Answer:
See the explanation below.
Explanation:
Density will remain the same since density is the relationship between mass and volume. As we can see in the equation below.
[tex]Ro=m/V[/tex]
where:
Ro = density = 2.5 [g/cm³]
m = mass [g]
V = volume [cm³]
In such a way that when the glass is broken the small fragments retain the same density ratio. That is, each fragment has a small mass and a small volume. That's why the density remains the same.
Which heat related illness requires immediate medical attention
Answer:
Heat stroke
Explanation:
Heat Stroke is the most serious heat-related illness and requires immediate medical attention.
Answer:
Heat Stroke is the most serious heat-related illness and requires immediate medical attention. Symptoms include: confusion, fainting, seizures, very high body temperature and hot, dry skin or profuse sweating. CALL 911 if a coworker shows signs of heat stroke.
Explanation:
I HOPE THIS HELPS * pls mark me brainliest }
A spherical balloon with a 36cm diameter is being deflated. Its volume V is a function of its radius r according to
Question:
A spherical balloon with a 36 cm diameter is being deflated. Its volume V is a function of its radius r according to [tex]V(r) = \frac{4}{3}\pi r^3.[/tex]
As it's deflating, it is much easier to measure its radius than its volume. Suppose the radius of the balloon is [tex]r(t) = 18 - 18e^{-0.24t}[/tex] cm at t seconds.
Determine the value of V'(4) (accurate to 3 decimal places). Write a complete sentence including the units of both 4 and V'(4) in this context. Determine the value of r(4) (accurate to 3 decimal places). Write a complete sentence including the units of both 4 and r(4) in this context.Answer:
[tex]V'(4) = 201.143[/tex]
[tex]r(4) = 11.106[/tex]
Explanation:
Given
[tex]V(r) = \frac{4}{3}\pi r^3.[/tex]
[tex]r(t) = 18 - 18e^{-0.24}[/tex]
Solving (a): V'(4)
First, we determine V('r)
[tex]V(r) = \frac{4}{3}\pi r^3.[/tex]
Differentiate w.r.t r
[tex]V'(r) = 3 * \frac{4}{3}\pi r^{3-1}[/tex]
[tex]V'(r) = 3 * \frac{4}{3}\pi r^2[/tex]
[tex]V'(r) = 4\pi r^2[/tex]
Substitute 4 for r and take [tex]\pi = \frac{22}{7}[/tex]
[tex]V'(4) = 4 * \frac{22}{7} * 4^2[/tex]
[tex]V'(4) = 4 * \frac{22}{7} * 16[/tex]
[tex]V'(4) = \frac{4 * 22* 16}{7}[/tex]
[tex]V'(4) = \frac{1408}{7}[/tex]
[tex]V'(4) = 201.143[/tex]
This means that the volume of the balloon when the radius is deflated to 4 seconds is 201.143 cm^3
Solving (b): r(4)
Substitute 4 for t in [tex]r(t) = 18 - 18e^{-0.24t}[/tex]
[tex]r(4) = 18 - 18e^{-0.24*4}[/tex]
[tex]r(4) = 18 - 18e^{-0.96}[/tex]
[tex]r(4) = 18 - 18*0.383[/tex]
[tex]r(4) = 11.106[/tex]
This means that the radius of the balloon when at 4 seconds of deflation is 11.106 cm
The force of water flowing over a dam spillway is to be measured in a geometrically similar small-scale laboratory model. The width of the prototype is 30 m and the width of the model is 2 m. If the force on the model spillway is measured to 10 kN, the expected force on the prototype spillway for dynamically similar conditions is:
Answer:
Jay shree krishna j to the same time kr rhe h u have UCC fm rob and studing a few minutes ago
Marvin Martian is standing on planet Potatoine.
Potatoine has the following physical data:
Radius: 6.4x106 m
Mass: 2.0x1024 kg
If Marvin's mass is 35 kg, calculate his weight on Potatoine.
