In a Broadway performance, an 83.0-kg actor swings from a R = 3.90-m-long cable that is horizontal when he starts. At the bottom of his arc, he picks up his 55.0-kg costar in an inelastic collision. What maximum height do they reach after their upward swing?

Answer:

a. The final velocity of the block of mass 2 kg is 3 m/s or 3.86 m/s. The final velocity of the block of mass 1.5 kg is 4 m/s or 2.86 m/s b. The kinetic energy change is 0 J or -12.235 J. Since the collision is elastic, we choose ΔK = 0

Explanation:

From principle of conservation of momentum,

momentum before impact = momentum after impact

Let m₁ = 2 kg, m₂ = 1.5 kg and v₁ = 3 m/s, v₂ = 4 m/s represent the masses and initial velocities of the first and second blocks of mass respectively. Let v₃ and v₄ be the final velocities of the blocks. So,

m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄

(2 × 3 + 1.5 × 4) = 2v₃ + 1.5v₄

6 + 6 = 2v₃ + 1.5v₄

12 = 2v₃ + 1.5v₄

2v₃ + 1.5v₄ = 12 (1)

Since the collision is elastic, kinetic energy is conserved. So

1/2m₁v₁² + 1/2m₂v₂² = 1/2m₁v₃² + 1/2m₂v₄²

1/2 × 2 × 3² +  1/2 × 1.5 × 4² = 1/2 ×2v₃² + 1/2 × 1.5v₄²

9 + 12 = v₃² + 0.75v₄²

21 = v₃² + 0.75v₄²

v₃² + 0.75v₄² = 21  (2)

From (1) v₃ = 6 - 0.75v₄ (3) . Substituting v₃ into (2)

(6 - 0.75v₄)² + 0.75v₄² = 21

36 - 9v₄ + 0.5625v₄² + 0.75v₄² = 21

36 - 9v₄ + 1.3125v₄² - 21 = 0

1.3125v₄² - 9v₄ + 15 = 0

Using the quadratic formula,

v₄ = [-(-9) ± √[(-9)² - 4 × 1.3125 × 15]]/(2 × 1.3125)

= [9 ± √[81 - 78.75]]/2.625

= [9 ± √2.25]/2.625

= [9 ± 1.5]/2.625

= [9 + 1.5]/2.625 or [9 - 1.5]/2.625

= 10.5/2.625 or 7.5/2.625

= 4 m/s or 2.86 m/s

Substititing v₄ into (3)

v₃ = 6 - 0.75v₄ = 6 - 0.75 × 4 = 6 - 3 = 3 m/s

or

v₃ = 6 - 0.75v₄ = 6 - 0.75 × 2.86 = 6 - 2.145 = 3.855 m/s ≅ 3.86 m/s

b. The kinetic energy change ΔK = K₂ - K₁

K₁ = initial kinetic energy of the two blocks =  1/2m₁v₁² + 1/2m₂v₂²

= 1/2 × 2 × 3² +  1/2 × 1.5 × 4² = 9 + 12 = 21 J

K₂ = final kinetic energy of the two blocks = 1/2m₁v₃² + 1/2m₂v₄². Using v = 3 m/s and v = 4 m/s

= 1/2 × 2 × 3² +  1/2 × 1.5 × 4² = 9 + 12 = 21 J.

ΔK = K₂ - K₁ = 21 - 21 = 0

Using v = 3.86 m/s and v = 2.86 m/s

K₂ = 1/2 × 2 × 3.86² +  1/2 × 1.5 × 2.86² = 14.8996 - 6.1347 = 8.7649 J ≅ 8.765 J

ΔK = K₂ - K₁ = 8.765  - 21 = -12.235 J

Since the collision is elastic, we choose ΔK = 0

Answer:

μ= F÷N

μ= 270/460= 0.587

Explanation:

The friction force always acts in the opposite direction of the intended or actual motion.

