An elevator is pulled up by a cable with a force of 65,000 N. The upward acceleration of the elevator is 1.8 m/s/s. What is the mass of the elevator?
a. 1.8 kg.
b. 1700 kg.
c. 54880 kg.
d. 5600 kg.
e. 36111 kg.

Answer:

The earliest known life-forms are putative fossilized microorganisms, found in hydrothermal vent precipitates, that may have lived as early as 4.28 Gya (billion years ago), relatively soon after the oceans formed 4.41 Gya, and not long after the formation of the Earth 4.54 Gya.

Explanation:

Is this what your looking for?

Answer:

Fall

Explanation:

When you turn off your bedroom lights, the kitchen lights can stay on. This is because your home is wired by using the series circuit. Hence, option B is correct.

The amount of current in the flow circuits of each element in a series circuit is the same. This is due to the recent flow having a single path. The flow rate (marble speed) at any point in a circuit (tube) at any given moment in time has to be equal because electric current moves through a conductor like marble in a duct.

The requirement that all values (voltage, current, resistance, and power) relate to one another in terms of the identical two places in a circuit is a significant exception to Ohm's law.

Therefore, the wiring of the home is in series combination. So, option B is correct.

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Answer:

a. Object A

Explanation:

The mass of an object implies the quantity of matter in it, while the weight is the amount of gravitational force applied on an object.

The object A has a mass of 25 lbs, but object B on the earth has a weight, W, of 25 N.

So that,

For object A on the moon, mass = 25 lbs

For object B on the earth, W = 25 N,

W = m x g

25 = m x 10                (g = 10 m/[tex]s^{2}[/tex])

m = [tex]\frac{25}{10}[/tex]

   = 2.5 lbs

Mass of object B is 2.5 lbs.

Therefore, the mass of the object A is more than that of B.

Answer:

15 m/s

Explanation:

The Speed Of The Car Does Not Add To The Speed Of The Bullet

Answer:

The mass of the cart is 150 kg.

Explanation:

Given that,

Mass of a boy, m₁ = 50 kg

Initial speed of boy, u₁ = 10 m/s

Initial speed of car, u₂ = 0 (at rest)

The speed of the cart with the boy on it is 2.50 m/s, V = 2.5 m/s

Let m₂ is the mass of the cart. Using the conservation of momentum as follows :

[tex]m_1u_1+m_2u_2=(m_1+m_2)V\\\\50(10)+m_2(0)=(50+m_2)(2.5)\\\\500=125+2.5m_2\\\\375=2.5m_2\\\\m_2=150\ kg[/tex]

So, the mass of the cart is 150 kg.

Answer:

Yes

Explanation:

I think this is because when you go out of the room and going to a hotter room you then get the heat from that room. It then becomes warmer in the room you are coming from because your body got the heat from the outside the room. I think it is because of body temperature.

HOPE THIS HELPED

We know, for single slit :

[tex]y =\dfrac{ n\lambda L}{a}\\\\\lambda = \dfrac{ya}{nL}[/tex]       ...1)

[tex]y = 1.4\ mm = 1.4 \times 10^{-3}\ m[/tex]

n = 2

L = 89 cm = 0.89 m

[tex]a=7.1\times 10^{-4}\ m[/tex]

Putting all these in equation 1), we get :

[tex]\lambda = \dfrac{ya}{nL}\\\\\lambda = \dfrac{1.4\times 10^{-3}\times 7.1\times 10^{-4}}{2\times 0.89 }\\\\\lambda = 5.584 \times 10^{-7}\ m[/tex]

Therefore, wavelength of the incident light is [tex]5.584 \times 10^{-7}\ m[/tex] or 558.4 nm.

Hence, this is the required solution.

Answer:

The total thermal energy required  is 8.93 Kcal

Explanation:

Given;

mass of the ice cube, m = 43 g

specific heat capacity of water, Cp = 4.18 J/gc

specific latent heat of fusion of ice, Cf = 334 J/g

First step, determine the heat needed to raise the temperature of the ice from -16.5 °C to 0° C

Q₁ = mCp[0 - (-16.5)]

Q₁ = 43 x 4.18(16.5)

Q₁ = 2965.71 J

Second step, determine the latent heat of fusion of ice at 0°C

Q₂ = mCf

Q₂ = 43 x 334

Q₂ = 14362 J

Third step, determine the quantity of heat required to raise the temperature of the water initially at 0°C to above 11.5 °C of boiling point of water.

