Answer:
163 N
Explanation:
The density of copper is about 8.96. The ratio of the weight in water to the weight in air is about 1-1/ρ, so is about 0.8884.
0.8884 × 184 N ≈ 163 N
The submerged weight is about 163 N.
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.
Learn more about work done here:
https://brainly.com/question/24230840
1.) There was an earthquake in Salt Lake City, Utah, on March 18, 2020, in the morning at 9 hours, 9 minutes, and 45 seconds Mountain Standard Time (9:9:45 MST). If the velocity of the p-wave is 7.3 km/sec and the velocity of the s-wave is 5.1 km/sec and the s-p time lag is 16 seconds, what is the distance in kilometers from Salt Lake City to the focus of the earthquake? Explain how you calculated the answer.
Answer:
7 because salt lake and Southis weat
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?
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
If you are riding your bike west to your friend’s house, and you ride the 1.25 miles in 5 minutes, what is your velocity, in miles per hour?
Answer:
velocity = 15 miles / hourExplanation:
distance = 1.25 mile
time traveled = 5 min.
find velocity in miles / hour
solution:
use the formula: velocity = distance / time
velocity = 1.25 mile x 60 min
5 min 1 hour
velocity = 15 miles / hour
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 :(
In football we see unbalanced forces. When 1 player exerts an unbalanced force on another player and causes a player to
Answer:
Fall
Explanation:
significance of practicals in the discipline of geography
A typical blood pressure is 120 mm Hg. How high would you need to hang an IV bag so that the fluid enters the blood stream at this pressure? Assume the IV fluid has a density of 1000 kg/m^3.
Answer:
1.63 m
Explanation:
The following data were obtained from the question given above:
Pressure (P) = 120 mmHg
Density (d) = 1000 Kg/m³
Height (h) =?
Next, we shall convert 120 mmHg to N/m². This can be obtained as follow:
760 mmHg = 101325 N/m²
Therefore,
120 mmHg = 120 mmHg × 101325 N/m² / 760 mmHg
120 mmHg = 15998.68 N/m²
Thus, 120 mmHg is equivalent to 15998.68 N/m².
Finally, we shall determine the height to which the IV bag should be placed so that the fluid can enter the blood stream. This is illustrated below:
Pressure (P) = 15998.68 N/m².
Density (d) = 1000 Kg/m³
Acceleration due to gravity (g) = 9.8 m/s²
Height (h) =?
P = dgh
15998.68 = 1000 × 9.8 × h
15998.68 = 9800 × h
Divide both side by 9800
h = 15998.68 / 9800
h = 1.63 m
Therefore, the the IV bag should be placed at a height of 1.63 m
A 30 kg box is being pulled with a force of 125 N. The coefficient of static friction between the box and the floor is 0.35. What is the minimum downward force on the box that will keep it from slipping?
Answer:
The minimum downward force on the box that will keep it from slipping is 63.14 N
Explanation:
Given;
mass of the object, m = 30 kg
applied force, f = 125 N
coefficient of static friction, μ = 0.35
Normal reaction (R) is acting upwards, weight of the box (mg) is acting downwards and the minimum downward force (F) on the box that will keep it from slipping is also acting downwards.
The net vertical forces on the box is given by;
R - mg - F = 0
F = R - mg
Now, determine normal reaction, R
f = μR
R = f / μ
R = 125 / 0.35
R = 357.14 N
Finally, determine the minimum downward force on the box that will keep it from slipping;
F = R - mg
F = 357.14 - (30 x 9.8)
F = 357.14 - 294
F = 63.14 N
Therefore, the minimum downward force on the box that will keep it from slipping is 63.14 N
How many stars are in the universe (approximately)? O 40 sixtillion 0 365 billion O 86.4 million O one
Answer:
a i belive
Explanation:
the univerce is VERY large so a, if im wrong i apologise :(
Fast and safe heart rate for workouts is called muscular strength? True or false
Answer:
False
Explanation:
Answer:
false :3
hope this helps
an a astronaut stands on the surface of a spherical asteroid that has a weak gravitational field but no atmosphere
Answer:
so what is your question that's not a question
An asteroid is a minor planet in the inner solar system. The asteroid orbits the sun and millions of asteroids exit as remains of planetesimals.
The weak gravitational field of the asteroid describes that there is no contact between the asteroid and the astronaut. As the asteroid has no atmosphere the astronaut jumps and drifts.Hence there is no contact as no gravitational force.
Learn more about the astronaut standing on the surface of a spherical asteroid.
brainly.com/question/13651731.
* 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 force of 1.50 N acts on a 0.20 kg trolley so as to accelerate it along an air track.
The track and force are horizontal and in line. How fast is the trolley going after acceleration from rest through 30 cm, if friction is negligible?
