What percentage of an iron anchor’s weight will be supported by buoyant force when submerged in salt water?

Answer:

divide the distance by the time and get average velocity in units of s. the detection to the left

Answer:

t = 0.029 s

Explanation:

We have,

Speed of sound is 343 m/s.

Dezeirey is positioned 5 m behind her.

It is required to find the time taken for the echo from the wall to reach her. The total distance covered by the echo when it reaches her is 2d or 10 m.

Time taken,

\(t=\dfrac{d}{v}\\\\t=\dfrac{10\ m}{343\ m/s}\\\\t=0.029\ s\)

So, it will take 0.029 seconds for the echo from the wall to reach her.

The absolute pressure at an ocean depth of 1.0 x 10^3 m is 1.002 x 10^8 Pa.

Hydrostatic pressure is the pressure that a fluid exerts on a surface due to the weight of the fluid above it. It is the result of the force of gravity acting on a column of fluid, and it is directly proportional to the height of the column of fluid and the density of the fluid.

The absolute pressure at an ocean depth of 1.0 x 10^3 m can be calculated using the hydrostatic pressure equation:

P = ρgh + Po

where:

P is the absolute pressure at the given depth

ρ is the density of the water

g is the acceleration due to gravity (assumed to be 9.81 m/s²)

h is the depth of the ocean

Po is the atmospheric pressure at the surface (assumed to be 1.01 x 10^5 Pa)

Substituting the given values, we get:

P = (1.025 x 10^3 kg/m³) x (9.81 m/s²) x (1.0 x 10^3 m) + 1.01 x 10^5 Pa

P = 1.025 x 9.81 x 10^6 Pa + 1.01 x 10^5 Pa

P = 1.002 x 10^8 Pa.

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The earthquake would occur 13 minutes before 10:05 a.m. which will be at 9.52 am.

The p-waves travel with a constant velocity of 7 km/s

The time can be calculated by using the formula

t = d / v

where

T1 =  10:05 a.m

d is the distance they take to travel from the epicenter

v is the speed of the p-waves

On average, the speed of p-waves is

v = 7 km/s

d = 5600 km (given)

Substituting the values in the formula;

t = d / v

t = 5600 ÷ 7

t = 800 seconds

Converting into minutes,

t = 800 ÷ 60

t = 13.3

≈ 13 mins

T1 -  13 mins = T2

10:05 - 13 mins = 9.52 am

It means the earthquake occurred prior 13 minutes, that is at 9.52 am.

Therefore, the earthquake occurred at 9.52 am.

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Answer:

whatttttttttttttttttttttttttttt are the items we have to classify into simple machines and compound machine

The moment of inertia of a rigid body with a cylindrical cross-section is (m*\(r^2\))/2.5, where m is mass and r is radius.

Given data:

Inclination angle, θ = 26 degrees

Distance traveled, s = 10.9 m

Time taken, t = 3.06 s

The acceleration of the body can be found by analyzing its motion down the incline.

Since the body is rolling, we need to consider both translational and rotational motion.

The net force acting on the body is the component of its weight parallel to the incline. Therefore:

Fnet = mgsinθ

The force causes both translational and rotational acceleration. The translational acceleration can be found using Newton's second law:

Fnet = ma

Therefore,

a = Fnet/m = gsinθ

The rotational acceleration can be found using the torque equation:

τ = Iα

where τ is the torque, I is the moment of inertia, and α is the angular acceleration. The torque is provided by the force of friction acting on the body, and it is given by:

τ = Fr

where r is the radius of the body.

The force of friction can be found using the relationship between the translational and rotational acceleration:

a = αr

Therefore

Fr = ma = mgsinθ

Substituting α = a/r and τ = Fr, we get:

I(a/r) = mgsinθ

Substituting a = gsinθ, we get:

I = (m*\(r^2\))/2.5

Therefore, the moment of inertia of the body in terms of its mass m and radius r is I = (m*\(r^2\))/2.5.

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No, the potential of a non-uniformly charged sphere cannot be the same as that of a point charge.