The final temperature of the water = 11.5 °C + 100° C = 111.5 °C

Q₃ = mCp Δθ

Q₃ = 43 x 4.18 (111.5 - 0)

Q₃ = 20041 J

Total thermal energy required = Q₁ + Q₂ + Q₃

Total thermal energy required =  2965.71 J + 14362 J +  20041 J

Total thermal energy required = 37,368.71 J

Total thermal energy required = 8.93 Kcal

We are given:

Homer the Astronaut:

Mass of Homer the astronaut(m1) = 200 kg

initial velocity of Homer the astronaut(u1) = 5 m/s

Larry the Pal:

Mass of Larry the Pal (m2) = 150 kg

initial velocity of Larry the Pal (u2) = 0 m/s

Since they will move together after the collision, they will have the same velocity:

v1 = v2 = V

Solving for the Final velocity:

from the law of conservation of momentum:

m1u1 + m2u2 = m1v1 + m2v2

since v1 = v2 = V:

m1u1 + m2u2 = V(m1 + m2)

replacing the variables with the given values

200 * 5 + 150 * 0 = V(200 + 150)

1000 = 350V

V  = 1000 / 350

V = 2.86 m/s

Answer:

How long do you jump (sec) ?

Explanation:

Answer:

Explanation:

a) centripetal acceleration is the acceleration of a body in a circular path. It is expressed as;

a = mv²/r

m is the mass of proton = 1.67×10^−27kg

v is the velocity = 3.00×10^8m/s

r is the radius

Since C = 2πr

7000m = 2πr

r = 7000/2π

r = 1114.08m

Substitute

a = 1.67×10^−27 (3.00×10^8)²/1114.08

a =  1.67×10^−27 * 9×10^16/1114.08

a = 15.03*10^-11/1114.08

a = 0.001346*10^-11

a = 1.346*10^-14m/s²

b) Force on the proton = mass * acceleration

Force = 1.67×10^−27kg * 1.346*10^-14

Force = 2.246*10^-41N

hence the force on the proton is 2.246*10^-41N

I think is True: Because...

when you throw a disc you can throw it in any direction, many people call it "Flying Saucer" I'll give you an example ... When you throw something, for example a paper, you want to throw it at your classmate. You already know what address you want to send it to, then I say it is: True ...

Sorry if it's wrong :(

Answer:

Th average force impact is [tex]F = 168.298 \ N[/tex]

Explanation:

From the question we are told that  

   The mass of the golf ball is  [tex]m_g = 26.7 \ g = 0.0267 \ kg[/tex]

    The angle made is [tex]\theta = 33.6 ^o[/tex]

    The range of the golf ball  is [tex]R = 190 \ m[/tex]

     The duration of contact is [tex]\Delta t = 7.13 \ ms = 7.13 *10^{-3} \ s[/tex]

Generally the range of the golf ball is mathematically represented as

       [tex]R = \frac{v^2 sin2(\theta)}{g}[/tex]

Here v  is the velocity with which the golf club propelled it with, making  v the subject

       [tex]v = \sqrt{\frac{R * g}{sin 2 (\theta)} }[/tex]

=>     [tex]v = \sqrt{\frac{190 * 9.8}{sin 2 (33.6)} }[/tex]

=>     [tex]v = 44.94 \ m/s[/tex]

Generally the change in momentum of the golf ball is mathematically represented as

      [tex]\Delta p = m * (v - u )[/tex]

here u  is the initial  velocity of the ball before being stroked and the value is 0 m/s

       [tex]\Delta p = 0.0267 * ( 44.94 - 0 )[/tex]

=>    [tex]\Delta p = 1.19996 \ kg \cdot m/s[/tex]

Generally the  average force of impact is mathematically represented as      

         [tex]F = \frac{\Delta p }{\Delta t}[/tex]

=>        [tex]F = \frac{1.19996 }{7.13 *10^{-3}}[/tex]

=>        [tex]F = 168.298 \ N[/tex]

Answer:

Dalton’s atomic theory was a scientific theory on the nature of matter put forward by the English physicist and chemist John Dalton in the year 1808. It stated that all matter was made up of small, indivisible particles known as ‘atoms’.

Explanation:

Atoms can be divided into their subatomic particles.

The statement that is part of Dalton's atomic theory is as follows: Matter is composed of small particles called atoms.

John Dalton is a scientist that first stated the theory of chemical combination in 1803.