Answer:
W = 113.98 N
Explanation:
Given that,
Radius of Potatoine[tex]R=6.4\times 10^6\ m[/tex]
Mass of Potatoine, [tex]M=2\times 10^{24}\ kg[/tex]
Mass of Marvin, m = 35 kg
We need to find his weight on Potatoine. Weight of an object is given by :
W = mg
g is acceleration due to gravity, [tex]g=\dfrac{GM}{R^2}[/tex]
So,
[tex]W=\dfrac{GMm}{R^2}\\\\W=\dfrac{6.67\times 10^{-11}\times 35\times 2\times 10^{24}}{(6.4\times 10^6)^2}\\\\W=113.98\ N[/tex]
So, his weight on Potatoine is 113.98 N.
A cruise ship can accelerate at 0.01 m/s2, if it starts from rest how long does it take
to reach a speed of 25 km/h? How long would it take if it started from a speed of
2.0 m/s?
Answer:
t = 8.24 [min]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f}=v_{o}+a*t[/tex]
where:
Vf = final velocity = 25 [km/h]
Vo = initial velocity = 2 [m/s]
a = acceleration = 0.01 [m/s²]
t = time [s]
Now we must convert the speed from kilometers per hour to meters per second.
[tex]25[\frac{km}{h} ]*[\frac{1h}{3600s}]*[\frac{1000m}{1km} ]=6.94[m/s][/tex]
Now replacing:
[tex]6.94=2+0.01*t\\t=494 [s]\\or\\t=8.24[min][/tex]
You are sitting on the beach and notice that a seagull floating on the water moves up and down 5 times in I minute. What is the frequency of the water waves
Answer:
f = 0.0833 Hz
Explanation:
Frequency is defined as the no. of cycles per unit time. The frequency of water waves can be given by the following formula:
[tex]Frequency\ of\ waves = f = \frac{No.\ of\ Waves}{time}\\[/tex]
here,
No. of Waves = No. of times seagull moves up and down = 5
Time = 1 minute = 60 sec
Therefore, using these values in the formula, we get:
[tex]f = \frac{5}{60\ sec}[/tex]
f = 0.0833 Hz
Different between rarer medium and denser medium at least 5 point each?? can anyone help me to solve this question
Answer:
A medium in which speed of light is more is known as optically rarer medium and a medium in which speed of light is less is said to be optically denser medium. For example in air and water, air is raer and water is a denser medium.
Explanation:
Answer:Rarer medium is less dense
Explanation:
Rarer medium is less dense than dense medium because rarer medium doesn't have a very high amount of particles packing it together. It's easier to see through a rarer medium, so you're able to get a clearer picture of what's on the other side of it. Denser medium is, simply put, more dense.
A 8.5 kg brick is dropped onto a 6.5 kg toy truck, which is moving across a level floor at 0.50 m/s. With what velocity do the truck and brick continue to move, after the brick has landed on the truck?
Answer:0.2167 m/s
Explanation:
Given
mass of brick M=8.5 kg
mass of toy truck m=6.5 kg
the velocity of truck u=0.5 m/s
Suppose v is the velocity after brick is landed on the truck
There is no external force acting so momentum is conserved
mu=(M+m)v
[tex]v=\dfrac{6.5}{15}\times 0.5\\v=0.2167\ m/s[/tex]
is a paper set on fire and burns to ashes chemical or physical change?
Answer:
Chemical
Explanation:
Find the distance between (-6, 1) and (2, 2). Round to the
nesest hundredth.
Answer:
d = 8.06 (units)
Explanation:
In this problem, we are given two points with X & y coordinates. therefore we must use the Pythagorean theorem with the respective coordinates of each point to determine the required distance.
[tex]A = (-6, 1)\\B = (2, 2)[/tex]
Now using the Pythagorean theorem.
[tex]d=\sqrt{(x_{1}-x_{0})^{2} +(y_{1}-y_{o})^{2} }\\d=\sqrt{(2-(-6))^{2} +(2-1)^{2} }\\d=\sqrt{64+1} \\d=8.06[/tex]
The dependent variable is always opposite of the independent variable? True or false?
Answer: The dependent variable is always the opposite of the independent variable. False ... Scientific theories are broad explanations that are widely accepted as true.
Explanation:
A 2000-kg car moving with a speed of 20 m/s collides with and sticks to a 1500-kg car at rest at a stop sign. Show that because momentum is conserved in the rest frame, momentum is also conserved in a reference frame moving with a speed of 10 m/s in the direction of the moving car.
