if one branch of a 120-v power lines is protected by a 20-A fuse, will the fuse carry an 8-Ώ load

Answer:

The highest point in the trajectory occurs at the midpoint of the path. This highest point increases as the angle increases. At a 75° launch angle, the maximum height is approximately 76 meters. However, a further increase in launch angle beyond this 75° angle will increase the peak height even more.

The time period of the spring is 2[tex]\pi[/tex][√(m/k)].

The spring constant of a spring is defined as the measurement of ratio of the force that is exerted on the spring to the displacement caused by it.

Here,

The mass of the block = m

Let F be the applied force on the spring and k be the spring constant.

When the mass attached to the spring is pulled to one side then released, it executes SHM.

Therefore we can write that, the applied force,

F = kx

Restoring force = -kx

According to Newton's law, we know that,

F = ma

So,

ma = -kx

Therefore, the acceleration,

a = (-k/m) x

For an SHM, the acceleration is given as,

a = -ω²x

Therefore, we can write that,

-ω²x = (-k/m) x

ω² = k/m

So, the time period of the spring,

T = 2[tex]\pi[/tex]/ω

T = 2[tex]\pi[/tex][√(m/k)]

Hence,

The time period of the spring is 2[tex]\pi[/tex][√(m/k)].

To learn more about spring constant, click:

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It does not change the chemical composition of water.

Answer:

D) the two spheres remain of equal size.

Explanation:

Answer:

[tex]$\frac{d}{\lambda} = 1.54$[/tex]

Explanation:

Given :

The first dark fringe is for m = 0

[tex]$\theta_1 = \pm 19^\circ$[/tex]

Now we know for a double slit experiments , the position of the dark fringes is give by :

[tex]$d \sin \theta=\left(m+\frac{1}{2}\right) \lambda$[/tex]

The ratio of distance between the two slits, d to the light's wavelength that illuminates the slits, λ :

[tex]$d \sin \theta=\left(\frac{1}{2}\right) \lambda$[/tex]     (since, m = 0)

[tex]$d \sin \theta=\frac{\lambda}{2}$[/tex]

[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin \theta}$[/tex]

[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin 19^\circ}$[/tex]

[tex]$\frac{d}{\lambda} = 1.54$[/tex]

Therefore, the ratio is [tex]$\frac{1}{1.54}$[/tex]  or 1 : 1.54

Answer:

Objects that are said to be undergoing free fall, are not encountering a significant force of air resistance; they are falling under the sole influence of gravity.

Explanation:

Under such conditions, all objects will fall with the same rate of acceleration, regardless of their mass.

Answer:

A = 26.875 rad/s²

Explanation:

Given the following data;

Initial angular speed, Uw = 150 rads/s.

Final angular speed, Vw = 580 rads/s.

Time = 16 seconds.

To calculate the angular acceleration;

From kinematics equation;

At = Vw - Uw

Where;

Substituting into the formula, we have;

A*16 = 580 - 150

16A = 430

A = 430/16

A = 26.875 rad/s²

Answer:

828 kg/m³ or 0.828 g/cm³

Explanation:

Applying,

D = m/V............. Equation 1

Where D = density of the liquid, m = mass of the liquid, V = volume of the liquid.

From the question,

Given: m = 77 g , V = 93 cm³

Substitute these values into equation 1

D = 77/93

D = 0.828 g/cm³

Converting to kg/m³

D = 828 kg/m³

Answer:

The velocities of the skaters are [tex]v_{1} = 3.280\,\frac{m}{s}[/tex] and [tex]v_{2} = 0.024\,\frac{m}{s}[/tex], respectively.

Explanation:

Each skater is not under the influence of external forces during process, so that Principle of Momentum Conservation can be used on each skater:

First skater

[tex]m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}[/tex] (1)

Second skater

[tex]m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}[/tex] (2)

Where:

[tex]m_{1}[/tex] - Mass of the first skater, in kilograms.