The potential of a charged sphere depends on the distribution of charge throughout the sphere, whereas the potential of a point charge depends only on its own charge and the distance from it. A non-uniformly charged sphere has different charges distributed at different distances from a point in space, so the potential at that point will vary depending on the distribution.

In contrast, a point charge has all its charge concentrated at a single point, so the potential at any given distance will be the same regardless of the distribution of charge in the surrounding space. Therefore, the potential of a non-uniformly charged sphere cannot be the same as that of a point charge.

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Answer:

\(3 \times 10 {}^6\)

Gravitational Potential Energy , Kinetic Energy and velocity of panther just before hitting the ground will be 790.41599 J , 990.41599 J , 5.6523m/s respectively.

Mass is 62 kg with a height 1.3m .

Gravitational Potential Energy means product of mass , gravitational acceleration and height . Mathematically, GPE= mgh

GPE= 62 x 1.3 x 9.8

= 790.41599 J

Energy lost = Final energy- GPE

Final Energy Δ =  790.41599 J + 200 J

=990.41599 J

To find Velocity of panther just before hitting the ground V:-

Δ⇒mV²/2

990.41599 J= 62 V²/2

V²= {990.41599 x 2}/62

= 31.9489029

V=5.6523m/s

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The acceleration with which the car uniformly increases its speed from zero to the given speed is 5m/s².

Given the data in the question

Since the car was initially at rest

Acceleration of the car; \(a = \ ?\)

Using the First Equation of Motion:

\(v = u + at\)

Where u is the initial velocity, v is the final velocity, a is the acceleration and t is the time taken.

We substitute our given values into the equation and solve for "a"

\(10m/s = 0m/s\ +\ ( a * 2s ) \\\\10m/s = a * 2s\\\\a = \frac{10m/s}{2s}\\\\a = 5m/s^2\)

Therefore, the acceleration with which the car uniformly increases its speed from zero to the given speed is 5m/s².

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Answer:

Explanation:

The mass of the car can be found by using the formula

\(m = \frac{f}{a} \\ \)

f is the force

a is the acceleration

From the question we have

\(m = \frac{11968}{8.8} \\ \)

We have the final answer as

Hope this helps you

Answer:

\(\displaystyle 1360\:kg.\)

Explanation:

\(\displaystyle am = F_{net} \\ \\ \frac{11968}{8,8} = \frac{8,8m}{8,8} \\ \\ \boxed{1360 = m}\)

I am joyous to assist you at any time.

The question is incomplete. The complete question is :

High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 60 m/s just before it strikes a  50 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40 m/s. Find the speed of the golf ball just after impact.

Solution :

We know that momentum = mass x velocity

The momentum of the golf club before impact = 0.200 x 60

                                                                             = 12 kg m/s

The momentum of the ball before impact is zero. So the total momentum before he impact is 12 kg m/s. Therefore, due to the conservation of momentum of the two bodies after the impact is 12 kg m/s.

Now the momentum of the club after the impact is = 0.2 x 40

                                                                                    = 8 kg m/s

Therefore the momentum of the ball is = 12 - 8

                                                                = 4 kg m/s

We know momentum of the ball, p = mass x velocity

                                                     4 = 0.050 x velocity

∴ Velocity =  \($\frac{4}{0.050}$\)

                 = 80 m/s

Hence the speed of the golf ball after the impact is 80 m/s.

The potential energy of the roller coaster is 176,400 J (joules).

The potential energy of an object is given by the formula PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical position of the object.

In this case, the roller coaster is at a peak of 20m and has a mass of 900kg. The acceleration due to gravity, g, is approximately 9.8 \(m/s^2\).

Using the formula, we can calculate the potential energy:

PE = mgh

= (900 kg)(9.8 \(m/s^2\))(20 m)

= 176,400 J

Therefore, the potential energy of the roller coaster is 176,400 J (joules).

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Answer:

add molecules

Answer:

It's important

Explanation:

Mental wellness is critical to overall physical health. Participating in recreational activities helps manage stress. Taking time to nurture oneself provides a sense of balance and self-esteem, which can directly reduce anxiety and depression.

