Question 12 of 15
The following equation is an example of
decay

Question 12 Of 15The Following Equation Is An Example Ofdecay

Answers

Answer 1

Answer:

beta decay

Explanation:

the decay above is known as beta decay because give electron after it decay.

Answer 2

Answer: It’s definitely “Beta” friend!


Related Questions

A 30° incline permanently sits on a 1.1 meter high table. Starting from rest a ball rolls off the incline with a velocity of 2m/s.
a. Calculate the length of the incline
b. calculate the horizontal and vertical components of the velocity at the end of the incline.​

Answers

It would be a good game for you but if I get a pic I don’t want you can you come to my crib I just

what is the answer to what does a 150 kg roller coaster traveling at 10m/s has how much kinetic energy

Answers

The answer to the question is 15 kinetic energy

If Nell was blocking a defensive lineman for the South Grand Prairie Warriors with a force of 126 Newtons and had a mass of 56.2 kg…what would his acceleration be equal to?

Answers

Answer:

2.24 m/s²

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

[tex]a = \frac{f}{m} \\ [/tex]

m is the mass

f is the force

From the question we have

[tex]a = \frac{126}{56.2} \\ = 2.241992...[/tex]

We have the final answer as

2.24 m/s²

Hope this helps you

Sugar crystals enter a dryer at the rate of 1000 kg h-1 and at 20% w.b. moisture content. They leave the dryer at 3% w.b. moisture content. If the drying process requires 3000 kJ kg-1 of water removed, estimate the amount of heat required per hour and the rate of dry crystals out of the dryer.

Answers

Answer:

The dryer evaporate 200 - 24.74 kg of water per hour

To remove 1kg of water it need 3000 K J

So to remove, 175.26 Kg

it need 5.257x [tex]10^{5}[/tex] KJ of heat per hour.

Explanation:

In one hour, the amount of sugar entering = 1000 kg

w.b moisture content is defined as,

weight of water / weight of water + weight of dry

[tex]W_{w}[/tex]/[tex]W_{w}[/tex] + [tex]W_{d}[/tex] x 100

[tex]W_{w}[/tex] + [tex]W_{d}[/tex] = 1000 kg when entering

it has 20% moisture content when entering

[tex]W_{w}[/tex] = 0.2 x 1000 = 200 kg

when leaving it has 3% moisture content then weight of dry material

[tex]W_{d}[/tex] = 1000 - 200 = 800 Kg

[tex]\frac{W_{w}^{'} }{W_{w}^{'} + W_{d}^{'} }[/tex] = 0.03

[tex]\frac{W_{w}^{'} }{W_{w}^{'} + 800 }[/tex] = 0.03

[tex]W_{w} ^{'}[/tex] = 0.03 x [tex]W_{w} ^{'}[/tex] + 0.03 x 800

[tex]W_{w} ^{'}[/tex] = 24.74 kg

When leaving the dryer, the crystals has total weight of the water = 24.74 kg per hour.

The dryer evaporate 200 - 24.74 kg of water per hour

To remove 1kg of water it need 3000 K J

So to remove, 175.26 Kg

it need 5.257x [tex]10^{5}[/tex] KJ of heat per hour.

A particle of mass m collides with a second particle of mass m. Before the collision, the first particle is moving in the x-direction with a speed 2v and the second particle is at rest. After the collision, the second particle is moving in the direction 45o below the x-axis and with a speed √2v.
a) Find the velocity of the first particle after the collision. (i.e. find the x-and y-components of the velocity.)
b) Find the total kinetic energy of the two particles before and after the collision.
c) Is the collision elastic or inelastic?

Answers

Answer:

a) v, v

b) 2mv^2

c) Elastic collion

Explanation:

(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v).  From momentum conservation in x-direction

Here x, y represent direction.They are not variable. 1 and 2 represent before and after.

2vm=v1xm+v2xm, we find v1x=v.

From momentum conservation in y-direction

0 =v1ym+v2ym, we findv1y=v.

(b) By energy conservation principle

Before: K=1/2m(2v)^2=2mv^2.

After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2

(c) The collision is elastic

4. You’re driving a car that can climb a maximum gradient of 500m/km. The hill in front of you starts at an elevation of 20m and reaches 100m. The total distance up the hill is 1.5km? What is the gradient of the hill and will your car make it?