Answer: The trolley is moving at 2.12m/s
Explanation:
Given from the question that the track and force are horizontal and inline, we have that
F= ma
where F= force= 1.50 N
m= mass = 0.2kg
therefore Acceleration , a = F/ m= 1.50/0.2 =7.5 m/s^2
To find how fast the trolley is going ie the Velocity, v
Having that
initial velocity at rest , u = 0,
acceleration a = 7.5 and
and distance, s = 30 cm = 30/100 = 0.30 m
we use motion equation that
v² = u² + 2 a s
v² = 0² + 2 x 7.5 x 0.30
v² = 4.5
v = [tex]\sqrt{4.5}[/tex]
v = 2.12 m/s
Describe which relationships in Ohm’s law are DIRECT and which are INVERSE. Use examples to support your answer. You can use calculations, drawings, or graphs to make your point more clear
Answer:
I like to memorize excerpt from articles to solve and answer questions like these. I hope this can help, it's from study.com: "The relationship between voltage, current, and resistance is described by Ohm's law. This equation, i = v/r, tells us that the current, i, flowing through a circuit is directly proportional to the voltage, v, and inversely proportional to the resistance, r."
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
Learn more: https://brainly.com/question/8155254
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]
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.
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.
You are a world-famous physicist-lawyer defending a client who has been charged with murder. It is alleged
that your client, Mr. Lawton, shot the victim, Mr. Cray. The detective who investigated the scene of the
crime, Mr. Dibny, found a second bullet, from a shot that missed Mr. Cray, that had embedded itself into a
chair. You arise to cross-examine the detective.
You: In what type of chair did you find the bullet?
Dinby: A wooden chair.
You: How massive was this chair?
Dinby: It had a mass of 20 kg.
You: How did the chair respond to being struck with a bullet?
Dinby: It slid across the floor.
You: How far?
Dinby: Three centimeters. The slide marks on the dusty floor are quite distinct.
You: What kind of floor was it?
Dinby: A wood floor.
You: What was the mass of the bullet you retrieved from the chair?
Dinby: Its mass was 10 g.
You: Have you tested the gun you found in Mr. Lawton's possession?
Dinby: I have.
You: What is the muzzle velocity of bullets fired from that gun?
Dinby: The muzzle velocity is 450 m/s.
With only slight hesitation, you turn confidently to the jury and proclaim, "My client's gun did not fire those
shots!"
(a) How are you going to convince the jury and judge?
(b) Choose one part of your solution and perform a sense-making analysis. Clearly state which sensemaking analysis you’ve chosen and why.
Answer:
It was not fired from the client's gun because the chair slid only 3 centimeters . If it had been fired from the client's gun the chair would slid 25.82 centimeters.
Explanation:
According to the law of conservation of momentum the momentum of the system before collision must be equal to the momentum of the system after the collision.
M1u1= m2u2
Let M1 = mass of the chair = 20kg
m2= mass of the bullet= 10g= 0.001kg
u1= velocity of the chair before collision = zero m/s
u2 = velocity of the bullet before collision = zero m/s
v1= velocity of the chair after collision = ? m/s
v2 = velocity of the bullet after collision = 450 m/s
After collision their velocities change from u1 to v1 and u2 to v2 so
M1v1= m2v2
v1= m2v2/M1
v1= 0.01 *450/ 20= 0.225 m/s
Now according to the law of conservation of energy the energy of the system before collision must be equal to the energy of the system after the collision.
The energy of the chair after the bullet is hit is
KE of the chair + KE of the bullet= 1/2 (M)(v1)²+ 1/2 m(v2)²=
1/2 ( 20) (0.225 )² + 1/2 (0.01) (450)²
= 0.50625 + 1012.5= 1013.00625 Joules
Frictional force = Coefficient of kinetic force of wood on wood ( M+m) g
= 0.2* ( 20.01) 9.8= 39.2196 N
Work done by friction = frictional force * distance
If law of conservation of energy is applied the KE must be equal to the work done
KE = W
W= f*d
KE= F*d
d = KE/f= 1013.00625/ 39.2196= 25.82 cm
The chair did not move 25.82 cm .
It only moved 3 centimeter.
Hence the bullet fired was not from the client's gun.
What is the highest order dark fringe, , that is found in the diffraction pattern for light that has a wavelength of 575 nm and is incident on a single slit that is 1410 nm wide?
Answer:
The highest order dark fringe is 2
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 575 \ nm = 575 *10^{-9} \ m[/tex]
The width is of the slit is [tex]d = 1410 nm = 1410 *10^{-9} \ m[/tex]
Generally the highest order dark fringe is mathematically represented as
[tex]m = \frac{w}{\lambda }[/tex]
=> [tex]m = \frac{ 1410 *10^{-9}}{ 575 *10^{-9} }[/tex]
=> [tex]m =2[/tex]
Find the density of a liquid, in lb/ft^3, that exerts a pressure of 0.400 lb/in^2 at a depth of 42.0 in.
Given :
Pressure, P = 0.4 lb/in².
Depth, h = 42 in.
To Find :
The density of a liquid.