Answer:

Mental wellness is critical to overall physical health. Participating in recreational activities helps manage stress. Taking time to nurture oneself provides a sense of balance and self-esteem, which can directly reduce anxiety and depression.

Explanation:

Answer:

Almost all machines require energy to offset the effects of gravity, friction, and air/wind resistance. Thus, no machine can continually operate at 100 percent efficiency.

Answer:

2.29 s

Explanation:

This problem is a description of an elastic collision (the two objects collide and effectively become one object).  The equation for an elastic collision is

\(m_1v_1+m_2v_2=m_3v_3\), where:

\(m_1v_1\) = the mass of the child (55.0 kg) times the velocity of the child (2.5\(\frac{m}{s}\))

\(m_2v_2\) =  the mass of the sled (12.0 kg) times the velocity of the sled (0.0\(\frac{m}{s}\))

\(m_3v_3\) = the combined mass of the child and the sled (67.0 kg) times the combined velocity of the child and the sled (\(v_3\))

First, rearrange the problem to solve for \(v_3\):

\(\frac{m_1v_1+m_2v_2}{m_3}=v_3\)

So,

\(\frac{(55.0\ kg)(2.5\ \frac{m}{s})+(12.0\ kg)(0.0\ \frac{m}{s})}{67.0\ kg}=v_3\\\frac{137.5\frac{kg*m}{s}}{67.0\ kg}=v_3\\2.05\frac{m}{s}=v_3\)

The force of friction for this problem is given as 60 N.  To stop the child and the sled, the force of friction must be equal and opposite to the force of the child and sled.  According to Newton's first law, force equals mass times acceleration (F=ma).  So, an equation to solve this portion of the problem can be given as -60 N=ma, where m=67.0 kg.  So,

\(-60\ N=(67.0\ kg)a\\\frac{-60\ \frac{kg*m}{s^2}}{67.0\ kg}=a\\-0.90\frac{m}{s^2}\)

Acceleration can then be used in kinematic equation #1 (\(v_f=v_i+at\)) to solve for time.  Rearrange the equation and let

\(v_f=0.0\frac{m}{s}\\v_i=2.05\frac{m}{s}\\a=-0.90\frac{m}{s^2}\)

So,

\(\frac{v_f-v_i}{a}=t\\\frac{0.0\frac{m}{s}-2.05\frac{m}{s}}{-0.90\frac{m}{s^2}}=t\\\frac{-2.05\frac{m}{s}}{-0.90\frac{m}{s^2}}=t\\2.29\ s=t\)

The factor that leads to loess deposit is when the wind carries fine sediment. That is option C.

The loess deposits are those deposits that are usually found at the edge of deserts.

The major factor that causes the formation of loess is the wind because they are entrained, transported, and deposited by the wind.

The fine particles carried by wind contains find grained sediments, organic particles that are capable of forming loess.

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Answer:

\(g=274\ m/s^2\)

Explanation:

Mass of the Sun, \(M=1.99\times 10^{30}\ kg\)

The radius of the Sun, \(r=6.96\times 10^8\ m\)

We need to find the acceleration due to gravity on the surface of the Sun. It is given by the formula as follows :

\(g=\dfrac{GM}{r^2}\\\\g=\dfrac{6.67\times 10^{-11}\times 1.99\times 10^{30}}{(6.96\times 10^8)^2}\\\\g=274\ m/s^2\)

So, the value of acceleration due to gravity on the Sun is \(274\ m/s^2\).

The acceleration due to gravity, in meters per second squared, on the surface of the Sun is \(296.88 \;m/s^2\).

Given the following data:

Gravitational constant = \(6.67 \times 10^{-11}\)

To calculate the acceleration due to gravity, in meters per second squared, on the surface of the Sun:

From the law of gravitational force, we have the formula:

\(g = \frac{Gm}{r^2}\)

Where:

Substituting the given parameters into the formula, we have;

\(g = \frac{6.67 \times 10^{-11} \times 1.99 \times 10^{30}}{(6.69 \times 10^8)^2} \\\\g= \frac{1.33 \times 10^{20} }{4.48 \times 10^{17}} \\\\g=296.88 \;m/s^2\)

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He was solely consuming high-fat foods until recently, when he began consuming nutritious foods.

