An automobile has a mass of 2100 kg and a velocity of +17 m/s. It makes a rear-end collision with a stationary car whose mass is 1900 kg. The cars lock bumpers and skid off together with the wheels locked. (a) What is the velocity of the two cars just after the collision? (b) Find the impulse (magnitude and direction) that acts on the skidding cars from just after the collision until they come to a halt. (c) Review: If the coefficient of kinetic friction between the wheels of the cars and the pavement is μk = 0.68, determine how far the cars skid before coming to rest.

To find the new orbital period of a satellite at 1500 miles above Earth with a velocity of 38,000 mph, we use a proportional relationship comparing it to a known orbit. We set up the ratio using the direct variation with the orbit's radius and inverse variation with the orbital velocity, then solve for the unknown orbital period.

The duration of a satellite's orbit around Earth can be described as varying directly with the orbit's radius and inversely with the orbital velocity. To calculate the orbital period for a new satellite trajectory, we need to compare it with a known situation using a proportional relationship. Given that a satellite orbits 660 miles (or 1062 kilometers) above Earth in 13 hours at 32,000 mph, we can set up a ratio to determine the orbital period at 1500 miles (or 2414 kilometers) above Earth with a velocity of 38,000 mph.

The ratio for the original orbit is T1 / (r1 / v1) = T2 / (r2 / v2), where T is time, r is radius (distance from Earth's center to the satellite), and v is velocity. Using the provided values, we get:

13 / ((3963 + 660) / 32000) = T2 / ((3963 + 1500) / 38000)

Solving for T2 gives us the new orbital period. Considering that we should convert the altitude to the same units and use Earth's radius in miles (3963 miles), the final calculation will provide us with the new orbital period in hours.

Answer:

39.8 m ≈ 40 m

Explanation:

power (P) = 50 W

sound intensity level ([tex]p[/tex]) = 94 dB

the distance (r) can be gotten from the equation I = [tex]\frac{power}{4nr^{2} }[/tex] (take not that π is shown as [tex]n[/tex])

making r the subject of the formula we have r = [tex]\sqrt{\frac{power}{4nI} }[/tex]   (take not that π is shown as [tex]n[/tex])

But to apply this equation we need to get the value of the intensity (I)

now that we have the value of the intensity (I)  we can substitute it into the formula for the distance (r)

distance (r) = [tex]\sqrt{\frac{power}{4nI} }[/tex]

r = [tex]\sqrt{\frac{50}{4x3.142x0.00251} }[/tex] = 39.8 m ≈ 40 m

Answer:

Because its dangerous.

Explanation:

During lighting strikes, there is discharge or energy transfer(electrons) from the clouds to the earth, these electrons flow through the path with least resistance between the cloud and the earth. Also the electric field around the tip of the leaves are strong, which makes trees a great target.

we as humans have lower resistance than trees, that is to say, the lighting may leave the tree and flow trough the body to the earth.

The tree and the ground around it are then raised to a high potential relative to the ground some distance away.

If you stand with your legs far apart, one leg on a higher-potential part of the ground than the other, or if you lie down with a potential difference between your head and your feet, you may find yourself a conducting path.

If it is also raining, the electricity may transfer down the wet tree to the wet ground and shock anyone standing near the tree.

During a thunderstorm, it is advised not to stand under a tree to avoid being struck by lightning, not to stand with legs far apart to reduce the electric potential difference, and not to lie down due to increased risk. Staying inside a car provides safety as it acts as a Faraday cage.

Standing under a tree during an electrical storm is dangerous since lightning tends to strike the tallest object in an area, which could be the tree you're under, potentially causing serious injury or death. As for not standing with legs apart, this is because, in the event of a ground strike, electricity can travel through the ground. If your legs are far apart, there could be a significant electric potential difference between them, which can result in a stronger current passing through your body, leading to severe harm. Lying down increases your contact with the ground, and consequently, the risk of current flowing through your body from a ground strike is greater.

