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skelet666 [1.2K]

3 years ago

5

A scientist wants to calculate the kinetic energy of a cheetah as it chases down its prey. What does the scientist need to measu

re in order to determine the kinetic energy of the cheetah?
Select all that apply.

the cheetah's amount of matter
the cheetah's speed
the amount of energy the cheetah absorbs from the sun
the amount of energy the cheetah releases into its surroundings
the cheetah's temperature

2 answers:

neonofarm [45]3 years ago

5 0

Answer:

the cheetah's amount of matter

the cheetah's speed

Explanation:

Kinetic energy is calculated according to the following equation

K.E = $\frac{mv^{2} }{2}$

where m represents the mass of the body in kilograms

and v represents the magnitude of the velocity which is know as speed of the body in m/s

Knowing this then, the scientist needs two things from that list:

the cheetah's amount of matter (also known as the mass)

the cheetah's speed

Feliz [49]3 years ago

4 0

Answer:

A + B

Explanation:

A) The cheetah's amount of matter

B) The cheetah's speed

(P.S. I just took a test on this and got it correct.)

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A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point

il63 [147K]

Answer:

31.1 N

Explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as

$T + mg =\frac{mv^{2}}{r}$

Inserting the values

$T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}$

T = 31.1 N

7 0

3 years ago

Read 2 more answers

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

7 0

3 years ago

3 Draw energy transfer diagrams

timofeeve [1]

Answer:

The outline of the energy transfer are;

a) Kinetic energy → Clockwork spring → Potential energy

b) Potential energy in clockwork car → Clockwork spring coil unwound → Clockwork car run

c) Chemical potential energy → Batteries in the car → Electric motors → Kinetic energy

Please find attached the drawings of the energy transfer created with MS Visio

Explanation:

The energy transfer diagrams are diagrams that can be used to indicate the part of a system where energy is stored and the form and location to which the energy is transferred

a) The energy transfer diagram for the winding up a clockwork car is given as follows;

Mechanical kinetic energy is used to wind up (turn) the clockwork car such that the kinetic energy is transformed into potential energy and stored in the wound up clockwork as follows;

Kinetic energy → Clockwork spring → Potential energy

b) Letting a wound up clockwork car run results in the conversion of mechanical potential energy into kinetic (energy due tom motion) energy as follows;

Potential energy in clockwork car → Clockwork spring coil unwound → Clockwork car run

c) The energy stored in the battery of a battery powered car is chemical potential energy. When the battery powered car runs, the chemical potential energy produces an electromotive force which is converted into kinetic energy as electric current flows from the batteries

Therefore, we have;

Chemical potential energy → Batteries in the car → Electric motors → Kinetic energy

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3 years ago

A car moving around a circular track with constant speed (Select all that apply.) has an acceleration with a direction that cann

BARSIC [14]

Answer:

Explanation:

Has an acceleration directed towards center of its path

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3 years ago

What is kinematics???<br>explain!!! -,-

Lena [83]

Answer:

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7 0

2 years ago

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