What is the acceleration of the object?
m/s²

What Is The Acceleration Of The Object?m/s

Answers

Answer 1

Answer:

-2.5m/s^2

Explanation:

10-40/12-0=-2.5


Related Questions

11. Trait theory claims that
O A. people from the same locations
share the same personality type.
B. you always behave the way your
personality type says you will.
O C. your personality is made up of a
number of traits.
O D. you have one characteristic that
defines your entire personality.

Answers

A. people from the same location share the same personality type.

The position of given object is s(t)= t^5 + cos 4t
Find the velocity and acceleration functions

Answers

Answer:

v(t)=s'(t)=5t⁴-sin(4).4, for velocity function at time t

a(t)=v'(t)=20t³, for acceleration function at time t

If the Moon had twice as much mass and still orbits Earth at the same distance, ocean bulges on Earth would be
smaller.

larger.

unequal in size.

not significantly different.

none of the available options

Answers

Ocean bulges on Earth would be bigger if the Moon had twice as much mass and yet orbited the planet at the same distance. Option B is correct.

What is ocean bludge?

The fluid and moveable ocean water are drawn towards the moon by the gravitational attraction between the moon and the Earth.

The ocean nearest to the moon experiences a bulge as a result, and as the Earth rotates, the affected seas' locations shift.

The Moon's bulges in the oceans would be larger if it had twice the mass and orbited Earth at the same distance.

Hence option B is corect.

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A box with a mass of 40 kg sits at rest on a frictionless tile floor. with your foot, you apply a 20 N force in a horizontal direction. what is the acceleration of the box?​

Answers

Answer:

0.5 m/s²

Explanation:

according to Newton's second law, we are goven a relationship between force, mass and acceleration, with the formula:

F = m×a

F for force

m for mass

a for acceleration

we use the given data and get:

20 = 40×a

we find a=20/40=0.5m/s²

acceleration of the box is 20m/s^2

given

mass (m) = 40 kg

force (f) = 20n

acceleration (a) =?

we know,

f = m × a

20 = 40 × a

40 - 20 = a

20m/s^2 = a

A car and a truck, starting from rest, have the same acceleration, but the truck moves twice the length of time. Compared with the car, the truck will travel

Answers

Compared with the car, the truck will travel four times as far.

The initial velocity of car and truck, u = 0

The acceleration of both the truck and car = a

The length of time for the acceleration = t

Let the time the truck accelerated be 2t

How to calculate the distance traveled by car and truck?

The distance traveled by car is calculated as;

s = ut + ¹/₂at²

s₁ = 0(t) + ¹/₂at²

s₁ = ¹/₂at²

The distance traveled by truck

s = ut + ¹/₂at²

s₂ = 0(2t) + ¹/₂a (2t)²

s₂ =  ¹/₂a x 4t²

s₂ = 4 (¹/₂at²)

s₂ = 4(s₁)

We can conclude that the Truck distance is four times the car distance.

Therefore, Compared with the car, the truck will travel four times as far.

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A skier is standing on a horizontal patch of snow. She is holding onto a horizontal tow rope, which is about to pull her forward. The skier’s mass is 59 kg, and the coefficient of static friction between the skies and the snow is 0.14. What is the magnitude of the maximum force that the tow rope can apply to the skier without causing her to move?

Answers

The magnitude of the maximum force that the tow rope can apply to the skier without causing her to move will be 81.03 N,

What is the friction force?

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

It is defined as the product of the coefficient of friction and normal reaction.

On resolving the given force and acceleration. Mathematically aion in the different components and balancing the equation gets. Components in the x-direction.

The normal force is balanced by weight;

N = mg

The magnitude of the maximum force that the tow rope can apply to the skier without causing her to move is;

F = μN

F= μmg

F=0.14 ×  59 kg × 9.81 m/s²

F = 81.03 N

Hence, the magnitude of the maximum force will be 81.03 N,

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A hydrogen bond forms by the electrostatic interaction of opposite charges in two molecules. If
the bond length is 2x10 -10 m and the magnitude of the charges involved is
approximately 1.60x10 -20C, what is the force between the molecules involved in the bond?