During thunderstorms, your car acts as a Faraday cage, which shields you from electric fields if a lightning strike occurs nearby. It's safest to remain inside the car with windows closed. However, it's critical to refrain from touching metal parts inside the car as lightning can transfer its charge through the car's metal frame.

Answer:

Ep= 3.8 10⁵ N/C

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

E = k*q/d²

E: Electric field in N/C

q: charge in Newtons (N)

k: electric constant in N*m²/C²

d: distance from charge q to point P in meters (m)

Equivalence

1nC= 10⁻⁹C

1cm= 10⁻²m

Data

k= 9*10⁹ N*m²/C²

q₁ =+7.5 nC = +7.5*10⁻⁹C  

q₂ =  -2.0 nC = -2.0*10⁻⁹C

d₁ =d₂ = 1.5cm = 1.5 *10⁻²m  = 0.015 m

Calculation of the electric fieldsat the midpoint (P) between the two charges

Look at the attached graphic:

E₁: Electric Field at point ;Due to charge q₁. As the charge q₁ is positive negative (q₁+), the field leaves the charge .

E₂: Electric Field at point : Due to charge q₂. As the charge q₂ is negative (q₂-) ,the field enters the charge

E₁ = k*q₁/d₁² = 9*10⁹ *7.5  *10⁻⁹/ ( 0.015 )² = 3*10⁵ N/C

E₂ = k*q₂/d₂²= 9*10⁹ *2*10⁻⁹/( 0.015 )² = 0.8*10⁵ N/C

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Ep= E₁ + E₂  

Ep= 3*10⁵ N/C +  0.8*10⁵ N/C

Ep= 3.8 10⁵ N/C

The electric field strength at the midpoint between a +7.5 nC point charge and a -2.0 nC point charge that are 3.0 cm apart is 1.56 *10^6 N/C (Newtons per Coulomb) away from the positive charge.

The question is asking for the electric field at the midpoint between two point charges. The formula for the electric field created by a point charge is given by E=kQ/r^2, where E is the electric field, k is Coulomb's constant (approximately 9.0 * 10^9 N*m^2/C^2), Q is the charge, and r is the distance from the charge. In this scenario we have two charges, so we calculate the electric field created by each charge and sum the results.

Applying this to both point charges and summing the electric fields we obtain: E_total = |E1| + |E2| = k*|Q1|/d^2 + k*|Q2|/d^2 = ((9.0*10^9 N*m^2/C^2) * 7.5 *10^-9 C/0.015m^2) + ((9.0*10^9 N*m^2/C^2) * 2 *10^-9 C/ 0.015m^2) = 1.8 *10^6 N/C and -0.24 * 10^6 N/C respectively. The total electric field strength at the midpoint is the sum which equals 1.56 *10^6 N/C away from the positive charge.

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Answer:

[tex]u_x=8.5454\ m.s^{-1}[/tex]

[tex]d=9.5782+16=25.5782\ m[/tex] was the her position ahead from her window when she caught the object.

Explanation:

Given:

[tex]r=3.2\times 5[/tex]

[tex]r=16\ m[/tex]

Since the projectile was thrown horizontally therefore the vertical component of velocity is zero.

Now the time taken for the object to reach to Henrietta ignoring the height where she catches:

[tex]h=u_x.t+\frac{1}{2} \times g.t^2[/tex]

[tex]43.9=0+\frac{1}{2} \times 9.8\times t^2[/tex]

[tex]t=2.9932\ s[/tex] is the time taken by the projectile to reach Henrietta.

Now the distance further walked by Henrietta in the above time from the point where she was when the projectile was launched:

[tex]\Delta d=v_w\times t[/tex]

[tex]\Delta d=9.5782\ m[/tex]

Now the total  horizontal distance from the window to Henrietta when the projectile reached her:

[tex]d=\Delta d+r[/tex]

[tex]d=9.5782+16=25.5782\ m[/tex] was the her position ahead from her window when she caught the object.

Now the initial horizontal velocity of launch of the projectile:

[tex]u_x=\frac{d}{t}[/tex]

[tex]u_x=\frac{25.5782}{2.9932}[/tex]

[tex]u_x=8.5454\ m.s^{-1}[/tex]

Answer:

b) Cells will pass through the G1/S checkpoint even if conditions are not ideal for cell division.