Answer:
13.33m/s
Explanation:
Given data
m1= 2000kg
u1= 20m/s
m2= 1500kg
u2= 0m/s
v1= 10m/s
Required
The speed of the sticks
We know that from the expression for the conservation of momentum
m1u1+m2u2= m1v1+m2v2
2000*20+1500*0=2000*10+1500*v2
40000=20000+1500v2
collect like terms
40000-20000= 1500v2
20000= 1500v2
v2= 20000/1500
v2= 13.33 m/s
Hence the velocity of the sticks is 13.33m/s
7. A runner runs on the track field with a velocity of 20 m/s and it is known that her momentum is
1,200 kg*m/s. What is the weight of the runner?
Answer:
Weight of the runner is 60 kilograms
Explanation:
Momentum is the product of weight and velocity.
The formula is : p=mv where m is mass in kg and v is velocity in m/s
Given in the question ;
v= 20 m/s and p=1200 kg*m/s
Applying the formula as;
p= mv
1200 = m * 20
1200/20 = m
60 kg = m
A prospector finds himself holding his bag of gold and standing in the middle of a large pond of frictionless ice. Use the Law of Conservation of Momentum to explain how he can get to the side before he freezes
Answer:
Explanation:
He can not walk on the ice because ice is frictionless . But he can move to shore or side using law of conservation of momentum . He will throw the bag of gold towards the side opposite to shore . This will create a motion in himself in opposite direction according to law of conservation of momentum.
Let m and M be mass of bag and prospector , v and V be velocity created in bag and prospector after throw of bag . Initially , both bag and prospector were at rest so initial momentum = 0 . Final momentum of both = mv - MV .
negative sign is due to their opposite motion .
According to law of conservation of momentum
MV - mv = 0
MV = mv
In this way he can create motion in himself to move to side .
A 2.34-kg gun has a recoil velocity of 5.2 m/sec. At what velocity does it fire its 95-gram bullets?
Explanation:
We need to use the conservation of energy to help us solve this one.
The energy imparted onto the bullet projectile is equal to the energy imparted into the recoiling gun. We can ignore potential energy and only consider the kinetic energy of the two masses (bullet and gun).
Remember that kinetic energy is expressed as [tex]K_E = \frac{1}{2}mv^2[/tex]
Conservation of energy means that both the gun and the bullet have the same amount of energy.
[tex]\frac{1}{2}m_gv_g^2 = \frac{1}{2}m_b v_b^2[/tex]
We'll want to put the mass terms into the same units. So the gun has a mass of 2340 grams.
Which stimulus increases blood glucose levels?
Time between meals.
Body cells taking up glucose.
Consuming food.
What determines whether a real or a virtual image is formed from a concave mirror?
Answer:
The distance of the object from the concave mirror determines if it is a virtual or real image
Explanation:
Concave mirrors form both real and virtual images. When the concave mirror is placed very close to the object, a virtual and magnified image is obtained and if we increase the distance between the object and the mirror, the size of the image reduces and real images are formed.
If a ball is thrown horizontally with a speed of 65 mph, how far will it fall while traveling 90 ft of horizontal distance?
Answer:
install socrati it give you all answers
Explanation:
If a ball is thrown horizontally with a speed of 65 miles per hour, then it would fall 4.32 meters while traveling 90 ft of horizontal distance.
What are the three equations of motion?There are three equations of motion given by Newton
v = u + at
S = ut + 1/2×a×t²
v² - u² = 2×a×s
Note that these equations are only valid for a uniform acceleration.
As given in the problem If a ball is thrown horizontally with a speed of 65 mph, then we have to find out how far will it fall while traveling 90 ft of horizontal distance,
1 mile = 1609 meters
65 miles = 65*1609 meters
=104585 meters
65 miles per hour = 104585 meters per hour
The speed in meters per second = 104585/3600
=29.05 meters / second
1 feet = 0.3048 meters
90 feets = 90*0.3048
=27.432 meters
The time is taken by the ball = 27.432/29.05
=0.945
S = ut + 1/2*a*t²
= 0 + 0.5*9.81*0.945²
S = 4.32 meters
Thus, the ball would fall 4.32 meters while traveling 90 ft of horizontal distance.