Answers

The force between the molecules involved in the bond is 6. 426 *10^-11 Newton

How to determine the force

Using the formula:

F = K[q1 x q2]/D^2

where K is coulombs constant =9 *10 ^9 Nm^2/C^2.

q1  and q2 = charges  =  1.60x10 -20C

d = distance between the charges = 2x10 -10 m

Substitute the values into the formula

F =  [tex]9 * 10^9\frac{ 1.60*10^ -20 * 1.60 *10^ -20}{2x10^ -10^{2} }[/tex]

F = [tex]9 *10^9\frac{2. 856* 10^-40}{4* 10^-20}[/tex]

F = [tex]9* 10^9 * 7. 14* 10^-21[/tex]

F = [tex]6. 426 * 10^-11[/tex] Newton

Thus, the force between the molecules involved in the bond is 6. 426 *10^-11 Newton

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the following equation shows the position of a particle in time t, x=at2i + btj where t is in second and x is in meter. A=2m/s2, b=1m/s. Find
A, the average velocity of the particle in the time interval t1=2sec and t2=3sec
B, the velocity and acceleration at any time t.
C, the average acceleration in the time interval given in part (a)

Answers

(a) The average velocity of the particle in the time interval t₁=2sec and t₂=3sec is 10 m/s.

(b) The velocity and acceleration at any time t is v =  (2ati + bj) m/s and a = 2ai m/s²

(c)  The average acceleration in the time interval given in part (a)​ is 4 m/s².

Position of the particle

x = at²i + btj

x = 2t²i + tj

Average velocity, at t₁=2sec and t₂=3sec

Δv = Δx/Δt

x(2) = 2(2)²i + 2j

x(2) = 8i + 2j

|x(2)| = √(8² + 2²) = 8.246

x(3) =  2(3)²i + 3j

x(3) = 18i + 3j

|x(3)| = √(18² + 3²) = 18.248

Δv = (18.248 - 8.246)/(3 - 2)

Δv =  10 m/s

Velocity and acceleration at any time, t

x = at²i + btj

v = dx/dt

v =  (2ati + bj) m/s

a = dv/dt

a = 2ai m/s²

Average acceleration

a = 2ai m/s²

a = 2(2)(1)

a = 4 m/s²

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Need help with this question

Answers

Answer:

1. 92, 138, 92

2. 7, 7, 7

3. 17, 18, 17

4. 29, 34, 29

Explanation:

the number on the bottom is always going to be the atomic number, or the number of protons.

the number on the top is going to be the atomic mass number, which is the sum of the number of protons and neutrons.

there is always going to be the same number of electrons as protons, because that was the overall charge of each atom is 0.

for example, for the first one:

there are 92 protons

there are 230 - 92 = 138 neutrons

there are 92 electrons


If the distance between two objects is cut in half, what happens to the
gravitational force between them?
A. It decreases to 1/2 its original magnitude.
B. It decreases to 1/4 its original magnitude.
O
C. It increases to 4 times its original magnitude.
D. It increases to 2 times its original magnitude.

Answers

C. Increase to 4 times its original magnitude

An upward force is applied to a 6.0–kilogram box. This force displaces the box upward by 10.00 meters. What is the work done by the force on the box?

Answers

Answer:

588J

Explanation:

Work done = Fd

=wh=mgh

=6×9.8×10

=558J

Three small but dense objects are located in the x-y plane as shown in the figure. The objects have the following masses: mA = 3.19 kg, mB = 2.55 kg and mC = 1.41 kg.

Determine the x and the y coordinates of the center of the mass of this system. The objects are small in size, they can be treated as point masses.


x coordinate: ?

y coordinate: ?

Answers

The x and the y coordinates of the center of the mass of this system will be 43.1 m and 3.12 m respectively.

What is the center of mass?

A location is established in relation to an object or set of objects in the center of mass. It is the system's average position across all of its components.

Given data;

There are three little objects that are densely spaced out in the x-y plane.;

The value of the masses

m₁=1.41 kg

m₂=2.55 kg

m₃=3.19 kg

[tex]\rm X_{cm} =\frac{ (W_1x_1 + W_2x_2 + W_3x_3)}{(W_1 + W_2 + W_3)} \\\\\ X_{cm} ==\frac{1.41 \times 2 + 2.55 \times 6 +3.19 \times 4}{1.41+2.55+3.19} \\\\ X_{cm} =4.31 \ m[/tex]

[tex]\rm Y_{cm} =\frac{ (W_1y_1 + W_2y_2 + W_3y_3)}{(W_1 + W_2 + W_3)} \\\\\ X_{cm} ==\frac{1.41 \times 2 + 2.55 \times 4 +3.19 \times 7}{1.41+2.55+3.19} \\\\ X_{cm} =3.12 \ m[/tex]

Hence, the x and the y coordinates of the center of the mass of this system will be 4.31 m and 3.12 m respectively.

