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myrzilka [38]
3 years ago
10

100 + 50 + BRAINLIST PLEASE HELP!!!

Physics
2 answers:
irina1246 [14]3 years ago
6 0

Answer:

Explanation:

1. What are the forces acting on the block when it is hanging freely from the spring scale? What is the net force on the block? What are the magnitudes of each of the forces acting on the block? Explain.

When a block is hanging freely, two forces are acting on it = tension force from the spring scale and gravity force on the block itself. The net force is zero as the block is not accelerating.  The magnitudes of tension and gravity force are the same but in opposite directions.

2. What are the forces that act on the block when it is placed on the ramp and is held in place by the spring scale? What is the net force acting on the block? Explain. (Assume that the ramps are frictionless surfaces.)

There are three forces acting on the block when it is placed on the ramp and is held in place by the spring scale: as in 1, there are tension and gravity but there is a third force - reaction force from the ramp surface on the block that is perpendicular to the surface. Again the block is not moving so the net force is zero.

3. What is the magnitude of normal force acting on the block when it is resting on the flat surface? How does the normal force change as the angle of the ramp increases? Explain. (Assume that the ramps are frictionless surfaces.)

On flat surface, the normal force is equal to the gravity force of the block i.e. its weight. On a vertical surface, the normal force is equal to zero. For the angle of ramp, θ, the normal force = weight * cos θ.

SOVA2 [1]3 years ago
3 0

Answer:

Explanation:

1. block is hanging freely - freely means there is no net force

from the spring scale - spring force

block - weight

magnitude of weight = magnitude of spring force as there is no net force

2. same thing - held in place means there is no net force

from the spring scale - spring force

block - weight

ramp - normal reaction

3. when its resting, there is no net force

magnitude of normal reaction force = magnitude of block weight

normal force reaction force decreases as angle of ramp increases from zero

reaction = weight x sin (ramp angle)

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N2 + O2 → 2NO N-N triple bond: 941 kJ/mol O-O double bond: 495 kJ/mol N-O bond: 201 kJ/mol
kiruha [24]

Answer:

\large \boxed{\text{761 kJ}}

Explanation:

You calculate the energy required to break all the bonds in the reactants.

Then you subtract the energy needed to break all the bonds in the products.

                        N₂  +   O₂   ⟶         2NO

                     N≡N  + O=O ⟶       2O-N=O

Bonds:         2N≡N   1O=O        2N-O + 2N=O

D/kJ·mol⁻¹:     941      495           201      607

\begin{array}{rcl}\Delta H & = & \sum{D_{\text{reactants}}} - \sum{D_{\text{products}}}\\& = & 2 \times 941 +1 \times 495 - (2 \times 201 + 2\times 607)\\&=& 2377 - 1616\\&=&\textbf{761 kJ}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{761 kJ}}$}.

3 0
3 years ago
The bigclaw snapping shrimp shown in (Figure 1) is aptly named--it has one big claw that snaps shut with remarkable speed. The p
leva [86]

1) 1.86\cdot 10^6 rad/s^2

2) 2418 rad/s

3) 27000 m/s^2

4) 36.3 m/s

Explanation:

1)

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

\theta=\omega_i t +\frac{1}{2}\alpha t^2

where:

\theta is the angular displacement

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

\theta=90^{\circ} = \frac{\pi}{2}rad is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

Solving for \alpha, we find:

\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2

2)

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

\omega_f = \omega_i + \alpha t

where

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

Therefore, the final angular speed is:

\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s

3)

The tangential acceleration is related to the angular acceleration by the following formula:

a_t = \alpha r

where

a_t is the tangential acceleration

\alpha is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2

4)

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

v=u+at

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

a=27900 m/s^2 (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

v=0+(27900)(0.0013)=36.3 m/s

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3 years ago
How are longitudinal and transverse waves different?
Natalija [7]

Answer:

A longitudinal wave is a wave where the movement of the medium is in the same direction as the wave. On the other hand, a transverse wave is a wave where the movement of the medium is at a right angle to the wave direction.

Explanation:i got this right on a quiz so i know its right

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Answer:

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Evgesh-ka [11]

Answer:

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Explanation:

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Examples of conductors are: <em>copper, aluminum, gold, silver, seawater, etc.</em>

The opposite of conductors are called "insulators." These do not allow the free movement of charges from one object to the other.

Examples of insulators: <em>plastic, rubber, paper, glass, wool, dry air, etc.</em>

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