Answer:
The answer is 2 cm³Explanation:
The volume of a substance when given the density and mass can be found by using the formula
[tex]volume = \frac{mass}{density} \\ [/tex]
From the question we have
[tex]volume = \frac{100}{50} \\ [/tex]
We have the final answer as
2 cm³Hope this helps you
What is an independent variable?
a 2 kg ball traveling at 5m/s collides with a 1 kg ball at rest. After the collision the 1 kg ball moves off with a speed of 7 m/s. Find the final speed of the 2 kg ball.
An image of the Earth-moon-sun system is shown.The moon remains in orbit around Earth because of the force of —
When a light ray traveling in a higher index of refraction material passes into a lower index of refraction material, the light ray Group of answer choices Travels in a straight line without changing direction Bends toward the axis perpendicular to the surface between the materials Bends away from the axis perpendicular to the surface between the materials
The light ray ; ( C ) Bends away from the axis perpendicular to the surface between the materials
Snell's lawSnell's law states that the ratio of the sines of the angles of incidence is equal to the ratio of the refractive indexes of the materials surface through which the light rays pass through. therefore
As the light ray travels from a material with a higher index of refraction into a material with a lower index of refraction at the axis that is perpendicular to the surface which is in between the materials the light ray will bend away due to Snell's law .
Hence we can conclude that The light ray Bends away from the axis perpendicular to the surface between the materials.
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Please I need help with this :(
three charged particals are located at the corners of an equil triangle shown in the figure showing let (q 2.20 Uc) and L 0.650
Kepler's work revealed that the Earth was at the center of the circular orbits of the planets at the center of the elliptical orbits of the planets orbiting the Sun in an elliptical orbit O at one focus of the elliptical orbit of the planets
Answer:
orbiting the Sun in an elliptical orbit
Explanation:
When orbits are circular, the center and the focus are the same point. For an elliptical orbit, the orbited object is at one focus of the ellipse. Kepler found planetary orbits to be elliptical with the sun at one focus.
__
As with a lot of scientific theories, it only explained some of the motion. As with a lot of scientific theories, explaining the discrepancies resulted in new discoveries.
Answer:
at one focus of the elliptical orbit of the planets.
During which time Interval does the object travel approximately 10 meters?
OA. O seconds to 3 seconds
OB. 3 seconds to 5 seconds
OC. 5 seconds to 7 seconds
OD. 7 seconds to 8 seconds
OE. 8 seconds to 10 seconds
Answer:
[tex]OA. \: O \: seconds \: to \: 3 \: seconds[/tex]
A man slides on snow without friction starting at 8.96m/s at the top of an inclined plane with height 8.21m. What is his speed at the bottom of a plane?
Answer:
V2 = 15.53 [m/s]]
Explanation:
In order to solve this problem we must use the principle of energy conservation, where potential energy is transformed into kinetic energy. At the bottom is taken as a reference level of potential energy, where the value of this energy is equal to zero.
Above the inclined plane we have two energies, kinetics and potential. While when the sled is at the reference level all this energy will have been transformed into kinetic energy.
[tex]E_{1}=E_{2}\\ m*g*h+(\frac{1}{2} )*m*v_{1} ^{2}=\frac{1}{2}*m*v_{2} ^{2} \\(9.81*8.21)+(0.5*8.96^{2} )=(0.5*v_{2}^{2} )\\(0.5*v_{2}^{2} )=120.68\\v_{2} ^{2}=241.36\\v_{2} =\sqrt{241.36}\\ v_{2} =15.53[m/s][/tex]
Describe the buoyant force and explain how
it relates to Archimedes principle.
Answer:
Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.
Which information did the Glomar Challenger study in 1968?
the rate of seafloor spreading
the direction of seafloor spreading
the age of rocks in various places in the ocean
the contents of rocks in various places in the ocean
Explanation:
Glomar Challenger studies about the "age of rocks in various places in the ocean" in 1968. EXPLANATION: Glomar Challenger was a "deep sea research vessel" for marine geology and oceanography studies.
I hope this helps you :)
Answer:
c no cappp :)
Explanation:
I need to know what the answer is to this
Answer:
i think its the top one
Explanation:
pls tell me if im wrong
48. Final Velocity Your sister drops your house keys down
to you from 4.3 m What is the velocity of the keys when you
catch them?