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What is the monthly (30 days) energy cost (in
$) of leaving a 40-W bulb on continuously if
electricity costs 21¢ per kWh? Round final
answer to the nearest cent.

Answers

Explanation:

Power= 40 W

cost=21¢/kWh

t=1 month=30 days = 720h

Energy= power*time

E= 40W* 720h = 28800kWh

Cost= Energy*rate

cost= 28800kWh*21¢/kWh

cost=604800¢

Suppose astronomers find an earth-like planet that is twice the size of Earth. What must be the mass of this planet such that the gravitational force at the surface would be identical to Earth’s?

Answers

This planet's mass must be four times that of Earth in order for the gravitational pull at its surface to be comparable to Earth's.

What is Newton's law of gravitation?

Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.

The gravitational force is proportional to the product of the masses of the two bodies and inversely proportional to the square of their distance.

The planet's mass is directly related to its mass at the surface and inversely proportional to the square of its radius;

When mass increases and distance reduces, gravity rises. Gravity also lowers when the distance between two points grows and the mass decreases.

The astronomers discovered a planet twice the size of Earth that resembles Earth.

This planet's mass must be four times that of Earth in order for the gravitational pull at its surface to be comparable to Earth's.

Hence the masses of this planet will be 4 times that of Earth.

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How much work does a football player do in the weight room when he squats 150 kg up a distance of 1 meter?
How much work does he do if he does 3 sets of 10 squats in a row?

Answers

Answer:

See below

Explanation:

The work done equals the increase in Potential Energy

PE = mgh

    = 150 * 9.81 * 1 = 1471.5 j

3 sets of 10 is 30 times this = 44145 j

The amount of work done by a football player in the weight room when he squats 150 kg will be equal to 1471.5 J.

What is potential energy?

Potential energy is a form of stored energy that is dependent on the relationship among different components. When a spring is compressed or stretched, its potential energy increases. If a steel ball is raised above the floor as opposed to falling to the ground, it has more potential energy. It is capable of carrying out additional work when raised.

Potential energy is a feature of systems rather than of particular bodies or particles; for instance, the system created up of Earth and the elevated ball has more energy stored as they become further apart.

Potential energy develops in systems components whose configurations, or spatial arrangement, determine the amount of the forces they apply to one another.

As per the instructions given in the question,

Work performed is equivalent to Potential Energy Increase,

PE = mgh

= 150 × 9.81 × 1

PE = 1471.5 J

Also,

3 sets of 10 are 30 times this = 44145 J

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5. A uniform solid ring of mass M = 10 kg and radius R = 2 m is initially rotating in mid-air with angular speed 20 rad/s [the Before state in figure (I) below]. It is then placed in contact with a surface (assume that the surface immediately feels the full weight of the ring). (1) M Before Wi (II) wf V Ө ə Ө After Ө wf H ? (a) What kind of frictional force acts on the ring upon contact with the surface? (b) If the ring begins to roll smoothly to the right with speed v = 10 m/s after 5 seconds of contact (the After state), find the coefficient of friction corresponding to the frictional force you mentioned in (a). (c) What is the increase in the thermal energy of the ring/surface system from right before the ring contacts the surface, to when the ring starts to roll smoothly? The ring then rolls smoothly up a ramp of 0 = π/6 rad and H = 5 m [see figure (II)] (d) What is the horizontal distance, from the end of the ramp, at which the ring lands?

Answers

(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.

(b) If the ring begins to roll smoothly to the right with speed v = 10 m/s after 5 seconds of contact (the After state), the coefficient of friction corresponding to the frictional force is 0.612

(c) The increase in the thermal energy of the ring/surface system from right before the ring contacts the surface, to when the ring starts to roll smoothly is 7500 J.

