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Paraphin [41]
3 years ago
6

What is the kinetic energy of a 74.0 kg skydiver falling at a terminal velocity of 52.0m/s?

Physics
1 answer:
skad [1K]3 years ago
8 0

Answer:

100,048

Explanation:

K.E = 1/2 m (v)^2

K.E = 1^/2 * 74 * (52)^2

K.E = 100,048J =100.048kJ

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A linear elastic spring can be compressed 10. 0 cm by an applied force of 5. 0 N. A 4. 5
Daniel [21]

The maximum compression of the spring is 0.6 m.

<h3>Spring constant</h3>

The spring constant of the spring is calculated as follows;

F = kx

k = F/x

k = (5)/(0.1)

k = 50 N/m

<h3>Maximum compression of the spring</h3>

The maximum compression of the spring is calculated as follows;

U = K.E

¹/₂kx²  = ¹/₂mv²

x² = (mv²)/(k)

x = √(mv²)/(k)

x = √(4.5 x 2²)/(50)

x = 0.6 m

Thus, the maximum compression of the spring is 0.6 m.

Learn more about spring constant here: brainly.com/question/1968517

4 0
2 years ago
A car starts from rest at the top of a hill with 45 J of gravitational
kherson [118]

Answer:

<em>The car will be moving at 5.48 m/s at the bottom of the hill</em>

Explanation:

<u>Principle of Conservation of Mechanical Energy</u>

In the absence of friction, the total mechanical energy is conserved. That means that

E_m=U+K is constant, being U the potential energy and K the kinetic energy

U=mgh

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

When the car is at the top of the hill, its speed is 0, but its height h should be enough to produce the needed speed v down the hill.

The Kinetic energy is then, zero. When the car gets enough speed we assume it is achieved at ground level, so the potential energy runs out to zero but the Kinetic is at max. So the initial potential energy is transformed into kinetic energy.

We are given the initial potential energy U=45 J. It all is transformed to kinetic energy at the bottom of the hill, thus:

\displaystyle \frac{mv^2}{2}=45

Multiplying by 2:

\displaystyle mv^2=90

Dividing by m:

\displaystyle v^2=\frac{90}{m}

Taking square roots:

\displaystyle v=\sqrt{\frac{90}{m}}

\displaystyle v=\sqrt{\frac{90}{3}}

v=\sqrt{30}

v = 5.48 m/s

The car will be moving at 5.48 m/s at the bottom of the hill

3 0
3 years ago
If a calorie is equivalent to 4.184 joules how many joules are contained in 250 kg calories slice of pizza
Andreyy89
In order to answer this, we will set up a simple ratio as such:

1 calorie = 4.184 joules

1 kilocalorie = 1000 calories

1 kilocalorie = 4,184 joules

250 kilocalories = x joules

Cross multiplying the second and third equations, we get:

x joules = 4,184 * 250

250 kilocalories are equivalent to 1,046 kJ
6 0
3 years ago
Assortative mating changes ______ frequencies but does not change ______ frequencies.
andrew-mc [135]

Allele frequencies are unaffected by assortative mating, but genotype frequencies .

<h3>Assortative mating: </h3>

Individuals with similar phenotypes and genotypes mate with others more frequently than is anticipated under a random mating pattern in assortative mating, which is a mating pattern and a type of sexual selection.

<h3>Frequencies of genotypes:</h3>

A population's genotype frequency is calculated by dividing the number of people having a particular genotype by the overall population size. The genotype frequency in population genetics is the frequency or ratio (i.e., 0 f 1) among genotypes inside a population.

<h3>The frequency for alleles in biology:</h3>

The term "allele frequency" describes the prevalence of an allele in a population. It is calculated by calculating the number of times the allele occurs in the population and dividing by the sum of all the gene copies.

To know more about Assortative mating visit:

brainly.com/question/28238408

#SPJ4

4 0
1 year ago
A rectangular coil with 50 turns of conducting wire and a total resistance of 10.0 Ω initially lies in the yz-plane at time t =
castortr0y [4]

Answer:

a) 43.20V

b) 2.71W/s

c) 40.25s

d) 7.77Nm

Explanation:

(a) The emf of a rotating coil with N turns is given by:

emf=NBA\omega sin(\omega t)

N: turns

B: magnitude of the magnetic field

A: area

w: angular velocity

the emf max is given by:

emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V

(b) the maximum rate of change of the magnetic flux is given by:

\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}

(c) emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V

(d) The torque is given by:

\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm

3 0
3 years ago
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