A rectangular field is of length 42 cm and breadth 25 m. Find the area of the field in SI unit.​

Answers

Answer 1

Answer:

(42/100) (25)=area

10.5 m^2 = area


Related Questions

A mass is suspended on a spring. The spring is compressed so that the mass is located 5 cm above its rest position. The mass is released at time t= 0 and allowed to oscillate. It is observed that the mass reaches its lowest point 1/2s after it is released.

Required:
Find an equation that describes the motion of the mass.

Answers

Answer:

y = 5 cos 2πt

Explanation:

We will use the formula for simple harmonic motion curve where;

y = a cos ωt

Where;

a is amplitude

t is period

ω is angular frequency with the formula; ω = 2π/t

We are told that when the spring is compressed, the mass is located 5 cm above its rest position.

Thus;

a = 5 cm

it's highest point is 5 cm, but we are told that after 1/2 second of being released, it reaches its lowest point.

Since highest point is 5, then lowest point will be -5.

The difference in time between the highest and lowest point is ½ s. Which is half of the period.

Thus;

t/2 = ½

Thus, t = 1 s

Now, we know that;

t = 1/f = 2π/ω

Since t = 1, then 1 = 1/f

f = 1

Thus;

2π/ω = 1

Thus, ω = 2π

Thus, the equation is;

y = 5 cos 2πt

The equation that describes the motion of the mass is y = 5 cos 2πt.

The given parameters;

maximum displacement of the spring, A = 5 cmtime taken for the mass to reach the lowest point (half period), t = 0.5 s

The general equation of the wave is given as;

[tex]y = A\ cos\ \omega t[/tex]

where;

A is the amplitude of the vibrationω is the angular speed of mass

The angular speed of the mass is calculated as;

[tex]\omega = 2\pi f\\\\[/tex]

The period of the oscillation is calculated as;

[tex]T = 2t \\\\T = 2(0.5 s) = 1 \ s[/tex]

The frequency of the wave is calculated as;

[tex]f = \frac{1}{T} \\\\f = \frac{1}{1} \\\\f = 1\ Hz[/tex]

The equation that describes the motion of the mass is calculated as;

[tex]y = A\ cos \ \omega t\\\\y = A\ cos \ 2\pi ft\\\\y = 5\ cos \ 2\pi (1) t\\\\y = 5 \ cos \ 2\pi t[/tex]

Thus, the equation that describes the motion of the mass is y = 5 cos 2πt.

Learn more here:https://brainly.com/question/14017869

A camera lens with focal length f = 50 mm and maximum aperture f>2
forms an image of an object 9.0 m away. (a) If the resolution is limited
by diffraction, what is the minimum distance between two points on the
object that are barely resolved? What is the corresponding distance
between image points? (b) How does the situation change if the lens is
“stopped down” to f>16? Use λ= 500 nm in both cases

Answers

Answer:

The minimum distance between two points on the  object that are barely resolved is 0.26 mm

The corresponding distance between the  image points = 0.0015 m

Explanation:

Given  

focal length f = 50 mm and maximum aperture f>2

s =  9.0 m

aperture = 25 mm = 25 *10^-3 m

Sin a = 1.22 *wavelength /D  

Substituting the given values, we get –  

Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m

Sin a = 2.93 * 10 ^-5 rad

Now  

Y/9.0 m = 2.93 * 10 ^-5

Y = 2.64 *10^-4 m = 0.26 mm

Y’/50 *10^-3 = 2.93 * 10 ^-5  

Y’ = 0.0015 m

The graph below shows the distance traveled by the skateboarder on each of the different road conditions. Using the graph, determine which of the roads was dry, wet, or muddy. Explain your answer using complete sentences.

Answers

Answer:

Road A- dry

Road B- mud

Road C- wet

Explanation:

Surface conditions do affect the ease and speed with which a skateboarder can move, on a muddy surface, the tyres of the skate boards finds it difficult to establish adequate fictional force between the skates trees and the traveling surface. Hence, the muddy surface presents a very slippery travel ground for the skate, hence leading the to skateboarder needing to apply caution.

