Does anyone knows how to do this ?

As heat is added to water, is it possible for the temperature measured by a thermometer in the water to remain constant?

Mazyrski [523]

Answer:

C. Yes, the water could be changing the phase.

Explanation:

6 0

3 years ago

Suppose that a car traveling to the west (the - x direction) begins to slow down as it approaches a traffic light. Which stateme

Amanda [17]

Answer:

(A) it's acceleration is negative but but it's velocity is positive

Explanation:

In the question it is given that begins to slow down so its speed is decreasing it is does not means that its speed is negative

For example let first the velocity of the car is 30 m/sec and when its velocity decreases it becomes 20 m/sec in 5 sec

So it is not negative at all

Now the acceleration is the rate of change of velocity

$a=\frac{dv}{dt}$

$a=\frac{v_2-v_1}{dt}=\frac{20-30}{5}=-2 m/sec^2$

So acceleration is negative here

So option (a) will be the correct option

4 0

3 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

Mariana [72]

a) $F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

Here the crate is moving at constant velocity, so no acceleration:

a = 0

Let's analyze the forces acting along the horizontal and vertical direction.

- Vertical direction: the equation of the forces is

$R-Fsin \theta - mg = 0$ (1)

where

R is the normal reaction of the floor (upward)

$F sin \theta$ is the component of the force F in the vertical direction (downward)

mg is the weight of the crate (downward)

- Horizontal direction: the equation of the forces is

$F cos \theta - \mu_k R = 0$ (2)

where

$F cos \theta$ is the horizontal component of the force F (forward)

$\mu_k R$ is the force of friction (backward)

From (1) we get

$R=Fsin \theta +mg$

And substituting into (2)

$F cos \theta - \mu_k (Fsin \theta +mg) = 0\\F cos \theta -\mu _k F sin \theta = \mu_k mg\\F(cos \theta - \mu_k sin \theta) = \mu_k mg\\F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

b) $\mu_s=cot(\theta)$

In this second case, the crate is still at rest, so we have to consider the static force of friction, not the kinetic one.

The equations of the forces will be:

$R-Fsin \theta - mg = 0$ (1)

$F cos \theta - \mu_s R = 0$ (2)

In this second case, we want to find the critical value of $\mu_s$ such that the woman cannot start the crate: this means that the force of friction must be at least equal to the component of the force pushing on the horizontal direction, $F cos \theta$ .

Therefore, using the same procedure as before,

$R=Fsin \theta +mg$

$F cos \theta - \mu_s (Fsin \theta +mg) = 0$

And solving for $\mu_s$ ,

$F cos \theta = \mu_s (Fsin \theta +mg) \\\mu_s = \frac{F cos \theta}{F sin \theta + mg}$

Now we analyze the expression that we found. We notice that if the force applied F is very large, $F sin \theta >> mg$ , therefore we can rewrite the expression as