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4 years ago

A 0.850 kg mass is placed on a

Physics

1 answer:

densk [106]4 years ago

5 0

Answer:

Picture

Explanation:

I am trying to solve this Q

but not sure if this a correct answer

or not.

Hope this helps.

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What is the first sign of dehydration? What is the first sign of dehydration?

Inga [223]

I think thirst or dark colored urine. :)

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3 years ago

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The density of an object is .835 g/cm3. If its volume is 34 cm3, what is the mass of the object?

zaharov [31]

D = m/v

Given:

D: .835 g/cm3

V: 34 cm3

M: ?

M= Dxv

M= .835 g/cm3 x 34 cm3

M= 28.39 g

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3 years ago

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As a motorcycle takes a sharp turn, the type of motion that occurs is called _______________ motion.

Darya [45]

Answer:

circular motion

Explanation:

As a motorcycle takes a sharp turn, the type of motion that occurs is called circular motion.

Circular motion is a movement of an object along a circular path. As this motorcycle makes the sharp turn, it is acted upon by a centripetal force which directs the motorcycle towards the center.

Therefore, circular motion is the correct answer to the question.

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4 years ago

A 9.0-cm-long spring is attached to the ceiling. when a 1.8 kg mass is hung from it, the spring stretches to a length of 18 cm .

Alex Ar [27]

The spring constant is computed by:
F = kx

Where: F is the force applied in newtons (N)

k is the spring constant measured in newtons per meter (N/m); and

x is the distance the spring is stretched (m)
and

F = mg

Where: F is the force pulling objects in the direction of the Earth.

m is the mass of the object.

g is the acceleration due to gravity;
So plugging our values in the formula:

F = mg

= (1.8) (9.81) = 17.658N

k =

F/x = 17.658 /0.09 = 196.2 N/meter

5 0

4 years ago

An object is projected at 25m/s from the top of a building of height 50m. At the same instant,another object is projected from t

docker41 [41]

A) The objects have the same vertical position after 2 seconds

B) The objects have same vertical position at y = 30.4 m (but they do not collide since they have different x-position)

Explanation:

The motion of the first object along the vertical direction is a uniformly accelerated motion, so we can write its position at time t using the following equation:

$y_1(t)=h+u_1 t + \frac{1}{2}gt^2$

where:

h = 50 m is the initial height

$u_1=0$ is the initial vertical velocity (the object is projected horizontally, so the vertical velocity is zero at the beginning)

$g=-9.8 m/s^2$ is the acceleration of gravity

So, its vertical position can be rewritten as

$y_1(t)=50-4.9t^2$

The position of object 2 instead can be written as

$y_2(t)=(u_2 sin \theta)t + \frac{1}{2}gt^2$

where

$u_2 sin \theta$ is the initial vertical velocity, where

$u_2 = 50 m/s$ is the initial velocity

$\theta=30^{\circ}$ is the angle of projection

Substituting, we get:

$y_2(t)=(50)(sin 30^{\circ})t+\frac{1}{2}(-9.8)t^2=25t-4.9t^2$

The two objects collide when their vertical position is the same, so:

$y_1(t)=y_2(t)\\50-4.9t^2 = 25t-4.9t^2$

And solving for t, we find:

$50=25t\\t= 2 s$

Note that this means that the two object at t = 2 s have the same vertical position: however, this is not true for the horizontal position.

In order to find the point where they collide, we have to substitute the time of the collision that we found in part A into one of the expressions of the vertical position.

Substituting into the expression of object 2, we find:

$y_2(t) = 25t-4.9t^2=25(2.0)-4.9(2.0)^2=30.4 m$