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2 years ago

NEED ANSWER ASAP PLEASE!!!!!!!

Physics

1 answer:

Usimov [2.4K]2 years ago

7 0

In both questions, the objects are falling from rest so their final velocities after time $t$ will be given by

$v_f=v_i+at=at$

where $a$ is the acceleration due to gravity given. Note that I take the downward direction to be positive.

23.

$v_f=\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)(7\,\mathrm s)\implies v_f=68.6\,\dfrac{\mathrm m}{\mathrm s}$

24.

$49\,\dfrac{\mathrm m}{\mathrm s}=\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\implies t=5.0\,\mathrm s$

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You are piloting a helicopter which is rising vertically at a uniform velocity of 14.70 m/s. When you reach 196.00 m, you see Ba

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Answer:

The ball reaches Barney head in $t = 8 \ s$

Explanation:

From the question we are told that

The rise velocity is $v = 14.70 \ m/s$

The height considered is $h = 196 \ m$

The horizontal velocity of the large object is $v_h = 8.50 \ m/s$

Generally from kinematic equation

$s = ut + \frac{1}{2} gt^2$

Here s is the distance of the object from Barney head ,

u is the velocity of the object along the vertical axis which is equal but opposite to the velocity of the helicopter

$u = -14.7 m/s$

$196 = -14.7 t + \frac{1}{2} * 9.8 * t^2$

= $4.9 t^2 - 14.7t - 196 = 0$

Solving the above equation using quadratic formula

The value of t obtained is $t = 8 \ s$

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2 years ago

A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar

notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2> $\omega = 0.23 rad/s$ </h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

$L_i = L_f$

now we can say

$m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega$

so we will have

$0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega$

$\omega = 0.23 rad/s$

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

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2 years ago

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sergiy2304 [10]

Answer:

The circular loop experiences a constant force which is always directed towards the center of the loop and tends to compress it.

Explanation:

Since the magnetic field, B points in my direction and the current, I is moving in a clockwise direction, the current is always perpendicular to the magnetic field and will thus experience a constant force, F = BILsinФ where Ф is the angle between B and L.

Since the magnetic field is in my direction, it is perpendicular to the plane of the circular loop and thus perpendicular to L where L = length of circular loop. Thus Ф = 90° and F = BILsin90° = BIL

According to Fleming's left-hand rule, the fore finger representing the magnetic field, the middle finger represent in the current and the thumb representing the direction of force on the circular loop.

At each point on the circular loop, the force is always directed towards the center of the loop and thus tends to compress it.

<u>So, the circular loop experiences a constant force which is always directed towards the center of the loop and tends to compress it.</u>

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2 years ago

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Total current flow in circuit=0.75
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2 years ago

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hodyreva [135]

That means the total of mass for the reactants needs to equal the total mass of their productions.

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2 years ago