Answer:
1.1 x 10⁻⁵m³
Explanation:
Given parameters:
Weight in air = 11.29N
Weight in water = 9.190N
Density of gold = 19320kg/m³
Unknown:
Volume of the secret cavity = ?
Solution:
To solve this problem, we need to find the mass of the secret cavity.
Weight of the cavity = Weight of sculpture in air - Weight of sculpture in water
Weight of cavity = 11.29N - 9.190N = 2.1N
Weight = mass x acceleration due to gravity
Mass =[tex]\frac{2.1}{9.8}[/tex] = 0.2kg
So;
Volume of substance can be determined from the density.
Density = [tex]\frac{mass}{volume}[/tex]
Volume = [tex]\frac{mass}{density}[/tex]
Now insert the parameters and solve;
Volume = [tex]\frac{0.2}{19320}[/tex] = 1.1 x 10⁻⁵m³
A small truck has a mass of 2085 kg. How much work is required to decrease the speed of the vehicle from 22.0 m/s to 13.0 m/s on a level road?
Answer:
Work required is 328387.5 Joules.
Explanation:
Given the following data;
Mass = 2085kg
Initial velocity, Vi = 13m/s
Final velocity, Vf =22m/s
To find the workdone;
We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.
Mathematically, it is given by the equation;
W = Kf - Ki
Where;
W is the work required.
Kf is the final kinetic energy possessed by the object.
Ki is the initial kinetic energy possessed by the object.
But Kinetic energy = ½MV²
W = ½MVf² - ½MVi²
Substituting into the equation, we have;
W = ½(2085)*22² - ½(2085)*13²
Simplifying the equation, we have;
W = 1042.5 * 484 - 1042.5 * 169
W = 504570 - 176182.5
W = 328387.5J
Therefore, the work required to decrease the speed of the vehicle is 328387.5 Joules.
NEED A PHYSICS EXPERT TO HELP ME PLS
I've been stuck on this for two hours:
Water flows steadily from an open tank as shown in the figure. (Figure 1)The elevation of point 1 is 10.0 m , and the elevation of points 2 and 3 is 2.00 m . The cross-sectional area at point 2 is 4.80×10−2 m2 ; at point 3, where the water is discharged, it is 1.60×10−2 m2 . The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe.
Assuming that Bernoulli's equation applies, compute the volume of water ΔV that flows across the exit of the pipe in 1.00 s . In other words, find the discharge rate ΔV/Δt.
Express your answer numerically in cubic meters per second.
The discharge rate of water (Q) that flows across the exit of the pipe in 1.00 seconds is 0.200 [tex]m^3/s[/tex].
Given the following data:
Elevation of point 1 = 10.0 mElevation of point 2 = 2.00 mElevation of point 3 = 2.00 mCross-sectional area at point 2 = [tex]4.80\times 10^{-2} \;m^2[/tex]Cross-sectional area at point 3 = [tex]1.60\times 10^{-2} \;m^2[/tex]To find the discharge rate of water (Q) that flows across the exit of the pipe in 1.00 seconds assuming that Bernoulli's equation applies:
In this exercise, we would apply Torricelli's Law, which states that the speed of the efflux in an apparatus, under the force of gravity is directly proportional to the square root of the height of the fluid above it.
Mathematically, Torricelli's Law is given by the formula:
[tex]V =\sqrt{2gh}[/tex]
Where:
g is the acceleration due to gravity.h is the height of the apparatus.V is the speed of liquid.Substituting the given parameters into the formula, we have;
[tex]V =\sqrt{2 \times 9.8 (10-2)}\\\\V =\sqrt{19.6 (8)}[/tex]
Speed, V = 12.52 m/s
Now, we can find the discharge rate of the pipe:
[tex]Q = V \times A\\\\Q = 12.52 \times 1.60\times 10^{-2}[/tex]
Discharge rate, Q = 0.200 [tex]m^3/s[/tex]
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I need help lol .........
Answer:
a
Explanation:
as an object falls:
a) both velocity and acceleration increase
b) velocity increases and acceleration decreases
c) velocity increases and acceleration is unchanged
Answer:
c it is not accelerating on it's on but gravity pulls it there for velocity increases.
As an object falls, velocity increases and acceleration is unchanged. Therefore, option C is correct.
What do you mean by velocity ?The term velocity is defined as the direction of the movement of the body or the object.
Speed is fundamentally a scalar quantity. Velocity is, in essence, a vector quantity. It is the rate at which distance changes. It is the displacement rate of change.
