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xxTIMURxx [149]

3 years ago

11

the gravitational force that earth exerts on the moon equals 2.03 x 10^20N. The moons mass is 7.35 x 10^22kg. What is the accele

ration of the moon due to earth’s gravitational pull?

1 answer:

Ghella [55]3 years ago

8 0

The first time we read this, we think that we'll have to go look up a bunch of stuff about the gravitational constant, centripetal force, orbital mechanics, the Moon's period of revolution, and who knows whut owl !

But you know whut ? The only thing we need is Newton's 2nd law:

F = m A

Divide each side by m :

Acceleration = Force / mass

Acceleration = (2.03 x 10²⁰ Newtons) / (7.35 x 10²² kg)

Acceleration = (2.03 x 10²⁰ / 7.35 x 10²² kg) (m/s²)

<em>Acceleration = 0.00276 m/s²</em>

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Nana76 [90]

Report this clown who put the first answer he’s trying to get your ip

4 0

3 years ago

What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use

marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force $F_{1}=18\ N$

Force $F_{2}=26\ N$

Force $F_{3}=14\ N$

We need to calculate the torque due to force F₁

Using formula of torque

$\tau_{1}=-F_{1}d_{1}$

$\tau_{1}=-F_{1}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{1}=-18\times\dfrac{0.2}{2}$

$\tau_{1}=-1.8\ N-m$

We need to calculate the torque due to force F₂

Using formula of torque

$\tau_{2}=F_{2}d_{2}$

$\tau_{2}=F_{2}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{2}=26\times\dfrac{0.2}{2}$

$\tau_{2}=2.6\ N-m$

We need to calculate the torque due to force F₃

Using formula of torque

$\tau_{3}=F_{3}d_{3}$

$\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{3}=0.1(14\sin45+14\cos45)$

$\tau_{3}=1.92\ N-m$

We need to calculate the net torque on the square plate

$\tau=\tau_{1}+\tau_{2}+\tau_{3}$

$\tau=-1.8+2.6+1.92$

$\tau=2.72\ N-m$

Hence, The net torque on the square plate is 2.72 N-m.

3 0

3 years ago

1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position

antiseptic1488 [7]

Answer:

The value of the distance is $\bf{14.52~cm}$ .

Explanation:

The velocity of a particle(v) executing SHM is

$v = \omega \sqrt{A^{2} - x^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)$

where, $\omega$ is the angular frequency, $A$ is the amplitude of the oscillation and $x$ is the displacement of the particle at any instant of time.

The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e., $x = 0$ .

The maximum velocity( $\bf{v_{m}}$ ) is

$v_{m} = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Divide equation (1) by equation(2).

$\dfrac{v}{v_{m}} = \dfrac{\sqrt{A^{2} - x^{2}}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)$

Given, $v = 0.25 v_{m}$ and $A = 15~cm$ . Substitute these values in equation (3).

$&& \dfrac{1}{4} = \dfrac{\sqrt{15^{2} - x^{2}}}{15}\\&or,& A = 14.52~cm$

6 0

3 years ago

A loudspeaker diaphragm is vibrating in simple harmonic motion with a frequency of 760 Hz and a maximum displacement of 0.85 mm.

Alchen [17]

Answer:(a) 4775.2Hz (b) 4.06m/s (c) 19382.15m/s²

Explanation: Given that the frequency of oscilation f, is 760Hz and the maximum displacement x, is 0.85mm= 0.00085m

(a) Angular frequency w= 2πf

w= 2π × 760 = 4775.2Hz

(b) Maximum speed v is given as the product of angular frequency and maximum displacement

V=wx

V= 4775.2 × 0.00085

V= 4.06m/s

(c) The maximum acceleration a

= w²x

= (4775.2)² × (0.00085)

a= 19382.15m/s².

5 0

4 years ago

If a runner has a speed of 8.66m/s and runs for 46.2s what distance is covered? tv = d

kati45 [8]

Answer:

$\text{Using the formula: }v=\frac{d}{t}\\\therefore vt=d\\\text{Plug and chug:}46.2(8.66)=400.092\text{ metres}$

6 0

3 years ago