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tankabanditka [31]
2 years ago
5

Help me please i really need this one

Physics
1 answer:
stiks02 [169]2 years ago
3 0

Answer:

T=0.0278

Explanation:

f=1/T

(36)=1/T

T=0.0278

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I'm not exactly sure but I'm thinking that it's the last one. Sorry if I'm wrong
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Explain why it is dangerous to jump from a fast moving train
Sergeu [11.5K]

Answer:

When you jump off a train, you jump off a certain height and your downwards (vertical) velocity is zero. But your forward (horizontal) velocity is not. You will hit the ground on split second with your horizontal velocity practically the same as the train.

Explanation:

you be in serious injury.

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2 years ago
A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angul
zepelin [54]

The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

i.e.

P.E = K.E + R.K.E

\mathbf{mgh = \dfrac{1}{2}m(r \times \omega)^2 + \dfrac{1}{2}\times I \times \omega^2}

\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}

\mathbf{2 \times \dfrac{gh}{r^2} =\omega^2 +  \omega^2}

\mathbf{2 \omega^2=2 \times \dfrac{9.81 \times 8 m }{2.0 ^2}  }

\mathbf{\omega^2=\dfrac{39.24 }{2}}

\mathbf{\omega=\sqrt{19.62 } \ rad/sec}

\mathbf{\omega=4.429 \ rad/sec}

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

Learn more about angular velocity here:

brainly.com/question/1452612

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2 years ago
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A 2011 Porsche 911 Turbo S goes from 0-27 m/s in 2.5 seconds. What is the car's acceleration?
Natalka [10]

Answer:

-10.8m/s^2

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a=change in velocity/change in time

-27 m/s/2.5=10.8m/s^2

or if its not negative

27m/s/2.5=10.8m/s^2

3 0
2 years ago
two forces 3N and 4N act on a body in a direction due north and due East respectively calculate their equivalent​
Lostsunrise [7]

Two forces 3N and 4N act on a body in a direction due north From East, the equilibrant's angle is given by \theta=\tan ^{-1} \frac{3}{4}=36.8^{\circ}.

<h3>What are equilibrium and resultant force?</h3>

The equilibrium force is the balanced force when the net force acting is zero and is the exact opposite of the consequent force. The resultant force is one single force replaced by numerous forces.

<h3>Briefing:</h3>

3N and 4N are the two forces pulling on a body.

The forces work along the North and the East, which are perpendicular to one another.

The resultant of the forces, which is provided by the equilibrant force,

R  = √(3)²+(4)²

R = 5N

From East, the equilibrant's angle is given by

\theta=\tan ^{-1} \frac{3}{4}=36.8^{\circ}

To know more about equilibrium force visit:

brainly.com/question/12582625

#SPJ9

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