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2 years ago

A flea jumps straight up to a maximum height of 0.540 m . what is its initial velocity v0 as it leaves the ground?

1 answer:

3 0

For an uniformly accelerated motion, we can write
$2ah=v_f^2-v_0^2$
where $a=g=-9.81 m/s^2$ is the acceleration of this motion, which in this problem is the gravitational acceleration, with a negative sign because it points downward, against the direction of the motion; h=0.540 m is the distance covered by the flea, and $v_0$ is the initial velocity.

At the maximum height, the velocity is zero, so $v_f =0$ . Therefore we can solve to find $v_0$ :
$v_0 = \sqrt{2ah}= \sqrt{2(9.81 m/s^2)(0.540 m)} =3.25 m/s$

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Part D- Isolating a Variable with a Coefficient In some cases, neither of the two equations in the system will contain a variabl

Shtirlitz [24]

Answer:

D) D = $\frac{5}{4} - \frac{3}{4} \ C$ , E) (C, D) = ( $\frac{17}{7}, \ \frac{-4}{7}$

Explanation:

Part D) two expressions are indicated

3C + 4D = 5

2C +5 D = 2

let's simplify each expression

3C + 4D = 5

4D = 5 - 3C

we divide by 4

D = $\frac{5}{4} - \frac{3}{4} \ C$

The other expression

2C +5 D = 2

2C = 2 - 5D

C = $1 - \frac{5}{2} \ D$

we can see that the correct result is 1

Part E.

It is asked to solve the problem by the substitution method, we already have

D = $\frac{5}{4} - \frac{3}{4} \ C$

we substitute in the other equation

2C +5 D = 2

2C +5 (5/4 - ¾ C) = 2

we solve

C (2 - 15/4) + 25/4 = 2

-7 / 4 C = 2 - 25/4

-7 / 4 C = -17/4

7C = 17

C = $\frac{17}{7}$

now we calculate D

D = $\frac{5}{4} - \frac{3}{4} \ \frac{17}{7}$

D = 5/4 - 51/28

D = $\frac{35-51}{28}$

D = - 16/28

D = $- \frac{4}{7}$

the result is (C, D) = ( $\frac{17}{7}, \ \frac{-4}{7}$ )

8 0

2 years ago

A body of mass 1kg is made to oscillate on a spring of force constant 16 n/m calculate 1 the angular frequency 2 the frequency o

Scilla [17]

Explanation:

Given that,

Mass of a body, m = 1 kg

Force constant, k = 16 N/m

We need to find the angular frequency and the frequency of oscillation.

(a) The angular frequency of a body is given by :

$\omega=\sqrt{\dfrac{k}{m}} \\\\=\omega=\sqrt{\dfrac{16}{1}} \\\\=4\ rad/s$

(b) The frequency of oscillation is given by :

$f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{4}{2\pi}\\\\=\dfrac{2}{\pi}\ Hz$

Hence, this is the required solution.

7 0

2 years ago

Help with these three

Elenna [48]

The first: alright, first: you draw the person in the elevator, then draw a red arrow, pointing downwards, beginning from his center of mass. This arrow is representing the gravitational force, Fg.
You can always calculate this right away, if you know his mass, by multiplying his weight in kg by the gravitational constant
$g = 9.81 \frac{m}{s {}^{2} }$
let's do it for this case:
$f_{g} = m \times g \\ f _{g} = 65kg \times 9.81 \frac{m}{s {}^{2} } = 637.65$
the unit of your fg will be in Newton [N]
so, first step solved, Fg is 637.65N
Fg is a field force by the way, and at the same time, the elevator is pushing up on him with 637.65N, so you draw another arrow pointing upwards, ending at the tip of the downwards arrow.
now let's calculate the force of the elevator
$f = m \times a \\ f = 65 \times 5 \frac{m}{s {}^{2} } \\ f = 325n$
so you draw another arrow which is pointing downwards on him, because the elevator is accelating him upwards, making him heavier
the elevator force in this case is a contact force, because it only comes to existence while the two are touching, while Fg is the same everywhere

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2 years ago

Question:

iogann1982 [59]

The correct answer is:
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2 years ago

A girl is floating in a freshwater lake with her head just above the water. If she weighs 610 N, what is the volume of the subme

Elden [556K]

Answer:

The volume of the submerged part of her body is $0.0622m^{3}$