Answers

Answer:

Gradient of the hill is 53.33m/km which is lesser than the 500 m/km. So, he will be able to climb the hill conveniently.

Explanation: Given that

Maximum gradient = 500 m/km

Total distance = 1.5 km

Starting elevation = 20 m

Final elevation = 100 m

Gradient = change in elevation/ total distance.

Now, substitute the values into the formula.

Gradient = (100m - 20m)/1.5km

= 80m/1.5km

= 53.33m/km

Gradient of the hill is 53.33m/km which is lesser than the 500 m/km. So, he will be able to climb the hill conveniently.

PLEASE HELP!!!!
Suppose a tractor was moving forward at 6.0 m/s. It runs into a pile of hay and is
brought to a stop in 1.1 s. If the tractor had a mass of 2550 kg, with what force did
the pile of hay stop the tractor?
A. -5.5N
B. -14000N
C. 15000N
D. -15000N
E. -0.0021N
F. -0.000072

Answers

Answer:

B )-14000N

Explanation:

F=mv-mu

t

F =2550(0)-2550(6)

1.1

F = -13909.09

approximately -14000N

Suppose a tractor was moving forward at 6.0 m/s. It runs into a pile of hay and is brought to a stop in 1.1 s. If the tractor had a mass of 2550 kg, the force did the pile of hay stop the tractor 14000N

What are the types of force ?

Force can be a unit of  pushing or pulling of any object which result   from the object’s interaction or movement, without applying force the objects can not be moved, can be stopped or change the direction.

Force is a  quantitative parameter between two physical bodies, means an object and its environment, there are various  types of forces in nature.

If an object present in its moving state will be either static or motion, the position of the object will be changed if it is pushed or pulled and The external push or pull upon the object is mainly called as Force.

The contact force  that occurs when we apply some effort on an object such as Spring Force, Applied Force, Air Resistance Force, Normal Force, Tension Force, Frictional Force

Non-Contact forces are the type of forces that occur from a distance  such as Electromagnetic Force, Gravitational Force, Nuclear Force

F=mv-mu

t

F =2550(0)-2550(6)

1.1

F = -13909.09

approximately -14000N

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baseball player hits a line drive estimated to have traveled 145 meters the ball leaves the bat with a horizontal velocity of 40.0 m/s . how much time was the ball in the air

Answers

Answer:

5800

Explanation:

caluculatior

The ball was in the air for 3.625 second.

What is velocity?

Velocity is the direction at which an object is moving and serves as a measure of the rate at which its position is changing as seen from a specific point of view and as measured by a specific unit of time (for example, 60 km/h northbound).

In kinematics, the area of classical mechanics that deals with the motion of bodies, velocity is a fundamental idea.

A physical vector quantity called velocity must have both a magnitude and a direction in order to be defined.

Given that: the a horizontal velocity of the ball: v = 40.0 m/s .

The ball  traveled 145 meters.

Hence, the time during which the ball is in the air:

= distance travelled/ horizontal velocity of the ball

= 145 meter /  40.0 m/s .

= 3.625 second.

Learn more about velocity here:

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A crane raises a 12,000 N marble sculpture at a constant velocity onto a pedestal 1.5 m above the ground outside an art museum. How much work is done by the crane

Answers

For the work, applicate formula:

[tex]\boxed{\boxed{\green{\bf{W = F\times d}}}}[/tex]

According our data:

Replacing:

W = 12000 N * 1,5 m

Resolving:

W = 18000 J

The work done is 18000 Joules.

A car is traveling at a constant speed along the road ABCDE shown in the drawing. Section Ab and DE are straight. Rank the acceleration in each of the four sections according to magnitude (smallest first).
-AB & DE tie, CD, BC
-all tie
-CD, BC, AB & DE tie
-AB & DE tie, BC , CD
-BC, CD, AB & DE tie

Answers

Answer:

AB = DE <CD <BC

Explanation:

This is an exercise in kinetics, the accelerations defined as the change in velocity over the time interval, therefore the accelerations of a vector.

Because the acceleration is a vector, it has two parts, the modulus that the numerical value of the magnitude and the direction, a change in any of them implies the existence of a relationship.

Let's apply these reasoning to our problem.

