A car is initially moving at 35 km/h along a straight highway. To pass another car, it speeds up to 135 km/h in 10.5 seconds at a constant acceleration.(a) how large was the acceleration in m/s ^2
(b)how large was the acceleration, in units go g= 9.80 m/s ^2

Answers

Answer 1

Answer: acceleration = (velocity final-velocity initial)/ time
where
velocity final = 135 km/hr x 1 hr /3600 s x 1000m/1km
= 37.5 m/s
velocity initial = 35 km/hr x 1hr /3600 s x 1000 m/1 km
= 9.72 m/s
a) acceleration = 2.646 m/s^2
b) acceleration in g units = (2.646m/s^2)/(9.8m/s^2)
= 0.27 units

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A block (0.50 kg) is attached to an ideal spring with a spring constant of 80 N/m, oscillating horizontally on a frictionless surface. The total mechanical energy is 0.12 J. (a) What is the greatest extension of the spring from its equilibrium length? (b) Now the block is traveling 2.00 m/s, and brought to rest by compressing a very long spring of spring constant 800.0 N/m. How much does the spring compress?
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Select all the statements regarding electric field line drawings that are correct. Group of answer choices:
1. Electric field lines are the same thing as electric field vectors.
2. Electric field line drawings allow you to determine the approximate direction of the electric field at a point in space.
3. The number of electric field lines that start or end at a charged particle is proportional to the amount of charge on the particle.
4. The electric field is strongest where the electric field lines are close together.

Answers

Answer:

All statement are correct.

Explanation:

1. Electric field lines are the same thing as electric field vectors, electric field are mathematically vectors quantity. These vectors point in the direction in which a positive test charge would move.

2. Electric field line drawings allow you to determine the approximate direction of the electric field at a point in space. Yes it is correct tangent drawn at any point on these lines gives the direction of electric filed at that point.

3. The number of electric field lines that start or end at a charged particle is proportional to the magnitude of charge on the particle, is a correct statement.

4.The electric field is strongest where the electric field lines are close together, again a correct statement as relative closeness of field lines indicate a stronger strength of electric field.

Hence we can say that all the statement are correct.

What is the volume of a cone with a radius of 3 feet and a height of 6 feet use 3.14 for pie round your answer to the nearest hundredth

Answers

Answer:

56.52 feet³ to the nearest hundredth

Explanation:

the volume of a cone is given as

V =

$(1)/(3) \pi r^(2) h$

the radius is 3 feet

height is 6 feet

substituting this given values in the formular

we have that, V = $(1)/(3)$ x 3.14 x $3^(2)$ x 6

dividing , we have the volume (V)

V= 3.14 x 3 x 6

= 3.14 x 18

= 56.52 feet³ to the nearest hundredth

A Hall-effect probe to measure magnetic field strengths needs to be calibrated in a known magnetic field. Although it is not easy to do, magnetic fields can be precisely measured by measuring the cyclotron frequency of protons. A testing laboratory adjusts a magnetic field until the proton's cyclotron frequency is 9.70 MHz . At this field strength, the Hall voltage on the probe is 0.549 mV when the current through the probe is 0.146 mA . Later, when an unknown magnetic field is measured, the Hall voltage at the same current is 1.735 mV .A) What is the strength of this magnetic field?

Answers

Answer:

The value of the magnetic field is 2.01 T when Hall voltage is 1.735 mV

Explanation:

The frequency of the cyclotron can help us find the magnitude of the magnetic field, thus then we can compare the effect of increasing Hall voltage on the probe.

Magnetic field magnitude at initial Hall voltage.

The cyclotron frequency can be written in terms of the magnetic field magnitude as follows

$f = \cfrac{qB}{2\pi m}$

Solving for the magnetic field.

$B = \cfrac{2\pi mf}q$

Thus we can replace the given information but in Standard units, also remembering that the mass of a proton is $m_p=1.67 * 10^(-27) kg$ and its charge is $q_p=1.6 * 10^(-19) C$ .

So we get

$B = \cfrac{2\pi * 1.67 * 10^(-27) kg * 9.7 * 10^6 Hz}{1.6 * 10^(-19)C}$

$B =0.636 T$

We have found the initial magnetic field magnitude of 0.636 T

Magnetic field magnitude at increased Hall voltage.

