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Brut [27]
2 years ago
14

Please I need help with this one please

Physics
2 answers:
mash [69]2 years ago
6 0
It’s correct :)! Ty
Svetach [21]2 years ago
4 0

12 Step 3

13 Step 2

14 Step 4

15 Step 4

16 Step 5

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Two blocks are suspended from opposite ends of a light rope that passes over a light, frictionless pulley. One block has mass m1
rewona [7]

Answer:

a)  m₁ = 1.41 kg , b)  m₂ = 2.65 kg

Explanation:

For this exercise we will use Newton's second law

Block 1

             T - W₁ = m₁ a

Block 2

           W₂ - T = m₂ a

We have selected the positive  block 1 rising and block two lowering, as the pulley has no friction does not affect the movement

Let's use kinematics to look for acceleration

         y = v₀ t + ½ a t²

As part of the rest the initial speed is zero

           a = 2 y / t²

           a = 2 6.00 / 2²

           a = 3 m / s²

Let's replace in the equation of block 1

a)   T = m₁ g + m₁ a

     m₁ = T / (g + a)

     m₁ = 18.0 / (9.8 + 3)

     m₁ = 1.41 kg

b) we substitute in the equation of block 2

     W₂ - T = m₂ a

     m₂ g - m₂ a = T

     m₂ = T / (g-a)

     m₂ = 18.0 / (9.8 -3)

    m₂ = 2.65 kg

4 0
3 years ago
A jet airliner moving initially at 693 mph (with respect to the ground) to the east moves into a region where the wind is blowin
Nesterboy [21]

Answer:

The new speed of the aircraft with respect to the ground is 1414.3 mph.

Explanation:

Given that,

Angle = 37°

Velocity of jet airliner = 693 mph

Velocity of wind = 798 mph

We know that,

The new velocity of the aircraft with respect to the ground

v=v_{a}+v_{w}

We need to calculate the new speed of the aircraft with respect to the ground

Using formula for velocity

v=\sqrt{(v_{a})^2+(v_{w})^2+2v_{a}v_{w}\cos\theta}

Put the value into the formula

v=\sqrt{(693)^2+(798)^2+2\times693\times798\cos37}

v=1414.3\ mph

Hence, The new speed of the aircraft with respect to the ground is 1414.3 mph.

7 0
3 years ago
A straight wire is aligned east-west in a region where Earth’s magnetic field has magnitude 0.0780 mT and direction 67.1° below
Sergio [31]

Explanation:

It is given that,

Earth's magnetic field, B=0.078\ mT=0.078\times 10^{-3}\ T

Direction of magnetic field, 67.1 degrees below the horizontal, with the horizontal component directed due north.

The magnetic force on the wire per unit length of wire, \dfrac{F}{L}=0.0300\ N/m

(a) The given scenario is shown in the attached figure as :

Using the right hand rule to find the direction of magnetic force on the wire. By using this rule, we get the magnetic force acting on the wire is in upward direction.

(b) Let I is the current flowing in the wire. The magnetic force is given by the following formula as:

F=ILB

I=\dfrac{F}{LB}

I=\dfrac{0.0300}{0.078\times 10^{-3}}

I = 384.61 A

Hence, this is the required solution.

7 0
3 years ago
Connecting many devices in a single socket does not affect the flow of current in a
eimsori [14]

<u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>:</u><u>-</u><em> </em><em>F</em><em>a</em><em>l</em><em>s</em><em>e</em>

<u>E</u><u>x</u><u>p</u><u>l</u><u>a</u><u>n</u><u>a</u><u>t</u><u>i</u><u>o</u><u>n</u><u>:</u><u>-</u>

<em>False because it can leads to overloading and further to short circut.</em>

<h2><em><u>H</u></em><em><u>o</u></em><em><u>p</u></em><em><u>e</u></em><em><u> </u></em><em><u>I</u></em><em><u>t</u></em><em><u> </u></em><em><u>W</u></em><em><u>i</u></em><em><u>l</u></em><em><u>l</u></em><em><u> </u></em><em><u>H</u></em><em><u>e</u></em><em><u>l</u></em><em><u>p</u></em><em><u> </u></em><em><u>Y</u></em><em><u>o</u></em><em><u>u</u></em><em><u>!</u></em></h2>
7 0
3 years ago
A ball is projected horizontally from a table 1.0 m high and hits the ground after falling from a time t=0.45s. If the ball trav
photoshop1234 [79]

The initial velocity is 6.7 m/s

Explanation:

The motion of the ball is a projectile motion, therefore it consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction  

In this problem, we just need to analyze the horizontal motion: the horizontal velocity is constant, therefore the horizontal distance travelled is given by

x=v_x t

where

v_x is the horizontal velocity

t is the time of flight

Here we have:

t = 0.45 s

x = 3.0 m

And so solving for v_x, we find

v_x = \frac{3.0}{0.45}=6.7 m/s

And since the ball was initially projected horizontally, this is also the initial velocity.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

3 0
3 years ago
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