AB Path  

this path is straight and as they indicate that the constant speed the acceleration is zero

DE path

This path is straight and since the velocity is constant the zero steps

BC path

This path is a curve and the velocity modulus is constant, but its directional changes therefore there is an acceleration called centripetal, given by the expression

         [tex]a_{c}[/tex] = v² / r

where r is the radius of the curve and the direction of acceleration is towards the center of the curve

CD path

This path is a curve and it also has centripetal acceleration, as can be seen in the drawing, the radius of the curve is greater than in section BC, therefore the acceleration is less

              [tex]a_{BC}[/tex] > [tex]a_{CD}[/tex]

In  summary lower accelerations are

 AB = DE <CD <BC

A sinusoidal wave with wavelength 0.500 m travels along a string. The maximum transverse speed of a point on the string is. 4.00 m/s and the maximum transverse acceleration is 1.00 x 105 m/s2. What is the propagation speed of the wave

Answers

Answer:

The velocity [tex]v = 1989.2 \ m/s[/tex]

Explanation:

From the question we are told that

    The wavelength is [tex]\lambda = 0.500 \ m[/tex]

     The maximum transverse speed is  [tex]v = 4.0 \ m/s[/tex]

      The maximum transverse acceleration is  [tex]a = 1.00 *10^{5} \ m/s^2[/tex]

Generally the frequency of the wave is mathematically represented as

                    [tex]f = \frac{w}{2 \pi }[/tex]

Here  w  is the angular speed which is mathematically evaluated as

      [tex]w = \frac{a}{v}[/tex]

=>    [tex]w = \frac{1.00 *10^{5}}{4}[/tex]

=>    [tex]w = 25000 \ rad/sec[/tex]

So

       [tex]f = \frac{ 25000 }{2 * 3.142 }[/tex]

=>    [tex]f = \frac{ 25000 }{2 * 3.142 }[/tex]

=>    [tex]f = 3978.4 \ Hz[/tex]

Gnerally the propagation speed of the wave is mathematically represented as

        [tex]v = f * \lambda[/tex]

=>      [tex]v = 3978.4 * 0.500[/tex]

=>      [tex]v = 1989.2 \ m/s[/tex]

An athlete swings a 6.90-kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.830 m at an angular speed of 0.680 rev/s.
(a) What is the tangential speed of the ball?
m/s
(b) What is its centripetal acceleration?
m/s2
(c) If the maximum tension the rope can withstand before breaking is 110 N, what is the maximum tangential speed the ball can have?
m/s

Answers

Answer:

a) 0.5644 m/s

b) ~0.384 m/s^2

c) ~3.638 m/s

Explanation:

a) Tangential speed is found be the radius*rotational speed, so it is 0.83*0.68 = 0.5644 m/s

b) Centripetal acceleration is found by v^2/r, so it is (0.5644^2)/0.83 = ~0.384 m/s^2

c) Let the tangential speed be v. The maximum centripetal force 110 N (as given). Centripetal force = mass*centripetal acceleration = mass*v^2/r (because centripetal acceleration is found by v^2/r). Inputting the values from the problem and solving for v, we get:

110 = 6.9*v^2/0.83

v = sqrt(110*0.83/6.9) = ~3.638 m/s

I hope this helps! :)

What is the beat frequency heard when two organ pipes, each open at both ends, are sounded together at their fundamental frequencies if one pipe is 48 cm long and the other is 63 cm long?

Answers

Answer:

Beat frequency = 125.5 Hz

Explanation:

Given:

Length of first organ pipes  = 43 cm = 0.43 m

Length of second organ pipes = 63 cm = 0.63 m

Computation:

Wavelength L = λ/2

length λ = 2L

Frequency f1 = v/λ = 340 / 2(0.43)

Frequency f1 = 395.34 Hz

Frequency f2 = v/λ2 = 340/2(0.63)

Frequency f2 = 269.84 Hz

Beat frequency  = Frequency f1 - Frequency f2

Beat frequency = 395.34 - 269.84

Beat frequency = 125.5 Hz

You heat up a 3.0 kg aluminum pot with 1.5 kg of water from 5 to 90o Celsius. The specific heat capacity of Aluminum is 900 J/kgK. (a) How much energy did it take to heat the pot of water

Answers

Answer:

765,000Joules or 765kJ

Explanation:

The Quantity of heat required is expressed as;

Q = (mcΔt)al + (mcΔt)water

m is the mass

c is specific heat capacity

Δt is the change in temperature

Q = (3(900)(90-5)) + (1.5(4200)(90-5))

Q = 2700*85 + 6300*85

Q = (2700+6300)85

Q = 9000*85

Q = 765,000

Hence the amount of energy needed is 765,000Joules or 765kJ

A ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally away from and below the point of release. What is the magnitude of its velocity just before it strikes the ground?​

Answers

Answer:

44.3 m/s

Explanation:

Given that a ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally away from and below the point of release.