Answer:

Power coming from solar radiations is 6.94 * 10^14 times higher that the power consumption of all humans.

Explanation:

Intensity of sunlight = I = 1380 w/m^2

Area of earth  = A = 4*pi*r^2 = 4*pi*(6.37*10^6)^2 = 5.09*10^14 m^2

he intensity is defined as the total power spread over the area of earth (Area of Sphere with radius equal to distance between earth and sun) and given by the following formula:

                        Intenity of sunlight = Power/Area of earth

                                                 I = P/A

                                                 P = IA

                                                 P = (1380)(5.09*10^14)

                                                 P =  7.036*10^17 W

if we take ratio:

                                                 7.036*10^17/1013 = 6.94 * 10^14

Hence, power coming from solar radiations is 6.94 * 10^14 times higher that the power consumption of all humans.

Answer:The coefficient of static friction between the turntable and the coin is 0.1

Explanation:

The coefficient of static friction is the friction force between two objects when neither of the objects is moving. ... A value of 1 means the frictional force is equal to the normal force. It is a misconception that the coefficient of friction is limited to values between zero and one.

Answer:

b) 8.

Explanation:

Below is an attachment containing the solution.

Answer:

It will require 14.715 N of force to hold the cartoon beneath the water.

Explanation:

Given the the volume of cartoon is 1.5 liters.

We need to find the force required to hold this cartoon beneath the water.

As we know from the Archimedes principle that the net force is equal to the volume of liquid displaced.

Given volume of the cartoon is 1.5 liters. So, 1.5 liters of water will be displaced.

And we know the density of the water is [tex]1000\ kg/m^3[/tex]. That is [tex]1\ kg/L[/tex]

And [tex]g=9.81\ m/s^2[/tex]

[tex]F_N=\rho Vg\\F_N=1\times 1.5\times 9.81\\F_N=14.715\ N[/tex]

So, it will require 14.715 N of force to hold 1.5 liter volume of cartoon beneath the water.

Answer:

Explanation:

Applying the exponential function of decay

M=Cexp(-kt)

At t =0 the mass is 13mmg

Therefore

13=Cexp(0)

C=13

M=13exp(-kt)

After 12mins, M=8mmg

8=13exp(-K×12)

8/13=exp(-12k)

0.615=exp(-12k)

Take In of both sides

In(0.615)=-12k

-0.4855=-12k

Then, k=0.0405

Then the equations become

M=13exp(-0.0405t)

We need to find t at M=2mmg

M=13exp(-0.0405t)

2=13exp(-0.0405t)

2/13=exp(-0.0405t)

0.1538=exp(-0.0405t)

Take In of both sides

In(0.1538)=-0.0405t

-1.872=-0.0405t

Then t=-1.82/-0.0405

t=46.22mintes

Answer:[tex]1.53g[/tex]

Explanation:

average deceleration= ?

inial velocity: [tex]u=0[/tex]

final velocity: [tex]v=30m/s[/tex]

time: [tex]t=2seconds[/tex]

The first law of kinematics :

[tex]v=u+at[/tex]

find a the subject of the formula

[tex]a=v-u/ t[/tex]

[tex]a=\frac{0-30} 2[/tex]

[tex]a=-30/2[/tex]

[tex]a=-15m/s^{2}[/tex]

The deceleration about g(acceleration due to gravity) will be:

[tex]15/9.8[/tex]

[tex]1.53g[/tex]

The bungee jumper's average deceleration in g's is approximately -1.53 g's.

To find the average deceleration, we need to calculate the change in velocity and divide it by the time it took to come to a halt. The change in velocity is the final velocity minus the initial velocity, which is 0 m/s minus 30 m/s, giving us -30 m/s. The time is given as 2 s. Plugging these values into the formula for average deceleration, we get:

Average Deceleration = (Change in Velocity) / (Time)

Average Deceleration = (-30 m/s) / (2 s) = -15 m/s²

To convert this to g's, we need to divide by the acceleration due to gravity (9.8 m/s²).

