/tex] × [tex]\frac{1.6\times (10^{-19})^{2}}{865 \times (10^{-9})^{2}}[/tex]    

force F = 1.66 × [tex]10^{-13}[/tex] N

Answer:

average speed is 1 m/s and average velocity is 0 m/s.

Explanation:

Given that :

Length of round trip = 50 m

Time taken = 100 seconds

The average speed :

Total distance / total time taken

Length of complete round trip :

(50 + 50) m, total. Distance = 100 m

100 / 100 = 1m/s

The average velocity :

Total Displacement / total time taken

Total Displacement of round trip = end point - start point = 0

0 / 100 = 0

Average speed is 1 m/s and average velocity is 0 m/s.

The average speed is defined as the ratio of distance to time. Speed is a scalar quantity hence it does not take direction into account while velocity is a vector quantity hence it takes direction into account.

The speed is obtained from;

Speed = Distance/time = 2(50 m)/100 s = 1 m/s.

The velocity is 0 m/s since it is complete round-trip lap.

Learn more about speed: https://brainly.com/question/7359669

Answer:

It's true I took the test on Edge.

Explanation:

Answer:

True

Explanation:

Got it right on edg

Answer:

58.8J

Explanation:

Given parameters;

Mass of ball  = 4kg

Height above the floor  = 1.5m

g  = 9.8n/kg

Unknown:

Potential energy  = ?

Solution:

The potential energy of a body is the energy due to the position of the body.

It is mathematically expressed as:

  Potential energy = mass x acceleration due to gravity x height

  Potential energy  = 4 x 9.8 x 1.5  = 58.8J

Answer:

it will option option A hope it helps

Answer:

A

Explanation:

Ke = 1/2 MV^2

Answer:

T = 10010 N

Explanation:

To solve this problem we must use the translational equilibrium relation, let's set a reference frame

X axis

       Fₓ-Fₓ = 0

       Fₓ = Fₓ

whereby the horizontal components of the tension in the cable cancel

Y Axis  

        [tex]F_{y} + F_{y} - W =0[/tex]

        2[tex]F_{y}[/tex] = W

let's use trigonometry to find the angles

        tan θ = y / x

        θ = tan⁻¹ (0.30 / 0.50 L)

        θ = tan⁻¹ (0.30 / 0.50 15)

        θ = 2.29º

the components of stress are

         F_{y} = T sin θ

we substitute

       2 T sin θ = W

       T = W / 2sin θ

        T = [tex]\frac{ 800}{ 2sin 2.29}[/tex]

        T = 10010 N

Distance = (speed) x (time)

Car A: Distance = (8 m/s) x (43 s)  =  344 meters

Car B: Distance = (7 m/s) x (50 s)  =  350 meters

350 meters is a longer distance than 344 meters.

Car-B traveled a longer distance than Car-A did.

Answer:

[tex]\boxed {\boxed {\sf Car \ B : 350 \ meters }}[/tex]

Explanation:

Distance is equal to the product of speed and time.

[tex]d=s*t[/tex]

1. Car A

Car A has a speed of 8 meters per second and travels for 43 seconds.

[tex]s= 8 \ m/s \\t= 43 \ s[/tex]

Substitute the values into the formula.

[tex]d= 8 \ m/s *43 \ s[/tex]

Multiply and note that the seconds will cancel out.

[tex]d= 8 \ m*43= 344 \ m[/tex]

2. Car B

Car B has a speed of 7 meters per second and travels for 50 seconds.

[tex]s= 7 \ m/s \\t= 50 \ s[/tex]

Substitute the values in and multiply.

[tex]d= 7 \ m/s * 50 \ s[/tex]

[tex]d= 7 \ m * 50 = 350 \ m[/tex]

350 meters is a longer distance than 344 meters, so Car B traveled the longer distance.

Without friction, NO.

The speed at D depends only on the difference in height between A and D. Whatever happens between them doesn't matter.

The speed of the coaster car at point D will be affected if  the height of point C is changed.

Potencial Energy:

It is the enrgy in a body due to the position of differnt part of the object or system.

As we increase the the hight of the car the potetial enrgy increase, the gravitational acceleration on car will be more due to the high of the point C.

Therefore, the speed of the coaster car at point D will be affected if  the height of point C is changed.

To know more about  speed of the coaster car,

https://brainly.com/question/9178285

Answer:

B

Explanation:

Answer:

The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.

Explanation:

Let suppose that RC car-slingshot-steelpellet is a conservative system, that is, that non-conservative forces (i.e. friction, air viscosity) can be neglected. The velocity of the steel pellet can be found by means of the Principle of Energy Conservation and under the consideration that change in gravitational potential energy is negligible and that the RC car travels at constant velocity:

[tex]\frac{1}{2}\cdot (m_{C}+m_{P})\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{C}\cdot v_{o}^{2} + \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]

[tex]\frac{1}{2}\cdot m_{P}\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{P}\cdot v^{2}[/tex]

[tex]m_{P}\cdot v_{o}^{2} + k\cdot x^{2} = m_{P}\cdot v^{2}[/tex]

[tex]v^{2} = v_{o}^{2} + \frac{k}{m_{P}}\cdot x^{2}[/tex]

[tex]v = \sqrt{v_{o}^{2}+\frac{k}{m_{P}}\cdot x^{2} }[/tex] (1)

Where:

[tex]v_{o}[/tex] - Initial velocity of the steel pellet, measured in meters per second.

[tex]v[/tex] - Final velocity of the steel pellet, measured in meters per second.

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]m_{P}[/tex] - Mass of the steel pellet, measured in kilograms.

[tex]m_{C}[/tex] - Mass of the RC car, measured in kilograms.

[tex]x[/tex] - Initial deformation of the spring, measured in meters.

If we know that [tex]v_{o} = 5.6\,\frac{m}{s}[/tex], [tex]k = 85\,\frac{N}{m}[/tex], [tex]m_{P} = 0.025\,kg[/tex] and [tex]x = 0.035\,m[/tex], then the velocity of the steel pellet relative to the ground when it leaves the sling shot is:

[tex]v = \sqrt{\left(5.6\,\frac{m}{s} \right)^{2}+\frac{\left(85\,\frac{N}{m} \right)\cdot (0.035\,m)^{2}}{0.025\,kg} }[/tex]

[tex]v \approx 5.960\,\frac{m}{s}[/tex]

The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.

Answer:

C. Is traveling on a curve in the road

    Hope this helps :3

The pick up truck has a changing velocity because, it is travelling on a curve in the road. A change in direction results in its change in velocity because, velocity is a vector quantity.

Velocity is a physical quantity that measures the distance covered by an object per unit time. It is a vector quantity, thus having magnitude as well direction.

The rate of change in velocity is called acceleration of the object. Like velocity, acceleration also is a vector quantity. Thus, a change in magnitude or direction or change in both for velocity make the object to accelerate.

Here, all the three vehicles  are travelling with the same velocity. But, the truck is moving to a curve on the road. The curvature in the path will make a change in its velocity.

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