Complete the following statement: The electromotive force is:______.a. the force that accelerates electrons through a wire when a battery is connected to it.b. the maximum potential difference between the terminals of a battery. c. the force that accelerates protons through a wire when a battery is connected to it.d. the maximum capacitance between the terminals of a battery.e. the potential difference between the terminals of a battery when the battery is not in use.

Answer:

how to solve this problem ???????

The magnitude of the resultant force is 919.6 N and the value of angle θ is 36.87⁰.

The resultant of the two forces is determined by resolving the force into x and y component as shown below;

[tex]F_1_x + F_2x_x = F_R_x \ --- (1) \\\\F_1_y + F_2_y = F_R_y\ ---(2)[/tex]

where;

The value of Angle θ is determined from equation (1)

-500sinθ + 600sin(30) = 0

500sinθ = 600sin(30)

500sinθ = 300

sinθ = 3/5

θ = 36.87⁰

The resultant of the forces is determined using the second equation;

500cosθ + 600cos(30) = R

500 x cos(36.87) + 600 x cos(30) = R

919.6 N = R

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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)].

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Answer:

................ftf6x

Answer:

when the mass of an object is decreased, the acceleration will increase

when mass is increased, acceleration decreases

Answer:

f = 1 m

Explanation:

The magnification of the lens is given by the formula:

[tex]M = \frac{q}{p}[/tex]

where,

M = Magnification = 4

q = image distance = 5 m

p = object distance = ?

Therefore,

[tex]4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m[/tex]

Now using thin lens formula:

[tex]\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\[/tex]

f = 1 m

Answer: I beleive A

Explanation:

Answer:

A

Explanation:

We can see the light being reflected off the mirror.

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:

2m

Explanation:

wavelength=speed/frequency

=340/170

=2m

Answer:

The change in entropy is 1.6 W/K.

Explanation:

Thickness, d = 0.5 cm

Area, A = 1 m^2

T = 25°C

T' = - 10°C

Coefficient of thermal conductivity of glass, K = 0.8 W/mK

The change in entropy is given by

S = Q/T

Here,

[tex]S =\frac{Q}{T}\\\\S = \frac{K A (T - T')}{d(T - T')}\\\\S = \frac{0.8\times 1}{0.5} = 1.6 W/K[/tex]

Answer:

True. A permanent magnet like the earth produces its own B field due to movement of the iron core. The earths magnetic field is the reason why we have an atmosphere and it also is the only defense against solar flares. A coil of wire or solenoid that has current have so much moving charge that the motion of the electrical charge can create a significant G b-field

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:

Physical properties are properties that can be measured or observed without changing the chemical nature of the substance. Some examples of physical properties are:

color (intensive)

density (intensive)

volume (extensive)

mass (extensive)

boiling point (intensive): the temperature at which a substance boils

melting point (intensive): the temperature at which a substance melts

Explanation:

Answer:

Given, Apparent weight(W₂)=4.2N

          Weight of liquid displaced (u)=2.5N

          Let weight of body in air = W₁

Solution,

             U=W₁-W₂

              W₁=4.2=2.5=6.7N

∴Weight of body in air is 6.7N

Answer:

The impact force is 98000 N.

Explanation:

mass = 10 tons

The impact force is the weight of the object.

Weight =mass x gravity

W = 10 x 1000 x 9.8

W = 98000 N

The impact force is 98000 N.

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:

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:

hello the answer is 47m/s

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:

Reflective

Explanation:

The radiation pressure of the wave that totally absorbed is given by;

[tex]P_{abs}= \frac{I}{C}[/tex]

and While the radiation pressure of the wave totally reflected is given by;

[tex]P_{ref}= \frac{2I}{C}[/tex]

Now compare the two-equation you can clearly see that the pressure due to reflection is larger than absorption therefore the sail should be reflective.

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.

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:

[tex]{ \tt{check \: in \: the \: pic}}[/tex]

The final speed of the ball is 91.72 m/s.

Given the following data:

To find the final speed of the ball, we would use the following formula:

[tex]F = \frac{M(V - U)}{t}[/tex]

Where:

Substituting the parameters into the formula, we have;

[tex]15000 = \frac{0.145(V \;- \;40)}{0.0005}\\\\15000(0.0005) = 0.145(V \;- \;40)\\\\7.5 = 0.145V - 5.8\\\\0.145V = 7.5 + 5.8\\\\0.145V = 13.3\\\\V = \frac{13.3}{0.145}[/tex]

Final speed, V = 91.72 m/s

Therefore, the final speed of the ball is 91.72 m/s.