Average Deceleration in g's = (-15 m/s²) / (9.8 m/s²) ≈ -1.53 g's

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Answer:

Time taken to reach your destination will be 10hours

Explanation:

Recall the formula for Speed;

speed=Total distance/Total time taken

Speed=2km/h

Total distance=20km

Time taken=x

let x be the unknown time taken

Input each values into the formula;

2=20/x

Making x subject of the equation

x=20/2

x=10

Total time taken =10hours.

Answer:

10 hours

Explanation:

The average speed of a body is given by;

[tex]v_{avg}=\frac{s}{t}..................(1)[/tex]

where s is the total distance travelled and t is the total time spent.

Given;

s = 20km

t = ?

[tex]v_{avg}=2km/h[/tex].

We substitute into equation (1) and then solve for t.

[tex]2=\frac{20}{t}\\2t=20\\t=\frac{20}{2}\\t=10hrs.[/tex]

Answer:

[tex]\frac{D_{jet}}{D_{prop}}=2.865[/tex]

Explanation:

Given data

Speed of jet Vjet=1190 km/h

Speed of prop driven Vprop=595 km/h

Height of jet 7.5 km

Height of prop driven transport 3.8 km

Density of Air at height 10 km p7.8=0.53 kg/m³

Density of air at height 3.8 km p3.8=0.74 kg/m³

The drag force is given by:

[tex]D=\frac{1}{2}CpAv^2\\[/tex]

The ratio between the drag force on the jet to the drag force  on prop-driven transport is then given by:

[tex]\frac{D_{jet}}{D_{prop}}=\frac{(1/2)Cp_{7.5}Av_{jet}^2}{1/2)Cp_{3.8}Av_{prop}^2} \\\frac{D_{jet}}{D_{prop}}=\frac{p_{7.5}v_{jet}^2}{p_{3.8}v_{prop}}\\\frac{D_{jet}}{D_{prop}}=\frac{(0.53)(1190)^2}{(0.74)(595)^2}\\ \frac{D_{jet}}{D_{prop}}=2.865[/tex]

Answer:

Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.

Explanation:

Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.

When this alpha particles were made to strike the aluminum foil, some passed through the foil, some were reflected and speed others changed.

The ones reflected encountered heavier particle known as the nucleus, preventing them from passing through it. The whole observations indicated that atom is not is uniformly charged sphere as proposed by J.J Thomson.

Rutherford proposed new model known as the Planetary model of atom, which described atom as containing a nucleus which is revolved by electron, just like planets revolve round the sun. And this nucleus contains opposite charge to electron which is proton, to balance the motion.

3. Newton's third law

5. Conservation of momentum

Explanation:

Conservation of momentum is mostly used for describing collisions between objects. Here, the type of collision is inelastic collision in which the object when collides with the pendulum bob sticks to it and moves as a combined object. In this process the momentum is conserved.

Let the mass of the pendulum be m1 moving with a velocity v1.

Let the mass of the object be m2 moving with a velocity v2.

Since the momentum is conserved during collision, the equation will be

[tex]m1 v1 + m2 v2 = (m1 + m2) v[/tex]

Where, v is the velocity of the combined system.

Conservation of momentum is actually a direct consequence of Newton's third law.

Consider a collision between two objects, object A and object B. When the two objects collide, there is a force on A due to B. However, because of Newton's third law, there is an equal force in the opposite direction, on B due to A

FAB = -FBA

The mechanical energy is not conserved due to the fact that the kinetic energy is not the same before and after the collision.

The conservation of momentum is crucial for analyzing collisions, while the conservation of mechanical energy is ideal for studying swinging motion.

Conservation of momentum can best be used to analyze the collision between the object and the pendulum bob, while Conservation of mechanical energy is best suited to analyze the resulting swing.

The physical model of the sun's interior has been confirmed by observations of neutrino and seismic vibrations.

