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sergiy2304 [10]

3 years ago

10

A student mixed two clear liquids together in a beaker and a solid precipitate formed, this data is shown in the table below . w

hat is the mass of the beaker?

2 answers:

klasskru [66]3 years ago

8 0

The mass is 10 +17 +32

Natasha2012 [34]3 years ago

4 0

Mass of liquid reactant A + mass of liquid reactant B = 10 g + 17 g = 27 g

Beaker + solid product = 32 g

Solve the second equation for Beaker:

Mass of beaker = 32 g - mass of solid product

Mass of beaker = 32 g - 27 g = 5 g

The beaker has a mass of 5 g.

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A stone was dropped off a cliff and hit the ground with a speed of 88 ft/s. What is the height (in feet) of the cliff

igor_vitrenko [27]

Answer:

the height (in feet) of the cliff is 121 ft

Explanation:

A stone hit the cliff with

speed, v = 88 ft/s

Acceleration, a= 32 ft/s^2

initial speed, u = 0 ft/s

height is h.

To solve this problem we will apply the linear motion kinematic equations, Equation of motion describes change in velocity, depending on the acceleration and the distance traveled

so, writing the formula of Equation of motion:

v^2 - u^2 = 2*a*h

substituting the appropriate values,

(88)^2 - 0 = 2*32* h

h=(88)^2 / 64

h= 121 ft

hence

the height (in feet) of the cliff is 121 ft

learn more about height of the cliff here:

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3 0

1 year ago

A small omnidirectional stereo speaker produces waves in all directions that have an intensity of 8.00 at a distance of 4.00 fro

slavikrds [6]

Answer:

A. We have that radius r = 4.00m intensity I = 8.00 W/m^

total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W

b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2

c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J

5 0

3 years ago

You are sitting on a merry-go-round of mass 200 kg and radius 2m that is at rest (not spinning). Your mass is 50 kg. Your friend

Bogdan [553]

Answer:

a. $\tau=200J$ b. $\alpha=0.44 \frac{rad}{s^2}$ c. $\alpha=0.33\frac{rad}{s^2}$ d. The angular acceleration when sitting in the middle is larger.

Explanation:

a. The magnitude of the torque is given by $\tau=rF\sin \theta$ , being r the radius, F the force aplied and $\theta$ the angle between the vector force and the vector radius. Since $\theta=90^{\circ}, \, \sin\theta=1$ and so $\tau=rF=2m100N=200Nm=200J$ .

b. Since the relation $\tau=I\alpha$ hols, being I the moment of inertia, the angular acceleration can be calculated by $\alpha=\frac{\tau}{I}$ . Since we have already calculated the torque, all left is calculate the moment of inertia. The moment of inertia of a solid disk rotating about an axis that passes through its center is $I=\frac{1}{2}Mr^2$ , being M the mass of the disk. If we assume that a person has a punctual mass, the moment of inertia of a person would be given by $I_p=m_pr_p^{2}$ , being $m_p$ the mass of the person and $r_p^{2}$ the distance from the person to the center. Given all of this, we have

$\alpha=\frac{\tau}{I}=\frac{\tau}{I_{disk}+I_{person}}=\frac{Fr}{\frac{1}{2}Mr^2+m_pr_p^{2}}=\frac{200Nm}{\frac{1}{2}200kg*4m^2+50kg*1m^2}=\frac{200\frac{kgm^2}{s^2}}{450Nm^2}\approx 0.44\frac{rad}{s^2}$ .

c. Similar equation to b, but changing $r_p=2m$ , so

$alpha=\dfrac{200\frac{kgm^2}{s^2}}{\frac{1}{2}200*4kg\,m^2+50*4 kg\,m^2}=\dfrac{200}{600}\dfrac{1}{s^2}\approx 0.33 \frac{rad}{s^2}$ .

d. The angular acceleration when sitting in the middle is larger because the moment of inertia of the person is smaller, meaning that the person has less inertia to rotate.