(d) The horizontal distance, from the end of the ramp, at which the ring lands is 7.78 m

What is frictional force?

When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.

Given is a uniform solid ring of mass M = 10 kg and radius R = 2 m is initially rotating in mid-air with angular speed 20 rad/s [the Before state in figure (I) below]. It is then placed in contact with a surface (assume that the surface immediately feels the full weight of the ring).

(a) The kind of frictional force acts on the ring upon contact with the surface is kinetic friction before rolling and afterwards it is static friction.

(b) angular frequency = velocity / radius

ωf = 10/2 = 5 rad/s and ωi = 20 rad/s

Angular acceleration, α = ωf - ωi /t

Put the values, we get

α = -15/5 = -3 rad/s²

Coefficient of friction, μ = a/g = rα/g

Plug the values, we get

μ = 0.612

Thus, the coefficient of friction corresponding to the frictional force is 0.612.

(c) The energy lost = heat generated

 energy lost = 1/2 Iω² + 1/2 Mv²

 energy lost = 1/2 MR²ω² + 1/2 Mv²

Plug the values, we get

 energy lost = 7500 J

Thus, the  increase in the thermal energy is 7500 J.

(d) The horizontal distance, from the end of the ramp, at which the ring lands

s= (v- 2gH) sin2θ /g

s = (10 - 2x (-9.81)x5 ) sin (2x π/6) / 9.81

s = 7.78m

Thus, the horizontal distance is  7.78 m

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A small sphere of mass 10 kg
is released from rest at a height of
15.0 m above the ground level.
The sphere experiences a constant
resistive force (due to air
resistance) of magnitude R = 10.0
N.
a) Calculate the speed of the
sphere after it has fallen
through a distance of 5.00 m

bCalculate the speed of the ball just before a it hits the gound.

Answers

Answer:

Approximately [tex]9.39 \; {\rm m\cdot s^{-1}}[/tex] after the sphere has travelled a distance of [tex]5\; {\rm m}[/tex].

Approximately [tex]16.3\; {\rm m\cdot s^{-1}}[/tex] right before touching the ground (a distance of [tex]15\; {\rm m}[/tex].)

Assumption: [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].

Explanation:

Weight of the sphere: [tex]m\, g = 9.81\; {\rm N \cdot kg^{-1}} \times 10\; {\rm kg} = 98.1\; {\rm N}[/tex], downwards.

Drag on the sphere: [tex]10.0\; {\rm N}[/tex] upwards.

Net force on the sphere: [tex]98.1\; {\rm N} - 10\; {\rm N} = 88.1\; {\rm N}[/tex] downwards.

Acceleration of the sphere: [tex]a = F_\text{net} / m = 88.1\; {\rm N} / (10\; {\rm kg}) = 8.81\; {\rm m\cdot s^{-2}}[/tex].

Apply the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex], where [tex]v[/tex] is the final velocity, [tex]u[/tex] is the initial velocity ([tex]0[/tex] in this case, as the sphere was released from rest,) and [tex]x[/tex] is the distance (displacement) that the sphere has travelled so far.

Rearrange this equation to obtain an expression for [tex]v[/tex]:

[tex]\displaystyle v = \sqrt{2\, a\, x + u^{2}}[/tex].

For example, after the ball travelled a distance of [tex]5.00\; {\rm m}[/tex], [tex]x = 5.00 \; {\rm m}[/tex]:

[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 5.0\; {\rm m} + 0} \\ &\approx 9.39\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Similarly, [tex]x = 15.0\; {\rm m}[/tex] right before landing, such that:

[tex]\begin{aligned} v &= \sqrt{2\, a\, x + u^{2}} \\ &= \sqrt{2 \times 8.81\; {\rm m\cdot s^{-2}} \times 15.0\; {\rm m} + 0} \\ &\approx 16.3\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Suppose the resistance of a copper wire is 1.05 N at 30°C.
Find the temperature at which the resistance becomes 1.50
Ω.