Explanation:

Sun's interior is composed of very high temperature and solar flares. So it is very difficult to understand the interior of the sun. But by using the vibrations of neutrino and seismic waves emitted by the solar waves, the physical model can be assumed.

As the interior of the sun performs continuous chain of hydrogen cycle. So the continuous emission of energy from the chain reaction releases neutrino. So these vibrations in neutrino and seismic vibrations, the physical model can be assumed easily.

Answer:

The answer to the question is

The luminosity of stars A is four times that of star B

Explanation:

Flux (F) produced by a source of light is directly proportional to the brightness or Luminosity (L), and varies inversely to its distance d, that is [tex]F \alpha \frac{L}{d^2}[/tex]

Therefore if the two stars present the same flux then we have

[tex]\frac{L_1}{d_1^2} = \frac{L_2}{(2d_1)^2}[/tex] then crossing out like terms gives [tex]\frac{L_1}{1} = \frac{L_2}{2^2}[/tex] or 4·L₁ = L₂

The luminosity of  star A is 4 times the that of star B

Final answer:

Star A, being twice as far away from Earth as Star B but appearing equally bright, must have a luminosity that is four times greater than Star B's luminosity, due to the inverse square law of light.

Explanation:

When comparing the luminosities of two stars that appear equally bright from Earth, but one is twice as far away as the other, we must account for the inverse square law of light. This law dictates that the intensity (brightness) of light from a source (in this case, a star) decreases proportionally to the square of the distance from the source.

Therefore, if Star A is twice as far away from Earth as Star B, yet they appear to have the same brightness, Star A must have a luminosity four times greater than that of Star B. This is because, to compensate for the increased distance, Star A must emit more light to be perceived as equally bright as Star B from Earth.

Answer:

4. Both points have the same instantaneous angular velocity

Explanation:

Angular velocity is a measure of the the number of rotations per unit time. This does not depend on the radius of the wheel. Hence, all points on the wheel have the same angular velocity. This invalidates option 1.

The centripetal acceleration is given by the product to the square of the angular velocity and the radius or distance from the centre. A and B are located at different distances from the centre. Hence, they have different centripetal acceleration. This invalidates option 2.

The tangential acceleration depends on the linear velocity which itself is a product of the angular velocity and the distance from the centre. Hence, it is different for both points because they are at different distances from the centre.

Since both A and B are fixed points on the wheel, they move through equal angles in the same time. In fact, for any other fixed point, they all move through the same angle in the same time. This invalidates option 5.

Final answer:

All points on a rotating wheel share the same instantaneous angular velocity; however, points farther from the axis will experience greater centripetal acceleration. The correct statement is that both points have the same instantaneous angular velocity.

Explanation:

The question concerns the properties of points located at different radii of a rotating wheel, specifically relating to angular velocity, centripetal acceleration, and tangential acceleration. We can address the situation by considering the principles of circular motion. When a rigid wheel is rotating about a fixed axis, all points on the wheel have the same instantaneous angular velocity since every point on the wheel rotates through the same angle in the same amount of time.

Centripetal acceleration is proportional to the radius and the square of the angular velocity. Since point A, being at the rim, is farther from the axis than point B, point A experiences a greater centripetal acceleration. On the other hand, tangential acceleration is related to the angular acceleration and the radius. If the wheel is rotating with decreasing angular velocity, both point A and B experience the same tangential acceleration because it is a property of the wheel's rotation, not the points' individual locations.

The correct statement in this scenario is that both points have the same instantaneous angular velocity, which makes option 4 the true statement.

Answer:

Option B

Explanation:

The orbital periods of Phobos and Deimos can be calculated using the Newton's form of Kepler's third law:  

[tex] T^{2} = \frac {4 \pi^{2}}{G*M_{m}} \cdot a^{3} [/tex]  

where T: is the period, G: is the gravitational constant = 6.67x10⁻¹¹ m³kg⁻¹s⁻², Mm: is the mass of Mars = 6.42x10²³ kg, [tex]a_{P}[/tex]: is the average radius of orbit for the satellite Phobos = 9376 km, and [tex]a_{D}[/tex]: is the average radius of orbit for the satellite Deimos = 23463 km.  

