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mafiozo [28]
3 years ago
7

2. Jessie studies seaweed and she usually takes large amounts of samples. She would take

Physics
1 answer:
Leni [432]3 years ago
7 0

Answer:

a

Explanation:

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The minimum cooking temperature for eggs that will be served immediately is
solong [7]
Cooking and Serving. Cook raw shell eggs that are broken for immediate preparation and service to heat all parts of the food to a temperature of 63°C<span> (</span>145°F<span>) for 15 seconds</span>
5 0
3 years ago
Read 2 more answers
A 94.7 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.75 r
Marta_Voda [28]

Answer:

The angular velocity of the platform is 1.114 rad/s.

Explanation:

Step 1:  Given data

Mass of the horizontal circular platform = 94.7 kg

Mass of the monkey = 21.1 kg

Initial angular velocity = 1.75 rad/s

A monkey drops a 9.25 kg bunch of bananas

They hit the platform at 4/5 of its radius from the center

Model the platform as a disk of radius 1.63 m

Step 2: Calculate the moment of inertia of the disk

I = ½ * m * r² = ½ * 94.7 * 1.63² = 125.80

Step 3: Calculate the initial angular momentum

I = 125.80 * 1.75 = 220.15

Step 4: Calculate the moment of inertia for the bananas

For the bananas, r = 4/5 * 1.63 = 1.304 m

I = 9.25 * 1.304² = 15.73

Step 5: Calculate Moment of inertia for the monkey

I = 21.1 * 1.63² = 56.06

Step 6: Total moment of inertia = 125.80 + 15.73 + 56.06 = 197.59

Step 7: Calculate final angular momentum = 197.59 * ω

197.59 * ω = 220.15

ω = 220.15 / 197.59

This is approximately 1.114 rad/s.

3 0
3 years ago
5/137 Under the action of its stern and starboard bow thrusters, the cruise ship has the velocity vB = 1 m/s of its mass center
quester [9]

The image is missing, so i have attached it;

Answer:

A) V_rel = [-(2.711)i - (0.2588)j] m/s

B) a_rel = (0.8637i + 0.0642j) m/s²

Explanation:

We are given;

the cruise ship velocity; V_b = 1 m/s

Angular velocity; ω = 1 deg/s = 1° × π/180 rad = 0.01745 rad/s

Angular acceleration;α = -0.5 deg/s² = 0.5 x π/180 rad = -0.008727 rad/s²

Now, let's write Velocity (V_a) at A in terms of the velocity at B(V_b) with r_ba being the position vector from B to A and relative velocity (V_rel)

Thus,

V_a = V_b + (ω•r_ba) + V_rel

Now, V_a = 0. Thus;

0 = V_b + (ω•r_ba) + V_rel

V_rel = -V_b - (ω•r_ba)

From the image and plugging in relevant values, we have;

V_rel = -1[(cos15)i + (sin15)j] - (0.01745k * -100j)

V_rel = - (cos15)i - (sin15)j - 1.745i

Note that; k x j = - i

V_rel = [-(2.711)i - (0.2588)j] m/s

B) Let's write the acceleration at A with respect to B in terms of a_b.

Thus,

a_a = a_b + (α*r_ba) + (ω(ω•r_ba)) + (2ω*v_rel) + a_rel

a_a and a_b = 0.

Thus;

0 = (α*r_ba) + (ω(ω•r_ba)) + (2ω*v_rel) + a_rel

a_rel = - (α*r_ba) - (ω(ω•r_ba)) - (2ω*v_rel)

Plugging in the relevant values with their respective position vectors, we have;

a_rel = - (-0.008727k * -100j) - (0.01745k(0.01745k * -100j)) - (2*0.01745k * [-(2.711)i - (0.2588)j])

a_rel = 0.8727i - (0.01745² x 100)j + 0.0946j - 0.009i

Note that; k x j = - i and k x i = j

Thus,simplifying further ;

a_rel = 0.8637i + 0.0642j m/s²

4 0
3 years ago
A concave mirror with a radius of curvature of 20 cm has a focal length of
xxTIMURxx [149]

Answer:

A concave mirror has a radius of curvature of 20 cm. What is it's focal length? If an object is placed 15 cm in front of it, where would the image be formed? What is it's magnification?

The focal length is of 10 cm, object distance is 30 cm and magnification is -2.

Explanation:

Given:

A concave mirror:

Radius of curvature of the mirror, as C = 20 cm

Object distance in-front of the mirror = 15 cm

a.

Focal length:

Focal length is half of the radius of curvature.

Focal length of the mirror =  \frac{C}{2} = 10 cm

According to the sign convention we will put the mirror on (0,0) point, of the Cartesian coordinate open towards the negative x-axis.

Object and the focal length are also on the negative x-axis where focal length and image distance will be negative numerically.

b.

We have to find the object distance:

Formula to be use:

⇒ \frac{1}{focal\ length}= \frac{1}{image\ distance} + \frac{1}{object\ distance}

⇒ Plugging the values.

⇒ \frac{1}{-10} =\frac{1}{image\ distance}+\frac{1}{-15}

⇒ \frac{1}{-10} -\frac{1}{-15}=\frac{1}{image\ distance}

⇒ \frac{1}{-10} + \frac{1}{15}=\frac{1}{image\ distance}

⇒ \frac{-3+2}{30} =\frac{1}{image\ distance}

⇒ \frac{-1}{30} =\frac{1}{image\ distance}

⇒ -30\ cm=image\ distance

Image will be formed towards negative x-axis 30 cm away from the pole.

c.

Magnification (m) is the negative ratio of mage distance and object distance:

⇒ m=-\frac{image\ distance}{object\ distance}

⇒ m=-\frac{(-30)}{(-15)}

⇒ m=-2

The focal length of the concave mirror, is of 10 cm, object distance is 30 cm and magnification is -2.

5 0
3 years ago
What is the frequency, in hertz, of a sound wave (v = 340 m/s) with a wavelength of 68 m?
hram777 [196]

Answer:

frequency = 5 Hz

Explanation:

F = v/wavelength

F = 340/68 =5Hz

5 0
3 years ago
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