The speed on a wet surfave is height as the amount of firece that will be applied in other to accelerate is very small. The surface is wet and hence serves as a lubricant between the contact surface.

The dry road also has a high speed but lower than a wet surface, frictional force is high here and this tend to slow the skateboarder down except in sloppy terrains.

Select the correct answer.
What are the directions of an object's velocity and acceleration vectors when the object moves in a circular path with a constant speed?
OA. The question is meanimgless, since the acceleration is zero.
ов.
The vectors point in opposite directions.
Oc.
Both vectors point in the same direction.
OD
The vectors are perpendicular,

Answers

Answer:

A

Explanation:

If the object is moving at a constant speed, the object isn't accelerating as the velocity doesn't change.

Answer: C.

Explanation:  plato users

Help help help HELP AAAAA

Answers

The answer is B

As seen on the graph, the bus maintains a 9m/s speed for a majority of the trip to school.

What is the efficiency of a machine that uses 102 kJ of energy to do 98 kJ of work?

Answers

Mark Brainliest please

Answer : 96.08 % efficiency

A 97.0 kg ice hockey player hits a 0.150 kg puck, giving the puck a velocity of 48.0 m/s. If both are initially at rest and if the ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal 14.5 m away

Answers

Answer:

s₁ = 0.022 m

Explanation:

From the law of conservation of momentum:

[tex]m_1u_1 + m_2u_2 = m_1v_1+m_2v_2[/tex]

where,

m₁ = mass of hockey player = 97 kg

m₂ = mass of puck = 0.15 kg

u₁ = u₂ = initial velocities of puck and player = 0 m/s

v₁ = velocity of player after collision = ?

v₂ = velocity of puck after hitting = 48 m/s

Therefore,

[tex](97\ kg)(0\ m/s)+(0.15\ kg)(0\ m/s)=(97\ kg)(v_1)+(0.15\ kg)(48\ m/s)\\\\v_1 = -\frac{(0.15\ kg)(48\ m/s)}{97\ kg} \\v_1 = - 0.074 m/s[/tex]

negative sign here shows the opposite direction.

Now, we calculate the time taken by puck to move 14.5 m:

[tex]s_2 =v_2t\\\\t = \frac{s_2}{v_2} = \frac{14.5\ m}{48\ m/s} \\\\t = 0.3\ s[/tex]

Now, the distance covered by the player in this time will be:

[tex]s_1 = v_1t\\s_1 = (0.074\ m/s)(0.3\ s)[/tex]

s₁ = 0.022 m

A ball is thrown horizontally at a speed of 24 meters per second from the top of a cliff. If the ball hits the ground 6.0 seconds later, approximately how high is the cliff? ( EASY QUESTION.. PLZZ HELPPP MEEE I WILL MARK YOU THE BRAINLIEST PLZZ)​

Answers

Answer:

144 meters

Explanation:

the ball is thrown with a speed of 24 meters per second right so if the ball reaches the ground in 6 seconds. the hight of the cliff must be S=v.t

S (height cliff)=24m/s×6s=144

what is electricity ?​

Answers

Answer:

it is energy resulting in charged particles

Electricity a form of energy resulting from the existence of charged particles (such as electrons or protons), either statically as an accumulation of charge or dynamically as a current

g the total mechanical energy of the satellite-Earth system when the satellite is in its current orbit is E. In order for the satellite to orbit Earth in a new stable circular orbit at an altitude of 12RE, the energy of the satellite-Earth system must be

Answers

Answer:

The correct answer is "[tex]\frac{4E}{3}[/tex]".