Answer:
There is a difference between the words 'drops' and 'throws'
If an object is dropped, it's initial velocity is 0 but if the same object is thrown, it has some initial velocity
Since 'the sister' DROPS the keys, the initial velocity will be 0 m/s
We are given:
u=0 m/s
s = 4.3 m
a = 9.8 m/s/s (gravity)
From the third equation of motion:
v² - u² = 2as
Replacing the known values
v² - (0)² = 2(9.8)(4.3)
v² = 84.28
v = 9.2 m/s (approx)
Hence, the final velocity of the keys is 9.2 m/s
A campus shuttle bus makes 1 revolution per second round a circular track of radius 100 cm. Determine its periodic time.
The periodic time of the campus shuttle is = 1 sec
What is periodic time ?The periodic time is also known as revolution per second is the time it takes object in motion ( revolving ) to pass through a point twice ( usually the starting position ) of the object and it calculated as :
For a simple pendulum = 2π√L/g
But a campus shuttle bus in motion
since it takes
1 revolution = 1 sec therefore the time period is = 1 sec
Hence we can conclude that the periodic time of the campus shuttle = 1 second.
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At low pressures and high temperatures, the density of a gas
Answer:
Higher denisty
Explanation:
High pressure=high denisty
is water wet? If water is not wet does that make it dry?
Answer:
Water isn't wet by itself, but it makes other materials wet when it sticks to the surface of them.
Explanation:
Answer:
water is wet
it is a liquid
Explanation:
A body, with a volume of 2 m3, weighs 40 kN. Determine its weight when
submerged in a liquid with SG = 1.59.
Answer:
8.8 kN
Explanation:
V = 2 m³, W = 40 kN, SG = 1.59
Bouyant force N = 1.59 * 1000 kg/m³ * 9.81 N/kg * 2 m³ = 31.2 kN
So the weight becomes 40 - 31.2 = 8.8 kN
The knot at the junction is in equilibrium under the influence of four forces acting on it. The F force acts from above on the left at an angle of α with the horizontal. The 5.7 N force acts from above on the right at an angle of 50◦ with the horizontal. The 6.2 N force acts from below on the right at an angle of 44◦ with the horizontal. The 6.7 N force acts from below on the left at an angle of 43◦ with the horizontal.
1. What is the magnitude of the force F?
2. What is the angle a of the force F in the figure above?
The knot is in equilbrium, so there is no net force acting on it. Starting with the unknown force and going clockwise, denote each force by F₁, F₂, F₃, and F₄, respectively. We have
F₁ + F₂ + F₃ + F₄ = 0
Decomposing each force into horizontal and vertical components, we have
F cos(180º - α) + (5.7 N) cos(50º) + (6.2 N) cos(-44º) + (6.7 N) cos(-137º) = 0
F sin(180º - α) + (5.7 N) sin(50º) + (6.2 N) sin(-44º) + (6.7 N) sin(-137º) = 0
Recall that cos(180º - x) = - cos(x) and sin(180º - x) = sin(x), so these equations reduce to
F cos(α) ≈ - 3.22 N
F sin(α) ≈ 4.51 N
(1) Recall that for all x, sin²(x) + cos²(x) = 1. Use this identity to solve for F :
(F cos(α))² + (F sin(α))² = F ² ≈ 30.73 N² → F ≈ 5.5 N
(2) Use the definition of tangent to solve for α :
tan(α) = sin(α) / cos(α) ≈ 1.399 → α ≈ 126º
or about 54º from the horizontal from above on the left of the knot.
(a) The magnitude of the force F acting on the knot is 5.54 N.
(b) The angle α of the force F is 54.4⁰.