Explanation:

In the given problem, if there exists a gain-of-function mutation for the given cell, there would not be the formation of cyclin E when there is the possibility of cells movement via the checkpoint of the G1/S, even when there are non-deal conditions for the division of cell. Thus, the correct option in the lists of options is the option b.

Answer:

A) Increasing,  Increasing,  Increasing

Explanation:

The greater the rate of fuel ejection higher will be the kinetic energy of the rocket. As rocket is fire upward its fuel rejection rate increases and hence it's kinetic energy increases.

Gravitational potential energy,

as the rocket moves further it's r from the Earth increases. Hence the Gravitational potential energy increases as its height from the Earth increases.

Therefore,  mechanical energy of the rocket must also increase as it is sum of kinetic and potential energy.

Hence Option A is correct.

For the first kilometer of the rocket's ascent, the kinetic energy, gravitational potential energy, and mechanical energy of the Earth-rocket system increase.

The changes in the kinetic energy, gravitational potential energy, and mechanical energy of the Earth-rocket system for the first kilometer of the rocket's ascent can be determined.

Since the mass of fuel ejected is small compared to the mass of the rocket, the rocket's kinetic energy will increase as it accelerates away from Earth. The gravitational potential energy of the Earth-rocket system will also increase as the rocket moves higher against the pull of gravity. Therefore, the changes in the three quantities are as follows:

Answer:

The Tension T is 42120N

The Horizontal force component is 18322.2N

The Vertical force component is - 4729N

Explanation:

First, you have to find the angle between the drawbridge and the cable using sine and cosine rule. This will result in angle 44.2°. Hence, the angle between the horizontal axis and the cable will be 64.2° (44.2° + 20°).

Having done that, you apply two conditions of equilibrium.

1. THE VECTOR SUM OF ALL FORCES EQUAL ZERO.

∑Fx = 0

∑Fx = Rx - Tcos64.2 = 0

Rx = 0.435T

∑Fy = 0

∑Fy = Ry + Tsin64.2 - W - w = 0

W = 2000kg × 9.8 = 19600N

w =1000kg × 9.8 = 9800N

Ry + 0.9T = 29400N

Ry = 29400 - 0.9T

2. THE SUM TOTAL OF TORQUES EQUALS ZERO

Rx: τ = 0

Ry: τ = 0

T: τ = 5 × Tsin44.2

= 3.49T m

W: τ = 4 × 19600sin90

= 78400Nm

w: τ = 7 × 9800sin9

= 68600Nm

Note:

Rx = x component of Reaction force

Ry = y component of Reaction force.

T = Tension

W = weight of bridge

w = weight of Sir Lance a Lost and his steed

τ = torque

Note: The torque of Tension is counter clockwise while that of the weights is clockwise.

Hence,

∑τccw = ∑τcw

3.49T = 78400 + 68600

3.49T = 14700Nm

T = 147000/3.49

T = 42120N

Rx = 0.435 × 42120

Rx = 18322.2N

Ry = 29400N - (0.9×42120)N

Ry = 29400 - 34129

Ry = -4729N

Note: Ry being negative means that the hinge of the drawbridge exerts a downward force.

In this exercise we have to use the knowledge of tension and force, so we can say that this will result in:

A)The Tension T is 42120N

B)The Horizontal force component is 18322.2N

C)The Vertical force component is - 4729N

Before starting the calculations we have to remember some concepts such as:

First we will add all the force vectors so that it results in zero and that means that the body is in equilibrium, so:

[tex]\sum F_x = 0\\ \sum F_x = R_x - Tcos(64.2) = 0\\ R_x = 0.435T\\ \sum F_y = 0\\ \sum F_y = R_y + Tsin(64.2) - W - w = 0\\ W = 2000kg * 9.8 = 19600N\\ w =1000kg * 9.8 = 9800N\\ R_y + 0.9T = 29400N\\ R_y = 29400 - 0.9T[/tex]

