All categories

3 years ago

The speed limit in a school zone is 25.mph what is the speed in m/s

Physics

2 answers:

shtirl [24]3 years ago

7 0

Answer:

11.176 m/s

Explanation:

Divide speed value by 2.237

drek231 [11]3 years ago

5 0

Answer: 25 mph in residential or school districts,  55 mph on rural highways, and  70 mph on rural Interstate highways. Posted speed limits (sometimes called regulatory speed limits) are those that are sign-posted along the road and are enforceable by law.

Explanation:

You might be interested in

A spring of force constant 285.0 N/m and unstretched length 0.230 m is stretched by two forces, pulling in opposite directions a

Kaylis [27]

Answer: W = 0.3853 J, e = 0.052 m

Explanation: Given that,

K =285.0N/M , L = 0.230m , F = 15N , e = ?

F = Ke

15 = 285 × e

e = 15÷ 285

e =0.052 m

e + L = 0.052 + 0.230

= 0.282m ( spring new length )

Work needed to stretch the spring

W = 1/2ke2

W = 1/2 × 285 x 0.052 × 0.052

W = 0.3853 J

8 0

3 years ago

A ball is thrown vertically upward from the edge of a bridge 22.0 m high with an initial speed of 16.0 m/s. The ball falls all t

KonstantinChe [14]

To solve this problem we will apply the concepts related to the kinematic equations of motion. We will start calculating the maximum height with the given speed, and once the total height of fall is obtained, we will proceed to calculate with the same formula and the new height, the speed of fall.

The expression to find the change in velocity and the height is,

$v_f^2-v_0^2 = -2gh$

Replacing,

$0^2-16^2 = -2(9.8)h$

$h = 13.0612m$

Thus the total height reached by the ball is

H = 22m+13.0612m

H = 35.0612m

Now calculate the velocity while dropping down from the maximum height as follows

$v_f^2-v_0^2 = 2gh$

Substituting the new height,

$v_f^2 - 0^2 = 2(9.8)(35.0612)$

$v = \sqrt{687.2}$

$v = 26.2145m/s$

3 0

3 years ago

A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with incompressib

Natasha_Volkova [10]

Answer:

F = 1958.4 N

Explanation:

By volume conservation of the fluid on both sides we can say that volume of fluid displaced on the side of the car must be equal to the volume of fluid on the other side

so we have

$L_1A_1 = L_2A_2$

$1.20(\pi 18^2) = L_2(\pi 5^2)$

$L_2 = 15.55 m$

so the car will lift upwards by distance 1.2 m and the other side will go down by distance 15.55 m

So here the net pressure on the smaller area is given as

$P = P_{atm} + \frac{12,000}{\pi (0.18)^2} + \rho g (1.2 + 15.55)$

excess pressure exerted on the smaller area is given as

$P_{ex} = \frac{12000}{\pi (0.18)^2} + 800(9.81)(16.75)$

$P_{ex} = 2.49\times 10^5 Pascal$

now the force required on the other side is given as

$F = P_ex (area)$

$F = (2.49 \times 10^5)(\pi (0.05)^2)$

$F = 1958.4 N$

3 0

3 years ago

A 5.09 × 1014-hertz electromagnetic wave is traveling through a transparent medium. The main factor that determines the speed of

sergiy2304 [10]

We are given an electromagnetic wave with a frequency of 5.09 x 10^14 Hz and travelling through a transparent medium. If the medium was vacuum, the speed of the wave would be equal to the speed of light. Otherwise, the main factor that would determine the speed of the wave is its wavelength.

6 0

3 years ago

Studen

Vesna [10]

Complete Question

A football coach walks 18 meters westward, then 12 meters eastward, then 28 meters westward, and finally 14 meters eastward.

From this motion what is the distance covered

What is the magnitude and direction of the displacement

Answer:

$D = 72 \ m$

Magnitude

$d = - 20 \ m$

Direction

West

Explanation:

From the question we are told that

The first distance covered westward is $d_w_1 = 18 \ m /tex] The first distance covered eastward is [tex]d_e1 = 12 \ m /tex] The second distance covered westward is [tex]d_w_2 = 28 \ m /tex] The second distance covered eastward is [tex]d_e2 = 14 \ m /tex] Generally the distance covered is mathematically represented as [tex]D = d_w1 + d_w2 + d_e1 + d_e2$

=> $D = 18 + 28 + 12 + 14$

=> $D = 72 \ m$

For the second question eastward is in the direction of the positive x-axis so it would be positive and westward is in the direction of the negative x-axis so it would be negative

The magnitude of the displacement is

$d = -d_w1 -d_w2 + d_e1 + d_e2$

=> $d = -18-28 + 12 + 14$

=> $d = - 20 \ m$

The direction is west

7 0

3 years ago