Question: A) A car is driving too fast on a flat non-banked curve. The car cannot stay on the road. What path will the car take as it leaves the road? See Image. (a) (b) B) The driver wants to go faster around the same curve mentioned above and not run off the road. He adds sandbags to his car to make it weigh more so that the friction force between the tires and the road will be increased. Will this work? Explain your answer.​

Answer:

0.5 m.

Explanation:

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

Force (F) applied = 5 N

Work done (W) = 2.5 J

Distance (s) travalled by the book =?

We can obtain the distance travelled by the book by applying the formula of work done. This is illustrated below:

Work done (W) = Force (F) × Distance (s)

W = F × s

Force (F) applied = 5 N

Work done (W) = 2.5 J

Distance (s) travalled by the book =?

2.5 = 5 × s

Divide both side by 5

s = 2.5/5

s = 0.5 m

From the above calculation, the book will travel 0.5 m when a force of 5 N is applied.

Answer:

8

Explanation:

Answer:

Chemical Properties of Oxygen

At standard temperature and pressure (STP), two atoms of the element bind to form dioxygen, a colorless, odorless, tasteless diatomic gas with the formula O2. Oxygen is a member of the chalcogen group on the periodic table and is a highly reactive nonmetallic element.

Explanation:

Answer:

Explanation:

The force needed to keep the elephant from slowing down is expressed as shown according to Newtons second law of motion.

Force = mass * acceleration due to gravity

Given

Mass of elephant = 500kg

acceleration due to gravity = 9.81m/s²

Force = 500*9.81

Force = 4905N

Hence the force needed to keep the elephant from slowing down is 4905N

Answer:

The value is [tex]U = 0.06 \ J [/tex]

Explanation:

From the question we are told that

The value of charge on each three point charge is

[tex]q_1 = q_2 = q_3 =q= 1.05 \mu C = 1.05 *10^{-6} \ C[/tex]

The length of the sides of the equilateral triangle is [tex]r = 0.500 \ [/tex]

Generally the total potential energy is mathematically represented as

[tex]U = k * [ \frac{q_1 * q_2}{r} + \frac{q_2 * q_3}{r} + \frac{q_3 * q_1}{r} ][/tex]

=> [tex]U = k * 3 * \frac{q^2}{r} [/tex]

Here k is coulomb constant with value [tex]k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]

=>    [tex]U  = 9*10^9 * 3 * \frac{(1.05 *10^{-6})^2}{0.5 }  [/tex]

[tex]U  = 0.06 \  J  [/tex]

Answer:

A) should be able to complete daily tasks without being physically taxed

Explanation:

B) you dont need to run that fast to have good physical fitness

C) you dont need to workout four times a week

D) you should always eat healthy as diet contributes to health

In order to have good physical fitness, you should be able to complete daily tasks without being physically taxed.

Physical fitness is the ability to be fit to do work without getting tired easily.

Exercise and workouts are necessary to build and improve physical fitness.

Exercises required include jogging and running.

Therefore, in order to have good physical fitness, you should be able to complete daily tasks without being physically taxed.

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Answer:

a

         [tex]q_1 = -1.389 *10^{-5} \  C [/tex]   , [tex]q_2 = -1.389 *10^{-5} \  C [/tex]

OR

           [tex]q_1 = 1.389 *10^{-5} \  C [/tex]   , [tex]q_2 = 1.389 *10^{-5} \  C [/tex]    

b

[tex]q_1 = 1.389 *10^{-5} \ C [/tex] and [tex]q_2 = 1.389 *10^{-5} \ C [/tex]

Explanation:

Generally the force exerted on the string is mathematically represented as

[tex]F = k * e[/tex]

substituting values 180 N/m for k and 0.024 m for e

[tex]F = 180 * 0.024[/tex]

[tex]F = 4.32 \ N[/tex]

This force can also equivalent to the electrostatic force between the charges i.e

[tex]F = k * \frac{q^2}{ r^2}[/tex]

substituting [tex]9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex] for k and ( 0.34 + 0.024 = 0.364 m) for r we have

[tex] 4.32= 9*10^{9} * \frac{q^2}{ (0.364)^2}[/tex]