What is the magnitude of its velocity just before it strikes the ground ?

The parameters given are:

Height H = 100m

Since the ball is thrown from a top of a building, initial velocity U = 0

Let g = 9.8m/s^2

Using third equation of motion

V^2 = U^2 + 2gH

Substitute all the parameters into the formula

V^2 = 2 × 9.8 × 100

V^2 = 200 × 9.8

V^2 = 1960

V = 44.27 m/s

Therefore, the magnitude of its velocity just before it strikes the ground is 44.3 m/s approximately

What is the potential energy of a spring that is compressed 0.65 m by a 25 kg block if the spring constant is 95 N/m?
A. 1.6J
B. 7.9J
C. 15J
D. 20J

Answers

Answer:

D. 20J

Explanation:

Answer:

20 J

Explanation:

yes

A stone is dropped into a deep well and is heard to hit the water 5 s after being dropped. Determine the depth of the well.

Answers

Answer:

122.5 m

Explanation:

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

Time (t) = 5 s

Depth (h) =?

The depth of the well can be obtained as:

NOTE: Acceleration due to gravity (g) = 9.8 m/s²

h = ½gt²

h = ½ × 9.8 × 5²

h = 4.9 × 25

h = 122.5 m

Thus, the depth of the well is 122.5 m

Fon of the assignment will be graded by your teacher Question: In a lab experiment, A square of black paper and a square of white paper are place directly under 100-watt lamps for 30 minutes. Predict what happens to the temperature of each paper during the experiment.​

Answers

Answer: OVERVIEW

Students will use black and white construction paper and a light source to learn that dark objects

absorb more light and reflect less light than bright objects. The activity also demonstrates the conversion

of radiant light energy into heat energy.

CONCEPTS

• Dark surfaces absorb more visible light energy than bright surfaces

• Dark surfaces reflect less visible light energy than bright surfaces

• Energy can change forms, in this case from radiant light energy to heat

• Clouds, being bright, reflect significant amounts of sunlight and help to regulate Earth’s temperature

MATERIALS

• 2 thermometers

• Flood lamp, desk lamp, or area in direct sunlight

• Ruler

• Construction paper, 1 piece white, 1 piece black, or 2 sheets photocopy paper

• Scissors

• Cellophane tape or rubber bands

• 2 empty metal food cans, same size (be sure rims are not jagged)

PREPARATION

The paper and the cans can be prepared beforehand or prepared as part of the activity (see Procedure).

Although two cans with their tops completely removed can be used, the experiment will be more effective

(have fewer external effects), if only holes are placed in the cans’ lids, e.g., two holes from a bottle opener

to empty material out of the can, and one center hole created with an awl for the thermometer. Only one

hole is actually needed for the experiment - for the thermometer. Cans can optionally be filled with water,

or this can be done as a separate experiment to demonstrate the higher heat capacity of water compared to

air.

You can either use a flood lamp or a desk lamp (light bulb) to simulate sunlight, as described here, or

you can place the cans on a windowsill (window closed) or other sheltered area in direct sunlight. A flood

lamp will be the most effective option, causing the largest temperature increases.

PROCEDURE

Engagement

Discuss whether dark surfaces (e.g., asphalt) or bright surfaces (e.g., concrete) tend to get hotter in

sunlight. Which would you rather walk on during the day in the summertime? What color are solar cells,

for example those found on some calculators or freeway call boxes?

Explanation: I hope this helps! ∧    ∧

                                                ⊂∵→ω←∵⊃

A 1.40 kg block is attached to a spring with spring constant 16.5 N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 45.0 cm/s . What are

Answers

Answer:

A = 0.13 m

Explanation:

Given that,

Mass of a block, m = 1.4 kg

Spring constant of the spring, k = 16.5 N/m

While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 45.0 cm/s or 0.45 m/s

We need to find the amplitude of the subsequent oscillations.

We can use the conservation of energy here. Initially, the kinetic energy of the block is maximum and then it gets converted to the potential energy of the spring.

