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patriot [66]
3 years ago
9

Tom has built a large slingshot, but it is not working quite right. He thinks he can model the slingshot like an ideal spring, w

ith a spring constant of 45.0 N/m. When he pulls the slingshot back 0.515 m from a non-stretched position, it just doesn't launch its payload as far as he wants. His physics professor "helps" by telling him to aim for an elastic potential energy of 14.5 Joules. Tom decides he just needs elastic bands with a higher spring constant. By what factor does Tom need to increase the spring constant to hit his potential energy goal?
During a followup conversation, Tom's physics professor suggests that he should leave the slingshot alone and try pulling the slingshot back further without changing the spring constant. How many times further than before must Tom pull the slingshot back to hit the potential energy goal with the original spring constant?
Physics
1 answer:
lora16 [44]3 years ago
4 0
I don’t know right now, but I’ll get back to you later this afternoon in comment
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A ball is dropped from a height of 10m. At the same time, another ball is thrown
soldi70 [24.7K]

5.1 m

Explanation:

Let's set the ground as our reference point. Let's also call the dropped ball to be ball #1 and its height above the ground at any time t is given by

y_1 = 10 - \frac{1}{2}gt^2 (1)

where 10 represents its initial height or displacement of 10 m above the ground. At the same time, the displacement of the second ball with respect to the ground y_2, is given by

y_2 = v_0t - \frac{1}{2}gt^2 (2)

At the instant the two balls collide, they will have the same displacement, therefore

y_1 = y_2 \Rightarrow 10 - \frac{1}{2}gt^2 = v_0t - \frac{1}{2}gt^2

or

v_0t = 10\:\text{m}

Solving for t, we get

t = \dfrac{10\:\text{m}}{v_0} = \dfrac{10\:\text{m}}{10\:\text{m/s}} = 1\:\text{s}

We can use either Eqn(1) or Eqn(2) to hind the height where they collide. Let's use Eqn(1):

y_1 = 10\:\text{m} - \frac{1}{2}(9.8\:\text{m/s}^2)(1\:\text{s})^2

\:\:\:\:\:\:\:= 5.1\:\text{m}

8 0
3 years ago
What is the strength of the electric field ep 0.90 mm from a proton?
dem82 [27]

The electric field strength is inversely related to the square of the distance.so  the strength of the electric field is given by

E=\frac{1}{4\pi \epsilon _{0}  } \frac{q}{r^{2} } = \frac{k q}{r^{2} }

Here,  \frac{1}{4\pi \epsilon _{0}  } = k  is constant depend upon medium and its value is 9.0 \times10^{9} \ N m^2/C^2 and q is charge and  r is the distance.

Given  r = 0.90 mm = 9.0 \times 10^{-4} m and we know the charge of proton, q = 1.6\times 10^{-19} \ C.

Therefore,

E=\frac{9.0 \times10^{9} \ N m^2/C^2 \times1.6\times 10^{-19} \ C  }{(9.0 \times 10^{-4} m)^2} = 0.177 \times 10^{-2} \ N/C \\\\  E= 1.77 \times 10^{-3}  N/C

4 0
3 years ago
Two equal quantities of water, of mass m and at temperatures T1 and T2, (T1 > T2) are mixed together with the pressure kept c
Ray Of Light [21]

Answer:

ΔS=2*m*Cp*ln((T1+T2)/(2*(T1*T2)^1/2))

Explanation:

The concepts and formulas that I will use to solve this exercise are the integration and the change in the entropy of the universe. To calculate the final temperature of the water the expression for the equilibrium temperature will be used. Similarly, to find the change in entropy from cold to hot water, the equation of the change of entropy will be used. In the attached image is detailed the step by step of the resolution.

4 0
3 years ago
The distance from the earth to a star that has identical apparent and absolute magnitudes is ____ .
PSYCHO15rus [73]
I think it's distance.
5 0
3 years ago
A ball is thrown upward at a 45° angle. Inthe absence of air resistance, the ballfollows aA. tangential curve.B. sine curve.C. p
Evgesh-ka [11]

As ball is projected up in air at an angle of 45 degree without any air resistance

Let the initial speed will be v

now we will have

In x direction

v_x = v cos45

in y direction

v_y = vsin45

now displacement in x direction

x = (vcos45)t + 0

displacement in y direction

y = (vsin45)t - \frac{1}{2}gt^2

now from above two equations we have

y = (vsin45)\frac{x}{vcos45} - \frac{1}{2}g(\frac{x}{vcos45})^2

y = xtan45 - \frac{1}{2v^2cos^245}gx^2

so above equation is a quadratic equation and hence it will be a parabolic curve

so correct answer will be

<em>C. parabolic curve.</em>

8 0
3 years ago
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