[tex]m_{2}[/tex] - Mass of the second skater, in kilograms.

[tex]v_{1,o}[/tex] - Initial velocity of the first skater, in meters per second.

[tex]v_{1}[/tex] - Final velocity of the first skater, in meters per second.

[tex]v_{b}[/tex] - Launch velocity of the meter, in meters per second.

[tex]v_{2}[/tex] - Final velocity of the second skater, in meters per second.

If we know that [tex]m_{1} = 70\,kg[/tex], [tex]m_{b} = 0.043\,kg[/tex], [tex]v_{b} = 32\,\frac{m}{s}[/tex], [tex]m_{2} = 58.5\,kg[/tex] and [tex]v_{1,o} = 3.30\,\frac{m}{s}[/tex], then the velocities of the two people after the snowball is exchanged is:

By (1):

[tex]m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}[/tex]

[tex]m_{1}\cdot v_{1,o} - m_{b}\cdot v_{b} = m_{1}\cdot v_{1}[/tex]

[tex]v_{1} = v_{1,o} - \left(\frac{m_{b}}{m_{1}} \right)\cdot v_{b}[/tex]

[tex]v_{1} = 3.30\,\frac{m}{s} - \left(\frac{0.043\,kg}{70\,kg}\right)\cdot \left(32\,\frac{m}{s} \right)[/tex]

[tex]v_{1} = 3.280\,\frac{m}{s}[/tex]

By (2):

[tex]m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}[/tex]

[tex]v_{2} = \frac{m_{b}\cdot v_{b}}{m_{2}+m_{b}}[/tex]

[tex]v_{2} = \frac{(0.043\,kg)\cdot \left(32\,\frac{m}{s} \right)}{58.5\,kg + 0.043\,kg}[/tex]

[tex]v_{2} = 0.024\,\frac{m}{s}[/tex]

Answer:

A (3 seconds)

Explanation:

Well here we have a type of motion called projectile motion and it is pretty similar to an upside down parabola. The top of the trajectory is the vertex of the parabola and is also when v=0.

Lets identify our givens.

Givens:

Horizontal speed= 30m/s

Vertical Speed= 30 m/s

Since the ball is in freefall after being launched ay=-g(take up to be positive) and ax=0

The ball is launched from the ground so y0=0

Final vertical velocity= 0

This problem is now relatively easy because we only need to find the vertical distance so we can ignore horizontal speed and use

vy=vy0+ayt

Plug in our givens

0=30-10t

solve for t

t=3 seconds

Answer:

Smaller and upside down

Explanation:

To answer the question, we must recognise that the characteristics of the image formed by a convex lens depends on the position of the object from the lens.

From the diagram given in the question above, the following data were obtained:

1. The image is smaller than the object.

2. The image is inverted i.e upside down.

3. The image is closer to the lens

4. The image between 2f and f

Now, considering the options given in question above, the correct answer to the question is:

The image is smaller and upside down.

Answer:

parametric representation: x = u, y = v - u ,  z = - v

Explanation:

Given vectors :

i - j ,  j - k

represent the vector equation of the plane as:

r ( u, v ) = r₀ + ua + vb

where:  r₀ = position vector

            u and v = real numbers

             a and b = nonparallel vectors

expressing the nonparallel vectors as :

a = i -j , b = j - k , r = ( x,y,z ) and r₀ = ( x₀, y₀, z₀ )

hence we can express vector equation of the plane as

r(u,v) = ( x₀ + u, y₀ - u + v,  z₀ - v )

Finally the parametric representation of the surface through (0,0,0) i.e. origin = 0

( x, y , z ) = ( x₀ + u,  y₀ - u + v,   z₀ - v )

x = 0 + u ,

y = 0 - u + v

z = 0 - v

∴ parametric representation: x = u, y = v - u ,  z = - v

Answer:

True.

Explanation:

If the sum of the external forces on an object is zero, then the sum of the external torques on it  must also be zero.