To learn more about equations of motion from here, refer to the link ;
brainly.com/question/5955789
#SPJ2
The weight of an object at the surface of Earth is 90 N. What is its weight at a distance 2R from the surface of Earth
Answer:
The weight at a distance 2R from the surface of Earth will be [tex]F'=10 N[/tex].
Explanation:
First of all, we need to find the acceleration of gravity at 2R. Using the gravitational force equation
[tex]F=G\frac{mM}{R^{2}}[/tex]
Where:
M is the mass of the earth
m is the mass of the object
G is the gravitational constant
R is the radius of the earth
We can equal the gravitational force with the second Newton's law (F=ma)
[tex]F=G\frac{mM}{R^{2}}[/tex]
[tex]mg=G\frac{mM}{R^{2}}[/tex]
We know the weight at the earth surface is 90 N, which means:
[tex]90=G\frac{mM}{R^{2}}[/tex]
Now, we have the same equation in the case of 2R as a distance from the surface of the object. Let's remember we need to use the distance from the center of the mass of the earth, so in this case, will be 3R.
[tex]F'=G\frac{mM}{(3R)^{2}}[/tex]
[tex]F'=G\frac{mM}{9R^{2}}[/tex]
[tex]F'=\frac{1}{9}G\frac{mM}{R^{2}}[/tex]
Using the above equation we have:
[tex]F'=\frac{1}{9}90[/tex]
[tex]F'=\frac{90}{9}[/tex]
Therefore, the weight at a distance 2R from the surface of Earth will be [tex]F'=10 N[/tex].
I hope it helps you!
Write down short notes on:
a. Electrical energy
Answer:
Explanation:
Electrical energy is energy derived from electric potential energy or kinetic energy.
Or,
Electrical energy is a form of energy resulting from the flow of electric charge. Lightning, batteries and even electric eels are examples of electrical energy.People use electricity for lighting, heating, cooling, and refrigeration and for operating appliances, computers, electronics, machinery, and public transportation systems.
Hope it helped you.
An asteroid with a mass of 3.5x10kg is 26,000m from a second asteroid with a
mass of 3.0x1012 kg.
Calculate the gravitational force between the two asteroids.
Answer:
F₁₋₂ = 1.036*10⁻⁵ [N]
Explanation:
In order to solve this problem, we must use the law of universal gravitation, which can be defined by means of the following expression.
[tex]F_{1-2}=G*\frac{m_{1}*m_{2}}{d^{2} }[/tex]
where:
F₁₋₂ = Force among the two bodies [N]
G = Universal gravity constant = 6.67 *10⁻¹¹ [N*m²/kg²]
m₁ = mass of the first body = 3.5*10 [kg]
m₂ = mass of the second body = 3*10¹² [kg]
d = separation distance = 26000 [m]
Now replacing:
[tex]F_{1-2}=6.67*10^{-11}*\frac{3.5*10*3*10^{12} }{26000^{2} } \\F_{1-2}=1.036*10^{-5}[N][/tex]
PLS HELP this is due in Jan 28
For this assignment I want you to define physical fitness. After I want you to name 3 physical fitness activities that you do at home or outside. Do you enjoy doing the activities? How long do you do them for? Write a paragraph explaining your answers. Use the Becoming physically fit notes to help you if you need.
Para esta tarea quiero que defina la aptitud física. Después, quiero que menciones 3 actividades de acondicionamiento físico que realizas en casa o al aire libre. ¿Disfrutas haciendo las actividades? ¿Por cuánto tiempo los haces? Escribe un párrafo que explique tus respuestas. Utilice las notas Cómo estar en buena forma física como ayuda si lo necesita
How does a hurricanes get its name
Answer:
The word "hurricane" is widely known and recognized, but its etymology is lesser-known. Named for Mayan God The English word "hurricane" comes from the Taino (the indigenous people of the Caribbean and Florida) word "Huricán," who was the Carib Indian god of evil. Their Huricán was derived from the Mayan god of wind, storm, and fire, "Huracán."
Explanation:
if that is not what you were looking for I can edit it
HELP!!!
How does Dr. Hewitt define Potential Energy?
