If an 800.0 Kg sports car slows to 8.0 m/s in the forward direction to check out an accident scene and a 1200.0 kg pick-up truck behind them continues traveling at 15.0 m/s forward, with what velocity will the two move if they lock bumpers after a rear-end collision

Answer:

d

Explanation:

Answer:

Question 1 the answer is 5 cm for the one pulley system, I just took the quiz

Question 2 is 10 cm

Explanation:

The rope only needs to move 5 cm for the mass to move from 15 to 20.

For the 2 pulley system the string has to move twice as far making it 10 cm. I took the quiz and got 100%

Answer:

Acceleration: -9.8 m/s^2

Velocity: -28.2 m/s

Displacement: 143.1 m

Explanation:

The acceleration of gravity for any object close to earth is approximately -9.8 m/s^2.

Now, to find the velocity after 9 seconds, we can use a kinematics formula, where x is the final velocity:

Final Velocity = Initial Velocity + Acceleration * Time

x = 60 + -9.8*9

x = 60 - 88.2

x = -28.2

The velocity is -28.2 m/s.

Lastly, to find the displacement, we can use another kinematics formula, where y is the displacement:

Displacement = (Final Velocity + Initial Velocity)/2 * Time

y = (-28.2 + 60)/2 * 9

y = 143.1

The displacement is 143.1 meters.

Answer:

Wavelength = 478.46 m

Explanation:

It is given that,

A local AM radio station broadcasts at a frequency of 627 kHz, f = 627000 Hz

We need to find the wavelength at which it is broadcasting. The wavelength is given by :

[tex]\lambda=\dfrac{c}{f}\\\\=\dfrac{3\times 10^8}{627\times 10^3}\\\\=478.46\ m[/tex]

So, the wavelength is 478.46 m.

Answer:

2.1 rad/s

Explanation:

Given that,

Mass of a tether ball, m = 0.546 kg

Length of a rope, l =  4.56 m

The maximum tension the rope can withstand before breaking is 11.0 N

We need to find the maximum angular speed of the ball. Let v is the linear velocity. The maximum tension is balanced by the centripetal force acting on it. It can be given by :

[tex]F=\dfrac{mv^2}{r}\\\\v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{11\times 4.56}{0.546}} \\\\=9.584\ m/s[/tex]

Let [tex]\omega[/tex] is the angular speed of the ball. The relation between the angular speed and angular velocity is given by :

[tex]v=r\omega\\\\\omega=\dfrac{v}{r}\\\\=\dfrac{9.584}{4.56}\\\\=2.1\ rad/s[/tex]

So, the maximum angular speed of the ball is 2.1 rad/s.

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is [tex]|v| = 6.93[/tex]

Explanation:

From the question we are told that

    The initial point is [tex](x_1 , y_1 , z_1 ) = (-1 , 7 , 4 )[/tex]

    The  terminal point is  [tex](x_2 , y_2 , z_2) = (-5 , 11, 8 )[/tex]

Generally the magnitude of the vector is mathematically represented as

     [tex]|v| = \sqrt{(x_2 -x_1 )^2 + (y_2 - y_1 )^2 + (z_2 -z_1)^2}[/tex]

=>   [tex]|v| = \sqrt{(-5 -(-1) )^2 + (11 - 7 )^2 + (8 -4)^2}[/tex]

=>   [tex]|v| = 6.93[/tex]

Answer:

The value of 96,745 joules in Btus is 91.6965673 or 91.7 Btus.

Explanation:

Btus stands for British thermal unit. It is the unit of energy. 96,745 Joules is equal to 91.7 Btus of energy.

Energy is the quantitative property of an object which is transferred to an object or to a physical system, that is recognizable in the performance of the work done and in the form of heat and light generated by the object or system. Energy is a conserved quantity, according to the law of conservation of energy which states that energy can be transformed, but not created or destroyed.

A British thermal unit (Btu) is a measure of the heat content or energy of fuels or energy sources. 1 BTU is roughly equal to the amount of energy which is required to raise the temperature of one pound of water by 1 degree Fahrenheit.

1 Joule = 0.000947817 Btu

Therefore, 96,745 Joules = 91.696 or 91.7 Btus

The value of 96,745 joules in Btus is 91.6965673 or 91.7 Btus.

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Answer:

Explanation:

reading of scale = reaction force of surface R

centripetal force = R - mg = m v² / R , m is mass , v is velocity and R is radius of the circular path .

