An example of a decomposition reaction is

A grocery cart of mass 16 kg is being pushed at a constant speed up a 12-degree ramp by a force of [{MathJax fullWidth='false' f

lawyer [7]

Answer:

If the cart is being pushed at a constant speed, then the acceleration in the direction of motion is zero. Hence, the force in the direction of the motion is zero, according to Newton's Second Law.

$\Sigma F_x = ma_x$

For simplicity, I will denote the direction along the inclined ramp as x-direction.

In the question the value of the force is not clearly given, so I will denote it as $F_P$

$\Sigma F_{net_x} = ma_x\\\Sigma F = F_{P}\cos(29^\circ) - mg\sin(12^\circ) = ma_x = 0\\F_{P}\cos(29^\circ) = mgsin(12^\circ)\\F_{P}\times 0.8746 = 16\times 9.8\times 0.2079\\F_{P} = 37.2740$

Here the angle between the applied force and the x-direction is 12° + 17° = 29°

The x-component of the weight of the cart is equal to sine component of the weight.

Since the cart is rolling on tires the kinetic friction does no work.

Work done by the applied force:

$W_{F_P} = F_P_x \cos(29^\circ)\times 7.5 = 244.5 ~J$

Work done by the weight of the cart:

$W_{mg} = -mg\sin(12^\circ)\times 7.5 = -16\times 9.8 \times 0.2079 \times 7.5 = - 244.5~J$

Since the x-component of the weight is in the -x-direction, its work is negative.

Conveniently, the total work done on the particle is zero, since its velocity is constant.

5 0

3 years ago

What are the disadvantages of using alcohol instead of mercury in a thermometer ?

vichka [17]

Answer:

alcohol thermometers are used rather than Mercury thermometers in very cold regions because alcohol has a lower freezing point than Mercury.

Explanation:

6 0

3 years ago

When a system fails it _____ our other systems causing us to be sick.

krok68 [10]

Answer:

c) affects

Explanation:

im like 90% sure

3 0

3 years ago

How to solve god initial speed

CaHeK987 [17]

Answer:

Velocity is a function of time and defined by both a magnitude and a direction. [1] Often in physics problems, you will need to calculate the initial velocity (speed and direction) at which an object in question began to travel. There are multiple equations that can be used to determine initial velocity. Using the information given in a problem, you can determine the proper equation to use and easily answer your question.

This is from a website btw.

4 0

4 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

$\mu_s \sim \frac{F cos \theta}{F sin \theta}\\\mu_s=cot(\theta)$

So, this is the critical value of the coefficient of static friction.

8 0

3 years ago