Suppose that the speedometer of a truck is set to read the linear speed of the truck, but uses a device that actually measures the angular speed of the tires. If larger-diameter tires are mounted on the truck, will the reading on the speedometer be correct? If not, will the reading be greater than or less than the true linear speed of the truck? Why?

Answers

Answer 1

Answer:

The measurement will be significantly affected.

Recall that the relationship between linear velocity and angular velocity is subject to the formula

$v = \omega r$ ,

Where r indicates the radius and $\omega$ the angular velocity.

As the radius increases, it is possible that the calibration is delayed and a higher linear velocity is indicated, that to the extent that the velocity is directly proportional to the radius of the tires.

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An electric field of 710,000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -6.00 C at this spot? (14C - 10 6C) Give your answer in Si unit rounded to two decimal places

Answers

Answer:

$4.26*10^6N$

Explanation:

A charge within an electric field E experiences a force proportional to the field whose module is F = qE, whose direction is the same, if the charge is negative, it experiences a force in the opposite direction to the field and if the charge is positive, experience a force in the same direction of the field.

In our case we are interested in the magnitude of the force, therefore the sign of the charge has no relevance

A person is making homemade ice cream. She exerts a force of magnitude 26 N on the free end of the crank handle on the ice-cream maker, and this end moves on a circular path of radius 0.26 m. The force is always applied parallel to the motion of the handle. If the handle is turned once every 2.0 s, what is the average power being expended?

Answers

Answer:

P = 31.83 W

Explanation:

Our data are,

Magnitude of the force F = 26 N

Radius of the circular path r = 0.26 m

The angle between force and handle $\theta = 0$ °

Time t = 2 s

We know that the formula to find the velocity is given by

Velocity $v = (2\pi r)/(t)$

$v= (2\pi r)/(t)$

$v=(2 \pi 0.26)/(2)$

$v= 0.8168m/s$

We know also that the formula to find the power is given by,

$P = F*v$

$P = (26)(0.8168)$

$P = 31.83 W$

Two narrow slits separated by 1.5 mm are illuminated by 514 nm light. Find the distance between adjacent bright fringes on a screen 5.0 m from the slits. Express your answer in meters using two significant figures.

Answers

The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M

Given data :

separation between slits ( d ) = 1.5 x 10⁻³ m

wavelength of light ( λ ) = 514 * 10⁻⁹ m

Distance from narrow slit ( D ) = 5.0 m

Determine the distance between the adjacent bright fringes

we apply the formula below

w = D * λ / d ---- ( 1 )

where : w = distance between adjacent bright fringes

Back to equation ( 1 )

w = ( 5 * 514 * 10⁻⁹ ) / 1.5 x 10⁻³

= 1.7 * 10⁻³ M

Hence we can conclude that The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M

Learn more about bright fringes calculations : brainly.com/question/4449144

Answer:

$1.7* 10^(-3)$ m

Explanation:

d = separation between the two narrow slits = 1.5 mm = 1.5 x 10⁻³ m

λ = wavelength of the light = 514 nm = 514 x 10⁻⁹ m

D = Distance of the screen from the narrow slits = 5.0 m

w = Distance between the adjacent bright fringes on the screen

Distance between the adjacent bright fringes on the screen is given as

$w = (D\lambda )/(d)$

$w = ((5.0)(514* 10^(-9)) )/(1.5* 10^(-3))$

$w = 1.7* 10^(-3)$ m

Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. (a) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time. (b) Find the velocity of each block after they have moved apart.

Answers

Answer:

av=0.333m/s, U=3.3466J

$v_(A2)=-1.333m/s,\n v_(B2)=0.667m/s$

Explanation:

a. let $m_A$ be the mass of block A, and $m_B=10.0kg$ be the mass of block B. The initial velocity of A, $\rightarrow v_A_1=2.0m/s$

-The initial momentum =Final momentum since there's no external net forces.

$pA_1+pB_1=pA_2+pB_2\n\nP=mv\n\n\therefore m_Av_A_1+m_Bv_B_1=m_Av_(A2)+m_Bv_(B2)$

Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):

$v_A_1-v_B_1=v_(B2)-v_(A2)$

-Applying the conservation of momentum. The blocks have the same velocity after collision:

$v_(B2)=v_(A2)=v_2\n\n2* 2+10* 0=2v_2+10v_2\n\nv_2=0.3333m/s$

#Total Mechanical energy before and after the elastic collision is equal:

$K_1+U_(el,1)=K_2+U_(el,2)\n\n#Springs \ in \ equilibrium \ before \ collision\n\nU_(el,2)=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\n\nU_(el,2)=0.5* 2* 2^2-0.5(2+10)(0.333)^2\n\nU_(el,2)=3.3466J$

Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s

b. Taking the end collision:

From a above, $m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0$

We plug these values in the equation:

$m_Av_A_1+m_Bv_B_1=m_Av_(A2)+m_Bv_(B2)$

$2*2+10*0=2v_A_2+10v_B_2\n\n2=v_A_2+5v_B_2\n\n#Eqtn 2:\nv_A_1-v_B_1=v_(B2)-v_(A2)\n\n2-0=v_(B2)-v_(A2)\n\n2=v_(B2)-v_(A2)\n\n#Solve \ to \ eliminate \ v_(A2)\n\n6v_(B2)=2.0\n\nv_(B2)==0.667m/s\n\n#Substitute \ to \ get \ v_(A2)\n\nv_(A2)=(4)/(6)-2=1.333m/s$

A 81.0 kg diver falls from rest into a swimming pool from a height of 4.70 m. It takes 1.84 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.

Answers

Explanation:

The given data is as follows.

height (h) = 4.70 m, mass = 81.0 kg

t = 1.84 s

As formula to calculate the velocity is as follows.

$\nu$ = 2gh

= $2 * 9.8 m/s^(2) * 4.70 m$

= 92.12 $s^(2)$

As relation between force, time and velocity is as follows.

F = $(m * \nu)/(t)$

Hence, putting the given values into the above formula as follows.

F = $(m * \nu)/(t)$

= $(81.0 kg * 92.12 s^(2))/(1.84 s)$

= 4055.28 N

Thus, we can conclude that the magnitude of the average force exerted on the diver during that time is 4055.28 N.

Sophia wanted to estimate the product 73 x 12 so she made an estimate of 70x 10. Would her estimate be greater or near the actual estimate?

Answers

her estimate would not be greater, so it would be near the actual estimate. the numbers 70 and 10 are smaller than the numbers 73 and 12, so it is not possible for 70 x 10 to be more than 73 x 12 is. i hope this helps!!

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