Secondly, we will add up all the torques so that it results in zero and that means that the body is in equilibrium, so:

[tex]T: T = 5 * Tsin(44.2) = 3.49T m\\ W: T = 4 * 19600sin(90)= 78400Nm\\ w: T = 7 * 9800sin(90) = 68600Nm\\ \sum Tccw = \sum Tcw\\ 3.49T = 78400 + 68600\\ 3.49T = 14700Nm\\ T = 147000/3.49\\ T = 42120N R_x = 0.435 * 42120\\ R_x = 18322.2N\\ R_y = 29400N - (0.9*42120)N\\ R_y = 29400 - 34129\\ R_y = -4729N[/tex]

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Answer:

1050 possible selections

Explanation:

Number of red balls = 7

Number of blue balls = 10

Let blue balls be B and red balls be R

four balls are to be selected. At least three must be blue.

the first combination is 4blue 0red

second combination is 3blue 1red

(10C4*7C0 + 10C3*7C1)

210 *1 +120 *7

210+840

=1050 possibilities

There are 71 different selections possible if at least 3 balls must be blue.

To find the number of different selections possible, we can consider the different scenarios where at least 3 balls are blue:

Summing up the number of selections from both scenarios, we get 70 + 1 = 71 different selections possible.

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Answer:

ANULAR ECLIPSE

Explanation:

ANULAR ECLIPSE. Because the moon is very far, only a portion of the sun would be obscured, and then only the moon's outer ring will be viewable; this is called the anular eclipse.The ring of fire marks the maximum stage of an annular solar eclipse.

The angular velocity of a solid uniform sphere rolling without slipping down an inclined plane does not depend on the mass or the radius of the sphere; it depends on the height from which it rolls down and gravitational acceleration only.

The question asks what determines the angular velocity of a solid uniform sphere as it rolls without slipping down an inclined plane. It specifically queries whether this angular velocity depends on the mass (M) of the sphere, the radius (R) of the sphere, both, or neither.

The answer is that the angular velocity of a rolling sphere at the bottom of the incline does not depend on the mass of the sphere or the radius of the sphere. According to the principles of conservation of energy and the dynamics of rolling motion, all objects regardless of their mass or radius, will roll down an incline and reach the bottom with the same angular velocity if they start from rest, provided that they do not slip and there is no air resistance. This is because the potential energy lost is completely converted into kinetic energy (both translational and rotational), and the mechanical energy of the system remains constant.

The angular velocity of the sphere at the bottom of the incline depends on d) neither the mass nor the radius of the sphere.

When the sphere rolls without slipping, there is a relationship between its linear velocity (v) and angular velocity (ω) given by the equation v = ωR.

Total mechanical energy of the sphere = translational kinetic energy + rotational kinetic energy + potential energy.

At the bottom of the incline, all potential energy will have converted into kinetic energy:

By energy conservation:

Substituting I and the relation v = ωR gives us:

Solving for ω:

From the above steps, it's evident that the angular velocity ω depends on the height h the sphere rolls down, and the acceleration due to gravity g, but not on the mass M or radius R of the sphere.

Therefore, the correct answer is: d) neither the mass nor the radius of the sphere.

Answer:

The milky way does not have an active galactic nuclei and active galactic nuclei reduce their activity as time pass.

Explanation:

An active galactic nucleus (AGN) is a region that is compact and at the galaxy's center. It has a extra high luminosity in which some observations indicate that the luminosity is not as a result of the presence of stars.

Explanation:

Formula for final volume of chamber if the partition is ruptured will be as follows.

        [tex]V_{2}[/tex] = 1.5 + 1.5

                   = 3.0 [tex]ft^{3}[/tex]

As mass remains constant then the specific volume at this state will be as follows.

             [tex]\nu_{2} = \frac{V_{2}}{m}[/tex]

                          = [tex]\frac{3.0}{2}[/tex]

                          = 1.5 [tex]ft^{3}/lbm[/tex]

Now, at final temperature [tex]T_{2}[/tex] = 300 F according to saturated water tables.

