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timurjin [86]
3 years ago
10

A box that has a weight of 40 N is hung from a spring. The spring constant of the spring is 400 N/m. How many centimeters will t

he spring stretch? 0.1 cm 1 cm 10 cm 100 cm
Physics
2 answers:
suter [353]3 years ago
7 0

the correct answer is c. 10 cm

Andrej [43]3 years ago
5 0
The mathematical relationship between force and extension for a spring is F = -kx , where F is the restoring force, k is the spring constant and x is the extension. If a box that weighs 40 N is hung from a spring of content 400 N/m, then the extension is equal to x = -F / k = -400 / 40 = -10 cm. The negative sign simply shows the extension and restoring force are in opposite directions. The answer is 10 cm. 
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30cm³ of brine of relative density 1.15 and 42cm³ of water are mixed. What is the density of the final solution​
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Answer:

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If Anisa swims 85.4 yards in five minutes, how many meters will she swim in 70.0 seconds? ( use the metric system for this quest
Ira Lisetskai [31]

Answer : 18.22 meters

Explanation:

1 yard. = 0.9144 meters

85.4 yards = 78.08976 meters

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5 minutes = 300 seconds

Speed of Anisa = distance / time

Speed of Anisa = 78.08976 meters / 300 seconds

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8 0
2 years ago
A car driver traveling at a speed of 108km per hour ,sees a traffic light and stopped after travelling for 20seconds .Find the a
navik [9.2K]

Answer:

– 2.5 m/s²

Explanation:

We have,

• Initial velocity, u = 180 km/h = 50 m/s

• Final velocity, v = 0 m/s (it stops)

• Time taken, t = 20 seconds

We have to find acceleration, a.

\longrightarrow a = (v ― u)/t

\longrightarrow a = (0 – 50)/20 m/s²

\longrightarrow a = –50/20 m/s²

\longrightarrow a = – 5/2 m/s²

\longrightarrow a = – 2.5 m/s² (Velocity is decreasing) [Answer]

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When designing a roller coaster what are the biggest considerations?
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3 0
3 years ago
Red light of wavelength 630 nm passes through two slits and then onto a screen that is 1.4 m from the slits. The center of the 3
VARVARA [1.3K]

Answer:

Part a)

f = 4.76 \times 10^{14} Hz

Part b)

d = 3.48 \times 10^{-4} m

Part c)

\theta = 0.311 degree

Explanation:

Part a)

As we know that the speed of light is given as

c = 3 \times 10^8 m/s

\lambda = 630 nm

now the frequency of the light is given as

f = \frac{c}{\lambda}

so we have

f = \frac{3 \times 10^8}{630 \times 10^{-9}}

f = 4.76 \times 10^{14} Hz

Part b)

Position of Nth maximum intensity on the screen is given as

y_n = \frac{n\lambda L}{d}

so here we know for 3rd order maximum intensity

y_3 = 0.76 cm

n = 3

L = 1.4 m

0.76 \times 10^{-2} = \frac{3(630 \times 10^{-9})(1.4)}{d}

d = 3.48 \times 10^{-4} m

Part c)

angle of third order maximum is given as

d sin\theta = 3 \lambda

3.48 \times 10^{-4} sin\theta = 3(630 \times 10^{-9})

\theta = 0.311 degree

8 0
2 years ago
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