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3 years ago

Which two statements are true?

Physics

1 answer:

Daniel [21]3 years ago

7 0

Answer:

the middle two.electric forces and magnetic forces are different aspects of a single electromagnetic force.

The operation of a motor depends on the relationship between electric and magnetic forces.

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A telephone pole has three cables pulling as shown from above, with F⃗ 1=(300.0iˆ+500.0jˆ) , F⃗ 2=−200.0iˆ , and F⃗ 3=−800.0jˆ .

Ray Of Light [21]

A) Net force in component form: $F=100.0i-300.0j$

B) Magnitude of the net force: 316.2, direction: $-71.6^{\circ}$

Explanation:

The three forces given in this problem are:

$F_1=300i+500j$

$F_2=-200i$

$F_3=-800 j$

The three forces are given in component form, where the components with unit vector i is the component along the x-direction, while the components with unit vector j is the component along the y-direction.

In order to find the net force in component form, we just need to add the components of the three forces along each direction. Therefore:

- Along the x-direction:

$F_x = F_{1x}+F_{2x}+F_{3x}=300+(-200)+0=100$

- Along the y-direction:

$F_y=F_{1y}+F_{2y}+F_{3y}=500+0+(-800)=-300$

So, the net force in component form is

$F=100.0i-300.0j$

The magnitude of a vector F is given by Pythagorean's theorem:

$|F|=\sqrt{F_x^2+F_y^2}$

where in this problem,

$F_x=100$ is the x-component

$F_y=-300$ is the y-component

Substituting,

$|F|=\sqrt{(100)^2+(-300)^2}=316.2$

The direction instead is given by

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{-300}{100})=-71.6^{\circ}$

where the negative sign means the direction is below the positive x-axis.

Learn more about vector addition:

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brainly.com/question/5892298

#LearnwithBrainly

8 0

3 years ago

"A 75 kg swimmer dives horizontally off a 500 kg raft. If the diver's speed immediately after leaving the raft is 4 m/s, what is

svet-max [94.6K]

Answer:

0.6 m/s

Explanation:

Let the raft speed be v.

Given:

Mass of the swimmer (m) = 75 kg

Mass of the raft (M) = 500 kg

Speed of swimmer after leaving the raft (u) = 4 m/s

Now, the given situation can be considered a problem of conservation of total momentum before and after collision.

Momentum is the product of mass and velocity.

Here, the swimmer and raft are the bodies in collision.

So, before the collision, both the bodies were at rest. So, initial momentum is 0. Now, from conservation of momentum, the final momentum of the system must be 0 after the collision. Therefore,

Final Momentum = 0

$mu+Mv=0\\\\Mv=-mu\\\\v=-\dfrac{mu}{M}$

Plug in the given values and solve for 'v'. This gives,

$v=-\frac{75\times 4}{500}\\\\v=-\frac{300}{500}=-0.6\ m/s$

The final velocity of the raft is -0.6 m/s. Now, speed is the magnitude of velocity.

Therefore, the corresponding raft speed is 0.6 m/s.

7 0

4 years ago

What detects the original stimulus ?

FinnZ [79.3K]

Wherever the change in the body happened will detect the stimuli and send the signals to the brain and Spinal does

8 0

4 years ago

A concave mirror produces a real image that is three times as large as the object. If the object is 20 cm in front of the mirror

TiliK225 [7]

Answer:

The image is produced 60 cm behind the mirror

The focal length of the mirror is 30 cm

Explanation:

u = Object distance = 20 cm

v = Image distance

f = Focal length

m = Magnification = 3

$m=-\frac{v}{u}\\\Rightarrow 3=-\frac{v}{20}\\\Rightarrow v=-3\times 20\\\Rightarrow v=-60\ cm$

The image is produced 60 cm behind the mirror

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{20}+\frac{1}{-60}\\\Rightarrow \frac{1}{f}=\frac{1}{30}\\\Rightarrow f=\frac{30}{1}=30\ cm$

The focal length of the mirror is 30 cm

8 0

3 years ago

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 71.0 m/s. Th

elena55 [62]

Answer:

Thunderbird is 995.157 meters behind the Mercedes

Explanation:

It is given that all the cars were moving at a speed of 71 m/s when the driver of Thunderbird decided to take a pit stop and slows down for 250 m. She spent 5 seconds in the pit stop.

Here final velocity $v=0 \ m/s$