Briefing:

Instead of helping you lose weight, hopping from one diet fad to another can make you put on weight. Your daily calorie intake can fluctuate significantly, which can cause your metabolism to slow down. To adjust to new food, your metabolism needs about three weeks.

Obesity is typically brought on by excessive eating and insufficient exercise. A large portion of the excess energy will be stored by the body as fat if you consume large amounts of energy, especially fat and sugars, without expelling it through exercise and physical activity.

Even if the food is good for you, the calories can still add up and result in a weight loss plateau or even weight gain.

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The answers are 1) The value of R2 is not relevant as it implies a downward force on the plank, 2) The reactions at C and D are 66.3 N and 90 N, respectively, and 3) The vertical force at B to lift the plank clear of D is 686.4 N. The reaction at C is zero, and the reaction at D is 61.4 kg.

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Answer:

D

Explanation:

trust me bro

Answer:

An object with no net forces acting on it which is initially at rest will remain at rest. If it is moving, it will continue to move in a straight line with constant velocity. Forces are "pushes" or "pulls" on the object, and forces, like velocity and acceleration are vector quantities.

Answer:

b gases

Explanation:

this is because gases spread out through out the entire container because they have no definite shape and are always moving.

Explanation:

You want  N/m

 N = 66 * 9.81

 m = 2.3 x 10^-2 m

66* 9.81 / 2.3 x 10^-2  = 28150 =  28 000 N/m    to two S D

The airplane will strike the ground at a horizontal distance of 490 meters.

To determine how far the airplane will strike on the ground, we need to consider the horizontal distance traveled by the airplane during its flight.

The horizontal distance traveled by an object can be calculated using the formula:

Distance = Speed × Time

In this case, the speed of the airplane is given as 100 meters per second and the time it takes to cover the distance of 490 meters is unknown. Let's denote the time as t.

Distance = 100 m/s × t

Now, to find the value of time, we can rearrange the equation as follows:

t = Distance / Speed

t = 490 m / 100 m/s

t = 4.9 seconds

Therefore, it takes the airplane 4.9 seconds to cover a horizontal distance of 490 meters.

Now, to calculate the distance on the ground where the airplane will strike, we can use the formula:

Distance = Speed × Time

Distance = 100 m/s × 4.9 s

Distance = 490 meters

It's important to note that this calculation assumes a constant speed and a straight flight path. In reality, various factors such as wind conditions, changes in speed, and maneuvering can affect the actual distance traveled by the airplane.

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Answer:

Have the same speed.

Explanation:

All electromagnetic waves travel at the same speed, which is 300,000,000 m/s.

Answer:

192.6N

Explanation:

Let's consider the forces acting on the beam:

Weight of the beam (W): It acts vertically downward and has a magnitude of W = mass * gravitational acceleration = 39.4 kg * 9.8 m/s^2.

Force exerted by the link on the beam (F_link): It acts at an angle of 90 degrees with respect to the beam and has two components: the vertical component and the horizontal component.

Tension in the cable (T): It supports the far end of the beam and acts at an angle of 90 degrees with respect to the beam. Since the angle between the beam and the cable is 90 degrees, the tension in the cable only has a vertical component.

Let's break down the forces acting on the beam:

Vertical forces:

W (weight of the beam) - T (vertical component of tension) = 0

T = W

Horizontal forces:

F_link (horizontal component of the force exerted by the link) = ?

To find the magnitude of the horizontal component of the force exerted by the link on the beam (F_link), we need to consider the equilibrium of forces in the horizontal direction.

Since the beam is inclined at an angle of θ = 33.1 degrees with respect to the horizontal, the horizontal equilibrium equation can be written as:

F_link = W * sin(θ)

Let's substitute the given values:

W = 39.4 kg * 9.8 m/s^2

θ = 33.1 degrees

F_link ≈ (39.4 kg * 9.8 m/s^2) * sin(33.1 degrees)

Using a calculator, we find that the magnitude of the horizontal component of the force exerted by the link on the beam (F_link) is approximately 192.6 N.