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3 years ago

I need help please

Physics

1 answer:

densk [106]3 years ago

4 0

Answer:

PLIS HELP PLSI PLSI PLIS

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Tamara begins with an empty toy truck, released it down the ramp and records its speed. She then adds 50 grams of mass to the to

Ksenya-84 [330]

Answer:

fortnite

Explanation: it is fortnite

6 0

3 years ago

PLZ HELP ILL GIVE RBAINLIEST

Rom4ik [11]

All of the above .........

3 0

3 years ago

The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

BARSIC [14]

A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$

So we have:

$F_{2x}=F_2 cos \theta = (15)(cos 120)=-7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 120)=13.0 N$

So, the components of the resultant this time are:

$F_x = F_{1x}+F_{2x}=20-7.5 = 12.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{13.5^2+13.0^2}=18.7 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{13.5})=43.9^{\circ}$

Learn more about vector addition:

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#LearnwithBrainly

7 0

3 years ago

Acceleration is zero if

tester [92]

It’s doesn’t change meaning it’s 0

4 0

3 years ago

A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of

34kurt

Answer:

The centripetal acceleration of the runner is $1.73\ m/s^2$ .

Explanation:

Given that,

A runner completes the 200 m dash in 24.0 s and runs at constant speed throughout the race. We need to find the centripetal acceleration as he runs the curved portion of the track. We know that the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

v is the velocity of runner

$v=\dfrac{200\ m}{24\ s}\\\\v=8.34\ m/s$

Centripetal acceleration,

$a=\dfrac{(8.34)^2}{40}\\\\a=1.73\ m/s^2$

So, the centripetal acceleration of the runner is $1.73\ m/s^2$ . Hence, this is the required solution.

5 0

4 years ago

I need help please ​

I need help please