The velocity of something is the rate at which it moves in a specific direction. For example, the speed of a car driving north on a highway or the speed of a rocket after launch. The scalar denotes that the absolute magnitude of the velocity vector is always the speed of motion.
Thus, it is not accelerating on it's on, but gravity pulls it there as velocity increases.
To learn more about velocity, follow the link:
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#SPJ2
how much heat is necessary to warm 500g of water from 20°c to 65°c
Explanation:
so sorry
don't know but please mark me as brainliest please
1. Synthesize Information You push your
younger sister on a swing in a park. Then you
give her a harder push. Explain what happens
in each case, in terms of the second and third
laws of motion
Answer:separate
Explanation:
feather is dropped from 13 m high after it has fallen 3 m what is its acceleration
Answer:
Explanation:
The constant acceleration of any object, neglecting air resistance and friction, is always -9.8. Proof:
[tex]s(t)=-4.9t^2+13[/tex] which is the position of this feather before it is dropped. The first derivative of position is velocity, so the velocity function for the feather is:
[tex]v(t)=-9.8t[/tex]. We could find the velocity that the feather is experiencing at 3 seconds, but that is not what we are being asked. What we are being asked is the acceleration of the feather at 3 seconds. So we find the acceleration function of the feather:
a(t) = -9.8
It turns out that the time doesn't matter to acceleration due to gravity because feathers and elephants all fall under the same pull.
Aria is riding her bike. She is bent slightly over, holding onto her handlebars. She wants to go faster. Which best explains what she should do? She should bend her hips so her head is closer to the handlebars. She should sit straight up on the bike, keeping her arms tight to her body. She should spread her arms wide while sitting straight in her seat. She should flap her arms up and down like a bird.
Answer:
She should bend her hips so her head is closer to the handlebars
Explanation:
Answer:
Yea its A
Explanation:
A car travels at a constant speed up a ramp making an angle of 28 degrees with the horizontal component of velocity is 40 kmh^-1, find the vertical and horizontal component of velocity.
Answer:
Vx = 35.31 [km/h]
Vy = 18.77 [km/h]
Explanation:
In order to solve this problem, we must decompose the velocity component by means of the angle of 28° using the cosine function of the angle.
[tex]v_{x} = 40*cos(28)\\V_{x} = 35.31 [km/h][/tex]
In order to find the vertical component, we must use the sine function of the angle.
[tex]V_{y}=40*sin(28)\\V_{y} = 18.77 [km/h][/tex]
Consider a trip to a lake. The sand warms up rapidly (and then burns your feet), while the water remains cool.
Which has the greater specific heat capacity, the sand or water?
List four safety measures one should take when there is a cyclone warning.
Answer:
I'm gonna give you more than 4 So u can choose.
Turn off all electricity, gas and water.
Unplug all electrical appliances.
Keep your emergency kit close at hand.
If the building starts to collapse, protect yourself with mattresses, rugs.
Stay tuned to the radio for updates.
Stay inside until told it is safe to go outside.
A periodic wave travels from one medium to another. Which pair of variables are likely to change in the process?
A.
velocity and wavelength
OB.
velocity and frequency
Ο Ο Ο Ο
OC.
frequency and wavelength
OD.
frequency and phase
E.
wavelength and phase
Answer:
Velocity and frequency
Explanation:
Which best describes her speed and velocity? (There are 60 seconds in 1 minute.) Her speed is 4.4 m/s, and her velocity is 0 m/s. Her speed is 1.1 m/s, and her velocity is 0 m/s. Her speed is 0 m/s, and her velocity is 2400 m/s. Her speed is 4.4 m/s, and her velocity is 4.4 m/s.
Answer:
A
Explanation:
Yw :)
Answer:
a
Explanation:
An object is pulled with a 85 N force inclined at 27° along a horizontal surface ABC
where AB is smooth but BC is rough.
a) Compute the mass of the object if the object moves with an acceleration of 5
ms-1 along AB.
Answer:
m = 15.15 kg
Explanation:
Newton's Second Law of motion states that when an unbalanced force is applied on a body, an acceleration is produced in it in the direction of force. The component of force along the horizontal direction here, will be given by the Newton's Second Law as:
Fx = ma
F Cosθ = ma
where,
F = Magnitude of Force = 85 N
θ = Angle with horizontal = 27°
m = mass of object = ?
a = acceleration of object = 5 m/s²
Therefore,
85 N Cos 27° = m(5 m/s²)
m = 75.73 N/5 m/s²
m = 15.15 kg
I WILL MARK YOU AS BRAINLIEST IF RIGHT The fastest pitch ever recorded in MLB was thrown by Aroldis Chapman at 5 points 105.1 mph . If the 0.145 kg baseball accelerated 96 m/s / s , how much force did Chapman apply to the ball? Round your answer to the nearest whole number
Answer: The force was 13.92 Newtons.