The given parameters:
F force at α 5.7 N force at 50⁰6.2 N force at 44⁰6.7 N force at 43⁰The net vertical force on the knot is calculated as follows;
[tex]F_y = Fsin(\alpha) + 5.7 sin(50) - 6.2 sin(44) - 6.7 sin(43)\\\\F_y = F sin(\alpha) -4.51\\\\Fsin(\alpha) = 4.51[/tex]
The net horizontal force on the knot is calculated as follows;
[tex]F_x = -F cos(\alpha) + 5.7 cos(50) + 6.2cos(44) - 6.7cos(43)\\\\F_x = -Fcos(\alpha) + 3.22\\\\Fcos(\alpha) = 3.22[/tex]
From the trig identity;
[tex]sin^2 \theta + cos^ 2 \theta = 1\\\\[/tex]
[tex](Fsin(\alpha))^2 + (Fcos(\alpha))^2 = (4.51)^2 + (3.22)^2\\\\F^2(sin^ 2\alpha + cos^2 \alpha) = 30.71\\\\F^2(1) = 30.71\\\\F = \sqrt{30.71} \\\\F = 5.54 \ N[/tex]
The angle α of the force F is calculated as follows;
[tex]Fsin(\alpha) = 4.51\\\\sin(\alpha) = \frac{4.51}{F} \\\\sin(\alpha ) = \frac{4.51}{5.54} \\\\sin(\alpha ) = 0.814\\\\\alpha = sin^{-1}(0.814)\\\\\alpha = 54.5 \ ^0[/tex]
Find the image uploaded for the complete question.
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Help real quick someone
Calculate the wave speed (in m/s) for the following waves:
a) A sound wave in steel with a frequency of 500 Hz and a wavelength of 3.0 meters. (2pts)
b) a ripple on a pond with a frequency of 2 Hz and a wavelength of 0.4 meters. (2pts)
Calculate the wavelength (in meters) for the following waves:
A wave on a slinky spring with a frequency of 2 Hz travelling at 3 m/s. (2pts)
An ultrasound wave with a frequency 40,000 Hz travelling at 1450 m/s in fatty tissue. (2pts)
Calculate the frequency (in Hz) for the following waves:
A wave on the sea with a speed of 8 m/s and a wavelength of 20 meters. (2pts)
A microwave of wavelength 0.15 meters travelling through space at 300,000,000 m/s. (2pts)
Answer: A : 250 is the answer
B; The frequency of a wave is the number of complete oscillations (cycles) made by the wave in one second.
Instead, the wavelength is the distance between two consecutive crests (highest position) or 2 troughs (lowest position) of the wave.
In this problem, we are told that the leaf does two full up and down bobs: this means that it completes 2 full cycles in one second. Therefore, its frequency is
where is called Hertz (Hz). So, the correct answer is
Explanation:
#Wavespeed
#1
[tex]\\ \rm\Rrightarrow v=\nu\lambda=500(3)=1500m/s[/tex]
#2
[tex]\\ \rm\Rrightarrow v=2(0.4)=0.8m/s[/tex]
#Wavelength
#1
[tex]\\ \rm\Rrightarrow \lambda=\dfrac{v}{\nu}=\dfrac{3}{2}=1.5m[/tex]
#2
[tex]\\ \rm\Rrightarrow \lambda= \dfrac{1450}{40000}=0.03625m[/tex]
#Frequency
[tex]\\ \rm\Rrightarrow \nu=\dfrac{v}{\lambda}=\dfrac{8}{20}=0.4Hz[/tex]
#2
[tex]\\ \rm\Rrightarrow \nu=\dfrac{3\times 10^8}{15\times 10^{-2}}=0.2\timee 10^{10}=2\times 10^9Hz[/tex]
The magnitude of the vertical velocity vector for an upwardly launched projectile _________. a stays constant b gets smaller and then larger c decreases throughout the flight d increases throughout the flight
Answer:
changes by 9.8 m/s each second.
what is acceleration due to gravity?
The acceleration due to gravity is 9.8 m/s^2.
When putting the ball on the tee you want half of the golf ball to _________.
Question 1 options:
be above the club
be below the club
be in front of the club
be behind the club
Answer:
be in front of the club
Explanation:
The ball should be highest off the ground for a driver. The general recommendation is that the bottom of the golf ball on a tee should be level with the top of the driver; for long and mid-irons, push the tee into the ground so that only about a quarter-inch is above ground.
You measured the length, diameter and mass of two different cylinders. In both cases, you found that the length had 3 significant figures and that length was the measurement with the fewest number of significant digits. If you found the weight densities to be 38119 N/m^3 and 38081 N/m^3 and you round these values to the correct number of significant figures, can you conclude the two cylinders are made of the same material (do they have the same weight density)?
a. Not enough information given.
b. Yes.
c. No.