Explanation:

According to the question,

Energy of satellite,

[tex]E_s=-\frac{GM_sM_E}{2r}[/tex]

For the very 1st case:

[tex]r = R_E+R_E[/tex]

  [tex]=2R_E[/tex]

or,

⇒ [tex]E=-\frac{GM_sM_E}{4R_E}[/tex]...(1)

For the new case:

[tex]r = R_E+\frac{R_E}{2}[/tex]

  [tex]=\frac{3R_E}{2}[/tex]

then,

⇒ [tex]E'=-\frac{GM_sM_E}{2 \frac{3R_E}{2} }[/tex]

        [tex]=-\frac{GM_sM_E}{3R_E}[/tex]...(2)

From equation (1) and (2), we get

⇒ [tex]E'=\frac{1}{3}(4E)[/tex]

        [tex]=\frac{4E}{3}[/tex]  

Kulsum’s TV uses 45 W. How much does it cost her to watch TV for one month (30 days). She watches TV for 4 hours/day during mid-peak time (10.4 cents/kWh).

Answers

Answer:

Total cost = 56.16 cents

Explanation:

Given the following data;

Power = 45 Watts

Time = 4 hours

Number of days = 30 days

Cost = 10.4 cents

To find how much does it cost her to watch TV for one month;

First of all, we would determine the energy consumption of the TV;

Energy = power * time

Energy = 45 * 4

Energy = 180 Watt-hour = 180/1000 = 0.18 Kwh (1 Kilowatts is equal to 1000 watts).

Energy consumption = 0.18 Kwh

Next, we find the total cost;

Total cost = energy * number of days * cost

Total cost = 0.18 * 30 * 10.4

Total cost = 56.16 cents

which two changes would make this reaction favor the formation of products

Answers

Answer:

C. ...

D. ...

Given.

Required.

Solution.

As the temperature decreases, the equilibrium will shift towards the exothermic reaction, so the reaction shifts to the right towards SO₃( products-favored)

And increasing the pressure, then the reaction shifts to the right SO₃( products-favored)⇒the number of coefficients is greater.

Answer:

decrease the temperature

and

raising the pressure

Explanation:

you are stowing items and come across an aerosol bottle of hairspray.what should you do?

Answers

Answer:

below

Explanation:

small plastic container, called the coolant reservoir, catches the radiator fluid that overflowswhen the automobile engine becomes hot. The radiator is made of copper, and the coolant has a coefficient of volume expansion of 410 x 10-6 [1/ oC]. If the radiator is filled to its 15 [quart] capacity when the engine is cold at 6.0 [oC], how much overflow from the radiator will spill into the reservoir when the coolant reaches its operating temperature of 92 [oC]

Answers

Answer:

0.53 quart

Explanation:

The volume expansion of the coolant is gotten from ΔV = VγΔθ where ΔV   = change in volume of the coolant, V = initial volume of coolant = 15 quart, γ = coefficient of volume expansion of coolant = 410 × 10⁻⁶ /°C and Δθ = temperature change = θ₂ - θ₁ where θ₁ = initial temperature of coolant = 6 °C and θ₂ = final temperature of coolant = 92 °C. So, Δθ = θ₂ - θ₁ = 92 °C - 6 °C = 86 °C

Since, ΔV = VγΔθ

substituting the values of the variables into the equation, we have

ΔV = VγΔθ

ΔV = 15 × 410 × 10⁻⁶ /°C × 86 °C

ΔV = 528900 × 10⁻⁶ quart

ΔV = 0.528900 quart

ΔV ≅ 0.53 quart

Since the change in volume of the coolant equals the spill over volume, thus the overflow from the radiator will spill into the reservoir when the coolant reaches its operating temperature of 92 °C is 0.53 quart.

what is measured by the ammeter

Answers

Answer:

amperes

Ammeter, instrument for measuring either direct or alternating electric current, in amperes. An ammeter can measure a wide range of current values because at high values only a small portion of the current is directed through the meter mechanism; a shunt in parallel with the meter carries the major portion.

