A) If the squid has 1.6 kg of water in its cavity, how fast should it expel this water so that it reaches 2.5 m/s and escapes the predator? What is the kinetic energy of the squid after this maneuver?

2 months ago

Guest #9800807

2 months ago

**Answer:**

see attachment

**Explanation:**

Question

A 4.3-kg block slides down an inclined plane that makes an angle of 30° with the horizontal. Starting from rest, the block slides a distance of 2.7 m in 5.8 s. Find the coefficient of kinetic friction between the block and plane.

Solution 1

**Answer:**

0.56

**Explanation:**

Let the coefficient of friction is μ.

m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s

By the free body diagram,

Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N

Friction force, f = μ N = 36.49 μ

Net force acting on the block,

Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ

Fnet = 21.07 - 36.49μ

Net acceleartion, a = Fnet / m

a = (21.07 - 36.49μ) / 4.3

use second equation of motion

s = ut + 1/2 a t^2

2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3

By solving we get

μ = 0.56

Question

An object with mass M is attached to the end of a string and is raised vertically at a constant acceleration of g 10 . If it has been raised a distance ℓ from rest, how much work has been done by the tension in the string?

Solution 1

The work by tension is **¹¹/₁₀ Mg***ℓ*

[tex]\texttt{ }[/tex]

__Complete Question:__

*An object with mass M is attached to the end of a string and is raised vertically at a constant acceleration of g/10 . If it has been raised a distance ℓ from rest, how much work has been done by the tension in the string?*

[tex]\texttt{ }[/tex]

__Given:__

Mass of the object = M

Acceleration of the object = g/10

Distance = *ℓ*

__Asked:__

Work by Tension = W = ?

__Solution:__

Let's find the magnitude of tension as follows:

[tex]\Sigma F = ma[/tex]

[tex]T - Mg = Ma[/tex]

[tex]T = Mg + Ma[/tex]

[tex]T = M(g + a)[/tex]

[tex]T = M(g + \frac{1}{10}g)[/tex]

[tex]T = M(\frac{11}{10}g)[/tex]

[tex]T = \frac{11}{10}Mg[/tex]

[tex]\texttt{ }[/tex]

[tex]W = T \times L[/tex]

[tex]W = \frac{11}{10}Mg \times L[/tex]

[tex]W = \frac{11}{10}MgL[/tex]

[tex]\texttt{ }[/tex]

The work by tension is **¹¹/₁₀ Mg***ℓ*

[tex]\texttt{ }[/tex]

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[tex]\texttt{ }[/tex]

**Grade: **High School

**Subject: **Physics

**Chapter: **Dynamics

[tex]\texttt{ }[/tex]

**Keywords:** Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

Solution 2

**Answer:**

The **work done** by tension in string is [tex]\dfrac{11}{10}MgL}[/tex].

**Explanation:**

**Given data:**

**Mass** of object is, **M**.

**Acceleration** of object is, **a = g/10**.

**Distance** covered vertically is, **L**.

The work done by **tension** in the string is given as,

[tex]W = T \times L[/tex] .......................................................... (1)

Here, **T **is the **tension** force on string.

Apply the **equilibrium of forces** on string as,

[tex]T- Mg=Ma[/tex]

Here, **g **is gravitational acceleration.

[tex]T- Mg=M(\dfrac{g}{10} )\\T=M(\dfrac{g}{10} )+Mg\\T=\dfrac{11}{10}Mg[/tex]

Substituting value in equation (1) as,

[tex]W = \dfrac{11}{10}Mg \times L\\W = \dfrac{11}{10}MgL}[/tex]

Thus, **the work done** by tension in string is [tex]\dfrac{11}{10}MgL}[/tex].