   [tex]\nu_{f} = 0.01745 ft^{3}/lbm[/tex]

   [tex]\nu_{fg} = 6.4537 ft^{3}/lbm[/tex]  

   [tex]\nu_{g} = 6.47115 ft^{3}/lbm[/tex]

Hence, we obtained [tex]\nu_{f} < \nu_{2} < \nu_{g}[/tex] and the state is in wet condition.

       [tex]\nu_{2} = \nu_{f} + x_{2}\nu_{fg}[/tex]

             1.5 = [tex]0.01745 + x_{2} \times 6.4537[/tex]

        [tex]x_{2}[/tex] = 0.229

Now, the final pressure will be the saturation pressure at [tex]T_{2}[/tex] = 300 F

and,   [tex]P_{2}[/tex] = [tex]P_{sat}[/tex] = 66.985 psia

Formula to calculate internal energy at the final state is as follows.

         [tex]U_{2} = m(u_{f}_{300 F} + x_{2}u_{fg_{300 F}}[/tex]

                   = [tex]2(269.51 + 0.229 \times 830.45)[/tex]

                   = 920.56 Btu

Therefore, we can conclude that the final pressure of water, in psia is 66.985 psia and total internal energy, in Btu, at the final state is 920.56 Btu.

Answer:

B. 2.5 W

Explanation:

Power: This can be defined as the rate at which energy used or it i the rate at which work is done. The S.I unit of power is Watt (W).

Mathematically,

Power = Energy or work/Time

P = W/t ........................Equation 1

Where P = power used to perform the task, W = work, t = time.

Given: W = 50 J, t = 20 s.

Substitute into equation 1

P = 50/20

P = 2.5 W.

Hence the power is 2.5 W.

The right option is B. 2.5 W

The power used to perform the work of 50J in 20 seconds is 2.5 watts. This much power should be generated by it. The correct answer is B.

Given that 50 J of work was performed in 20 seconds,

The power used to perform a task, you can use the formula:

Power = Work / Time

Power = 50 J / 20 s

Power = 2.5 W

Therefore, the power used to perform the task is 2.5 W.

Therefore, the correct answer is B) 2.5 W.

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Answer:

Explanation:

Given

mass of children [tex]m_1=20\ kg[/tex]

[tex]m_2=30\ kg[/tex]

distance between two children [tex]L=3\ m[/tex]

suppose  small child is at a distance of x m from  pivot point

so torque of small child and heavier child must be equal

[tex]20\times (x)=30\times (3-x)[/tex]

[tex]2x=9-3x[/tex]

[tex]5x=9[/tex]

[tex]x=1.8\ m[/tex]    

The distance between the center of mass of the molecule and carbon atom is 0.65 × 10⁻¹⁰ m

To answwer the question, we need to know the center of mass of CO molecule

The center of mass of the CO molecule is given by

x = m₁x₁ + m₂x₂/(m₁ + m₂) where

Substituting the values of the variables into the equation, we have

x = (m₁x₁ + m₂x₂)/(m₁ + m₂)

x = (12 u × 0 m + 16 u × 1.13 × 10⁻¹⁰ m)/(12 u  + 16 u)

x = (0 + 16 u × 1.13 × 10⁻¹⁰ m)/28 u

x = 18.08 × 10⁻¹⁰ um/28 u

x = 0.646 × 10⁻¹⁰ m

x ≅ 0.65 × 10⁻¹⁰ m

Since the carbon atom is at x₁ = 0 m and the center of mass is at x = 0.65 × 10⁻¹⁰ m, the distance between the carbon atom and the center of mass of the molecule is d = x - x₁

= 0.65 × 10⁻¹⁰ m - 0 m

= 0.65 × 10⁻¹⁰ m

So, the distance between the center of mass of the molecule and carbon atom is 0.65 × 10⁻¹⁰ m

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Answer:

2 in

Explanation:

Area of a sheet = 3.5 x 14 in²

Volume of ream, V = 98 in³

Let t be the thickness of ream

So, Volume of ream  Area of a sheet x thickness of ream

98 = 3.5 x 14 x t

t = 2 in

Thus, the  thickness of ream is 2 in.