[tex]q = \sqrt{1.929 *10^{-10}}[/tex]

[tex]q = 1.389 *10^{-5} \ C [/tex]

Given the spring was stretched it means that the force between the charges is a repulsive for which tell us that both charge are of the same sign thus the possible  algebraic signs  of the charges are

         [tex]q_1 = -1.389 *10^{-5} \  C [/tex]   , [tex]q_2 = -1.389 *10^{-5} \  C [/tex]

OR

[tex]q_1 = 1.389 *10^{-5} \ C [/tex] , [tex]q_2 = 1.389 *10^{-5} \ C [/tex]

Answer:

the y axis, every moment is subjected to a vertical acceleration directed towards the center of the Earth, which is called the acceleration of gravity

Explanation:

In projectile launching, the movement is separated into two movements, one on the x-axis and the other on the y-axis, related through time.

In horizontal movement, the speed is constant, because there is no acceleration in this axis, the effect of air friction is almost always eliminated.

In the other movement on the y axis, every moment is subjected to a vertical acceleration directed towards the center of the Earth, which is called the acceleration of gravity.

This value of the eceleration of gravity is constant for small distances

compared to the radius of the Earth, for higher altitudes an expansion in beings of the distance is used, giving a linear dependence.

Answer:

 a = √ (a_t² + a_c²)

a_t = dv / dt ,    a_c = v² / r  

Explanation:

In a two-dimensional movement, the acceleration can have two components, one in each axis of the movement, so the acceleration can be written as the components of the acceleration in each axis.

            a = aₓ i ^ + a_y j ^

Another very common way of expressing acceleration is by creating a reference system with a parallel axis and a perpendicular axis. The axis called parallel is in the radial direction and the perpendicular axis is perpendicular to the movement, therefore the acceleration remains

         a = √ (a_t² + a_c²)

where the tangential acceleration is

           a_t = dv / dt

the centripetal acceleration is

          a_c = v² / r

Answer:

Electric current in amperes = 1.1808 A

Explanation:

Given:

Intercept protons rate = 1800 protons per second

Area = 41 × 10⁷ km²

Find:

Electric current in amperes

Computation:

Current density = Intercept protons rate × 1.6 × 10⁻¹⁹

Current density = 1800 × 1.6 × 10⁻¹⁹

Current density = 2.88 × 10⁻¹⁶

1 km² = 10⁶m²

So,

Electric current in amperes =  2.88 × 10⁻¹⁶ × 41 × 10⁷ × 10⁶

Electric current in amperes = 1.1808 A

Answer:

(a) The velocity of the car before the brakes were applied is 77.46 ft/s

(b) The time required for the car to stop is 7.8 s

Explanation:

Given;

acceleration of the car, a = 10 ft/s²

distance traveled by the car, d = 300 ft

(a) the velocity of the car before the brakes were applied is given;

v² = u² + 2ad

v² = 0 + 2(10 x 300)

v² = 6000

v = √6000

v = 77.46 ft/s

(b) the time required for the car to stop

d = ut + ¹/₂at²

d = 0 +  ¹/₂at²

d = ¹/₂at²

t² = 2d / a

t = √ ( 2d / a)

t = √ ( 2 x 300 / 10)

t = 7.8 s

Therefore, the time required for the car to stop is 7.8 s

The speed of the car immediately before the brakes were applied was 77.46 ft/s and it took 7.75 s for the car to stop.

Using the formula:

v² = u² + 2as

Where S is the distance = 300 ft, u is the initial velocity, a is the acceleration = -10 ft/s², v is the final velocity = 0 ft/s (stops)

0² = u² + 2(-10)(300)

0 = u² - 6000

u² = 6000

u = 77.46 ft/s

b)

v = u + at

0 = 77.46 - 10t

t = 7.75 s

The speed of the car immediately before the brakes were applied was 77.46 ft/s and it took 7.75 s for the car to stop.

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Answer:

The distance traveled by the second block is 0.197 m

Explanation:

Given;

mass of the small block, m₁ = 0.2 kg

distance traveled by the block, d₁ =0.394 m

time of travel, t = 2 s

mass of the second block, m₂ = 0.4 kg

distance traveled by the second block, d₂ = ?

