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3 years ago

14

I WILL GIVE BRAINLIEST!!!

1 answer:

AnnZ [28]3 years ago

8 0

Answer:

7.27 m/s

Explanation:

40/5.5 = 7.27m/s

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A clam of mass 0.12 kg dropped by a seagull takes 3.0 s to hit the ground. [Neglect friction.]

inna [77]

This is a defective, misleading question, and should never be asked in a Physics class.

There is no such thing as the force due to the impact.

If you know how long it takes the clam to stop once it begins to hit the dirt,
then you can calculate the impulse transferred to it, and tease a force out
of that. But the question doesn't give us the time.

It depends on the material of the surface. Was the clam dropped onto dirt ?
Into a dumpster ? Onto grass ? Concrete ? Styrofoam ? Mud ? The answer
is different in each case, and we still need to know the short length of time
AFTER it first encountered whatever surface brought it to rest.

I would kick this question back to the Physics teacher. It's meaningless,
and the longer you try to work on it, the more nonsense you'll plant into
your head that'll need to be dug out later.

8 0

3 years ago

A 2-kg object is moving 6 m/s in a horizontal direction.

german

Answer: A) O N

Explanation:

An object in motion will maintain its state of motion. The presence of an unbalanced force changes the velocity of the object.

4 0

3 years ago

Please help me with this i dont know it

Inessa05 [86]

Answer:

no it would not. that is an open circuit and it would need to be closed at the switch for current to flow.

5 0

2 years ago

What part of the hammer acts as the fulcrum when the hammer is used to remove a nail

Natali5045456 [20]

The end of it with the dent

6 0

3 years ago

A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

Artist 52 [7]

Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

$y = 20 + 7t - 9t^2$

Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

$a_x = -18t$

$a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)$

$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

now magnitude of the force will be

$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

$v_x = \frac{dx}[dt}$

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

now at t = 1.1 s

$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

7 0

3 years ago