Explanation:

hope it helps

What is the potential difference per unit length between two infinitely long concentric cylindrical shells with inner radius 1.5 cm and outer radius 5.6 cm, if the charge on the inner and outer cylinders are 7.0 nC and -7.0 nC, respectively

Answers

Answer:

165.8 V/m

Explanation:

The capacitance of a long concentric cylindrical shell of length, L and inner radius, a and outer radius, b is C = 2πε₀L/㏑(b/a)

Since the charge on the cylindrical shells, Q = CV where V = the potential difference across the capacitor(which is the potential difference between the concentric cylindrical shells)

V = Q/C

V = Q ÷ 2πε₀L/㏑(b/a)

V = Q㏑(b/a)/2πε₀L

So, the potential difference per unit length V' is

V' = V/L = Q㏑(b/a)/2πε₀

Given that a = inner radius = 1.5 cm, b = outer radius = 5.6 cm and Q = 7.0 nC = 7.0 × 10⁻⁹ C and ε₀ = 8.854 × 10⁻¹² F/m substituting the values of the variables into the equation, we have

V' = Q㏑(b/a)/2πε₀

V' = 7.0 × 10⁻⁹ C㏑(5.6 cm/1.5 cm)/(2π × 8.854 × 10⁻¹² F/m)

V' = 7.0 × 10⁻⁹ C㏑(3.733)/(55.631 × 10⁻¹² F/m)

V' = 7.0 × 10⁻⁹ C × 1.3173/(55.631 × 10⁻¹² F/m)

V' = 9.2211 × 10⁻⁹ C/(55.631 × 10⁻¹² F/m)

V' = 0.16575 × 10³ V/m

V' = 165.75 V/m

V' ≅ 165.8 V/m

A 55 kg pole vaulter falls from rest from a height of 5.4 m onto a foam rubber pad. The pole vaulter comes to rest 0.24 s after landing on the pad.

a. Calculate the athlete's velocity just before reaching the pad
b. Calculate the constant force exerted on the pole vaulter due to the collision.

Answers

Answer:

a) 10.3 m/s

b) 566 N

Explanation:

[tex]v {}^{2} = {u}^{2} + 2as \\ v {}^{2} = 0 {}^{2} + 2(9.81)(5.4) \\ v = 10.3 \: ms {}^{ - 1} [/tex]

[tex]force \: = \frac{d(mv)}{dt} \\ = 55(10.293) \\ = 566 \: newtons[/tex]

The athelete velocity will be 10.3 and constant force 566 N.

What is velocity?

The displacement that an object or particle experiences with respect to time is expressed vectorially as velocity. The meter per second (m/s) is the accepted unit of velocity magnitude (also known as speed).

Alternately, the magnitude of velocity can be expressed in centimeters per second (cm/s). Depending on how many dimensions are included, there are numerous ways to indicate the direction of a velocity vector.

The car's velocity in relation to your body is zero when you are driving. The speed of the car in relation to you if you were to stand by the side of the road is 20 m/s northward.

Therefore, The athelete velocity will be 10.3 and constant force 566 N.

To learn more about velocity, refer to the link:

https://brainly.com/question/18084516

#SPJ5

An airplane, starting from rest, moves down the runway at constant acceleration for 23 s and then takes off at a speed of 66 m/s. What is the average acceleration of the plane (in m/s2)?

Answers

Answer:

46

Explanation:

I really don’t know the answer for this

Answers

The correct answer is the reverse wave I took the test

plz answer the question

Answers

Answer:

Ray A - incident ray

Ray B - reflected ray

A particle of unit mass moves so that displacement after t seconds is given by x = 2 cos (t - 2). Find the acceleration and kinetic energy at the end of 3 seconds. (K.E = (1/2) m v²)

Answers

Answer:

  a₃ = -1.08 m/s²,    K = 1.42 J

Explanation:

The particle is in a periodic motion, so the general expression is

           x = A cos (wt + Ф)

let's compare the terms with the expression they give us

           x = 2 cos (t - 2)

the amplitude of motion is A = 2 m, the angular velocity w = 1 rad / s, and the phase is Ф = - 2.

to find the acceleration we use its definition

          v = dx / dt

          a = dv / dt

          a = [tex]\frac{ d^2x}{dt^2}[/tex]

let's perform the derivative

          v = - A w sin (wt + Ф)

          a = - A w² cos wt + Ф)

substituting the values

          a = - 2 1² cos (t-2)