•Extra energy
•Energy that might or might not be there
•Energy of motion
•Energy of position
•Potent energy
Answer:
energy of position
Explanation:
I think that is the answer
Suppose that, while lying on a beach near the equator watching the Sun set over a calm ocean. you start a stopwatch just as the top of the Sun disappears. You then stand, elevating your eyes by a height H=1.70 m, and stop the watch when the top of the Sun again disappears. If the elapsed time is t=11.1 s, what is the radius r of Earth?
Answer:
R = 1.699 × [tex]10^{4}[/tex]
Explanation:
given data
height H = 1.70 m
time is t = 11.1 s
to find out
radius r of Earth
solution
we get here cosθ so
cos(θ) = R ÷ (R+1.70) ................1
here θ = 360 × 11.1 ÷ (24×60×60)
θ = 0.04625
so put here value in eq 1
cos(0.04625) = R ÷ (R+1.70)
so radius R will be
0.9999 = R ÷ (R+1.70)
1.6999 + 0.999 R = R
R = 1.699 × [tex]10^{4}[/tex]
Two cars, initially at rest and 5 km apart at t=0 , simultaneously move toward each other. Car A travels at a constant speed of 55 km/hour, while car B travels at a constant speed of 12 km/hour. How long will it take care from t=0 for car A to meet car B?
Select one:
a. 24.5 s
b. 490 s
c. 245 s
d. 268 s
e. None of the above.
Answer:
d. 268 s
Explanation:
Constant Speed Motion
An object is said to travel at constant speed if the ratio of the distance traveled by the time taken is constant.
Expressed in a simple equation, we have:
[tex]\displaystyle v=\frac{d}{t}[/tex]
Where
v = Speed of the object
d = Distance traveled
t = Time taken to travel d.
From the equation above, we can solve for d:
d = v . t
And we can also solve it for t:
[tex]\displaystyle t=\frac{d}{v}[/tex]
Two cars are initially separated by 5 km are approaching each other at relative speeds of 55 km/h and 12 km/h respectively. The total speed at which they are approaching is 55+12 = 67 km/h.
The time it will take for them to meet is:
[tex]\displaystyle t=\frac{5}{67}[/tex]
t = 0.0746 hours
Converting to seconds: 0.0746*3600 = 268.56
The closest answer is d. 268 s
PLEASE HELP PHYSICS QUIZ DUE TODAY!
WILL GIVE BRAINLIEST TO THE CORRECT ANSWER
A block of mass 4kg is pushed up against a wall by a force (P) that makes the 50 degree angle with the horizontal.
A. Calculate the force P needed if the coefficient of static friction between the block and the wall is 0.
B. Calculate the the force or forces needed if the coefficient 0.25.
Please answer in Newton’s with two 2 significant figures.
I suppose you mean to say that the wall is frictionless in the first scenario? Also, I assume the block is to held in place. Construct a free body diagram for the block. There are 3 (in part A) or 4 (in part B) forces acting on it.
• its weight w, pulling it downward
• the normal force (magnitude n), pushing outward from the wall to the left
• the push as described, with magnitude p
• static friction (mag. f ), opposing the upward net force and thus pointing downward.
The static friction force has a magnitude proportional to that of the normal force. If the coefficient of static friction is µ, then
f = µ n
so if the wall is frictionless with µ = 0, then f = 0 and does not need to be considered.
(A) If µ = 0, then by Newton's second law we have
• net vertical force:
∑ F = p sin(50°) - w = 0
and we don't need to consider the net horizontal force to determine p from here. We get
p = w / sin(50°) = (4 kg) (9.8 m/s²) / sin(50°) ≈ 51 N
(B) If µ = 0.25, then Newton's second law gives
• net vertical force:
∑ F = p sin(50°) - w - f = 0
p sin(50°) - f = (4 kg) (9.8 m/s²)
p sin(50°) - f = 39.2 N
• net horizontal force:
∑ F = p cos(50°) - n = 0
p cos(50°) - f /0.25 = 0
[since f = 0.25 n]
p cos(50°) - 4f = 0
Multiply the first equation by -4, then add it to the second equation to eliminate f and solve for p :
-4(p sin(50°) - f ) + (p cos(50°) - 4f ) = -4 (39.2 N) + 0
p (cos(50°) - 4 sin(50°)) = -156.8 N
p = (156.8 N) / (4 sin(50°) - cos(50°)) ≈ 65 N
What are the 4 basic skills of Badminton. Explain.