The work done per unit time on the inclined plane is given by;

[tex]\frac{f_1d_1}{t} = \frac{f_1d_1}{t}\\\\f_1d_1 = f_2d_2\\\\m_1gd_1 = m_2gd_2\\\\m_1d_1 = m_2d_2\\\\d_2 = \frac{m_1d_1}{m_2} \\\\d_2 = \frac{0.2*0.394}{0.4}\\\\d_2 = 0.197 \ m[/tex]

Therefore, the distance traveled by the second block is 0.197 m

Answer:

Acceleration /Speed

Explanation:

An objects Velocity can be determined by acceleration,

Please pay attention in your middle school class, speed and velocity quiz.

Answer:

[tex]x=8m[/tex]

Explanation:

Hello,

In this case, the most suitable kinematic equation for the calculation of the squirrel's displacement is:

[tex]x=v_ot+\frac{1}{2}at^2[/tex]

Whereas the initial velocity is zero as it starts from rest, thus, we obtain:

[tex]x=\frac{1}{2}*1.0\frac{m}{s^2}*(2.0s)^2\\ \\x=8m[/tex]

Best regards.

Explanation:

They speed up, slow down, or change direction.

Answer:In order for object to be accelerating, the net force (sum of all forces) acting on the object has to be non-zero. According to Newton's Second Law of Motion, Force = Mass x Acceleration. ... If the object is sitting motionless on the ground, then the net force (sum of all forces) acting on it is zero.

Explanation:

Answer:

F = ma - with this you will get 2.6 m/s^2

Explanation:

Use this formula to get the acceleration...

where F is force, M is mass and A is acceleration

by using this we get...

65 = 25 * a

so, a = 65/25

Therefore, the acceleration is 2.6 m/s^2

Hope that helped :)

The acceleration of a lawn mower will be 2.6 m/s². It is the ratio of force and the mass.

The rate of change of velocity with respect to time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.

The given data in the problem is;

The force act on a lawn mower,[tex]\rm F = 65 \ N[/tex]

Mass of  lawn mower,[tex]\rm m = 25 kg[/tex]

The acceleration of a lawn mower is,[tex]\rm a = ?[/tex]

Acceleration, is found as the ratio of force and the mass.

[tex]\rm F= ma \\\\ a = \frac{F}{m} \\\\ a= \frac{65}{25} \\\\ a = 2.6 \ m/s^2[/tex]

Hence, the acceleration of a lawn mower will be 2.6 m/s².

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Answer: 34 km, 21 km  61 degrees north of east

Explanation: distance = 13 + 3 + 18 = 34

   displacement = 13 - 3 = 10

               10^2 + 18^2 = 424

              find the square root of 424 ( 20.5 rounded to 21 )

The total distance walked is 34 km and the resulting displacement is 20.6 km.

a) The total distance is gotten by summing up all the distance.

Total distance = Distance moved east + distance moved north + distance moved west

Total distance = 13 km + 18 km + 3 km = 34 km

b) The displacement is the distance from the beginning point to end point.

Displacement² = 18² + (13 - 3)² = 18² + 10²

Displacement² = 424

Displacement = 20.6 km

Therefore the total distance walked is 34 km and the resulting displacement is 20.6 km.

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Answer:

F = 0.012 N

Explanation:

Given that,

Mass of the object, m = 6 g

Acceleration, a = 2 m/s²

1 kg = 1000 grams

6 g = 0.006 kg

Force, F = ma

So,

[tex]F=0.006\ kg\times 2\ m/s^2\\F=0.012\ N[/tex]

So, the force is 0.012 N.

Answer:

(C) no change in the temperature of the gas.

Explanation:

isothermal process of a gas, can result to compression of the gas or expansion of the gas at a constant temperature (ΔT = 0).

For isothermal gas expansion, work is done in reducing the pressure of the gas by increasing its volume at a constant temperature. The change in the internal energy of this process is zero (ΔU = 0).

For isothermal gas compression, work is done in decreasing the volume of the gas by increasing its pressure at a constant temperature. The change in the internal energy of this process is also zero (ΔU = 0).

Therefore, when a gas undergoes an isothermal process, there is no change in the temperature of the gas.

Answer:

Alright the other day.