Mathematically,

[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2[/tex]

A is the amplitude of subsequent oscillations.

[tex]A=\sqrt{\dfrac{mv^2}{k}} \\\\A=\sqrt{\dfrac{1.4\times (0.45)^2}{16.5}} \\\\A=0.13\ m[/tex]

So, the amplitude of subsequent oscillations is 0.13 m.

Two identical blocks fall a distance H. One falls directly down, the other slides down a frictionless incline. Which has the larger speed at the bottom?

Answers

The question incomplete, the complete question is;

Two identical blocks fall a distance H. One falls directly down, the other slides down a frictionless incline.

(A)the one falling vertically

(B)the one on the incline

(C)Both have the same speed.

(D)cannot be determined

Answer:

(C)Both have the same speed.

Explanation:

When we consider the question closely, we will discover that an object falling down a frictionless incline is comparable to an object falling freely under gravity.

In both instances, the acceleration of objects is just the same irrespective of mass.

Hence, the object falling vertically and the object sliding down a frictionless plane will have the same speed at the bottom.

A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal force of 98.0 N for 1.40 s.(a) Determine the magnitude of each force acting on the block of ice while you are pulling.Fpull = NFg = NFN = N(b) With what speed is the ice moving after you are finished pulling?m/s

Answers

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block ([tex]W[/tex]), measured in newtons, is:

[tex]W = m\cdot g[/tex] (1)

Where:

[tex]m[/tex] - Mass of the block of ice, measured in kilograms.

[tex]g[/tex]  - Gravitational acceleration, measured in meters per square second.

If we know that [tex]m = 14\,kg[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the magnitudes of the weight and normal force of the block of ice are, respectively:

[tex]N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]

[tex]N = W = 137.298\,N[/tex]

And the pull force is:

[tex]F_{pull} = 98\,N[/tex]

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

[tex]m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f}[/tex] (2)

Where:

[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds of the block, measured in meters per second.

[tex]\Sigma F[/tex] - Horizontal net force, measured in newtons.

[tex]\Delta t[/tex] - Impact time, measured in seconds.

Now we clear the final speed in (2):

[tex]v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}[/tex]

If we know that [tex]v_{o} = 0\,\frac{m}{s}[/tex], [tex]m = 14\,kg[/tex], [tex]\Sigma F = 98\,N[/tex] and [tex]\Delta t = 1.40\,s[/tex], then final speed of the ice block is:

[tex]v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}[/tex]

[tex]v_{f} = 9.8\,\frac{m}{s}[/tex]

The final speed of the block of ice is 9.8 meters per second.

Calculate the escape velocity
the moon's surface given that a man on the moon has 1/6 his weight on earth​

Answers

Answer:

v = 2.38 × 10³ m/s

Explanation:

Escape velocity, v = √(2gR) where g = acceleration due to gravity on planet and R = radius of planet.

Since it is given that the weight of the man on the moon is 1/6 his weight on earth, and g' = acceleration due to gravity on moon and g = acceleration due to gravity on earth and m = mass of man,

mg' = mg/6

g' = g/6

Since g = 9.8 m/s²,

g'= 9.8 m/s² ÷ 6

g' = 1.63 m/s²

The escape velocity of the moon is thus  v = √(2g'R) where R = radius of moon = 1.737 × 10⁶ m.

Substituting these into v, we have

v = √(2g'R)

v = √(2 × 1.63 m/s² × 1.737 × 10⁶ m)

v = √[5.663 × 10⁶ (m/s)²]

v = 2.38 × 10³ m/s

An object has mass 4 kg. What is its weight (in newton) on earth?​

Answers

i think it’s 78 carrots

Answer:

Should be -39.2 N

Explanation:

w=mg

w=4 x -9.8 m/s2

= -39.2 N

Do field forces exist in nature?

Answers

Answer:

Yes, field forces exist in nature.

Explanation:

A field force is a force experience due to an interaction with fields. In this case, a contact is not required before the force can be felt.

The three major field forces are: magnetic field, electric field and gravitational field. The magnetic fields are produced due to the interaction between the north and south poles of a magnet, the electric field is one from charges, while gravitational field is a force of attraction due to gravity on the earth. All these fields occur in nature, therefore making field forces to exist in nature.