The net external force and the net external torque acting on the object have to be zero for an object to be in mechanical equilibrium.

Hence, the given statement is true.

Answer:

1.67 m

Explanation:

The potential energy change of the small ball ΔU equals the work done by the buoyant force, W

ΔU = -W

Now ΔU = mgΔh where m = mass of small ball = ρV where ρ = density of small ball and V = volume of small ball. Δh = h - h' where h = final depth of small ball and h' = initial height of small ball = 5 m. Δh = h - 5

ΔU = mgΔh

ΔU = ρVgΔh

Now, W = ρ'VgΔh'   where ρ = density of water and V = volume of water displaced = volume of small ball. Δh' = h - h' where h = final depth of small ball and h' = initial depth of small ball at water surface = 0 m. Δh' = h - h' = h - 0 = h

So, ΔU = -W

ρVgΔh = -ρ'VgΔh'

ρVg(h - 5) = -ρ'Vgh

ρ(h - 5) = -ρ'h

Since the density of the small ball equals 1/2 the density of water,

ρ = ρ'/2

ρ(h - 5) = -ρ'h

(ρ'/2)(h - 5) = -ρ'h

ρ'(h - 5)/2 = -ρ'h

(h - 5)/2 = -h

h - 5 = -2h

h + 2h = 5

3h = 5

h = 5/3

h = 1.67 m

So, the maximum depth the ball reaches is 1.67 m.

Answer:

R/2

Explanation:

The potential at a distance r is given by :

[tex]V=\dfrac{kq}{r}[/tex]

Where

k is electrostatic constant

q is the charge

The potential (relative to infinity) due to a point charge is V at a distance R from this charge. So,

[tex]\dfrac{V_1}{V_2}=\dfrac{r_2}{r_1}[/tex]

Put all the values,

[tex]\dfrac{V}{2V}=\dfrac{r_2}{R}\\\\\dfrac{1}{2}=\dfrac{r_2}{R}\\\\r_2=\dfrac{R}{2}[/tex]

Answer:

false.

Explanation:

We know that for a wave that moves with velocity V, with a wavelength λ, and a frequency f, we have the relation:

V = λ*f

So, if the velocity is constant and we increase the frequency to:

f' > f

we will have a new wavelength λ'

Such that:

V = f'*λ'

And V = f*λ

Then we have:

f'*λ' = f*λ

Solvinf for λ', we get:

λ' =(f/f')*λ

And because:

f' > f

then:

(f/f') < 1

Then:

λ' =(f/f')*λ < λ

So, if we increase the frequency, we need to decrease the wavelength.

So, for higher frequency waves, we must have proportionally shorter wavelengths.

Then we can conclude that the given statement:

"or waves moving through the atmosphere at a constant velocity, higher frequency waves must have proportionally longer wavelengths"

is false.

Let m be the mass of the second car, so the first car's mass is 2m.

Let K be the kinetic energy of the second car, so the first car's kinetic energy would be K/2.

Let u and v be the speeds of the first car and the second car, respectively. At the start,

• the first car has kinetic energy

K/2 = 1/2 (2m) u ² = mu ²   ==>   K = 2mu ²

• the second car starts with kinetic energy

K = 1/2 mv ²

It follows that

2mu ² = 1/2 mv ²

==>   4u ² = v ²

When their speeds are both increased by 2.76 m/s,

• the first car now has kinetic energy

1/2 (2m) (u + 2.76 m/s)² = m (u + 2.76 m/s)²

• the second car now has kinetic energy

1/2 m (v + 2.76 m/s)²

These two kinetic energies are equal, so

m (u + 2.76 m/s)² = 1/2 m (v + 2.76 m/s)²

==>   2 (u + 2.76 m/s)² = (v + 2.76 m/s)²

Solving the equations in bold gives u ≈ 1.95 m/s and v ≈ 3.90 m/s.