Answer:
Weight Density 1 = 38100 N/m³
Weight Density 2 = 38100 N/m³
b. Yes
Explanation:
The formula for volume of cylinder is:
V = πr²l
where,
V = Volume
r = radius
l = length of cylinder
So, if length has the 3 significant figures which is least in all values, Then the volume must also be in 3 significant figures. The formula for weight density is:
Weight Density = Weight/Volume
Here, the volume has the least significant figures of 3, therefore, the weight densities must also have 3 significant figures:
Weight Density 1 = 38119 N/m³
Weight Density 1 = 38120 N/m³
Weight Density 1 = 38100 N/m³
Weight Density 2 = 38119 N/m³
Weight Density 2 = 38120 N/m³
Weight Density 2 = 38100 N/m³
Hence, the answer is:
b. Yes
what is the order of magnitude of the final velocity of an object that begins from rest and accelerates at a rate of 20 meters per second2 for 5.0 seconds
The order of magnitude of the final velocity of an object that begins from rest and accelerates at a rate of 20m/s² for 5.0 seconds is 100m/s
The acceleration of a body is the change in velocity with respect to time as shown:
[tex]a=\frac{v-u}{t}[/tex]
Given the following parameters
a is the acceleration = 20m/s²
u is the initial velocity = 0m/s
t is the time taken = 5.0seconds
Required
Final velocity "v"
Substitute the given parameters into the formula:
[tex]20=\frac{v-0}{5} \\20 =\frac{v}{5}\\v = 20 \times 5\\v =100m/s[/tex]
Hence the order of magnitude of the final velocity of an object that begins from rest and accelerates at a rate of 20m/s² for 5.0 seconds is 100m/s
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A boy of mass 40kg eats bananas contains of 980 joule. If this energy is used to lift him up from ground,the height to which he can climb is
Answer:
h = 2.49 [m]
Explanation:
In order to solve this problem we must use the definition of potential energy, which tells us that energy is equal to the product of mass by gravity by height.
The potential energy can be calculated by means of this equation:
Ep = m*g*h
where:
Ep = potential energy = 980 [J]
m = mass = 40 [kg]
g = gravity acceleration = 9.81 [m/s^2]
h = elevation [m]
Now replacing:
980 = 40*9.81*h
h = 2.49 [m]
Harry is pushing a car down a level road at 2.0 m/s with a force of 243 N. The
total force acting on the car in the opposite direction, including road friction and
air resistance, is which of the following?
a. Slightly more than 243 N.
b. Exactly equal to 243 N.
c. Slightly less than 243 N.
Answer:
C, slightly less than 243 N
Explanation:
Road friction and air resistance aren't that much on a force. Try pushing something and see how much friction there is. Not that much.
a car moving at a speed shows that the force applied to the car is greater than the frictional force and air resistance
c. Slightly less than 243 N.
A, B, or C for this question?
Answer:
A. right before it hits the ground.
Explanation:
because all the way down, it is building kinetic energy.
Can someone help pleaseeee
Answer:
Motion with constant velocity of magnitude 1 m/s (uniform motion) for 4 seconds in a positive direction and then for 2 seconds uniform motion with constant velocity of magnitude 3 m/s in reverse direction .
Explanation:
The graph shows a constant velocity of 1 m/s for 4 seconds in the positive direction. After that, between 4 seconds and 6 seconds, the object reverses its motion with constant velocity of magnitude 3m/s.
how does gravity change as it nears an object
2. Calculate the work done by a 47 N force pushing a 0.025 kg pencil 0.25 m against a force of 23 N.
Answer: 6 J
Explanation:
Total force applied = 47 N Assuming that direction of movement of pencil and applied force is same. Work done by force in moving the pencil W = Force × Distance through which force moves ⇒ W = 47 × 0.25 = 11.75 J .-.-.-.-.-.-.-.-. In case we are asked useful work done then we calculate net force used for pushing the pencil: Net force used for pushing the pencil = 47 − 23 = 24 N Assuming that direction of movement of pencil and net force is same. Useful Work done by force in moving the pencil W u = Force × Distance ⇒ W u = 24 × 0.25 = 6 J
The work done the pushing force is required.
The work done by the pushing force is [tex]6\ \text{J}[/tex]
[tex]F_1[/tex] = Pushing force = 47 N
[tex]F_2[/tex] = Opposing force = 23 N
m = Mass of object = 0.025 kg
s = Displacement = 0.25 m
Work done is given by
[tex]W=F_ns[/tex]
[tex]\Rightarrow W=(47-23)\times 0.25[/tex]
[tex]\Rightarrow W=6\ \text{J}[/tex]
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