The orbital period of Phobos is:

[tex] T = \sqrt {\frac {4 \pi^{2}}{6.67 \cdot 10^{-11} m^{3} kg^{-1} s^{-2}*6.42 \cdot 10^{23} kg} \cdot (9.376 \cdot 10^{6} m)^{3}} = 2.75 \cdot 10^{4} s = 7 hours 36 min [/tex]        

The orbital period of Deimos is:

[tex] T = \sqrt {\frac {4 \pi^{2}}{6.67 \cdot 10^{-11} m^{3} kg^{-1} s^{-2}*6.42 \cdot 10^{23} kg} \cdot (2.35 \cdot 10^{7} m)^{3}} = 1.09 \cdot 10^{5} s = 1 day 6 hours [/tex]      

Therefore, the approximate orbital periods of Phobos and Deimos are 7 hours 35 minutes and 1 day 6 hours, respectively, so the correct answer is option B.    

I hope it helps you!      

Answer:

[tex]4.08cm/s^2[/tex]

Explanation:

The second equation of a uniformly accelerated motion could be used to solve this problem. This is given by equation (1);

[tex]s=ut+\frac{1}{2}at^2....................(1)[/tex]

where u is the particle's initial velocity, t is the time taken, a is the acceleration and s is the distance travelled.

Given;

u = 20cm/s

t = 7s

a = ?

s = ?

The particle moved from one point [tex]x_1[/tex] to another point [tex]x_2[/tex] along the x-axis, where [tex]x_1=10cm[/tex] and [tex]x_2=-30cm[/tex]. This information could be used to find the distance s covered by the object as follows;

[tex]s=x_1-x_2.................(2)\\s=10-(-30)\\s=10+30\\s=40cm[/tex]

We the make appropriate substitutions into equation (1) and then solve for the acceleration.

[tex]40=(20*7)+\frac{1}{2}*a*7^2\\40=140+\frac{1}{2}*a*49\\40=140+24.5a\\40-140=24.5a\\hence\\24.5a=-100\\a=\frac{-100}{24.5}\\a=-4.08cm/s^2[/tex]

The negative sign is an indication that the particle is decelerating.

Answer:

7.347 cm / s²

Explanation:

Using equation of linear motion

S = ut + 1/2 at²

where total displacement = final displacement - initial displacement

S = - 30 - 10 = - 40 cm

- 40 cm = (20 cm /s × 7 s) + 1/2 a (7²)

- 40 cm = 140 cm + 1/2 49 a

- 40 cm - 140 cm =  1/2 × 49 a

- 180 cm × 2 / 49 s² = a

a = -7.347 cm / s²

It is probably decelerating.

Final answer:

To calculate the flashlight's power usage, we use power and current formulas based on given charge and voltage values. Similarly, to estimate battery life for a flashlight bulb, we divide the battery's Amp-Hour capacity by the bulb's current draw.

Explanation:

The question is about determining the average current used by a flashlight bulb over a time period and how long the battery would last. To calculate the power usage, we apply the formula P = IV, where P is the power in watts, I is the current in amperes, and V is the voltage in volts. Given that 600 C (coulombs) of charge passes through the flashlight in 0.500 hours (which is 1800 seconds) and the voltage is 3.00 V, we can find the average current using the formula I = Q/t, where Q is the charge in coulombs and t is the time in seconds.

To find out how long a battery will keep a flashlight bulb burning, we divide the battery capacity in ampere-hours (Ah) by the current drawn by the bulb. If a 1.00-W bulb is used and the battery is rated at 1.00 Ah and 1.58 V, we first need to use the power formula P = IV to find the current drawn by the bulb.