Answer:

[tex]c.) \: systematic \: error \\ \\ = > it \: is \: the \: error \: caused \: \\ \\ due \: to \: the \: procedure \\ \\ \: and \: used \: apparatuses \\ \\ \huge\mathfrak\red{Hope \: it \: helps}[/tex]

Answer:

Electromechanical transducer and Electrical receiver.

Explanation:

Electromechanical transducer is the part of a communication system which converted electrical waves or electrical energy into sound waves. The most common example loudspeaker while on the other hand, Electrical receiver is a device that converts electrical energy into another form of energy, except thermal. Examples are cell phones, computers and television.

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass [tex]m_A[/tex] as

[tex]x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)[/tex]

[tex]y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)[/tex]

Substituting (2) into (1), we get

[tex]\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)[/tex]

where [tex]f_N= \mu_kN[/tex], the frictional force on [tex]m_A.[/tex] Set this aside for now and let's look at the forces on [tex]m_B[/tex]

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on [tex]m_B[/tex] as

[tex]x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)[/tex]

[tex]y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)[/tex]

From (5), we can solve for N as

[tex]N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)[/tex]

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

[tex]T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)[/tex]

Substituting (7) into (4) we get

[tex]m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba[/tex]

Collecting similar terms together, we get

[tex](m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag[/tex]

or

[tex]a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)[/tex]

Putting in the numbers, we find that [tex]a = 1.4\:\text{m/s}[/tex]. To find the tension T, put the value for the acceleration into (7) and we'll get [tex]T = 21.3\:\text{N}[/tex]. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get [tex]N = 50.9\:\text{N}[/tex]

Answer:

978.6084 Newton

Explanation:

Given the following data;

To find the weight in Newtown;

Conversion:

1 lb = 4.448220 N

220 lb = 220 * 4.448220 = 978.6084 Newton

220 lb = 978.6084 Newton

Therefore, the weight of the astronaut in Newton is 978.6084.

Weight can be defined as the force acting on a body or an object as a result of gravity.

Mathematically, the weight of an object is given by the formula;

Weight = mg

Where;

Note:

Answer:

64.20m

Explanation:

As we can see from the image I have attached below, the route that the chipanzee makes forms a right triangle. In this case, the shortest distance is represented by x in the image, which is the hypotenuse. To find this value we use the Pythagorean theorem which is the following.

[tex]a^{2} +b^{2} = c^{2}[/tex]

where a and b are the length of the two sides and c is the length of the hypotenuse (x). Therefore, we can plug in the values of the image and solve for x

[tex]51^{2} +39^{2} =x^{2}[/tex]

2,601 + 1,521 = [tex]x^{2}[/tex]

4,122 = [tex]x^{2}[/tex]   ... square root both sides

64.20 = x

Finally, we see that the shortest distance is 64.20m

Answer:

increased because as you run into each sound wave the time between each sound decreases meaning the period of each wave decreases to your years and since f=1/T and T is decreasing by greater than 0, f must increase.

Explanation:

Answer:

7.46×10⁻⁶ m

Explanation:

Applying,

F = kqq'/r²............ Equation 1

make r the subject of the equation

r = √(F/kqq').......... Equation 2

From the question,

Given: F = 8 N, q' = q= 4 C

Constant: k = 8.98×10⁹ Nm²/C²

Substitute these values into equation 2

r = √[8/(4×4×8.98×10⁹)]

r = √(55.7×10⁻¹²)

r = 7.46×10⁻⁶ m

Answer:

mercury and alcohol

ii) used to test temperatures

i) It is a good conductor of heat and therefore the whole liquid reaches the temperature of the surroundings quickly.

ii) It does not wet (cling to the sides of) the tube.

Alcohol:

i) Alcohol has greater value of temperature coefficient of expansion than mercury.

ii) it's freezing point is below –100°C.

Answer:

C

Explanation:

20 cm = 0.2m

since uncertainty is 0.1 cm (0.001 m), should be recorded to same number of decimal place as uncertainty

therefore it's 0.200m

Answer:

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