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snow_tiger [21]
3 years ago
12

Use Hooke's Law to determine the work done by the variable force in the spring problem. A force of 250 newtons stretches a sprin

g 30 centimeters. How much work is done in stretching the spring from 30 centimeters to 60 centimeters? N-cm
Physics
1 answer:
diamong [38]3 years ago
5 0

Answer:

112.5 J

Explanation:

F = 250 N, x = 30 cm = 0.3 m

Let the spring constant be K.

By using the Hooke's law

F = k x

250 = k x 0.3

k = 833.3 N / m

xi = 30 cm = 0.3 m, xf  60 cm = 0.6 m

Work done = 1/2 k (xf^2 - xi^2)

Work done = 0.5 x 833.33 x (0.6^2 - 0.3^2)

Work done = 112.5 J

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A fish in an aquarium with flat sides looks out at a hungry cat. To the fish, does the distance to the cat appear to be less tha
lakkis [162]

Answer:

p = -q  

he distance is equal to the current distance, so the distance does not change

Explanation:

For this exercise we can solve it using the equation of the constructor

            1 / f = 1 / p + 1 / q

where f is the focal length, p the distance to the object and q the distance to the image

For a flat surface the radius is at infinity, therefore 1 / f = 0, which implies

          1 / p = - 1 / q

           p = -q

Therefore the distance is equal to the current distance, so the distance does not change

7 0
3 years ago
What is the the acceleration of a proton that is 4.0 cm from the center of the bead? input positive value if the acceleration is
QveST [7]
Missing detail in the text:
"<span>A small glass bead has been charged to + 25 nC "

Solution
The force exerted on a charge q by an electric field E is given by
</span>F=qE
<span>Considering the charge on the bead as a single point charge, the electric field generated by it is
</span>E=k_e  \frac{Q}{r^2}
with k_e = 8.99\cdot 10^9 Nm^2/C^2, Q=+25 nC=25 \cdot 10^{-9}C is the charge on the bead. We want to calculate the field at r=4.0 cm=0.04 m:
E=(8.99\cdot 10^9) \frac{25\cdot 10^{-9}}{(0.04)^2}=1.4\cdot 10^5 V/m
The proton has a charge of q=1.6\cdot 10^{-19}C, therefore the force exerted on it is
F=qE=1.6\cdot 10^{-19}C \cdot 1.4\cdot 10^5 V/m=2.25\cdot 10^{-14} N

And finally, we can use Newton's second law to calculate the acceleration of the proton. Given the proton mass, m=1.67\cdot 10^{-27} kg, we have
F=ma
a= \frac{F}{m}= \frac{2.25\cdot 10^{-14} N}{1.67\cdot 10^{-27} kg}=1.35 \cdot 10^{13} m/s^2

The charge on the bead is positive, and the proton charge is positive as well, therefore the proton is pushed away from the bead, so:
a=-1.35 \cdot 10^{13} m/s^2
6 0
3 years ago
The brakes on your automobile are capable of creating a deceleration of 5.0 m/s2. If you are going 135 km/h and suddenly see a s
poizon [28]

Answer:

3.33 seconds

Explanation:

We can use the velocity formula [ v = u + at ] to solve.

Find the value "u".

135km/h -> 135km*1000m/3600s -> 37.5m/s

Find the value "v".

75km/h -> 75km*1000m/3600s -> 20.83m/s

Keep in mind we are dealing with "deceleration" so when we input 5.0m/s into the formula, it will be a negative value.

Now, find "t" which is the value we aren't given with the values we're given in the question.

20.83 = 37.5 - 5t

-16.67 = -5t

3.33 = t

Best of luck!

8 0
3 years ago
An object, accelerating from rest at a constant rate, travels over 28 m in 11 s. What is its final velocity?
Paul [167]

Answer:

answer is 3

Explanation:

by using s= [(v+ u)/2] x t

28= (v+ 0)/2 x 11

v= 5.09 ms^-1

4 0
3 years ago
Please help ASAP I have to turn this in soon
Allisa [31]

Answer:

See in explanation

Explanation:

Scientific use: The Einstein's THEORY of relativity states that "Time Is Absolute".

Everyday use: Einstein's LAW of relativity says that time is not the same at all places and events.

5 0
3 years ago
Read 2 more answers
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