The components of these theory are as follows:

Therefore, according to this question, the statement that is part of Dalton's atomic theory is as follows: Matter is composed of small particles called atoms.

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Answer:

Explanation:

The total work done by the wave is expressed as;

Workdone = Potential energy + Kinetic energy

Workdone = mgh + 1/2mv²

m is the mass = 77kg

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

v is the velocity = 8.2m/s

h is the height = 1.65m

Substitute into the formula;

Workdone = 77(9.8)(1.65) + 1/2(77)8.2²

Workdone = 1245.09 + 2588.74

Workdone = 3833.83Joules

Hence the amount of non conservative work done on the sofa is 3833.83Joules

Given:

We know,

→ [tex]Work \ done = Potential \ energy +Kinetic \ energy[/tex]

or,

                     [tex]= mgh +\frac{1}{2} mv^2[/tex]

By putting the values,

                     [tex]= 77\times 9.8\times 1.65+\frac{1}{2}\times 77\times (8.2)^2[/tex]

                     [tex]= 1245.09+2588.74[/tex]

                     [tex]= 3833.83 \ Joules[/tex]

Thus the above approach is right.

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Answer:

Explanation:

To get the focal length, we will use the lens formula;

1/f = 1/u + 1/v

f is the focal length

u is the object distance

v is the image distance

Given

since the physicist's left eye is myopic, it will be corrected using concave lens and the image distance is negative.

u = 35cm

v = -2.0

1/f = 1/35-1/2

1/f = 2-35/70

1/f = -33/70

f = -70/33

f = -2.12 cm

f = -0.0212m

Power of a lend is the reciprocal of its focal length

Power of the lens = 1/f

P = 1/-0.0212

P = -47.17dioptres

The power of the lens is -47.17D

Answer:

(a) V₂ = 1.86 m/s

(b) Q = 5.1 x 10⁻⁴ m³/s

Explanation:

(a)

The formula derived for Venturi tube is as follows:

P₁ - P₂ = (ρ/2)(V₂² - V₁²)

where,

P₁ - P₂ = Difference in Pressure of Inlet and Outlet = 1.2 KPa = 1200 Pa

ρ = Density of Gasoline = 7.4 x 10² kg/m³

V₂ = Exit Velocity = ?

V₁ = Inlet Velocity

Therefore,

1200 Pa = [(7.4 x 10²kg/m³)/2](V₂² - V₁²)

V₂² - V₁² = (1200 Pa)/(3.7 x 10² kg/m³)

V₂² - V₁² = 3.24 m²/s²   ------------------- equation (1)

Now, we will use continuity equation:

A₁V₁ = A₂V₂

where,

A₁ = Inlet Area = πd₁²/4 = π(0.0374 m)²/4 = 1.098 x 10⁻³ m²

A₂ = Exit Area = πd₂²/4 = π(0.0187 m)²/4 = 2.746 x 10⁻⁴ m²

Therefore,

(1.098 x 10⁻³ m²)V₁ = (2.746 x 10⁻⁴ m²)V₂

V₁ = (2.746 x 10⁻⁴ m²)V₂/(1.098 x 10⁻³ m²)

V₁ = 0.25 V₂  

using this value in equation (1):

V₂² - (0.25 V₂)² = 3.24 m²/s²

0.9375 V₂² = 3.24 m²/s²

V₂² = (3.24 m²/s²)/0.9375

V₂ = √(3.456 m²/s²)

V₂ = 1.86 m/s

(b)

For fluid flow rate we use the following equation:

Flow Rate = Q = A₂V₂ = (2.746 x 10⁻⁴ m²)(1.86 m/s)

Q = 5.1 x 10⁻⁴ m³/s

The formula for finding variables in a  Venturi tube is shown below:

P₁ - P₂ = (ρ/2)(V₂² - V₁²)

where, P₁ - P₂ is difference in pressure of Inlet and outlet, ρ = density, V₂ = exit velocity and V₁ is inlet velocity

P₁ - P₂ = 1.2 KPa = 1200 Pa

ρ = 7.4 x 10² kg/m³

V₂ = Exit Velocity = ?

V₁ = Inlet Velocity

We then substitute the variables into this equation.