R = mg + m v² / R

given ,

m v² / R = .80 mg

v² = .80 x g x R

= .8 x 9.8 x 9 = 70.56

v = 8.4 m /s

Answer:

a. stay the same for very long

Explanation:

It is rare for any motion to stay the same for a very long time. The force applied on a body causes changes in the magnitude of motion.

Answer:

hello your question is incomplete below is the missing part of the question

Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and then injected into a region of constant magnetic field with a field strength of 0.55 T.

answer :   ≈ 42

Explanation:

Ratio of the Radii of curvature for a proton and an electron travelling through this apparatus

= [tex]\frac{Rp}{Re}[/tex] = 0.025 m / 5.94 * 10^-4 m = 42.08

Rp = 0.025 m ( calculated )

Re = 5.94 * 10^-4 ( calculated )

Answer:

69

Explanation:

Answer:

620nm

Explanation:

According to the question,

As we know,

→ [tex]t = \frac{\lambda}{2n}[/tex]

or,

→ [tex]\lambda = 2tn[/tex]

By putting the values, we get

      [tex]= 2\times 233\times 10^{-9}\times 1.33[/tex]

      [tex]= 620 \ nm[/tex]

Thus the above response is correct.

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Answer:

a

 [tex]\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s[/tex]

b

 [tex]I = 0.106 \ A[/tex]

Explanation:

From the question we are told that

  The current is  [tex]I = 0.106 \ A[/tex]

   The length of one side of the square [tex]a = 4.60 \ cm = 0.046 \ m[/tex]

    The separation between the plate is  [tex]d = 4.0 mm = 0.004 \ m[/tex]

Generally electric flux is mathematically represented as

       [tex]\phi_E = \frac{Q}{\epsilon_o}[/tex]

differentiating both sides with respect to t is  

       [tex]\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} * \frac{d Q}{ dt}[/tex]

=>     [tex]\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} *I[/tex]

Here [tex]\epsilon_o[/tex] is the permitivity of free space with value  

        [tex]\epsilon _o = 8.85*10^{-12} C/(V \cdot m)[/tex]

=>   [tex]\frac{d \phi_{E}}{dt} = \frac{0.106}{8.85*10^{-12}}[/tex]

=>   [tex]\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s[/tex]

Generally the displacement current between the plates in A

    [tex]I = 8.85*10^{-12} * 1.1977 *10^{10}[/tex]

=>  [tex]I = 0.106 \ A[/tex]

Answer:

The astronaut is moving at a speed of 0.238 m/s in a direction opposite the direction of the water shot out.

Explanation:

We are given;

Mass of astronaut; m1 = 84 kg

Mass of water shoot out; m2 = 2 kg

Initial speed of astronaut; u1 = 0 m/s

Initial speed of water shoot out; u2 = 0 m/s

Final speed of shoot out; v2 = 10 m/s

From law of conservation of momentum, we can say that;

Initial momentum = final momentum

Thus;

m1•u1 + m2•u2 = m1•v1 + m2•v2

Where v1 is the final speed of the astronaut

Plugging in the relevant values, we get;

(84 × 0) + (2 × 0) = (84 × v1) + (2 × 10)

0 = 84v1 + 20

-20 = 84v1

v1 = -20/84

v1 = -0.238 m/s

The negative sign indicates that the astronaut is moving 0.238 m/s in a direction opposite the direction of the water shot out.

Answer:

The tangential speed at Livermore is approximately 284.001 meters per second.

Explanation:

Let suppose that the Earth rotates at constant speed, the tangential speed ([tex]v[/tex]), measured in meters per second, at Livermore (37.6819º N, 121º W) is determined by the following expression:

[tex]v = \left(\frac{2\pi}{\Delta t}\right)\cdot R \cdot \sin \phi[/tex] (1)

Where:

[tex]\Delta t[/tex] - Rotation time, measured in seconds.

[tex]R[/tex] - Radius of the Earth, measured in meters.

[tex]\phi[/tex] - Latitude of the city above the Equator, measured in sexagesimal degrees.

If we know that [tex]\Delta t = 86160\,s[/tex], [tex]R = 6.371\times 10^{6}\,m[/tex] and [tex]\phi = 37.6819^{\circ}[/tex], then the tangential speed at Livermore is:

[tex]v = \left(\frac{2\pi}{86160\,s} \right)\cdot (6.371\times 10^{6}\,m)\cdot \sin 37.6819^{\circ}[/tex]

[tex]v\approx 284.001\,\frac{m}{s}[/tex]

The tangential speed at Livermore is approximately 284.001 meters per second.

