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crimeas [40]
3 years ago
5

You throw a rock upward. The rock is moving upward, but it is slowing down. If we define the ground as the origin, the position

of the rock is _____ and the velocity of the rock is _____. You throw a rock upward. The rock is moving upward, but it is slowing down. If we define the ground as the origin, the position of the rock is _____ and the velocity of the rock is _____. negative, negative positive, negative negative, positive positive, positive
Physics
1 answer:
Alecsey [184]3 years ago
3 0

Answer:

positive, positive

You throw a rock upward. The rock is moving upward, but it is slowing down. If we define the ground as the origin, the position of the rock is positive and the velocity of the rock is positive

Explanation:

Given that the ground is defined as the origin.

The position of the rock is positive since the rock is thrown upward, the position also increases with time until it reaches the maximum height. Also, since the rock is thrown upward with the ground as the origin, the velocity of the rock is positive but the velocity reduces with time (change in height per unit time as the rock moves up is positive)

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List two of the number of strategies you have learned
jolli1 [7]

Answer:

what?

Explanation:

does that means also i learned Ratios and multiplication i keared them like 5 years ago?

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A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica
Ugo [173]

Answer:

f_{e}  = 1.9 cm

Explanation:

The magnification of a microscope is the product of the magnification of the eyepiece by the magnifier with the objective

    M = M₀ m_{e}

Where M₀ is the magnification of the objective and  m_{e} is the magnification of the eyepiece.

The eyepiece is focused to the near vision point (d = 25 cm)

       m_{e} = 25 /  f_{e}

The objective is focused on the distances of the tube (L)

     M₀ = -L / f₀

Substituting

     M = - L/f₀    25/f_{e}  

1) Let's look for the focal length of the eyepiece (faith)

     f_{e}  = - L 25 / f₀ M

     M = 400X = -400

     f_{e}  = - 12 25 /0.40 (-400)

     f_{e}  = 1.875 cm

Let's approximate two significant figures

    f_{e}  = 1.9 cm

8 0
4 years ago
What is square rot of 80
guajiro [1.7K]

The square root of 80 is: 8.944

3 0
3 years ago
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What is the kinetic energy of a 478 kg object that is moving with the speed of 15 m/s
Tpy6a [65]
Kinetic energy = (1/2)*mass*velocity^2
KE = (1/2)mv^2
KE = (1/2)(478)(15)^2
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5 0
3 years ago
A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coe
andrey2020 [161]

Answer:

The magnitude of force is 1593.4N

Explanation:

The sum of the horizontal components of the friction and the normal force will be equal to the centripetal force on the car. This can be represented as

fcostheta + Nsintheta = mv^2/r

Where F = force of friction

Theta = angle of banking

N = normal force

m = mass of car

v = velocity of car

r = radius of curve

The car has no motion in the vertical direction so the sum of forces = 0

The vertical component of the normal force acts upwards whereas the weight of the car and the vertical component friction acts downwards.

Taking the upward direction to be positive,rewrite the equation above to get:

Ncos thetha = mg - fsintheta =0

Ncistheta = mg + fain theta

N = mg/cos theta + sintheta/ costheta

fcostheta +[mg/costheta + ftan theta] sin theta = mv^2/r

Substituting gives:

f = (1/(costheta + tanthetasintheta) + mgtantheta = mv^2/r - mgtantheta)

Substituting given values into the above equation

f = 1/(cos25 + tan 25 )(sin25)[ 600×30/120 - (600×9.81)tan

f = 1593.4N25

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3 years ago
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