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3 years ago

The y component of a vector is 36, and the angle between the vector and the x axis is 27 what is the magnitude of the vector

Physics

1 answer:

xz_007 [3.2K]3 years ago

3 0

Answer:

Magnitude of Vector = 79.3

Explanation:

When a vector is resolved into its rectangular components, it forms two vector components. These components are named as x-component and y-component, they are calculated by the following formulae:

x-component of vector = (Magnitude of Vector)(Cos θ)

y-component of vector = (Magnitude of Vector)(Sin θ)

where,

θ = angle of the vector with x-axis = 27°

Therefore, using the values in the equation of y-component, we get:

36 = (Magnitude of Vector)(Sin 27°)

Magnitude of Vector = 36/Sin 27°

<u>Magnitude of Vector = 79.3</u>

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100 g of Ice at -10°C is added into a

Andrei [34K]

Answer:

The mass of the juice responsible for melting the ice is 949.043 grams.

Explanation:

By the First Law of Thermodynamics, we understand that juice releases heat to the ice, which turns into water under the assumption that interactions between the ice-juice system and surroundings are negligible and energy processes are done in steady-state. Since juice is done with water, its specific heat will be taken as of the water. The process is described by the following formula:

$m_{i} \cdot [c_{i}\cdot (T_{1}-T_{2}) - L_{f} + c_{w}\cdot (T_{2}-T_{3})] + m_{w} \cdot c_{w}\cdot (T_{4}-T_{3}) = 0$ (1)

Where:

$m_{i}$ - Mass of ice, in grams.

$m_{w}$ - Mass of the juice, in grams.

$c_{i}$ - Specific heat of ice, in joules per gram-degree Celsius.

$c_{w}$ - Specific heat of water, in joules per gram-degree Celsius.

$L_{f}$ - Latent heat of fusion, in joules per gram.

$T_{1}$ - Initial temperature of ice, in degrees Celsius.

$T_{2}$ - Melting point of water, in degrees Celsius.

$T_{3}$ - Final temperature of the ice-juice system, in degrees Celsius.

$T_{4}$ - Initial temperature of the juice, in degrees Celsius.

If we know that $m_{i} = 100\,g$ , $c_{i} = 2.090\,\frac{J}{g\cdot ^{\circ}C}$ , $c_{w} = 4.18\,\frac{J}{g\cdot ^{\circ}C}$ , $L_{f} = 334\,\frac{J}{g}$ , $T_{1} = -10\,^{\circ}C$ , $T_{2} = 0\,^{\circ}C$ , $T_{3} = 10\,^{\circ}C$ and $T_{4} = 20\,^{\circ}C$ , then the mass of the juice is:

$m_{w} = \frac{m_{i}\cdot [c_{i}\cdot (T_{1}-T_{2}) - L_{f} + c_{w}\cdot (T_{2}-T_{3})]}{c_{w} \cdot (T_{3}-T_{4})}$

$m_{w} = \frac{(100\,g)\cdot \left[\left(2.090\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (-10\,^{\circ}C) - 334\,\frac{J}{g} +\left(4.18\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (-10\,^{\circ}C) \right]}{\left(4.180\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (-10\,^{\circ}C)}$

$m_{w} = 949.043\,g$

The mass of the juice responsible for melting the ice is 949.043 grams.

5 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

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

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

8 0

3 years ago

Artemon [7]

Answer:

constant volicty of the pumper when they hit ground 7.03-/s

8 0

3 years ago

A man works on a striaght road from his home to market 2.5km away dwith a speed of 5km/h.Finding the market closed, he instantly

umka21 [38]

Answer:

5.625km/h

Explanation:

We are given that

Distance between home and market, d=2.5 km

Speed, v1=5km/h

Speed, v2=7.5 km/h

We have to find the average speed of the man over the interval of time 0 to 40min.

Time, $t=\frac{distance}{speed}$

Using the formula

$t_1=\frac{2.5}{5}=0.5 h=0.5\times 60=30 min$

1 hour =60 min

$t_2=\frac{2.5}{7.5}=\frac{1}{3}hour=\frac{60}{3}=20min$

Distance traveled by man in 10 min with speed 7.5 km/h= $\frac{2.5}{20}\times 10=2.5/2km$

Therefore,

Total distance covered=2.5+2.5/2=3.75 km

Time=40 min=40/60=2/3 hour

Average speed= $\frac{total\;distance}{total\;time}$

Average speed= $\frac{3.75}{2/3}=5.625km/h$

7 0

3 years ago

Does the sun sun traces shortest path across local sky on june solstice

viktelen [127]

The June solstice in the Northern hemisphere is the summer solstice. The June Solstice in the Southern hemisphere is the winter solstice. The summer solstice is equivalent to the longest day while the winter solstice is equivalent to the shortest day. Therefore on the local sky, when is the June solstice we have have the longest day (longest path of sun in the sky) in the Northern hemisphere and the shortest day (shortest path of sun in the sky) in the Southern hemisphere.

7 0

4 years ago