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Marizza181 [45]

2 years ago

11

Two toddlers are fighting over a toy. Joey pulls the toy with a force of 8 N while Tommy pulls the toy with a force of 7.5 N. Wh

at is the net force of the toy? Who is going to get the toy?

1 answer:

Firdavs [7]2 years ago

5 0

Answer:

15.5 and Joey will get the toy

Explanation:

Net force is all the forces acting on the object and since Joey is pulling .5 newtons more he will get the toy.

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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 disk with a diameter of 0.05 m is spinning with a constant velocity about an axle perpendicular to the disk and running throug

Maksim231197 [3]

Answer:

f= 7.8Hz

Explanation:

a= 12g= 120m/s²

a= ω²r

120= ω²×0.05

ω= 48.99 rad/s

2πf= 48.99

f= 7.8Hz

8 0

2 years ago

A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radi

Mamont248 [21]

Answer:

$a=368.97\ m/s^2$

Explanation:

Given that,

Initial angular velocity, $\omega=0$

Acceleration of the wheel, $\alpha =7\ rad/s^2$

Rotation, $\theta=14\ rotation=14\times 2\pi =87.96\ rad$

Let t is the time. Using second equation of kinematics can be calculated using time.

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s$

Let $\omega_f$ is the final angular velocity and a is the radial component of acceleration.

$\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s$

Radial component of acceleration,

$a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2$

So, the required acceleration on the edge of the wheel is $368.97\ m/s^2$ .

6 0

2 years ago

A string of length L, fixed at both ends, is capable of vibrating at 309 Hz in its first harmonic. However, when a finger is pla

Scorpion4ik [409]

Answer:

i = 0.3326 L

Explanation:

A fixed string at both ends presents a phenomenon of standing waves, two waves with the same frequency that are added together. The expression to describe these waves is

2 L = n λ n = 1, 2, 3…

The first harmonic or leather for n = 1

Wave speed is related to wavelength and frequency

v = λ f

λ = v / f

Let's replace in the first equation

2 L = 1 (v / f₁)

For the shortest length L = L-l

2 (L- l) = 1 (v / f₂)

These two equations form our equation system, let's eliminate v

v = 2L f₁

v = 2 (L-l) f₂

2L f₁ = 2 (L-l) f₂

L- l = L f₁ / f₂

l = L - L f₁ / f₂

l = L (1- f₁ / f₂)

.

Let's calculate

l / L = (1- 309/463)

i / L = 0.3326

4 0

2 years ago

As a bullet shot vertically upward rises, the kinetic energy of the bullet

algol [13]

The kenitc energy of the bullet lowers as it keeps going up.

Because gravity is pushing the bullet down as the bullet goes up.

I'm pretty sure that the way to put this answer.

3 0

1 year ago