P₁ - P₂ = (ρ/2)(V₂² - V₁²)

1200 Pa = [(7.4 x 10²kg/m³)/2](V₂² - V₁²)

V₂² - V₁² = (1200 Pa)/(3.7 x 10² kg/m³)

V₂² - V₁² = 3.24 m²/s²  ------ equation (1)

The  continuity equation A₁V₁ = A₂V₂ is then used

where,A₁ = Inlet area = πd₁²/4 = π(0.0374 m)²/4 = 1.098 x 10⁻³ m²

A₂ = Exit Area = πd₂²/4 = π(0.0187 m)²/4 = 2.746 x 10⁻⁴ m²

(1.098 x 10⁻³ m²)V₁ = (2.746 x 10⁻⁴ m²)V₂

V₁ = (2.746 x 10⁻⁴ m²)V₂/(1.098 x 10⁻³ m²)

V₁ = 0.25 V₂  

We then substitute the value into equation 1

V₂² - (0.25 V₂)² = 3.24 m²/s²

0.9375 V₂² = 3.24 m²/s²

V₂² = (3.24 m²/s²)/0.9375

V₂ = √(3.456 m²/s²)

V₂ = 1.86 m/s

This can be calculated using the formula

Flow Rate = Q = A₂V₂

                  = (2.746 x 10⁻⁴ m²)(1.86 m/s)

                 = 5.1 x 10⁻⁴ m³/s

From a flying plane a body should dropped in advance to hit the target,Why? ... The body should be dropped in advance as when the body is dropped it has the velocity of the plane. So, in air the body moves forward which we have to take into consideration in order to hit the target.

Answer:

Explanation:

Frequency is oscillations per second.

So we have to find the Number of seconds she paced.

2 minutes = 2 X 60

               = 120 seconds

Therefore,

Her frequency = 10 / 120

                       = 1/12 Hertz

I would say food but like they take weight too

Answer:

False

Explanation:

Answer:

False

Explanation:

Hope this helped, Have a Wonderful Day/Night!!

Answer:

Watt

Explanation:

Power is defined as the work done by an object per unit time. Its mathematical form is given by :

[tex]P=\dfrac{W}{t}[/tex]

The SI unit of work done is Joules (J) and that of time is seconds (s).

J/s is equal to watts. Watt is the metric unit of power.

Answer:

[tex]\mid \vec C\mid=31.9[/tex]

Explanation:

Consider the vectors:

[tex]\vec A=9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}}[/tex]

[tex]\vec B=-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}}[/tex]

Calculate the magnitude of

[tex]\vec C=-2\vec B-\vec A[/tex]

Substitute the values of the vectors:

[tex]\vec C=-2(-9.5\mathbf{\hat{i}}-13.4\mathbf{\hat{j}})-(9.4\mathbf{\hat{i}}-3.6\mathbf{\hat{j}})[/tex]

Operate and remove parentheses:

[tex]\vec C=19\mathbf{\hat{i}}+26.8\mathbf{\hat{j}}-9.4\mathbf{\hat{i}}+3.6\mathbf {\hat{j}}[/tex]

Operating both components separately:

[tex]\vec C=9.6\mathbf{\hat{i}}+30.4\mathbf{\hat{j}}[/tex]

Now find the magnitude of C:

[tex]\mid \vec C\mid=\sqrt{9.6^2+30.4^2}[/tex]

[tex]\mid \vec C\mid=\sqrt{1016.32}[/tex]

[tex]\mathbf{\mid \vec C\mid=31.9}[/tex]

Answer:

0.130

Explanation:

From the given data, the coefficient of static friction for each trial are:

1. 0.053

2. 0.081

3. 0.118

4. 0.149

5. 0.180

6. 0.198

The sum of the coefficient of static friction = 0.053 + 0.081 + 0.118 + 0.149 + 0.180 + 0.198

                                              = 0.779

So that;

the average coefficient of static friction = [tex]\frac{sum of coefficient of static friction}{number of trials}[/tex]

                                              = [tex]\frac{0.779}{6}[/tex]

                                              = 0.12983

The average coefficient of static friction is 0.130

The average coefficient of static friction is 0.13.

The coefficient of static friction is obtained using the formula; μ = F/R

Where;

F = force acting on the body

R = reaction

μ = coefficient of static friction

The average of measurements is given as; ∑summation of measurements/number of measurements

We can see from the question that there were 6 measurements of the coefficient of static friction. Hence, the average coefficient of static friction is obtained from;

0.053 + 0.081 + 0.118 + 0.149 + 0.180 +  0.198/6

= 0.13

The average  coefficient of static friction is 0.13

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