           

for t = 3 s

          a₃ = 2 cos (3-2)

remember angles are in radians

          a₃ = -1.08 m/s²

To calculate kinetic energy, let's find the velocity for t = 3 s

          v = - 2 sin (t-2)

          v = -2 sin (3-2)

          v = - 1.683 m / s

body mass is m = 1 kg

we calculate

          K = ½ m v²

          K = ½ 1 (-1.683) ²

          K = 1.42 J


A bike and rider of total mass 75.0 kg moving at a velocity of 30 m/s to the right
collides with a skater of mass 60,0 kg moking at a velocity of 6.0 m/s to the left.
After they collide, the skater has a velocity of O m/s. What is the final velocity of the
bike and rider?
O A. O m/s
O B. 1.8 m/s to the left
O C. 3.0 m/s to left
O D. 1.8 m/s to the right

Answers

Answer:

[tex](75.0 \times 30) + (60.0 \times 6.0) = (75.0 \times V) + (60.0 \times 0) \\ 2250 + 360 = 75V \\ 75V = 2610 \\ V = 34.8 \: m {s}^{ - 1} [/tex]

On topographic maps, contour lines that are farther apart indicate what ?

Answers

Answer:

if I am correct, they indicate less steep terrain. think of it as the steeper the terrain the closer together the lines would be. hope that makes sense for you guys.

Answer:

gentle slopes

Explanation:

A 2890-lb car is traveling with a speed of 58 mi/hr as it approaches point A. Beginning at A, it decelerates uniformly to a speed of 18 mi/hr as it passes point C of the horizontal and unbanked ramp. Determine the total horizontal force F exerted by the road on the car just after it passes point B.

Answers

Answer:

4592.57 lb

Explanation:

The missing diagram for this question is attached in the image below.

Given that:

the weight of the car = 2890 lb

At point A, the speed of the car [tex](V_A)[/tex] = 58 mi/hr

At point C, the speed of the car [tex](V_C)[/tex] = 18 mi/hr

To ft/s:

[tex](V_A)[/tex]  = 58 mi/hr × 5280 ft/1 mi × 1 hr/3600 s

[tex](V_A)[/tex]  = 85.07 ft/s

[tex](V_C)[/tex] = 18 mi/hr × 5280 ft/1 mi × 1 hr/3600 s

[tex](V_C)[/tex] = 26.4 ft/s

Between A to C, the total distance is;

[tex]S_{AC} = S_{AB}} + S_{BC} \\ \\ S_{AC} = 331 + \dfrac{\pi r}{2} \\ \\ S_{AC}= 331 + \dfrac{\pi \times 207}{2} \\ \\ S_{AC} = 656.154 \ ft[/tex]

Now, we need to determine the deceleration of the car using the formula:

[tex]V_C^2 = V_A^2 + 2 aS_{AC}[/tex]

[tex]26.4^2 = 85.07^2 + 2 a (654.154)[/tex]

[tex]696.96 = 7236.9049+ 2 a (654.154)[/tex]

[tex]696.96-7236.9049 = 2 a (654.154)[/tex]

[tex]-6539.9449 = 2 a (654.154)[/tex]

[tex]a= \dfrac{-6539.9449} {2(654.154)}[/tex]

a = -4.99 ft/s²

The velocity of the car as it passes via B

[tex]v_B^2 = v_A^2 + 2aS_{AB}[/tex]

[tex]v_B^2 = 85.07^2 + 2(-4.99 \times 331)[/tex]

[tex]v_B =\sqrt{ 85.07^2 + 2(-4.99 \times 331)}[/tex]

[tex]v_B =\sqrt{ 85.07^2 +3303.38}[/tex]

[tex]v_B =\sqrt{ 10540.2849}[/tex]

[tex]v_B =102.67 \ ft/s[/tex]

Along B, the car's acceleration is:

[tex]a_B = \sqrt{a^2 + (\dfrac{v_B^2}{r})^2}[/tex]

[tex]a_B = \sqrt{(-4.99)^2 + \dfrac{102.67^2}{207}^2 }[/tex]

[tex]a_B = 51.17 \ ft/s^2[/tex]