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Answer:

The pressure is 2505 Pa.  

Explanation:

Height, h = 1.5 m

density of blood, d = 1000 kg/cubic meter

Gravity, g = 1.67 m/s^2

let the pressure is P.  

The pressure due to the fluid is given by

P = h d g

P = 1.5 x 1000 x 1.67

P = 2505 Pa

Answer:

Liquid volume is usually measured using either a graduated cylinder or a buret. As the name implies, a graduated cylinder is a cylindrical glass or plastic tube sealed at one end, with a calibrated scale etched (or marked) on the outside wall.

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:

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:

Explanation:

The energy for an isothermal expansion can be computed as:

[tex]\mathsf{Q_H =nRTIn (\dfrac{V_b}{V_a})}[/tex] --- (1)

However, we are being told that the volume of the gas is twice itself when undergoing adiabatic expansion. This implies that:

[tex]V_b = 2V_a[/tex]

Equation (1) can be written as:

[tex]\mathtt{Q_H = nRT_H In (2)}[/tex]

Also, in a Carnot engine, the efficiency can be computed as:

[tex]\mathtt{e = 1 - \dfrac{T_L}{T_H}}[/tex]

[tex]e = \dfrac{T_H-T_L}{T_H}[/tex]

In addition to that, for any heat engine, the efficiency e =[tex]\dfrac{W}{Q_H}[/tex]

relating the above two equations together, we have:

[tex]\dfrac{T_H-T_L}{T_H} = \dfrac{W}{Q_H}[/tex]

Making the work done (W) the subject:

[tex]W = Q_H \Big(\dfrac{T_H-T_L}{T_H} \Big)[/tex]

From equation (1):

[tex]\mathsf{W = nRT_HIn(2) \Big(\dfrac{T_H-T_L}{T_H} \Big)}[/tex]

[tex]\mathsf{W = nRIn(2) \Big(T_H-T_L} \Big)}[/tex]

If we consider the adiabatic expansion as well:

[tex]PV^y[/tex] = constant

i.e.

[tex]P_bV_b^y = P_cV_c^y[/tex]

From ideal gas PV = nRT

we can have:

[tex]\dfrac{nRT_H}{V_b}(V_b^y)= \dfrac{nRT_L}{V_c}(V_c^y)[/tex]

[tex]T_H = T_L \Big(\dfrac{V_c}{V_b}\Big)^{y-1}[/tex]

From the question, let us recall  aw we are being informed that:

If the volumes changes by a factor = 5.7

Then, it implies that:

[tex]\Big(\dfrac{V_c}{V_b}\Big) = 5.7[/tex]

∴

[tex]T_H = T_L (5.7)^{y-1}[/tex]

In an ideal monoatomic gas [tex]\gamma = 1.6[/tex]

As such:

[tex]T_H = T_L (5.7)^{1.6-1}[/tex]

[tex]T_H = T_L (5.7)^{0.67}[/tex]

Replacing the value of [tex]T_H = T_L (5.7)^{0.67}[/tex] into equation [tex]\mathsf{W = nRIn(2) \Big(T_H-T_L} \Big)}[/tex]

[tex]\mathsf{W = nRT_L In(2) (5.7 ^{0.67 }-1}})[/tex]

From in the question:

W = 930 J and the moles = 1.90

using 8.314 as constant

Then:

[tex]\mathsf{930 = (1.90)(8.314)T_L In(2) (5.7 ^{0.67 }-1}})[/tex]

[tex]\mathsf{930 = 15.7966\times 1.5315 (T_L )})[/tex]

[tex]\mathsf{T_L= \dfrac{930 }{15.7966\times 1.5315}}[/tex]

[tex]\mathbf{T_L \simeq = 39 \ K}[/tex]

From [tex]T_H = T_L (5.7)^{0.67}[/tex]

[tex]\mathsf{T_H = 39 (5.7)^{0.67}}[/tex]

[tex]\mathbf{T_H \simeq 125K}[/tex]