Do you think antiseptic creams and lotions play an important part in our day
to day life? ​

Answers

Answer:

no, otherwise they just cause a side effect on our lives

yes, it keeps your skin from getting burnt when you play out side so much.

What is the relationship between the type of energy sublevels present and the
principal energy level?​

Answers

Each principal energy level has one sublevel containing one orbital, an s orbital, that can contain a maximum of two electrons. Electrons in this orbital are called s electrons and have the lowest energy of any electrons in that principal energy level.

One small speaker is placed 3m to the east of a second speaker, and a listener stands 4m directly south of one of the speakers. That listener finds that if they move in any direction, the sound gets louder. What is the longest possible wavelength of the sound from the speakers

Answers

Answer:

The value is [tex]\lambda = 2 \ m[/tex]

Explanation:

From the question we are told that

     The distance of the speaker  from the  second speaker  to the east is  [tex]d = 3 \ m[/tex]

      The distance of the speaker  from the listener  to the south is     [tex]a = 4 \ m[/tex]

Generally given that if the speaker move in any direction, their sound become  louder , it then mean that the position of the listener of minimum sound (i.e a position of minima ) ,

Generally the path difference of the sound produce by both speaker at a position of minima is mathematically represented as

              [tex]y = \frac{\lambda}{2}[/tex]

Generally considering the orientation  of the speakers and applying Pythagoras theorem we see that  distance from the second speaker to the listener  is mathematically represented as

             [tex]b = \sqrt{d^ 2 + a^2 }[/tex]

=>           [tex]b = \sqrt{3^ 2 + 4^2 }[/tex]

=>           [tex]b = 5[/tex]

Generally the path difference between the two speaker with respect to the  listener is  

              [tex]y = b - a[/tex]

=>           [tex]y = 5 - 4[/tex]

=>           [tex]y = 1[/tex]

So  

              [tex]1 = \frac{\lambda}{2}[/tex]

=>           [tex]\lambda = 2 \ m[/tex]

A powered winch is used to pull a sailboat to shore. The winch uses a 900 W motor.

If the motor is used for 30 s, how much work does it do? (Power: P = W/t)

0.03 J
30 J
960 J
27,000 J

Answers

Answer: 27,000 J :)

Explanation:

If the winch uses a 900 W motor and the motor is used for 30 s to pull a sailboat then work needed to do this work is 27000 J. hence option D is correct

What is Power?

Power is the rate of doing work. Power is also defined as work divided by time. i.e. Power = Work ÷ Time. Its SI unit is Watt denoted by letter W. Watt(W) means J/s or J.s-1. Something makes work in less time, it means it has more power. Work is Force times Displacement. Dimension of Power is [M¹ L² T⁻³]

Given,

Power P = 900W

Time = 30s

Work in joule =?

By using formula,

P = Work ÷ Time

Work = Power × time

W= 900 × 30

W= 27000 J

Hence Work of 27000J is needed to pull a sailboat to shore in 30s.

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A woman exerts a horizontal force of 113 N on a crate with a mass of 31.2 kg.

Required:
a. If the crate doesn't move, what's the magnitude of the static friction force (in N)?
b. What is the minimum possible value of the coefficient of static friction between the crate and the floor?

Answers

Answer:

a) 113N

b) 0.37

Explanation:

a) Using the Newton's second law:

\sum Fx =ma

Since the crate is not moving then its acceleration will be zero. The equation will become:

\sum Fx = 0

\sumFx = 0

Fm - Ff = 0.

Fm is the moving force

Ff is the frictional force

Fm = Ff

This means that the moving force is equal to the force of friction if the crate is static.

Since applied force is 113N, hence the magnitude of the static friction force will also be 113N

b) Using the formula

Ff = nR

n is the coefficient of friction

R is the reaction = mg

m is the mass of the crate = 31.2kg

g is the acceleration due to gravity = 9.8m/s²

R = 31.2 × 9.8

R = 305.76N

Recall that;

n = Ff/R

n = 113/305.76

n = 0.37

Hence the minimum possible value of the coefficient of static friction between the crate and the floor is 0.37

In the attachment there is a density column where there is colour

Question: tell me why is the red at the bottom of the density column if it is the least dense

Answers

because it’s on the bottom, and to get to that part of it, think of it like that scene from moana when they go down into the realm of monsters they go through the thick parts to get to the thin parts, soo the cube wouldn’t be nearly able to reach that level
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