Explanation:
First, let's recall the second Newton's law:
The net force is equal to the mass times the acceleration, or:
F = m*a
where:
F = force
m = mass
a = acceleration.
When the player hits the ball with the bat, he applies a force that accelerates the ball for a small period of time, that increases greatly the speed of the ball.
In this case, we know that:
the mass of the ball is 0.145 kg
The acceleration of the ball is 96m/s^2
Then we can input those values in the above equation to find the force.
F = 0.145kg*96m/s^2 = 13.92 N
The force was 13.92 Newtons.
A barge is pulled by two tugboats. If the resultant of the forces exerted by the
tugboats is 5kN force directed along the axis of the barge, determine the tension in
each of the ropes for α = 45
Answer:
Approximately [tex]3.5\; \rm kN[/tex] in each of the two ropes.
Explanation:
Let [tex]F_1[/tex] and [tex]F_2[/tex] denote the tension in each of the two ropes.
Refer to the diagram attached. The tension force in each rope may be decomposed in two directions that are normal to one another.
The first direction is parallel to resultant force on the barge.
The component of [tex]F_1[/tex] in that direction would be [tex]\displaystyle F_1\cdot \cos(45^\circ) = \frac{F_1}{\sqrt{2}}[/tex].Similarly, the component of [tex]F_2[/tex] in that direction would be [tex]\displaystyle F_2\cdot \cos(45^\circ) = \frac{F_2}{\sqrt{2}}[/tex].These two components would be in the same direction. The resultant force in that direction would be the sum of these two components: [tex]\displaystyle \frac{F_1 + F_2}{\sqrt{2}}[/tex]. That force should be equal to [tex]5\; \rm kN[/tex].
The second direction is perpendicular to the resultant force on the barge.
The component of [tex]F_1[/tex] in that direction would be [tex]\displaystyle F_1\cdot \sin(45^\circ) = \frac{F_1}{\sqrt{2}}[/tex].Similarly, the component of [tex]F_2[/tex] in that direction would be [tex]\displaystyle F_2\cdot \sin(45^\circ) = \frac{F_2}{\sqrt{2}}[/tex].These two components would be in opposite directions. The resultant force in that direction would be the difference of these two components: [tex]\displaystyle \frac{F_1 - F_2}{\sqrt{2}}[/tex]. However, the net force on the barge is normal to this direction. Therefore, the resultant force should be equal to zero.
That gives a system of two equations and two unknowns:
[tex]\displaystyle \frac{F_1 + F_2}{\sqrt{2}} = 5\; \rm kN[/tex], and[tex]\displaystyle \frac{F_1 - F_2}{\sqrt{2}} = 0[/tex].The second equation suggests that [tex]F_1 = F_2[/tex]. Hence, replace the [tex]F_2[/tex] in the first equation with [tex]F_1[/tex] and solve for [tex]F_1\![/tex]:
[tex]F_1 = \displaystyle \frac{5\; \rm kN}{2\, \sqrt{2}} \approx 3.5\; \rm kN[/tex].
Because [tex]F_1 = F_2[/tex] (as seen in the second equation,) [tex]F_2 = F_1 \approx 3.5\; \rm kN[/tex].
In other words, the tension in each of the two ropes is approximately [tex]3.5\; \rm kN[/tex].
The tension in each of the rope is 3,535.5 N
The given expression:
the resultant force, R = 5 kN = 5000 N
the angle in between the forces, α = 45
To find:
the tension in each of the ropesThe tension in each of the ropes is calculated as follows;
The tension in vertical direction
[tex]F_y = F \times sin(\alpha)\\\\F_y = 5000 \times sin(45)\\\\F_y = 5000 \times 0.7071\\\\F_y = 3,535.5 \ N[/tex]
The tension in horizontal direction;
[tex]F_x = F \times cos(\alpha)\\\\F_x = 5000 \times cos(45)\\\\F_x = 5000 \times 0.7071\\\\F_x = 3,535.5 \ N[/tex]
Thus, the tension in each of the rope is 3,535.5 N
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Superman strikes a golf ball on the ground at a 38 degree angle above the horizontal at 147 m/s. What is the maximum height the golf ball reaches?
255 m
418 m
687 m
1103 m
The ball is hit with an initial vertical velocity of
(147 m/s) sin(38º) ≈ 90.5 m/s
Recall that
v ² - u ² = 2 a ∆y
where u and v are initial and final velocities, respectively; a is acceleration; and ∆y is displacement.