Answer:
As the service marks the start of every rally and subsequently dictates its flow, it is a crucial aspect of the game to get right in badminton.
These are the four main types of services in badminton and most can be executed with either your forehand or backhand.
1. Low serve
This low serve is almost a gentle tap over the net with the shuttle, with the aim of flying just over the net, yet falling just over the front line of his service court. It must not be too high or predictable, otherwise it would be easy for your opponent to do an outright smash or net kill.
2. High serve
The high serve is a powerful strike upwards with the shuttle, that aims to travel a great distance upwards and fall deep at the rear end of the court.
Although it is a strong serve and the popular choice of beginner players, its a serve that isn't so easy to disguise especially since you're using a forehand grip. Your opponent will already expect the shuttlecock to land at the back of the court.
Do remember that shuttlecocks have to fall within the corresponding service areas and this is different in singles and doubles.
3. Flick serve
This flick serve is also played upwards but at a much lesser altitude. It is most common for players to use their backhand to execute the flick serve and the trajectory is lower as this grip has less power.
The whole point of the backhand flick serve is deception, by mixing your serves up and making it look like you're doing a low serve. For this reason, serving with your backhand is thus very popular with competitive players.
It becomes hard for your opponent to predict if you are going to do a flick or a low serve as your stroke will look exactly the same until the point of contact.
4. Drive Serve
This is an attacking serve that is used by top badminton players like Lin Dan. The idea is to hit the shuttle directly at your opponent, limiting their return options and catching them off guard, winning you easy points. It's a good change of pace but it is also risky as if your opponent is prepared, he could just smash the shuttlecock back at you.
This serve is executed with your forehand through underarm action and following through. The shuttle should be dropped a bit sideways rather than in front of your body and hit flatter.
Now that you've determined the type of serve you want to make, here are a four tips on how to execute these serves well.
1. Keep your feet still
During the service, some part of both your feet must be in contact with the ground for it to be a legal serve.
2. Disguise your shots
Make sure your stroke is the same up to the point of contact with the shuttle. This will make your serve possible to predict only at the last possible second. Advanced players can try to trick their opponent by making it deliberately look like you're leaning back and about to do a high serve when you're really going to do a low serve.
3. Observe your opponents position
Is your opponent leaning towards the back already anticipating a high serve to the rear-court? In that case, you may want to execute a low serve to catch him off-guard. Always be aware of the position of your opponent. Try to imagine what he's expecting and do the opposite to gain an advantage.
4. Mix up your serves
Using just one type of service will make you too easy to predict. Make sure you incorporate at least two types of serves into your play. Once you've mastered the basic high and low serves, you can learn the flick and drive serves to add more dimension to your play.
In a nutshell, executing a service well allows you to start the rally strong and dictate its flow.
Explanation:
I hope it's help
Tech a says that when testing a relay whining with an ohmmeter it should show the winding is open Tech B says that the solenoid closes a set of heavy contacts that send current flow to the starter motor
Complete question is:
Tech A says that when testing a relay winding with an ohmmeter, it should show the winding is open. Tech B says that the solenoid closes a set of heavy contacts that send current flow to the starter motor. Who is correct?
Answer:
Tech B is correct
Explanation:
First we need to know the basic working principle or relays. It is basically an electromagnetic switch controlled by a coil. When voltage is applied to the coil, the generated current produces an electromagnetic pulling the internal contacts and activating the switch.
Now, lets review each statement:
Tech A says that when testing a relay winding with an ohmmeter, it should show the winding is open.This is False. The winding is made of a conductive filament, like copper, so testing with an ohmmeter will show a closed circuit, usually with a low resistance value.
Tech B says that the solenoid closes a set of heavy contacts that send current flow to the starter motor.This is True. As the explanation above, the solenoid produces an electromagnetic field that activates the contacts, switching on or off the desired circuit.
Nancy walks 100 m west and then 60 m east. What is Nancy's displacement?
Answer:
40m to the East
Explanation:
Displacement is the distance moved in a specific direction. When writing the displacement value of a moving body, the direction must be put in the description.
Displacement takes into account the start and finish position of a body.
100m
Start -------------------------------------------------------------- →
60m
Final ←----------------------------------
Displacement = 100m - 60m = 40m
Therefore, the displacement is 40m due east
