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USPshnik [31]
3 years ago
8

This is for 6.02 in comprehensive science for flvs

Physics
1 answer:
Katen [24]3 years ago
4 0

Answer:

Please provide an image to help clarify

Explanation:

thanks :)

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Need help transforming one formula to another!
Sindrei [870]

Hello! I don't know if this will help you any but I worked it out, got nothing on it. So, I just had my twin help me and he had nothing either. But, this is just how to find the transformation. Seems like this is already complete to me.

To find the transformation, compare the equation to the parent function and check to see if there is a horizontal or vertical shift, reflection about the x-axis or y-axis, and if there is a vertical stretch.

Parent function : y=√a

Horizontal shift : None

Vertical shift : none

Reflection about the x-axis : none

Vertical stretch : Stretched.

I am so sorry if this doesn't help you but to me, in both my eyes and my brother's eyes, this already looks complete. I hope this helps you out. Again, so sorry if it doesn't.

-Karleif ☺

5 0
4 years ago
What speed would a fly with a mass of 0.72 g need in order to have the same kinetic energy as a 1250 kg automobile traveling at
Elena L [17]
Kinetic of automobile

Mass m = 1,250 Kg;         V = 11 m/s

Formula: K.E = 1/2 mV²

               K.E = 1/2(1,250 Kg)(11 m/s)²

               K.E = 75,625 J

Speed required for insect to have the same kinetic energy as automobile

Mass of insect = 0.72 g convert to Kg   m = 7.2 x 10⁻⁴ Kg

K.E = 1/2 mV²  Derive V =?

 V = 2 K.E/m

 V = √2(75,625 J)/7.2 x 10⁻4 Kg

 V = √2.1 x 10⁸ m²/s²

 V = 14,491.34 m/s  (velocity of insect)









5 0
3 years ago
An anchor with a density of 700 lb/ft3 tied to a rope is submerged in water (but is not touching the bottom). It has a volume of
Ivanshal [37]

Answer:

Explanation:

Given

Density of anchor \rho _a=700\ lb/ft^3

Volume of anchor V=2.8\ ft^3

Density of water \rho _w=62.4\ lb\ft^3

Buoyancy force on submerged object is given by

F_B=\rho _W\times V\times g

F_B=62.4\times 2.8\times 32.2=5625.98\approx 5626\ approx lbf          

3 0
3 years ago
The net external force on the propeller of a 3.0 kg model airplane is 6.8 N forward.What is the acceleration of the airplane? An
solong [7]

The correct answer to the question is 2.27 m/s^2 i.e the acceleration of the body is 2.27 m/s^2 along the forward direction.

CALCULATION:

As per the question, the net external force on the propeller of model airplane F =  6.8 N.

The mass of the model air plane m = 3.0 kg

We are asked to calculate the acceleration of the air plane.

From Newton's second law of motion, we know that the net external force acting on a body is equal to the product of mass with acceleration of that body.

Mathematically force F = m × a

                                  ⇒  a=\ \frac{F}{m}

                                          =\ \frac{6.8\ N}{3.0\ kg}

                                          =\ 2.27\ m/s^2                  [ans]

The direction of acceleration is along the direction of force. Hence, the acceleration of the propeller is 2.27 m/s^2 along forward direction.

8 0
4 years ago
Haley noticed a small spot on her skin that turned out to be skin cancer. Which treatment is her doctor most likely to use?
Igoryamba

Answer:

I'm going to guess D, or B

Explanation:

Medical oncologist A physician who treats cancer with medication, such as chemotherapy, immunotherapy, or targeted therapy. Radiation oncologist These doctors treat cancer with radiation therapy.

Simple excision, Mohs micrographic surgery, curettage and electrodesiccation, and cryosurgery are used to treat basal cell carcinoma and squamous cell carcinoma of the skin. Laser surgery is rarely used to treat basal cell carcinoma

7 0
3 years ago
Read 2 more answers
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