Solution :
Pressure at height h of a liquid of density [tex]\rho[/tex] is given by :
[tex]P = \rho g h[/tex] ....1)
Here, g = 386.04 in/s²( acceleration due to gravity )
Putting all values in equation 1, we get :
[tex]P = \rho g h\\\\0.4 = \rho\times 386.04 \times 42\\\\\rho=\dfrac{0.4}{42\times 386.04}\ lb/in^3\\\\\rho=\dfrac{0.4}{42\times 386.04}\times 12^3\ lb/ft^3\\\\\rho=0.0426\ lb/ft^3[/tex]
Hence, this is the required solution.
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)
A Spring is pulled to 10cm and held in place with a force of 500N .what is the spring constant
We are given:
extension of the spring (x) = 10 cm OR 0.1 m
Force applied to keep it in position (F) = 500 N
Solving for the spring constant:
We know that F = kx (when the spring is extended)
replacing the variables with the given values
500 = k * 0.1
k = 5000 N/m
This means that to extend the spring by 1 m, we have to apply a force of 5000 N
Calculate the average induced voltage between the tips of the wings of a Boeing 747 flying at 970 km/hr above East Lansing. The downward component of the earth's magnetic field at this place is
This question is incomplete, the complete question is;
Calculate the average induced voltage between the tips of the wings of a Boeing 747 flying at 970 km/hr above East Lansing. The downward component of the earth's magnetic field at this place is 0.7 × 10⁻⁴ T. (DATA: Assume that the wingspan is 60 meters.)
Answer:
the average induced voltage between the tips of the wings of a Boeing 747 flying is 1.132 Volts
Explanation:
Given that;
Speed S = 970 KM/hr
downward component of the earth's magnetic field B = 0.7 × 10⁻⁴
wingspan 1 = 60min
Velocity V = (970 × 10³) / 3600 = 269.44 m/s
So Average Induced Voltage E = VBI
we substitute
E = 269.44 × (0.7 × 10⁻⁴) × 60
E = 1.132 Volts
Therefore the average induced voltage between the tips of the wings of a Boeing 747 flying is 1.132 Volts
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.
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.
Find the magnitude of the gravitational force (in N) between a planet with mass 6.00 X 10^24 kg and its moon, with mass 2.50 X 10^22 kg, if the average distance between their centers is 2.70 X 10^8 m. What is the moon's acceleration (in m/s2) toward the planet? (Enter the magnitude.) What is the planet's acceleration (in m/s2) toward the moon? (Enter the magnitude.)
Answer:
Magnitude of gravitational force between this planet and its moon: approximately [tex]1.37 \times 10^{20}\; \rm N[/tex].
Acceleration of this moon towards the planet: approximately [tex]5.49 \times 10^{-3}\; \rm m \cdot s^{-2}[/tex].
Acceleration of this planet towards its moon: approximately [tex]2.29 \times 10^{-5}\; \rm m \cdot s^{-2}[/tex].
Explanation:
Look up the gravitational constant, [tex]G[/tex]:
[tex]G \approx 6.67 \times 10^{-11}\; \rm m^3\cdot kg^{-1} \cdot s^{-2}[/tex].
Assume that both this planet and its moon are spheres of uniform density. When studying the gravitational interaction between this planet and its moon, this assumption allows them to be considered as two point masses.
The formula for the size of gravitational force between two point masses [tex]m_1[/tex] and [tex]m_2[/tex] with a distance of [tex]r[/tex] in between is:
[tex]\displaystyle F = \frac{G \cdot m_1 \cdot m_2}{r^2}[/tex],
where [tex]G[/tex] is the gravitational constant.
Let [tex]m_1[/tex] and [tex]m_2[/tex] denote the mass of this planet and its moon, respectively.
Calculate the size of gravitational force between this planet and its moon:
[tex]\begin{aligned} F &= \frac{G \cdot m_1 \cdot m_2}{r^2} \\ &\approx \frac{6.67 \times 10^{-11}\; \rm m^3 \cdot kg^{-1} \cdot s^{-2} \times 6.00 \times 10^{24}\; \rm kg \times 2.50 \times 10^{22}\; \rm kg}{{\left(2.70 \times 10^{8}\; \rm m\right)}^2} \\ &\approx 1.37 \times 10^{20}\; \rm N\end{aligned}[/tex].
Assume that other than the gravitational force between this planet and its moon, all other forces (e.g., gravitational force between this planet and the star) are negligible. The magnitude of the net force on the planet and on the moon should both be approximately [tex]1.37 \times 10^{20}\; \rm N[/tex].
Apply Newton's Second Law of motion to find the acceleration of this planet and its moon:
[tex]\displaystyle \text{acceleration} = \frac{\text{net force}}{\text{mass}}[/tex].
For this moon:
[tex]\displaystyle \frac{1.37 \times 10^{20}\; \rm N}{2.50 \times 10^{22}\; \rm kg} \approx 5.49\times 10^{-3}\; \rm m \cdot s^{-2}[/tex].
For this planet:
[tex]\displaystyle \frac{1.37 \times 10^{20}\; \rm N}{6.00 \times 10^{24}\; \rm kg} \approx 2.29 \times 10^{-5}\; \rm m \cdot s^{-2}[/tex].
Forces that are equal in size but opposite in direction are called what ?
Answer:
I think they are called balanced forces
Explanation: