Answer:
36111 kg
Explanation:
Given
Force = 65000N
Acceleration = 1.8m/s²
Required
Determine the mass of the elevator
This question will be answered using the following Force formula.
Force = Mass * Acceleration
Substitute values for Force and Acceleration
65000N = Mass * 1.8m/s²
Make Mass the subject
Mass = 65000N/1.8m/s²
Mass = 36111.11 kg
From the list of given options, option E answers the question.
The metric unit of power is _____.
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.
A spring gun fires a ball horizontally at 15 m/s. It is mounted on a flat car moving in a straight line at 25 m/s. relative to the ground, what is the horizontal speed of the ball when the gun is aimed forward?
Answer:
15 m/s
Explanation:
The Speed Of The Car Does Not Add To The Speed Of The Bullet
* WILL GIVE BRAINLIEST TO CORRECT ANSWER *
Name a product that people commonly purchase by mass and not by weight.
I would say food but like they take weight too
A group of students collected the data shown below while attempting to measure the coefficient of static friction (of course, it looks like this group varied the amount of mass sitting on the block with each trial - this is not recommended). Nonetheless, what is their average coefficient of static friction?
Trial Mass of block (g) Hanging mass (kg)
1 105 0.053
2 165 0.081
3 220 0.118
4 280 0.149
5 315 0.180
6 385 0.198
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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Why does a watched pot never boil?
significance of practicals in the discipline of geography
Catching a wave, a 77 kg surfer starts with a speed of 1.3 m/s, drops through a height of 1.65 m, and ends with a speed of 8.2 m/s. How much non-conservative work was done on the surfer?
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:
Velocity, v = 8.2 m/sHeight, h = 1.65 mMass, m = 77 kgWe 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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2. Heather and Matthew walk with an average velocity of +0.87 m/s eastward.
If it takes them 27 minutes to walk to the store, what is their
displacement? (include direction)
(5 points)
If you jump upward with a speed of 1.70 m/s how high will you be when you stop rising?
Answer:
How long do you jump (sec) ?
Explanation:
The second-order dark fringe in a single-slit diffraction pattern is 1.40 mm from the center of the central maximum. Assuming the screen is 89.0 cm from a slit of width 0.710 mm and assuming monochromatic incident light, calculate the wavelength of the incident light.
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.
The CERN particle accelerator is circular with a circumference of 7.0 km.
Required:
a. What is the acceleration of the protons (m=1.67×10^−27kg) that move around the accelerator at of the speed of light? (The speed of light is v=3.00×10^8m/s.)
b. What is the force on the protons?
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
from a flying aeroplane abody should be dropped in advance to hit the target why
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.
Which statement is part of Dalton's atomic theory?
Matter is composed of small particles called atoms.
Atoms can be divided into their subatomic particles.
Atoms are able to be seen with proper spectroscopy equipment.
Chemical reactions can change atoms from one type to another.
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.
What is Dalton's atomic theory?John Dalton is a scientist that first stated the theory of chemical combination in 1803.
The components of these theory are as follows:
Elements consist of indivisible small particles called atoms.All atoms of the same element are identical i.e. different elements have different types of atom. Atoms can neither be created nor destroyedTherefore, 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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When you turn off your bedroom lights, the kitchen lights can stay on. This is because your home is wired using which of the following? (AKS 10b / DOK 1)
A.
Fuses
B.
Series circuits
C.
Batteries
D.
Parallel circuits
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.
What is a Series circuit?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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In ultimate the disc may be passed in any direction.
True
False
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 :(
Hmmm... Who can answer this question?
How did life begin?
Just a practice
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?
A golf ball (m=26.7g) is struck a blow that makes an angle of 33.6 degrees with the horizontal. The drive lands 190m away on a flat fairway. The acceleration of gravity is 9.8 m/s^2 . If the golf club and ball are in contact for 7.13 ms, what is the average force of impact?
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]
A physicist's left eye is myopic (i.e., nearsighted). This eye can see clearly only out to a distance of 35 cm.Find the focal length and the power of a lens that will correct this myopia when worn 2.0 cm in front of the eye.
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
An oil refinery uses a Venturi tube to measure the flow rate of gasoline. The density of the gasoline is
ρ = 7.40 ✕ 102 kg/m3,
the inlet and outlet tubes, respectively, have a radius of 3.74 cm and 1.87 cm, and the difference in input and output pressure is
P1 − P2 = 1.20 kPa.
a) find the speed of the gasoline as it leaves the hose
b) find the fluid flow rate in cubic meters per second
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:
The speed of the gasolineP₁ - 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
The fluid flow rate we use the following equation: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
In football we see unbalanced forces. When 1 player exerts an unbalanced force on another player and causes a player to
Answer:
Fall
Explanation:
Fast and safe heart rate for workouts is called muscular strength? True or false
Answer:
False
Explanation:
Answer:
False
Explanation:
Hope this helped, Have a Wonderful Day/Night!!
Can someone help plzzzzzzz I need it ASAP thank you
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
A 2.0 cm thick brass plate (k_r = 105 W/K-m) is sealed to a glass sheet (kg = 0.80 W/K m), and both have the same area. The exposed face of the brass plate is at 80°C, while the exposed face of the glass is at 20 °C. How thick is the glass if the glass brass interface is at 65 C? Ans. 0.46 mm
How much thermal energy (in kcal) is required to change a 43 g ice cube from a solid at - 16.5 oC to steam at 11.5 oC above boiling
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
C4. A 50.0 kg boy runs at 10.0 m/s, jumps on a cart and rolls off at 2.50 m/s. What is the mass of the cart
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.
we feel cold in winter when we come out from the quilt but the same room becomes warmer after coming back from outside the room
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
Objects are lighter on the moon than they are on earth. if an object A weighs 25lbs on the Moon and another object B weighs 25 Newtons on earth, which has more mass?
a. Object a
b. Object b
c. Same mass
d. Other
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.
If the velocity of Homer the astronaut (mass =200 kg) is 5 m/s and he runs into and grabs his stationary pal Larry (mass = 150 kg), what is the new velocity of the astronauts after the collision?
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
Help me guys please with this question
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]
Assume you are in the car and the car is moving at a certain speed to
school. Are you at rest or in motion with respect to the school? With
respect to the car?