Answer:

learning new activities like riding a bike and playing sports. and also having the ability to collect information and remember them (memory). <3

Answer:

The answer is "[tex]\bold{10.6 \ \ \frac{m}{s^2}}[/tex]"

Explanation:

The  Formula of acceleration:

[tex]\bold{a= \frac{v-v_0}{t}}[/tex]

Given value:

[tex]t= 50\\v= 856\\v_0= 324[/tex]

Calculating the acceleration:

[tex]\to a = \frac{(856 - 324)}{50}\\\\[/tex]

       [tex]= \frac{(532)}{50}\\\\= \frac{(106.4)}{10}\\\\= 10.64 \ \ \frac{ m}{s^2}[/tex]

i cant understand your question define briefly

Answer:

The mechanical efficiency of the pump is 91.8 %

Explanation:

Given;

input power, p = 44 kw

density of oil, ρ = 860 kg/m³

motor efficiency, η = 90 %

inlet diameter, d₁ = 8 cm

outlet diameter, d₂ = 12 cm

volume flow rate, V = 0.1 m³/s

pressure rise, P = 500kPa

output power  = motor efficiency x input power

output power  = 0.9 x 44 = 39.6 kW

Thus, the mechanical input power = 39.6 kW

The mechanical output power  is given by change in mechanical energy;

[tex]E = mgh + \frac{m}{2} (v_2^2 - v_1^2) \\\\E = \rho V g h + \frac{\rho V}{2} [(\frac{V_2}{\pi r_2^2} )^2 - (\frac{V_1}{\pi r_1^2})^2]\\\\E = PV + \frac{\rho V^3}{2\pi^2} [\frac{1}{ r_2^4} - \frac{1}{ r_1^4}]\\\\E = (500 *10^3)(0.1) + \frac{(860)(0.1)^3}{2\pi^2} [\frac{1}{ 0.06^4} - \frac{1}{ 0.04^4}]\\\\E = 50000 -13653.51\\\\E = 36346.48 \ W\\\\E = 36.347 \ kW[/tex]

The mechanical efficiency is given by

η = mechanical output power /  mechanical input power

η = 36.347 / 39.6

η = 0.918

η = 91.8 %

Therefore, the mechanical efficiency of the pump is 91.8 %

diagonal line means the velocity is constant.

Answer:

the mean

Explanation:

The mean (informally, the “average“) is found by adding all of the numbers together and dividing by the number of items in the set: 10 + 10 + 20 + 40 + 70 / 5 = 30. The median is found by ordering the set from lowest to highest and finding the exact middle. The median is just the middle number: 20.

Answer:

W= 0.319992 J  and distance is 3.75 cm

Explanation:

The energy needed to stretch the spring from 30 cm to 45 cm = 3 J

Now we required to find the requirement of energy when the spring is stretched from 35 cm to 37 cm.

So first find the work done to stretch the spring from 35 cm to 45 cm.

Work done, w = (1/2) kx^2

3 = (1/2)k(0.45 – 0.30)^2

k = 266.66 N/m

now, x1 = 0.35 – 0.30 = 0.05m

x2 = 0.37 – 0.30 = 0.07m

Now the amount of work done to stretch from 35cm to 37.

w = (1/2) k (x2^2 – x1^2)

w = (1/2) (266.66) (0.07^2 – 0.05^2 )

w= 0.319992 J

(b). Given F = 10 N

F = kx

x = F / k

x = 10 / 266.66

x = 0.0375m

x = 3.75 cm

Thus, distance is 3.75cm

Answer:

0.3

Explanation: when you average out the number sthat is the answer you get

[tex]\left(95\dfrac{\rm gal}{\rm min}\right)\left(\dfrac{231}{1.00}\dfrac{\mathrm{in}^3}{\rm gal}\right)\left(\dfrac{1.0}{12}\dfrac{\rm ft}{\rm in}\right)^3\left(\dfrac{1.0}{60}\dfrac{\rm min}{\rm s}\right)\approx0.211661\dfrac{\mathrm{ft}^3}{\rm s}[/tex]

Multiplying 95 gal by 231 converts it to in^3. Dividing by 12^3 converts in^3 to ft^3. Dividing by 60 converts min to s.