Vertically, the ball is in freefall, so it is only subject to acceleration due to gravity, with magnitude g = 9.80 m/s² in the downward direction. At its maximum height, the ball has zero vertical velocity (v = 0) and the displacement is equal to the maximum height, so
0² - (90.5 m/s)² = 2 (-g) ∆y
∆y = (90.5 m/s)² / (2g)
∆y ≈ 418 m
A piano string having a mass per unit length equal to 4.70 10-3 kg/m is under a tension of 1 400 N. Find the speed with which a wave travels on this string.
Answer:
The speed of the sound wave on the string is 545.78 m/s.
Explanation:
Given;
mass per unit length of the string, μ = 4.7 x 10⁻³ kg/m
tension of the string, T = 1400 N
The speed of the sound wave on the string is given by;
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
where;
v is the speed of the sound wave on the string
Substitute the given values and solve for speed,v,
[tex]v = \sqrt{\frac{T}{\mu} }\\\\v = \sqrt{\frac{1400}{4.7*10^{-3}} }\\\\v = \sqrt{297872.34}\\\\v = 545.78 \ m/s[/tex]
Therefore, the speed of the sound wave on the string is 545.78 m/s.
A ball rotates on a string as shown below.
If the string in the image above breaks at point 1, which of the following images best shows the direction the ball will travel?
Question 8 options:
Diagram A
Diagram B
Diagram C
Diagram D
Diagram E
Answer:
Diagram A
Explanation:
When an object is travelling in a circular path it has an acceleration called centripetal acceleration, even if it is moving with uniform speed. This acceleration is attributed to the object due to change in the direction of velocity at every instant during the motion.
The direction of the velocity of the object is tangent to the circular path at every instant and if the object beaks free of the circular motion, such as breaking of the string, then it will go in the direction tangent to the circle at that instant. Hence, the diagram that best shows the direction ball will travel is:
Diagram A
Sometimes in the winter, ice crystals grow directly from the moisture in the air. As the water changes from a gas to a solid, the molecules
Group of answer choices
slow down as they gain energy.
slow down as they lose energy.
speed up as they lose energy.
speed up as they gain energy.
Answer: B. slow down as they lose energy
What power is supplied to the starter motor of a large truck that draws 290 A of current from a 20.4 V battery hookup?
Answer:
Power = 5.916 Kw
Explanation:
Given:
Current = 290 A
Volt = 20.4 V
Find:
Power
Computation:
Power = IV
Power = (290)(20.4)
Power = 5916 w
Power = 5.916 Kw
А A pool of water of refractive index
4/3! is 60cm deep. Find its apparent
depth when viewed vertically through
air.
Answer:
Apparent depth = 45 cm
Explanation:
The refractive index of water in a pool, n = 4/3
Real depth, d = 60 cm
We need to find its apparent depth when viewed vertically through air. The ratio of real depth to the apparent depth is equal to the refractive index of the material. Let the apparent depth is d'. So,
[tex]n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{60}{\dfrac{4}{3}}\\\\d'=45\ cm[/tex]
So, the apparent depth is 45 cm.
can someone help im not sure of my answer
Answer:
Yes
Explanation:
But I feel also 2 is correct but your answer is right
An electron (m = 9.1 × 10−31 kg, q = −1.6 × 10−19 C) starts from rest at point A and has a speed of 5.0 × 106 m/s at point B. Only electric forces act on it during this motion. Determine the electric potential difference VA − VB.a) −71 Vb) +71 Vc) −26 Vd) +26 Ve) −140 V
Answer:
-71 V
Explanation:
Given that,
An electron starts from rest at point A and has a speed of 5.0 × 10⁶ m/s at point B.
We need to find the electric potential difference [tex]V_A-V_B[/tex]. It can be calculated by using the conservation of charges as follows :
[tex]qV=\dfrac{1}{2}mv^2[/tex]
m and q are mass and charge on electrons
[tex]V=\dfrac{mv^2}{2(-q)}\\\\V=-\dfrac{9.1\times 10^{-31}\times (5\times 10^6)^2}{2\times 1.6\times 10^{-19}}\\\\V=-71.09\ V[/tex]
So, the electric potential difference is (-71 V). So, the correct option is (a).
Lauren walks 100m in half a minute. What must her speed have been to travel this distance?
Answer:
3.33 m/s
Explanation:
Using the formula: s = d / t (where s is speed, d is distance, and t is time)
[Convert minutes to seconds then solve]
Half a minute is 30 seconds. Therefore:
s = 100 / 30 = 3.33 m/s