The flashlight bulb that draws 0.60 A can be lit for approximately 1.67 hours using a 1.00 Ah alkaline battery. This is calculated by dividing the battery capacity by the current drawn by the bulb.

To determine how long a flashlight bulb that draws 0.60 A can be lit, we must consider the battery's capacity and voltage. Here is a step-by-step explanation:

First, identify the battery capacity. Assume we have an alkaline battery rated at 1.00 Ah.

Next, understand that 1.00 Ah means the battery can supply 1.00 ampere for 1 hour.

Since the flashlight draws 0.60 A, we calculate the time the battery can light the bulb using the formula:

Time (hours) = Battery Capacity (Ah) / Current (A)

Substitute the values:

Time = 1.00 Ah / 0.60 A = 1.67 hours

Thus, the flashlight bulb can be lit for approximately 1.67 hours or 1 hour and 40 minutes before the battery is depleted.

Answer:

128 N

Explanation:

Using

F = Gm'm/r²....................... Equation 1

Where F = Force, G = Universal constant, m = mass of the human, m' = mass of the moon, r = radius of the moon

Given: m = 75 kg, m' = 7.4×10²² kg, r = 1.7×10⁶ m

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute into equation 1

F =  (6.67×10⁻¹¹ )(75)(7.4×10²²)/(1.7×10⁶)²

F = (3.7×10¹⁴)/(2.89×10¹²)

F = 1.28×10²

F = 128 N

Explanation:

Below is an attachment containing the solution.

Answer:

37057.9V

Explanation:

Electric potential is defined as the work done in moving a unit positive charge from infinity to a point.

Electric potential (E) = Potential Difference (V)/distance between plates(d)

Given; electric field of 5.71 x 10^5 N/C; distance between plates =6.49cm = 0.0649m

Since E = V/d

V = Ed

V = 5.71×10^5×0.0649

V = 37057.9Volts

The potential difference (in V) between the plates is 37057.9V

Answer:

V = 3.71×10⁴ V

Explanation:

Potential difference: This can be defined as the work done in moving a positive charge from infinity to any point in an electric field.

The S.I unit of potential difference is Volt (V).

The expression for potential difference is

V = E×d.............................. Equation 1

Where V = potential difference between the plates, E = Electric field , d = distance of separation between the plates

Given: E = 5.71×10⁵ N/C, d = 6.49 cm = 0.0649 m.

Substitute into equation 1

V = 5.71×10⁵×0.0649

V = 3.71×10⁴ V

Answer:

Mammography is the process in which low energy radiations are used to diagnose and screening. The purpose of this process is the early detection of the breast cancer. These low energy radiations may have some risks like damaging and burning of cells.  

In the current scenario, woman is apprehensive because she has read about the risks of using ionizing radiations. The radiographer should tell her the benefits of the mammography will outweigh its potential consequences. Screening, for instance, will let her know if she is suffering from breast cancer. Cancer is very dangerous disease as compare to very small burning.

In this way radiographer should handle the situation.

Answer:

Explanation:

a) To get the resistivity ρ at 50 Celsius, given the resitstivity at 20 Celsisus, use:

ρ = ρo(1 + α(T - To))

where To = 20 Celsius

b) Knowing the resistivity at 50 Celsius, and the (uniform) electric field E, you can determine the current density J using:

E = ρJ

(which is actually a density-averaged version of V = IR)

c) Assuming the current is uniform (which is should be in a uniform electric field and constant-diameter wire), the current i can be calculated using:

J = i/A --> i = JA

where A is the cross-sectional area of the wire (given by πr2); make sure to convert the given diameter to a radius, and the radius to base units

d) Since the electric field is given in volts per meter, and you have two meters of length in the wire, you can determine directly from that how many volts difference you need at the ends of the wire to get 0.2 volts per meter.