5 0

3 years ago

A 25.0 kg box of textbooks rests on a loading ramp that makes an angle α with the horizontal. The coefficient of kinetic frictio

Alekssandra [29.7K]

Answer:

The minimum angle at which the box starts to slip (rounded to the next whole number) is α=19°

Explanation:

In order to solve this problem we must start by drawing a sketch of the problem and its corresponding fre body diagram (See picture attached).

So, when we are talking about friction, there are two types of friction coefficients. Static and kinetic. Static friction happens when the box is not moving no matter what force you apply to it. You get to a certain force that is greater than the static friction and the box starts moving, it is then when the kinetic friction comes into play (kinetic friction is generally smaller than static friction). So in order to solve this problem, we must find an angle such that the static friction is the same as the force applie by gravity on the box. For it to be easier to analyze, we must incline the axis of coordinates, just as shown on the picture attached.

After doing an analysis of the free-body diagram, we can build our set of equations by using Newton's thrid law:

$\sum F_{x}=0$

we can see there are only two forces in x, which are the weight on x and the static friction, so:

$-W_{x}+f_{s}=0$

when solving for the static friction we get:

$f_{s}=W_{x}$

We know the weight is found by multiplying the mass by the acceleration of gravity, so:

W=mg

and:

$W_{x}=mg sin \alpha$

we can substitute this on our sum of forces equation:

$f_{s}=mg sin \alpha$

the static friction will depend on the normal force applied by the plane on the box, static friction is found by using the following equation:

$f_{s}=N\mu_{s}$

so we can substitute this on our equation:

$N\mu_{s}=mg sin \alpha$

but we don't know what the normal force is, so we need to find it by doing a sum of forces in y.

$\sum F_{y}=0$

In the y direction we got two forces as well, the normal force and the force due to gravity, so we get:

$N-W_{y}=0$

when solving for N we get:

$N=W_{y}$

When seeing the free-body diagram we can determine that:

$W_{y}=mg cos \alpha$

so we can substitute that in the sum of y-forces equation, so we get:

$N=mg cos \alpha$

we can go ahead and substitute this equation in the sum of forces in x equation so we get:

$mg cos \alpha \mu_{s}=mg sin \alpha$

we can divide both sides of the equation into mg so we get:

$cos \alpha \mu_{s}=sin \alpha$

as you may see, the angle doesn't depend on the mass of the box, only on the static coefficient of friction. When solving for $\mu_{s}$ we get:

$\mu_{s}=\frac{sin \alpha}{cos \alpha}$

when simplifying this we get that:

$\mu_{s}=tan \alpha$

now we can solve for the angle so we get:

$\alpha= tan^{-1}(\mu_{s})$

and we can substitute the given value so we get:

$\alpha= tan^{-1}(0.350)$

which yields:

α=19.29°

which rounds to:

α=19°

8 0

3 years ago

Lasers can be constructed that produce an extremely high intensity electromagnetic wave for a brief time—called pulsed lasers. T

alex41 [277]

Answer:

30643 J

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

t = Time taken = 1 ns

c = Speed of light = $3\times 10^8\ m/s$

$E_0$ = Maximum electric field strength = $1.52\times 10^{11}\ V/m$

A = Area = $1\ mm^2$

Magnitude of magnetic field is given by

$B_0=\dfrac{E_0}{c}\\\Rightarrow B_0=\dfrac{1.52\times 10^{11}}{3\times 10^8}\\\Rightarrow B_0=506.67\ T$

Intensity is given by

$I=\dfrac{cB_0^2}{2\mu_0}\\\Rightarrow I=\dfrac{3\times 10^8\times 506.67^2}{2\times 4\pi \times 10^{-7}}\\\Rightarrow I=3.0643\times 10^{19}\ W/m^2$

Power, intensity and time have the relation

$E=IAt\\\Rightarrow E=3.0643\times 10^{19}\times 1\times 10^{-6}\times 1\times 10^{-9}\\\Rightarrow E=30643\ J$

The energy it delivers is 30643 J

4 0

3 years ago