The relation given by Hall voltage with the magnetic field is:

$V_H =\cfrac{R_HI}t B$

Thus if we keep the same current we can write for both cases:

$V_(H1) =\cfrac{R_HI}t B_1\nV_(H2) =\cfrac{R_HI}t B_2$

Thus we can divide the equations by each other to get

$\cfrac{V_(H1) }{V_(H2)}=\cfrac{\cfrac{R_HI}t B_1}{\cfrac{R_HI}t B_2}$

Simplifying

$\cfrac{V_(H1) }{V_(H2)}=\cfrac{ B_1}{ B_2}$

And we can solve for $B_2$

$B_2 =B_1 \cfrac{V_(H2)}{V_(H1)}$

Replacing the given information we get

$B_2= 0.636 T * \left(\cfrac{1.735 mV}{0.549 mV} \right)$

We get

$\boxed{B=2.01\, T}$

Thus when the Hall voltage is 1.735 mV the magnetic field magnitude is 2.01 T

Two particles are traveling through space. At time t the first particle is at the point (−1 + t, 4 − t, −1 + 2t) and the second particle is at (−7 + 2t, −6 + 2t, −1 + t). (a) (5 Points) Do the paths of the two particles cross? If so, where?

Answers

Answer:

Yes, the paths of the two particles cross.

Location of path intersection = ( 1 , 2 , 3)

Explanation:

In order to find the point of intersection, we need to set both locations equal to one another. It should be noted however, that the time for each particle can vary as we are finding the point where the paths meet, not the point where the particles meet themselves.

So, we can name the time of the first particle $T_F$ , and the time of the second particle $T_S$ .

Setting the locations equal, we get the following equations to solve for $T_F$ and $T_S$ :

$(-1 + T_F) = (-7 + 2T_S)$ Equation 1

$(4 - T_F) = (-6 + 2T_S)$ Equation 2

$(-1 + 2T_F) = (-1 + T_S)$ Equation 3

Solving these three equations simultaneously we get:

$T_F =$ 2 seconds

$T_S =$ 4 seconds

Since, we have an answer for when the trajectories cross, we know for a fact that they indeed do cross.

The point of crossing can be found by using the value of $T_F$ or $T_S$ in the location matrices. Doing this for the first particle we get:

Location of path intersection = ( -1 + 2 , 4 - 2 , -1 + 2(2) )

Location of path intersection = ( 1 , 2 , 3)

Captain Kiddo is trapped inside a submarine. In order to escape, she first needs to open a heavy revolving door. a. Would be easier for her to apply all her force close to the axle of the door or as far away as possible from it?
b. To finally get the hatch open she needs to grab hold of the wheel with her hands on opposite sides and rotate the wheel by exerting a force toward the top of the door on one side while exerting the same force towards the bottom of the door on the opposite side. Is there a net force applied on the wheel? Is there a net torque applied? Explain your answers

Answers

Answer:

a) It will be easier to apply all her force as far as possible from the axle

b) There is a net torque applied

Explanation:

a) Applying a larger force far away from the axle of the door produces a larger force on the torque than pushing it near the axle. This is because Increasing the lever arm between the axle and the point at which you push the door increases the torque on the door

b)The door is turning faster and faster because you are exerting a torque on it and its undergoing angular acceleration. There is a net torque which is the addition of the torque applied on the opposite sides of the door

For a short time the position of a roller-coaster car along its path is defined by the equations r=25 m, θ=(0.3t) rad, and z=(−8 cosθ) m, where t is measured in seconds, Determine the magnitudes of the car's velocity and acceleration when t=4s .

Answers

Answer:

Velocity = v = 2.24m/s

Acceleration = a = 0.20m/s²

Explanation:

Please see attachment below.

Given

z=(−8 cosθ) and θ = 0.3t

z = -8Cos (0.3t)

V = dz/dt

a = v²/R.

Please see full solution below.

Final answer:

The roller coaster's velocity and acceleration at t=4 seconds is 7.64 m/s and 0.57 m/s² respectively.

Explanation:

The question is about understanding kinematics in cylindrical coordinates to analyze the motion of a roller coaster car. First, we need to understand that in polar coordinates, θ is changing with time t. So, the velocity vector v will have two components, one in the θ direction (rθ') and another in the z direction (z'). Given θ = 0.3t, we differentiate θ with respect to time to get θ' or dθ/dt = 0.3 rad/sec. Then, the z component of the velocity can be calculated by differentiating the equation of motion in the z-direction, z = -8 cos(θ), with respect to time. This gives z' = 8(0.3)sin(0.3t). So, at t=4s, z' = 8(0.3)sin(1.2) = 1.89 m/s. We then calculate rθ' = r*dθ/dt = 25*0.3 = 7.5 m/s.

The magnitude of velocity can then be calculated using the Pythagorean theorem: √((rθ')² + (z')²) = √((7.5)² + (1.89)²) = 7.64 m/s .

In a similar way, we can find the acceleration components. Given that r=25 m and is constant, radial acceleration is zero ( ar = r*(θ')²). The tangential acceleration is at = r*θ'' = r*d²θ/dt² =0 m/s² and z'' = dz'/dt = 8*0.3²*cos(0.3t). So, at t = 4s, z'' = 8(0.09)cos(1.2) = 0.57 m/s². The magnitude of the acceleration is given by √((ar)² + (at)² +(z'')²) = √((0)² + (0)² +(0.57)²)= 0.57 m/s².