0.2 = V/d

with d = 2 m. This corresponds to a uniform electric field being related to voltage by V = Ed, where d is distance along the field line.

Explanation:

Below are attachments containing the solution.

Answer:

The total velocity of the ball will be 14.14 m/s.

Explanation:

Horizontal Velocity component = 10 m/s

Vertical Velocity component = -10 m/s

Total velocity of the ball will be found from the following equation:

(Total velocity) ^2 = (Horizontal Velocity) ^2 + (Vertical Velocity) ^2

Total Velocity ^2 = 10^2 + (-10)^2

Total Velocity^2 = 100 + 100

Total Velocity = [tex]\sqrt{200}[/tex]

Total Velocity = 14.14 m/s

Answer: Two marbles separated by 300 kilometers

Explanation: Hope i helped have a great day and please mark brainliest i would appreciate it!

Answer:

[tex]A = 1.056\,m^{2}[/tex]

Explanation:

The portion of solar energy converted into electric energy is given by the following equation:

[tex]\dot E = \eta \cdot I\cdot A[/tex]

The area needed to produce energy is derived by clearing the corresponding variable:

[tex]A = \frac{\dot E}{\eta \cdot I}[/tex]

[tex]A = \frac{75\,W }{(0.1)\cdot (710\,\frac{W}{m^{2}} )}[/tex]

[tex]A = 1.056\,m^{2}[/tex]

Final answer:

To double the velocity of the wooden box, you need to double the force applied to it. Use the equation Force = coefficient of kinetic friction x normal force to calculate the force required to push the crate at a constant speed. Then, double this force to find the force required to double the velocity.

Explanation:

To double the velocity of the crate to 2.0 m/s, you need to double the force applied to the crate. The force required to push the crate at a constant speed can be calculated using the equation:

Force = coefficient of kinetic friction x normal force

The normal force is equal to the weight of the crate, which is given by:

Normal force = mass x gravity

With the given values of the mass of the crate (45 kg), the coefficient of kinetic friction (0.25), and the acceleration due to gravity (9.8 m/s²), you can calculate the force required to push the crate at a constant speed. Then, you can double this force to find the force required to double the velocity.

Answer:

he failed thousands of times

Explanation:

There is no known number for his failings. Edison may have failed in many of his experiments and in his schooling, but he had something better working in his favor. He had great determination and persistence.

He failed thousands of times in an attempt to develop an electric light, the great Edison simply viewed each unsuccessful experiment as the elimination of a solution that wouldn’t work, thereby moving him that much closer to a successful solution.

The given question is incomplete as the options are missing. The options related to this question are as follows-

(A) clear dry weather

(B) hot dry weather

(C) cloudy wet weather

(D) cold dry weather

Answer:

Option (C)

Explanation:

The surface temperature often increases because of increased absorption of solar radiation, the air present at the surface gets heated up more readily, as a result of which the air becomes less dense, and eventually rises up. This gives rise to the creation of a low air pressure system. It often forms clouds comprising of increase relative humidity, and generates wind and thereby causes precipitation. It also causes heavy storms when the atmospheric conditions are too intense.

Thus, the type of weather associated with this is wet and cloudy weather.

Hence, the correct answer is option (C).

Answer:

D

Explanation:

Cloudy wet weather

Answer:

It's true.

Explanation:

It's true. When we connect two resistors in parallel the current is divided between the two in such a way that the sum of the currents on each resistor should be equal to the current on that branch. By finding the equivalent resistance we can use Ohm's law to determine the voltage drop across the resistors. This voltage drop is the same for both, since they're connected in parallel.

This statement is true. In a series-parallel circuit, the voltage across the branches in the parallel portion can be found by multiplying the sum of the branch currents by the equivalent resistance of the resistors in the parallel portion.

In a series-parallel circuit, the voltage across the branches in the parallel portion can be found by multiplying the sum of the branch currents by the equivalent resistance of the resistors in the parallel portion. This statement is true.

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