Answer:

0.2Nm

Explanation:

F1 = 30.0N

F2 = 20.0N

R = 2.0cm = 0.02m

F1R1 - F2R2 = net Torque

30(0.02) - 20(0.02) = net Torque

0.6-0.4 = net Torque

Therefore net Torque = 0.2Nm

The figure below shows two forces [tex]F_1[/tex] = 30.0 N and [tex]F_2[/tex] = 20.0 N acting on an object that can rotate around a centre axle. The net torque in 0.2 N-m

The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is a torque.

Torque is defined as Γ=r×F=r F sin(θ). In other words, torque is the cross product of the force vector, where 'a' is the angle between r and F, and the distance vector (the distance between the pivot point and the place where force is applied).

Torque due to [tex]F_1[/tex] = 30.0 N is 0.6 N-m outside the plane and due to   [tex]F_2[/tex] = 20.0 N is 0.4 N-m inside the plane so net torque is 0.2 N-m outside the plane.

The figure below shows two forces [tex]F_1[/tex] = 30.0 N and  [tex]F_2[/tex] = 20.0 N acting on an object that can rotate around a centre axle. The net torque in 0.2 N-m outside the plane.

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Answer: The displacement is 18.8 meters.

Explanation:

Let's define the North as the positive y-axis and the East as the positive x-axis.

We also can assume that if the point (x, y) represents the position of the boy, the initial position is (0 paces, 0 paces).

"A boy on a treasure hunt walks 12 paces West,"

West would be the negative x-axis in this case, then:

the position will be (-12 paces, 0 paces)

"then 21 paces North"

Now the position will be (-12 paces, 21 paces)

"then 25 paces East."

Now the position will be (-12 paces + 25 paces, 21 paces)

                                       = (13 paces, 21 paces)

Now we have the initial and final positions of the boy,  (0 paces, 0 paces) and (13 paces, 21 paces) respectively.

We want to find the displacement, which is defined as the distance between the final position and the initial position.

Remember that for two vectors (a, b) and (c, d), the distance is:

D = √( (a - c)^2 + (b  - d)^2)

In this case, the displacement will be:

D = √( ( 13 paces - 0 paces)^2 + (21 paces - 0 paces)^2) = 24.7 paces.

But we want this quantity in meters, so we can use the relation:

1 pace = 0.762 meters.

Then 24.7 paces, are 24.7 times 0.762 meters.

D = 24.7*0.762 m = 18.8 meters.

Answer:

21.21 m/s

Explanation:

Let KE₁ represent the initial kinetic energy.

Let v₁ represent the initial velocity.

Let KE₂ represent the final kinetic energy.

Let v₂ represent the final velocity.

Next, the data obtained from the question:

Initial velocity (v₁) = 15 m/s

Initial kinetic Energy (KE₁) = E

Final final energy (KE₂) = double the initial kinetic energy = 2E

Final velocity (v₂) =?

Thus, the velocity (v₂) with which the car we travel in order to double it's kinetic energy can be obtained as follow:

KE = ½mv²

NOTE: Mass (m) = constant (since we are considering the same car)

KE₁/v₁² = KE₂/v₂²

E /15² = 2E/v₂²

E/225 = 2E/v₂²

Cross multiply

E × v₂² = 225 × 2E

E × v₂² = 450E

Divide both side by E

v₂² = 450E /E

v₂² = 450

Take the square root of both side.

v₂ = √450

v₂ = 21.21 m/s

Therefore, the car will travel at 21.21 m/s in order to double it's kinetic energy.

For the car to be able to double its kinetic energy, it would need to travel at a speed of approximately 21.21m/s.

Given the data in the question;

Speed of the car; [tex]v_1 = 15m/s[/tex]

Speed for the car to double its kinetic energy; [tex]v_2 = \ ?[/tex]

Using the expression for kinetic energy:

[tex]K_E = \frac{1}{2}mv^2[/tex]

Where m is the mass and v is the velocity

Now, Initial Kinetic Energy will be;

[tex]K_E_1 = \frac{1}{2}mv_1^2 \\\\K_E_1 = \frac{1}{2}*m*(15)^2\\\\K_E_1 = \frac{1}{2}*m*225\\\\K_E_1 = 112.5*m[/tex]

For the kinetic energy to become double

[tex]K_E_2 = 2 * K_E_1\\\\\frac{1}{2}mv_2^2 = 2 * ( 112.5 * m)\\\\\frac{1}{2}mv_2^2 = 2m( 112.5)\\\\v_2^2 = 4( 112.5)\\\\v_2^2 = 450\\\\v_2 = \sqrt{450}\\\\v_2 = 21.21m/s[/tex]

Therefore, for the car to be able to double its kinetic energy, it would need to travel at a speed of approximately 21.21m/s.