For more details, refer to the link:

https://brainly.com/question/14871010?referrer=searchResults

Question

A tank holds a 1.44-m thick layer of oil that floats on a 0.98-m thick layer of brine. Both liquids are clear and do not intermix. Point O is at the bottom of the tank, on a vertical axis. The indices of refraction of the oil and the brine are 1.40 and 1.52, respectively. A ray originating at O crosses the brine-oil interface at a point 0.7 m from the axis. The ray continues and emerges into the air above the oil. What is the angle that the ray in the air makes with the vertical axis

Solution 1

**Answer:**

Angle of ray makes with the vertical is 62.1 degree

**Explanation:**

As per the ray diagram we know that the angle of incidence on oil brine interface will be given as

[tex]tan\theta_i = \frac{0.7}{0.98}[/tex]

[tex]\theta_i = 35.5^0[/tex]

now by Snell'a law at that interface we have

[tex]\mu_1 sin\theta_i = \mu_2 sin \theta_r[/tex]

now we will have

[tex]1.52 sin35.5 = 1.40 sin\theta_r[/tex]

[tex]\theta_r = 39.12^0[/tex]

now this is the angle of incidence for oil air interface

so now again by Snell's law we will have

[tex]\mu_2 sin\theta_i' = \mu_{air} sin\theta[/tex]

[tex]1.40 sin39.12 = 1 sin\theta[/tex]

[tex]\theta = 62.1^0[/tex]

Question

A rope attached to a load of 175 kg bricks Ilifts the bricks with a steady acceleration of 0.12.m/s^2 straight up. What is the tension in the rope? (a)2028N (b)1645 N (c) 1894 N (d) 1976 N (e) 1736 N (f) 1792 N

Solution 1

**Answer:**

Tension, T = 1736 N

**Explanation:**

It is given that,

Mass of bricks, m = 175 kg

A rope is attached to a load of 175 kg bricks lifts the bricks with a steady acceleration of 0.12 m/s² in vertically upwards direction. let T is the tension in the rope. Using second equation of motion as :

T - mg = ma

T = ma + mg

T = m(a + g)

T = 175 kg ( 0.12 m/s² + 9.8 m/s² )

**T = 1736 N**

**Hence, the tension in the wire is 1736 N.**

Solution 2

**Answer:**

The tension in the rope is 1736 N.

**(e) is correct option.**

**Explanation:**

**Given that,**

Mass of bricks = 175 kg

Acceleration = 0.12 m/s²

**Let T is the tension in the rope.**

A rope attached to a load of 175 kg bricks lifts the bricks with a steady acceleration of 0.12.m/s^2 in vertically upward direction.

Using equation of balance

[tex]T-mg=ma[/tex]

[tex]T=mg+ma[/tex]

[tex]T=175(9.8+0.12)[/tex]

[tex]T= 1736\ N[/tex]

**Hence, The tension in the rope is 1736 N.**

Question

Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x = +2.0 cm respectively. Where should a third charge of +3.0-μC be placed on the +x-axis so that the potential at the origin is equal to zero? (k = 1/4πε0 = 8.99 × 109 N · m2/C2)

Solution 1

**Answer:**

see attachment

**Explanation:**

Solution 2

The** electric potential **varies inversely with the **distance**. So, the third charge should be placed at a distance of** 3 cm** from the origin on the x-axis.

The **work done** on an electric charge to shift it from infinity to a point is known as electric potential at that point. And its expression is,

[tex]V = \dfrac{kq}{r}[/tex]

here, **k** is the **coulomb's constant**.

**Given data:**

The magnitude of two point **charges** are, [tex]+2.0 \;\rm \mu C[/tex] and [tex]-6.0 \;\rm \mu C[/tex].

The **location** of each charge on the x-axis is** -1.0 cm** and **+2.0 cm**.

Let the third charge ( [tex]+3.0 \;\rm \mu C[/tex] ) be placed at a **distance** of** x**. Then the **electric potential** at origin is,

[tex]V = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times -6.0}{0.02} +\dfrac{k \times 3.0}{x}[/tex]

Since, **potential** at **origin** is **zero** **(V = 0)**. Then,

[tex]0 = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times -6.0}{0.02} +\dfrac{k \times 3.0}{x}\\\\\dfrac{k \times 6.0}{0.02} = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times 3.0}{x}\\\\\dfrac{6.0}{0.02} = \dfrac{2.0}{0.01} +\dfrac{3.0}{x}\\\\x = 0.03 \;\rm m =3 \;\rm cm[/tex]

Thus, we can conclude that the **third charge** should be placed at a distance of **3 cm** from the origin on the **x-axis**.