Answer:

t = 2 in

Explanation:

given,

width of sheet, W = 3.5 in

Length of the sheet, L = 14 in

thickness of the ream = ?

volume of paper = 98 in³

we now,

Volume = Length x width x thickness

V = l w t

98 =  14 x 3.5 x t

[tex]t = \dfrac{98}{14\times 3.5}[/tex]

t = 2 in

thickness of the ream is equal to 2 in.

Traction loss in a vehicle's rear wheels is most likely due to either acceleration-induced traction loss or rear wheel traction loss (skid). The former is caused by rapid acceleration, while the latter can be due to turning or braking.

The traction loss described in your question occurs in the rear wheels of a vehicle and is likely due to acceleration-induced traction loss or rear wheel traction loss (skid). Acceleration-induced traction loss happens when rapid acceleration causes the tires to lose grip on the road. This is common in high-powered, rear-wheel drive vehicles. Rear wheel traction loss or a rear wheel skid, on the other hand, generally occurs when the rear wheels lose grip during turning or braking. In this scenario, the vehicle's rear end can swing out in either direction, causing the vehicle to spin.

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Final answer:

Rear wheel traction loss (skid) occurs when the rear tires lose grip on the road surface. In a front-wheel-driven car, friction at the rear wheel is in the opposite direction of motion. This concept is important for analysis in vehicle dynamics and skid prevention.

Explanation:

The traction loss referred to in the question occurs in the rear wheels of a vehicle and is specifically known as rear wheel traction loss (skid). This is a type of skid that happens when the rear tires of a vehicle lose grip on the road surface, often as a result of oversteering, sudden acceleration, or slippery conditions. It's essential to distinguish this from other types of traction loss, such as braking-induced traction loss, which occurs when the tires can no longer grip the road during heavy braking, or driver-induced skids, which are the result of driver error.

With respect to the problem provided, in a front-wheel-driven car, the friction at the rear wheel is generally in the opposite direction of motion. This is because the rear wheels are not powered in this configuration; instead, they follow the rotation induced by the car's movement, which is primarily driven by the front wheels. Therefore, while the front wheels are pulled forward by the engine, creating friction in the direction of motion, the rear wheels experience a kind of rolling resistance and the friction present there acts in the opposite direction of the car's movement.

Understanding the dynamics of friction and traction is critical for analyzing cases like where a vehicle skids and comes to rest after traveling a certain distance or determining the force required to maintain a constant speed in the presence of kinetic friction. These analyses involve concepts like travel reduction ratio, torque transferred to the wheel axle, tractive efficiency of the wheel, and the actual velocity of the wheel.

Answer:

a. for any force, there is an equal and opposite reaction force.

Explanation:

Newton third law of motion states that " for every action, there is an equal and opposite reaction.

Action force= reaction force =

M₁ * a₁ = M₂ * a₂

where M₁ = mass of object 1

          a₁ = acceleration due to object 1

          M₂ = mass of object 2

          a₂ = corresponding acceleration due to object 2

this happen in everyday life of an individual. an example is seen in a groups of three man exerting a push force on a static car, the push force on the car make the car to exhibit some motion equivalent to the force applied by the three man.

Answer:

v' = -0.0906 m/s

Explanation:

given,

mass of cannon, M = 2240 Kg

mass of the ball, m = 15.5 Kg

speed of cannon ball, v = 131 m/s

speed of he cannon = ?

initial speed of cannon and the cannon ball is equal to 0 m/s

using conservation of energy

(M+m)V = M v' + m v

(M+m) x 0= 22400 v' + 15.5 x 131

22400 v' = -2030.5

v' = -0.0906 m/s

negative sign represent the canon will move in opposite direction of the ball.

hence, speed of cannon is equal to 0.0906 m/s

Using the law of conservation of momentum and the given values, the recoil velocity of the cannon is found to be approximately 0.9058 m/s after firing a cannonball.

