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3 years ago

Help i feel stressful in work i feel like giving up

Physics

2 answers:

Phantasy [73]3 years ago

8 0

Answer:

Keep going you are doing great, if you need a break go take a short walk

Schach [20]3 years ago

5 0

Just try your best, or ask your teacher for help! You could get tutoring just keep yourself positive you can do it!

You might be interested in

Determine the acceleration that results when a 12 N net force is applied to a 3 kg object.

Drupady [299]

A I hope it helpsss youuu ;:))))

8 0

3 years ago

As you are cycling to the Q center one spring day, you pass the construction site at Gant and stop to watch a few minutes. A cra

Rama09 [41]

Answer:

30.63 m

Explanation:

Using y = ut + 1/2gt² where u = initial speed of block = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time of fall = 2.5 s and y = height of fall.

So, substituting the values of the variables into the equation, we have

y = ut + 1/2gt²

y = 0 m/s × 2.5 s + 1/2 × 9.8 m/s² × (2.5 s)²

y = 0 m + 4.9 m/s² × 6.25 s²

y = 0 m + 30.625 m

y = 30.625 m

y ≅ 30.63 m

So, the brick fell 30.63 m

7 0

3 years ago

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

7 0

3 years ago

PLEASE HELP

GREYUIT [131]

Answer:

h = 1.8 m

Explanation:

The initial velocity of the glove, u =- 6 m/s

We need to find the maximum height of the glove. Let it is equal to h. Using equation of kinematics. At the maximum height v = 0

$v^2-u^2=2ah$ , h is the maximum height and a = -g

$0^2-(6)^2=2\times (-10)\times h\\\\h=\dfrac{36}{20}\\\\h=1.8\ m$

Hence, it will go up to a height of 1.8 m.

4 0

3 years ago

1. A rocket is forced forward by the ______ force of its engines, expelling gases out the rear of the rocket.

AURORKA [14]

There are two forces acting on a rocket at the moment of lift off: Thrust pushes the rocket upwards by pushing gases downwards in the opposite direction.Weight is the force due to gravity pulling the rocket downwards towards the centre of the earth.So I'm thinking the answer is THRUST.

6 0

3 years ago