Answers

Answer:

42.9°C

Explanation:

Resistance of a conducting material is directly proportion to Temperature, that is, when being heated, their is more of collisions between the molecules of the material which as a result offers more opposition to the flow of current through it.

therefore R=kT

where R=resistance of the wire

T=Temperature

k= proportionality constant

when R=1.05 T=30°C

so k=1.05/30

k=0.035

when R=1.5 k=0.035

so T=1.5/0.035

T=42.9°C. or 43°C approximated

What’s the allowable ampacity of a No. 12 copper wire with type THHN insulation installed in a raceway with 35 wires? Question 3 options: A) 10 A B) 15 A C) 8 A D) 12 A

Answers

The allowable ampacity of a No. 12 copper wire with type THHN insulation installed in a raceway with 35 wires is approximately 12 amperes. This is best represented by option D.

According to the National Electrical Code Table (NEC), a derating factor must be applied for any raceway with more than three current-carrying conductors (in our case, copper wires). In a raceway with 31-40 current-carrying conductors, the derating factor is 50%. Because the ampacity of a No. 12 copper wire usually is 25 amperes (A), we can multiply this by our derating factor to find that:

25 • 0.50 = 12.5

The exact allowable ampacity of this copper wire is 12.5 amperes; however, we can round down to find that the closest answer is 12 A.

Hence, option D represents the best answer.

Answer:

  D)  12 A

Explanation:

You want to know the allowable ampacity of AWG 12 type THHN copper wire in a raceway with 35 wires.

Nominal ampacity

Table 310.15(B)(16) of the 2017 National Electrical Code gives the nominal rating of AWG 12 type THHN copper wire as 30 A when installed in an environment with a temperature of 30 °C or less.

Derating

Table 310.15(B)(3)(a) gives the adjustment factor when there are more than three conductors in a raceway or cable. For 31–40 conductors, the ampacity must be derated to 40% of its nominal value.

In a raceway with 35 wires, the allowable ampacity of THHN AWG 12 wire is ...

  (30 A)(0.40) = 12 A

__

Additional comment

For environmental temperatures above 30 °C, Table 310.15(B)(2)(a) gives additional derating factors. The adjustment factor for temperatures up to 50 °C (122 °F) is 0.82, which would bring the allowable ampacity down to about 9.8 A.

3. Sarah is a world-class diver in the women's 3-m springboard competition. Her height
h (in metres), above the water t seconds after she leaves the board is given by
h = -4.9t2 + 8.8t + 3. How long is Sarah in the air before she reaches the water?
Make sure to round your answer to the nearest tenth.
Answer:
2011, 2013, 2016 Alberta Education

Answers

Answer:

1.163 seconds

Explanation:

see the attached. solve the quadratic equation.

How many seconds are in 28 hours?

Answers

Answer:

1680 seconds

Explanation:

[tex]28 hrs * \frac{60 s}{1 hr} =1680s[/tex]

A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 n. What must be done to find the acceleration of the sled? check all that apply.

Answers

The acceleration of this sled can be calculated by doing: option A, B, E and F.

How to calculate the acceleration?

The acceleration of this sled can be calculated by using Newton's Second law of motion. Mathematically, the acceleration of an object is given by this formula:

Net force = Mass × acceleration

Deductively, the acceleration of this sled can be calculated by doing the following:

The force of gravity must be broken into its parallel and perpendicular components.Acceleration can be found by dividing the net force by mass.Trigonometry can be used to solve for the magnitude of the force components.Net force must be found before acceleration can be found.

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Complete Question:

A sled is moving down a steep hill. The mass of the sled is 50 kg and the net force acting on it is 20 n. What must be done to find the acceleration of the sled? check all that apply.

A. The force of gravity must be broken into its parallel and perpendicular components.

B. Acceleration can be found by dividing the net force by mass.

C. Acceleration can be found by multiplying the net force by mass.

D. The magnitude of the force components must be multiplied by gravity.

E. Trigonometry can be used to solve for the magnitude of the force components.

F. Net force must be found before acceleration can be found.

Answer:

To find the acceleration, you do 20N/50kg = 0.4 m/s^2

Explanation:

example: a = F/m = 10/2 = 5 m/s2

the 3kg object in figure is released from rest at height of on a curved frictionless ramp. at the foot of the ramp is a spring of force constant 400N/m. the object slides down the ramp and into the spring,compressing it a distance x before coming momentarily to rest.
a) find x
b) describe the motion of the object (if any) after the block momentarily comes to rest?

Answers

(a) The compression of the spring is 7.4 cm.