According to the law of conservation of momentum, the total momentum before an event must equal the total momentum after the event, if no external forces act on the system. In this case, the cannon and the cannonball system is isolated and free to recoil, so the momentum before the cannon fires must be equal to the momentum after the cannon fires.

Momentum is the product of mass and velocity. Before the cannon fires, both the cannon and the cannonball are at rest and have a momentum of zero. After the cannon fires, the momentum of the cannonball is mcannonball × vcannonball, which must be equal and opposite to the momentum of the cannon. Therefore, mcannonball × vcannonball = mcannon × vcannon.

Using the given values for mass and velocity of the cannonball (15.5 kg and 131 m/s), and the mass of the cannon (2240 kg), we can calculate the cannon's recoil velocity using the equation:

mcannonball × vcannonball = mcannon × vcannon

(15.5 kg × 131 m/s) / 2240 kg = vcannon

The calculated recoil velocity of the cannon is approximately 0.9058 m/s.

Answer:

57.19461 m/s²

-9.20833 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Equation of motion

[tex]v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{\dfrac{663}{3.6}-0}{3.22}\\\Rightarrow a=57.19461\ m/s^2[/tex]

The acceleration during the race is 57.19461 m/s²

[tex]v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-\dfrac{663}{3.6}}{20}\\\Rightarrow a=-9.20833\ m/s^2[/tex]

The acceleration while stopping is -9.20833 m/s²

The assumptions made here are:

The initial velocity of the dragster at the start of the race is zero

The acceleration is constant in both cases of acceleration.

The final velocity when it stops is zero

Motion is in a straight line path

Answer:

The answer is 199920 meters.

Explanation:

First we need to find how many seconds they walked to the store:

[tex]34*60=2040[/tex]

If they walked 2040 seconds to eastward with velocity of 89 m/s, the displacement will be:

[tex]2040*98=199920[/tex]

Answer:

I think the answer is c

Explanation:

But it is kinda of opinionated

Statements C best describe a leader. A leader is a person who inspires other team members to reach the organizational goals

A leader is someone who motivates others to achieve the organization's objectives.

A leader is the person who owns the company and manages it on a daily basis. Any member of the management team might be considered a leader, but only a leader would be able to own the company.

A leader is a person who inspires other team members to reach the organizational goals

Hence,statements C best describe a leader.

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Answer: 0.006in/s

Explanation:

Let the rate at which air is being blown into a spherical balloon be dV/dt which is 1.68in³/s

Also let the rate at which the radius of the balloon is increasing be dr/dt

Given r = 4.7in and Π = 3.14

Applying the chain rule method

dV/dt = dV/dr × dr/dt

If the volume of the sphere is 4/3Πr³

V = 4/3Πr³

dV/dr = 4Πr²

If r = 4.7in

dV/dr = 4Π(4.7)²

dV/dr = 277.45in²

Therefore;

1.68 = 277.45 × dr/dt

dr/dt = 1.68/277.45

dr/dt = 0.006in/s

Answer:

We say that the object's spectrum is shifted.

Explanation:

Spectral lines will be shifted to the blue part of the spectrum1 if the source of the observed light is moving toward the observer, or to the red part of the spectrum when is moving away from the observer (that is known as the Doppler effect).

That shift can be used to find the velocity of the object by means of the Doppler velocity.

[tex]v = c\frac{\Delta \lambda}{\lambda_{0}}[/tex]  (1)

Where [tex]\Delta \lambda[/tex] is the wavelength shift, [tex]\lambda_{0}[/tex] is the wavelength at rest, v is the velocity of the source and c is the speed of light.

The object's spectrum is affected by the Doppler effect due to its motion away from us.

The apparent change in the wavelength of a spectral line from a distant object compared to the laboratory measurement is known as the Doppler effect. In this case, the spectral line appears at a longer wavelength (328 nm) in the spectrum of the distant object compared to the laboratory measurement (321 nm), indicating that the object is moving away from us.

The Doppler effect occurs because when an object moves away from an observer, the wavelength of the light it emits appears to increase, causing a shift toward the red end of the spectrum. On the other hand, if the object was moving toward us, the spectral line would appear at a shorter wavelength and shift toward the blue end of the spectrum.