Learn more about the **electric potential** here:

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Question

what net force is needed to give a 48.0kg grocery cart an acceleration of 3.39m/s^2 ?

Solution 1

you use the formula F=ma to get the answer

Question

A nuclear power plant operates at 40.0% efficiency with a continuous production of 1192 MW of usable power in 1.00 year and consumes 1.22×106 g of uranium-235 in this time period. What is the energy in joules released by the fission of a single uranium-235 atom?

Solution 1

**Answer:**

the answer is in the picture below......

Question

Consider two isolated, charged conducting spheres: a large sphere and a second smaller sphere with a radius 6 times smaller than that of the large sphere, but with 3 times as much charge.

(a) Calculate the ratio of the electric potential at the surface of the large sphere to that of the small sphere.

(a) Calculate the ratio of the electric potential at the surface of the large sphere to that of the small sphere.

Solution 1

Let the bigger sphere be sphere 1 and the let the smaller sphere be sphere 2. Rest of the answer is in the picture.

Question

Consider a solid sphere of radius R = 0.4 m that is uniformly charged with ? = -11 ?C/m3. What is the electric potential a distance 5 m from the center of the sphere?

Solution 1

**Answer:**

V=-5304.6V

**Explanation:**

*(I set the charge density unit to **m**i**c**r**o**c**o**u**l**o**m**b**s** **p**e**r** **m**e**t**r**e** **c**u**b**e**d**)*

**A****n****s****w****e****r****:**

In the picture.

Question

A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 50.4 m/s and returns the shot with the ball traveling horizontally at 37.0 m/s in the opposite direction. (Take the direction of the ball's final velocity (toward the net) to be the +x-direction.)

(a) What is the impulse delivered to the ball by the racket?

(b) What work does the racket do on the ball? (Indicate the direction with the sign of your answer.)

(a) What is the impulse delivered to the ball by the racket?

(b) What work does the racket do on the ball? (Indicate the direction with the sign of your answer.)

Solution 1

(a) The impulse delivered to the ball by the racket is **5.24 kg.m/s**

(b) The work that the racket does on the ball is **-35.1 Joule**

__Given :__

mass of ball = m = 0.06 kg

initial velocity = v₁ = -50.4 m/s

final velocity = v₂ = 37.0 m/s

__Unknown :__

(a) Impulse = I = ?

(b) Work = W = ?

__Solution :__

In this question , we could use the formula from Second Law of Newton :

[tex]I = \Delta p[/tex]

[tex]I = p_2 - p_1[/tex]

[tex]I = m \times v_2 - m \times v_1[/tex]

[tex]I = m \times (v_2 - v_1)[/tex]

[tex]I = 0.06 \times (37.0 - (-50.4))[/tex]

[tex]I = 0.06 \times (87.4)[/tex]

[tex]I = 5.244~kg.m/s[/tex]

[tex]\large { \boxed {I \approx 5.24~kg.m/s} }[/tex]

[tex]W = F \times d[/tex]

[tex]W = (\frac{I}{\Delta t})(\frac {v_1 + v_2}{2} \Delta t)[/tex]

[tex]W = \frac{I(v_1 + v_2)}{2}[/tex]

[tex]W = \frac{5.244(-50.4 + 37)}{2}[/tex]

[tex]W = \frac{5.244(-13.4)}{2}[/tex]

[tex]W = -35.1348~Joule[/tex]

[tex]\large { \boxed {W \approx -35.1~Joule}}[/tex]

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Example of Newton's Law: https://brainly.com/question/498822

**Grade:** High School

**Subject:** Physics

**Chapter:** Dynamics

**Keywords:** Newton, Law, Impulse, Work

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