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katen-ka-za [31]
3 years ago
11

two objects were lifted by a machine. One object had a mass of 2 kilograms, was lifted at a speed of 2m/sec . The other had a ma

ss of 4 kilograms and was lifted at 3m/sec. Which object had more kinetic energy while it was being lifted? Show all calculations
Physics
1 answer:
Lubov Fominskaja [6]3 years ago
5 0

Object 2 has more kinetic energy

Explanation:

The kinetic energy of an object is the energy possessed by the object due to its motion, and it is given by

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

In this problem, for object 1:

m = 2 kg

v = 2 m/s

So its kinetic energy is

K_1 = \frac{1}{2}(2)(2)^2=4 J

For object 2,

m = 4 kg

v = 3 m/s

So its kinetic energy is

K_2 = \frac{1}{2}(4)(3)^2=18 J

Therefore, object 2 has more kinetic energy.

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

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The mass of a colony of bacteria, in grams, is modeled by the function P given by P(t)=2+5tan−1(t2), where t is measured in days
ololo11 [35]

Answer:

<em>P'(1.015)=4.93 grams</em>

Explanation:

<u>Instantaneous Rate Of Change </u>

If one variable y is a function of another variable x, the rate of change can be measured by changing x by a small amount (dx) and computing the change in y (dy). The rate of change is

\displaystyle \frac{dy}{dx}

When dx tends to zero, we call it the instantaneous rate of change and is easily computed as the first derivative of y

Note: We'll be using the standart notation atan for the inverse tangent

We are given the relation between the mass of a colony of bacteria in grams P(t) and the time t in days

P(t)=2+5atan(t^2)

Let's find its first derivative, recalling that

[atan(u)]'=\displaystyle \frac{u'}{1+u^2}

P'(t)=\displaystyle 5\frac{(t^2)'}{1+(t^2)^2}

P'(t)=\displaystyle 5\frac{2t}{1+t^4}

P'(t)=\displaystyle \frac{10t}{1+t^4}

We need to know the value of t, so we use the provided condition P=6 gr

P(t)=2+5atan(t^2)=6

\displaystyle atan(t^2)=\frac{4}{5}

\displaystyle (t^2)=tan\left ( \frac{4}{5} \right )

\displaystyle t=\sqrt{tan\left ( \frac{4}{5} \right )}

\displaystyle t=1,015\ days

We use this value in the derivative:

P'(1.015)=\displaystyle \frac{10(1.015)}{1+(1.015)^4}

P'(1.015)=4.93\ grams

4 0
3 years ago
Use a variation model to solve for the unknown value. Use as the constant of variation. The stopping distance of a car is direct
svlad2 [7]

Complete question is;

Use a variation model to solve for the unknown value.

The stopping distance of a car is directly proportional to the square of the speed of the car.

a. If a car traveling 50 mph has a stopping distance of 170 ft, find the stopping distance of a car that is traveling 70 mph.

b. If it takes 244.8 ft for a car to stop, how fast was it traveling before the brakes were applied?

Answer:

A) d = 333.2 ft

B) 60 mph

Explanation:

Let the stopping distance be d

Let the speed of the car be v

We are told that the stopping distance is directly proportional to the square of the speed of the car. Thus;

d ∝ v²

Therefore, d = kv²

Where k is constant of variation.

A) Speed is 50 mph and stopping distance of 170 ft.

v = 50 mph

d = 170 ft = 0.032197 miles

Thus,from d = kv², we have;

0.032197 = k(50²)

0.032197 = 2500k

k = 0.032197/2500

k = 0.0000128788

If the car is now travelling at 70 mph, then;

d = 0.0000128788 × 70²

d = 0.06310612 miles

Converting to ft gives;

d = 333.2 ft

B) stopping distance is now 244.8 ft

Converting to miles = 0.046363636 miles

Thus from d = kv², we have;

0.046363636 = 0.0000128788(v²)

v² = 0.046363636/0.0000128788

v² = 3599.99658

v = √3599.99658

v ≈ 60 mph

6 0
3 years ago
A car with a mass of 860 kg sits at the top of a 32 meter high hill. Describe the transformations between potential and kinetic
I am Lyosha [343]

Answer:

Kinetic energy at the bottom equals potential energy at the top hence 269971.2  J

Explanation:

Assuming no friction and neglecting air resistance, from the law of conservation of energy, the potential energy that the car posses at 32 m is all converted to kinetic energy hence

PE= mgh

KE= 0.5mv^{2}

Here m denotes the mass of the car, g is acceleration due to gravity, h is the height and v is the speed of the car

The potential energy will be PE= 860*9.81*32=269971.2  J

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What type of machine are wire cutter pliers?
love history [14]
It a compound machine
4 0
3 years ago
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When light strikes an object, what happens to some of the light?
svp [43]

Answer:

The light wave could be reflected by the object

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3 years ago
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