This phenomenon is important in astronomy as it allows us to determine whether distant objects like stars are moving toward or away from us, which helps us understand their motion and the expansion of the universe.

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Answer:

The magnitude of charge on each sphere is [tex]q=2.50\times 10^{-8}\ C[/tex].

Explanation:

Given that,

Force of repulsion between the charges, F = 22 mN

The distance between spheres, r = 16 mm = 0.016 m

It is mentioned that both the spheres carry equal charges. The force between charges is given by :

[tex]F=\dfrac{kq^2}{r^2}[/tex]

[tex]q=\sqrt{\dfrac{Fr^2}{k}}[/tex]

[tex]q=\sqrt{\dfrac{22\times 10^{-3}\times (0.016)^2}{9\times 10^9}}[/tex]

[tex]q=2.50\times 10^{-8}\ C[/tex]

So, the magnitude of charge on each sphere is [tex]q=2.50\times 10^{-8}\ C[/tex]. Hence, this is the required solution.

Answer:

You have to apply force of 78.43 N to pull the sled or person across the snow at constant speed

Explanation:

Given data

Sled force F₁=58 N

Person weight F₂=655 N

Coefficient of kinetic friction u=0.11

To find

Friction Force

Solution

To pull the sled across the snow means the force you apply must be equal to and opposite the force of frictional.  

First we need to find the Total normal force

So

[tex]F_{Normal-Force}=F_{1}+F_{2}\\F_{Normal-Force}=58N+655N\\F_{Normal-Force}=713N[/tex]

Now the force of friction

[tex]F_{Friction-Force}=u_{coefficient}*F_{Normal-Force}\\F_{Friction-Force}=0.11*713N\\F_{Friction-Force}=78.43N[/tex]

So you have to apply force of 78.43 to pull the sled or person across the snow at constant speed

Answer:

Range, [tex]R = MV²/2QE[/tex]

Explanation:

The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.

Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses  all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.

So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops

Therefore {tex}R = MV²/2QE{/tex}

Answer:

14.4 m/s

Explanation:

mass of Anna (Ma) = 68 kg

speed of Anna (Va) = 17 m/s

mass of SandraDay (Ms) = 76 kg

speed of SandraDay (Vs) = 12 m/s

We can find their speed (V) immediately after collision from the conservation of momentum where

(Ma x Va) + (Ms + Vs) = (Ma + Ms) x V

where V = speed immediately after collision

(68 x 17) + (76 + 12) = (68 + 76) x V

2068 = 144 V

V = 2068 / 144 = 14.4 m/s

Answer:

Weber's Law

Explanation:

Weber’s law is an important law in psychology.

This law quantifies the perception of change within a provided stimulus.

This law is also known as Weber-Fechner law as it relates two hypotheses in the field of psychophysics which are Weber law and the Fechner law.

Weber’s law describes that an observed change in a stimulus is a constant ratio of the original stimulus.

Final answer:

Weber's Law describes the relationship between a stimulus and the resulting sensation, stating that the Just Noticeable Difference (JND) is a constant fraction of stimulus intensity. It is a key principle in psychophysics, useful for experiments such as determining the JND for different weights of rice in bags.

Explanation:

The concept you're inquiring about is Weber's Law, which is a fundamental principle in the field of psychophysics, a branch of psychology. Weber's Law describes the relationship between a stimulus and its resulting sensation by proposing that the Just Noticeable Difference (JND) is a constant fraction of the stimulus intensity.

To put this into practice, for example, if one were testing the JND of various weights of rice in bags, they could use incremental percentage increases of the weight to determine the smallest difference that can be perceived. By choosing increments like 10 percent or 20 percent between certain weight thresholds, a researcher can apply Weber's Law to determine the JND in a standardized way.

This law helps explain why when you have a cup of coffee with little sugar, adding a teaspoon can make a noticeable difference, but in a cup with much more sugar, that same teaspoon becomes less detectable, requiring a proportionally larger amount of sugar to notice a change in sweetness.