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nexus9112 [7]
2 years ago
15

A fisherman casts his bait toward the river at an angle of 25° above the horizontal. As the line unravels, he notices that the b

ait and hook reach a maximum height of 2.9 m. What was the initial velocity he launched the bait with? Assume that the line exerts no appreciable drag force on the bait and hook and that air resistance is negligible. A fisherman casts his bait toward the river at an angle of 25° above the horizontal. As the line unravels, he notices that the bait and hook reach a maximum height of 2.9 m. What was the initial velocity he launched the bait with? Assume that the line exerts no appreciable drag force on the bait and hook and that air resistance is negligible. 6.3 m/s 18 m/s 7.9 m/s 7.6 m/s
Physics
1 answer:
Butoxors [25]2 years ago
6 0

Answer:

v_{o}=18m/s

Explanation:

According to the exercise we know the angle which the bait was released and its maximum height

\beta =25\\y=2.9m

To find the initial y-component of velocity we need to do the following steps:

v_{y}^{2} =v_{oy}^{2}+2g(y-y_{o})

At maximum height the y-component of velocity is 0 and from the exercise we know that the initial y position is 0

0=v_{oy}^{2}-2(9.8m/s^2)(2.9m)

v_{oy}=\sqrt{2(9.8m/s^{2} )(2.9m)} =7.53m/s

Since the bait is released at 25º

v_{oy}=v_{o}sin(25)

v_{o}=\frac{7.53m/s}{sin(25)}=18m/s

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Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched o
IRINA_888 [86]

Answer:

a) W=-0.0103125\ J

b) W=0.0059375\ J

c) Compressing is easier

Explanation:

Given:

Expression of force:

F=kx-bx^2+cx^3

where:

k=100\ N.m^{-1}

b=700\ N.m^{-2}

c=12000\ N.m^{-3}

x when the spring is stretched

x when the spring is compressed

hence,

F=100x-700x^2+12000x^3

a)

From the work energy equivalence the work done is equal to the spring potential energy:

here the spring is stretched so, x=-0.05\ m

Now,

The spring constant at this instant:

j=\frac{F}{x}

j=\frac{100\times (-0.05)-700\times (-0.05)^2+12000\times (-0.05)^3}{-0.05}

j=-8.25\ N.m^{-1}

Now work done:

W=\frac{1}{2} j.x^2

W=0.5\times -8.25\times (-0.05)^2

W=-0.0103125\ J

b)

When compressing the spring by 0.05 m

we have, x=0.05\ m

<u>The spring constant at this instant:</u>

j=\frac{F}{x}

j=\frac{100\times (0.05)-700\times (0.05)^2+12000\times (0.05)^3}{0.05}

j=4.75\ N.m^{-1}

Now work done:

W=\frac{1}{2} j.x^2

W=0.5\times 4.75\times (0.05)^2

W=0.0059375\ J

c)

Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.

8 0
1 year ago
The δe of a system that releases 12.4 j of heat and does 4.2 j of work on the surroundings is __________ j.
Vedmedyk [2.9K]

The answer for this problem is clarified through this, the system is absorbing (+). And now see that it uses that the SURROUNDINGS are doing 84 KJ of work. Any time a system is overshadowing work done on it by the surroundings the sign will be +. So it's just 12.4 KJ + 4.2 = 16.6 KJ.

5 0
2 years ago
An alloy is made of a material of specific gravity 7.87 and another material of specific gravity 4.50. The alloy of mass 750g ha
julsineya [31]

Answer:

13.9

Explanation:

Apparent weight is the normal force.  Sum of the forces on the alloy when it is submerged:

∑F = ma

N + B − W = 0

N + ρVg − mg = 0

6.6 + (0.78 × 1000) V (9.8) − (0.750) (9.8) = 0

V = 9.81×10⁻⁵

If x is the volume of the first material, and y is the volume of the second material, then:

x + y = 9.81×10⁻⁵

(7.87×1000) x + (4.50×1000) y = 0.750

Two equations, two variables.  Solve with substitution:

7870 (9.81×10⁻⁵ − y) + 4500 y = 0.750

0.772 − 7870 y + 4500 y = 0.750

0.0222 = 3370 y

y = 6.58×10⁻⁶

x = 9.15×10⁻⁵

The ratio of the volumes is:

x/y = 13.9

8 0
2 years ago
A man weighing 750 n and a woman weighing 500 n have the same momentum. what is the ratio of the man's kinetic energy km to that
miss Akunina [59]
Because weight W = M g, the ratio of weights equals the ratio of masses.

(M_m g)/ (M_w g) = [ (p^2 Man )/ (2 K_man)] / [ (p^2 Woman )/ (2 K_woman)

but p's are equal, so

K_m/K_m = (M_w g)/(M_m g) = W_woman / W_man = 450/680 = 0.662
4 0
2 years ago
In the context of energy transfers with hot and cold reservoirs, the sign convention is that _______________.
Likurg_2 [28]

Answer:

B. QC > 0; QH < 0

Explanation:

Given that there are two reservoir of energy.

Sign convention for heat and work :

1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

From hot reservoir heat is going out that is why it is taken as negative

Q_H

From cold reservoir heat is coming inside the reservoir that is why it is taken as positive

Q_C>0

That is why the answer will be

Q_H ,Q_C>0

8 0
2 years ago
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