(b) When the object comes to rest its potential energy from the given height will be converted into elastic potential energy of the spring.

Compression of the spring [ x ]

The compression of the spring is calculated by applying Hooke's law as follows;

F = kx

x = F/k

x = (mg)/k

x = (3 x 9.8)/400

x = 0.074 m

x = 7.4 cm

When the object comes to rest its potential energy from the given height will be converted into elastic potential energy of the spring.

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the question in photo

Answers

Answer:

a

Explanation:

A cup of water is warmed from 21 °C to 85 °C. What is the difference between these two temperatures, in kelvins?

Answers

Answer:

337k

Explanation:

First, let us find the difference between the given two temperatures.

Difference = 85°C - 21°C

                  = 64°C

And now we have to write the temperature in kelvins.

To convert Celcius to Kelvins you can add 273 to the temperature in Celcius.

Let us find it now.

64°C + 273 = 337k

Therefore,

64°C ⇒ 337k

A house has a roof (colored gray) with the dimensions shown in the drawing. Determine the magnitude of the net force that the
atmosphere applies to the roof when the outside pressure drops suddenly by 13.2 mm of mercury, before the pressure in the attic can
adjust.

Answers

The magnitude of the net force that the atmosphere applies to the roof when the outside pressure drops suddenly is  1.9 x 10⁵ N.

What is pressure?

The pressure is the amount of force applied per unit area.

Pressure p = Force/Area

Given is a house has a roof (colored gray) with the dimensions shown in the drawing. The outside pressure drops suddenly by 13.2 mm of mercury, before the pressure in the attic can adjust.

The pressure difference ΔP = 13.2 mm of Hg

The length of the roof l = 14.5m

the breadth of the roof h = 4.21m

The force exerted by pressure is

Force, F = P x A

         = (13.2 mm of Hg) [(133 N/m²) /1 mm of Hg ](14.5 x 4.21)

         = 107,170.6 N

Then the net vertical force

    Fnet = 2F cos30

    Fnet = 2 ( 107,170.6) cos30

    Fnet = 185625 N

    Fnet =1.9 x 10⁵ N

The direction of the force is downwards, since the horizontal components of the forces cancel each other.

Hence,  magnitude of the net force is 1.9 x 10⁵ N

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A boat moves with a velocity of 15m/s, N in a river which flows with a velocity of 8.0m/s, Calculate the boat’s resultant velocity with respect to North

Answers

The boat’s resultant velocity with respect to North is determined as 17 m/s.

Resultant velocity of the boat

The boat’s resultant velocity with respect to North is calculated as follows;

R² = Va² + Vb²

where;

R is resultant velocityVa is velocity of boatVb is velocity of river

R = √(Va² + Vb²)

R = √(15² + 8²)

R = 17 m/s

Thus, the boat’s resultant velocity with respect to North is determined as 17 m/s.

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During World War II, it was found that r, the radius of the shockwave produced during an atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density, .

Answers

The atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.

What is an atomic bomb?

An atomic bomb is type of nuclear weapon that releases a vast amount of energy upon explosion in the form of fission reaction.

During an explosion of atomic bomb, a cloud of mushroom fire is formed which vaporises anything found within it. The extent of destruction depends on:

energy released, W,

the elapsed time, t, and

the air density.

Therefore, the atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.

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A room has a 60 Watt, a 100 Watt and a 150 Watt light bulb. How much does it
cost to use all the lamps for 2.5 hours at $0.08/ kWh?

Answers

The cost to use all lamps is $62.

room has 60 watts, 100 watts, and 150 watt light bulbs.Duration of using bulbs=2.5 hours power =$0.08kWh

           power=energy/time

          (60+ 100+150)=energy/2.5

          energy=775 joule

cost for 1 hour=$0.08

                      = 775 * 0.08

                     = $62

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Answer:

6.2 cents

Explanation:

(60+100+150/1000)*2.5*8=6.2 cents

A model rocket with a mass of 5 kg has a downward thrust of 100 newtons. Assuming there are no other forces acting on the rocket, what is the rocket’s acceleration upward?
a. 20 m/s2
b. 100 m/s2
c. 5 m/s2
d. 105 m/s2

Answers

Answer:

A

Explanation:

F = m * a

100 = 5 * a

a = 100/5 = 20 m/s^2

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