Finally, the total horizontal force F exerted = m[tex]a_B[/tex]

[tex]= (\dfrac{2890}{32.2}) \times 51.17[/tex]

= 4592.57 lb

An arrow is shot from a height of 1.5 m toward a cliff of height H. It is shot with a velocity of 30 m/s at an angle of 60 degrees above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff

Answers

Answer:

The height of the cliff is 27.02 m

Explanation:

Given;

height above the ground from which the arrow is shot, h₁ = 1.5 m

initial velocity of the projectile, v = 30 m/s

angle of projection, θ = 60⁰

time taken to reach top of the cliff, t = 4.0 s

The vertical component of the velocity is calculated as;

[tex]v_y = v \times sin(\theta)\\\\v_y = 30 \times sin(60)\\\\v_y = 25.98 \ m/s[/tex]

The height attained by the projectile at the given time is calculated as;

[tex]h_2= v_y t - \frac{1}{2}gt^2\\\\h_2 = 25.98\times 4 \ - \ \frac{1}{2} \times 9.8\times 4^2\\\\h_2= 103.92 \ - \ 78.4 \\\\h_2 = 25.52 \ m[/tex]

The height of the cliff is calculated as;

H = h₁ + h₂

H = 1.5 m  +  25.52 m

H = 27.02 m

On Ramesh’s13th birthday, his father invited all his friends and their relatives. It was a big party with lots of food and DJs. Ramesh didn’t like the loud sound of DJs and asked his father to play it in a low volume so that their neighbours do not get much disturbed and people at the party can also enjoy the music. Ramesh’s father felt good for his wisdom and did as he said.
→Do you think when loud music is played at a party is acceptable to all the people living in, neighbourhood? Give a reason for your answer
→How can you control noise pollution at your end?

Answers

Answer:

The first fact to understand which carries the most weight is we share this small world. We all have such similar characteristics that if you listed them all in one column and then list traits that make each person an absolute individual, one of the lists would be liken to an encyclopedia where as the other more akin to a doodle on a napkin in comparison. Now to the question.

Explanation:

We all know inherently the rules, so to say. yes its acceptable to do as people do normally. to go outside those bounds on purpose to be a nuisance intentionally is not. like 3am loud pounding music or downright disrespect is not acceptable. because when its done that way, since we all know the unspoken rules, its is being done on purpose to annoy.

When it comes to the movement of air, friction
A. increases with altitude.
B. is greater near the ground surface.
C. diminishes turbulence.
D. is responsible for weaker winds aloft.

Answers

Answer: When it comes to the movement of air, friction is greater near the ground surface.

Explanation:

A resistance in motion observed by an object while on another object is called friction.

For example, a vehicle moving on road will have friction between its tires and the road.

Friction is more near the ground surface rather than away from the ground surface.

Thus, we can conclude that when it comes to the movement of air, friction is greater near the ground surface.

an object that has momentum must have

Answers

Answer:

Mechanical energy

Explanation:

If an object does not have any momentum, then it doesn't have mechicanl energy.

12. What type of circuit is the diagram below?
series circuit
parallel circuit

Answers

Answer:

parallel circuit

Explanation:

An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.

Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.

Basically, the components of an electric circuit can be connected or arranged in two forms and these includes;

I. Series circuit

II. Parallel circuit: it's an electrical circuit that has the same potential difference (voltage) across its terminals or ends. Thus, its components are connected within the same common points so that only a portion of current flows through each branch.

Hence, the type of circuit that the above diagram above represents is a parallel circuit.

Answer:

parallel circuit

Explanation:

I got it right on my exam

For each value of the principal quantum number n, what are the possible values of the electron spin quantum number m_s?

a. 0
b. -3/2
c. +3/2
d. +1/2
e. -1/2

Answers

Answer:

d) and e)

Explanation:

For any value of the principal quantum n, the only possible values of the electron  spin quantum number m_s are + 1/2 and -1/2.Since this number is related with the angular momentum of the electron, it has a magnitude (1/2) and an orientation given by the sign (+ or -)So, the only right answers are d) and e).
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