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adell [148]
2 years ago
6

Cassidy Carver's family child care center has an activities table with four chairs an easel that can accommodate a child on each

side, a play kitchen area with room for four children and a building blocks area that can hold three children. How many play spaces are in Carver's environment
cannot determine from information prividrd

thirteen

three

four​
Physics
1 answer:
Akimi4 [234]2 years ago
6 0

Answer:4

Explanation:

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Assuming the same current is running through two separate coils, why is it easier to thrust a magnet into a wire coil with one l
s2008m [1.1K]
The magnetic field strength in a coil is directly proportional to the number of turns, or loops, in the coil.
Therefore, when there are four loops instead of one, the magnetic field strength has increased four times, making it harder to push the magnet in.
6 0
2 years ago
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When jumping, a flea reaches a takeoff speed of 1.0 m/s over a distance of 0.47 mm .What is the flea's acceleration during the j
garri49 [273]
We can use kinematics here if we assume a constant acceleration (not realistic, but they want a single value answer, so it's implied). We know final velocity, vf, is 1.0 m/s, and we cover a distance, d, of 0.47mm or 0.00047 m (1m = 1000mm for conversion). We also can assume that the flea's initial velocity, vi, is 0 at the beginning of its jump. Using the equation vf^2 = vi^2 + 2ad, we can solve for our acceleration, a. Like so: a = (vf^2 - vi^2)/2d = (1.0^2 - 0^2)/(2*0.00047) = 1,064 m/s^2, not bad for a flea!
8 0
2 years ago
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A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm li
babunello [35]

Answer:

0.00001266 m

Explanation:

D = Distance from source to screen

m = Order

d = Slit separation

The distance from a point on the screen to the center line

y=\frac{m\lambda D}{d}

At m = 0

y_0=0

y_1-y_0=35\ cm\\\Rightarrow y_1=35\ cm

At m = 1

y_1=\frac{1\times 633\times 10^{-9}\times 7}{d}\\\Rightarrow d=\frac{1\times 633\times 10^{-9}\times 7}{0.35}\\\Rightarrow d=0.00001266\ m

The slit separation is 0.00001266 m

3 0
2 years ago
In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th
qaws [65]

Answer:

The distance between both cars is 990 m

Explanation:

The equations for the position and the velocity of an object moving in a straight line are as follows:

x = x0 + v0 * t + 1/2 * a * t²

v = v0 + a * t

where:

x = position of the car at time "t"

x0 = initial position

v0 = initial speed

t = time

a = acceleration

v = velocity

First let´s find how much time it takes the driver to come to stop (v = 0).  We will consider the origin of the reference system as the point at which the driver realizes she must stop. Then x0 = 0

With the equation of velocity, we can obtain the acceleration and replace it in the equation of position, knowing that the position will be 250 m at that time.

v = v0 + a*t

v-v0 / t = a

0 m/s - 71.0 m/s / t =a

-71.0 m/s / t = a

Replacing in the equation for position:

x = v0* t +1/2 * a * t²

250 m = 71.0 m/s * t + 1/2 *(-71.0 m/s / t) * t²

250 m = 71.0 m/s * t - 1/2 * 71.0 m/s * t

250m = 1/2 * 71.0m/s *t

<u>t = 2 * 250 m / 71.0 m/s = 7.04 s</u>

It takes the driver 7.04 s to stop.

Then, we can calculate how much time it took the driver to reach her previous speed. The procedure is the same as before:

v = v0 + a*t

v-v0 / t = a      now v0 = 0 and v = 71.0 m/s

(71.0 m/s - 0 m/s) / t = a

71.0 m/s / t =a

Replacing in the position equation:

x = v0* t +1/2 * a * t²      

390 m = 0 m/s * t + 1/2 * 71.0 m/s / t * t²       (In this case, the initial position is in the pit, then x0 = 0 because it took 390 m from the pit to reach the initial speed).

390m * 2 / 71.0 m/s = t

<u>t = 11.0 s</u>

In total, it took the driver 11.0s + 5.00 s + 7.04 s = 23.0 s to stop and to reach the initial speed again.

In that time, the Mercedes traveled the following distance:

x = v * t = 71.0 m/s * 23.0 s = 1.63 x 10³ m

The Thunderbird traveled in that time 390 m + 250 m = 640 m.

The distance between the two will be then:

<u>distance between both cars = 1.63 x 10³ m - 640 m = 990 m.  </u>

3 0
2 years ago
The drag force F on a boat varies jointly with the wet surface area A of the boat and the square of the speed s of the boat. A b
Advocard [28]

Answer:

Wet surfaces areaA=+25.3ft^2

Explanation:

Using F= K×A× S^2

Where F= drag force

A= surface area

S= speed

Given : F=996N S=20mph A= 83ft^2

K = F/AS^2=996/(83×20^2)

K= 996/33200 = 0.03

1215= (0.03)× A × 18^2

1215=9.7A

A=1215/9.7=125.3ft^2

7 0
2 years ago
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