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Answer: C. Body fat percentage

Explanation:

Answer:

your answer is C

Explanation:

Answer:

Explanation:

separation between slit d = 1.2 x 10⁻⁴ m

distance of screed D = 3.5 m

distance of second order right fringe

= 2λD / d where λ is wavelength of light .

2λD / d  = .0375

2 x λ x 3.5 / 1.2 x 10⁻⁴ = .0375

λ = 6428 x 10⁻¹⁰ m

= 643 nm .

Answer:

θ₁ = 5.4°

θ₂ = 10.86°

Explanation:

The angle ca be found by using grating equation:

mλ = d Sinθ

where,

m = order of diffraction

λ = wavelength = 405.3 nm = 4.053 x 10⁻⁷ m

d = grating element = 1/230 lines/mm = 0.0043 mm/line = 4.3 x 10⁻⁶ m/line

θ = angle = ?

FOR m = 1:

(1)(4.053 x 10⁻⁷ m) = (4.3 x 10⁻⁶ m/line) Sin θ₁

Sin θ₁ = 0.09425

θ₁ = Sin⁻¹(0.09425)

θ₁ = 5.4°

FOR m = 2:

(2)(4.053 x 10⁻⁷ m) = (4.3 x 10⁻⁶ m/line) Sin θ₁

Sin θ₂ = 0.1885

θ₂ = Sin⁻¹(0.1885)

θ₂ = 10.86°

Answer:91% I think

Explanation:

The typical engine efficiency for a modern petrol combustion engine is from 20 to 30 percent. The remaining 70 to 80 percent of the heat energy of the petrol is released from the engine as friction loss, mechanical sound energy, or exhaust heat. Thus, option B is correct.

An engine's efficiency is measured by the heat produced divided by the amount of useful work done. Is the task complete. Is the task finished. Please be aware that the phrase “work done” refers to the force applied to the clutch or driveshaft.

Utilizing the following formula, efficiency may be represented as a ratio: output input. Output, often called work output, is the entire quantity of productive work completed, excluding waste and spoilage.

You may determine the efficiency by figuring out what proportion of the energy input results in the desired output: Efficiency is defined as Energy Out/Energy In 100%.

Therefore, This implies that the work produced by thermodynamic expansion is reduced by the friction and other losses.

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Answer:

It's mass would be 4.486768kg..

Answer:

because formula of kinetic energy is K=1/2mv^2 that means more the velocity then more kinetic energy

Answer:true

Explanation:

Answer:

The right approach will be "1.3". A further explanation is given below.

Explanation:

As we know,

In Decibels, the change in sound volume will be:

= [tex]10log\frac{I_1}{I_2}[/tex]

Now,

According to the question,

⇒ [tex]1=10 log \frac{I_1}{I_2}[/tex]

By applying cross multiplication and putting the value of log, we get

⇒ [tex]10^{\frac{1}{10} }=\frac{I_1}{I_2}[/tex]

⇒ [tex]1.26=\frac{I_1}{I_2}[/tex]

⇒ [tex]\frac{I_1}{I_2}=1.3[/tex]

Answer:

B

Explanation:

Answer:

B.)Static

Hope I helped

Answer:

Answer.

Explanation:

Explanation.

Answer:

They both hit the ground at the same time.

Explanation:

if you drop a 100 g and 200 g mass simultaneously from the same height they’ll hit the ground at the same time.

Yes, there is a role of a very simple equation of kinematics i.e.

s=ut+1/2(at^2)

so, s means the height is the same for both masses(h), u(initial velocity) is obviously zero cause you are dropping it, a (acceleration) is due to the gravity here only ‘g’.

time is taken for a particular object to hit the ground,

t= (2h/g)^1/2 .

We can see that there is no term for the mass of objects, meanwhile, the mass of the object has nothing to do with the time taken in hitting the ground in this case.