Answers
Answer:
English please?
Explanation:
Related Questions
a boulder with a mass of about 1.5 x 10^5 kg falls and strikes the ground at 70 m/s how much kinetic energy dies the boulder deliver to the ground PLEASE HELP NEED HELP ASAP
Answers
Explanation:
K.E=1/2mv^2
m=1.5×10^5
v=70
K.E=1/2×1.5×70
K.E=52.5×10^5
Was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s? Why?/Why Not?
(was not able, but I don't know the reason)
Answers
Answer:
It will not hit.
Explanation:
In this problem, we need to tell that was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s.
When it will hit the target, final velocity, v = 0
Using third equation of motion as follows :
[tex]v^2-u^2=2as[/tex]
Here, a =-g
[tex]s=\dfrac{u^2}{2g}\\\\s=\dfrac{(20)^2}{2\times 9.8}\\\\=20.40\ m[/tex]
The target is in a straight line 50m away on the x-axis. As 50m is far from 20.40 m and that’s why it won’t hit.
An object increases its velocity from 22 m/s to 36 m/s in 5 s. What is the acceleration of the
object?
Add process please
Answers
Explanation:
Using Kinematics,
we have a = (v - u) / t.
Therefore a = (36m/s - 22m/s) / 5s = 2.8m/s².
a 150kg roller coaster SITTING ON THE TOP OF A 200M HILL HAS HOW MUCH POTETNTIAL ENERGY
Answers
Answer:
Epot = 294300 [J]
Explanation:
Potential energy is defined as the product of mass by height by gravitational acceleration. The height is measured with respect to the reference level. At this reference level the potential energy is equal to zero.
[tex]E_{pot}=m*g*h\\[/tex]
where:
m = mass = 150 [kg]
g = gravity acceleration [m/s²]
h = elevation = 200 [m]
[tex]E_{pot}=150*9.81*200\\E_{pot}=294300 [J][/tex]
PLEASE HELP
If your mass is 120 Kg how many Newtons do you weigh on the moon? (gravity on moon is 1.63m/s2)
Answers
The correct answer is 195.6 N
Explanation:
Different from the mass (total of matter) the weight is affected by gravity. Due to this, the weight changes according to the location of a body in the universe as gravity is not the same in all planets or celestial bodies. Moreover, this factor is measured in Newtons and it can be calculated using this simple formula W (Weight) = m (mass) x g (force of gravity). Now, leps calculate the weigh of someone whose mass is 120 kg and it is located on the moon:
F = 120 kg x 1.63 m/s2
F= 195.6 N
A box is pulled 6 meters across the ground at a constant velocity by a horizontally applied force of 50 newtons. At the same time, kinetic friction acts on the box as it slides so that it eventually stops when the pulling force stops. What is the magnitude of the frictional force
Answers
Answer:
The magnitude of the frictional force is 50 newtons.
Explanation:
The frictional force can be found as follows:
[tex] \Sigma F = ma [/tex]
[tex] F - F_{\mu} = ma [/tex]
Where:
F: is the applied force = 50 N
[tex]F_{\mu}[/tex]: is the frictional force
m: is the box's mass
a: is the acceleration
Since the box is pulled at a constant velocity, a = 0, so:
[tex] F - F_{\mu} = 0 [/tex]
[tex] F = F_{\mu} = 50 N [/tex]
Therefore, the magnitude of the frictional force is equal to the applied force, that is to say, 50 newtons.
I hope it helps you!
If the Earth were flat, then the shadows of two towers at two different places on the Earth would:__________.
A. be different lengths with the one further South being longer
B. be different lengths with the one further North being longer
C. be different lengths with the one further East being longer
D. be the exact same length
Answers
I believe answer
D.
I’m not exactly sure go ahead and let me know if I can be of any assistance thank u happy new year
What is 6 Fahrenheit in celcious
Answers
Answer:
6 Fahrenheit, converted over to Celsius, would be -14.444444
An elastic conducting material is stretched into a circular loop of 11.2 cm radius. It is placed with its plane perpendicular to a uniform 0.880 T magnetic field. When released, the radius of the loop starts to shrink at an instantaneous rate of 68.8 cm/s. What emf is induced in volts in the loop at that instant?
Answers
Answer:
0.426 volts
Explanation:
It is given that,
The radius of a circular loop, r = 11.2 cm = 0.112 m
An elastic conducting material is stretched into a circular loop.
It is placed with its plane perpendicular to a uniform 0.880 T magnetic field.
The radius of the loop starts to shrink at an instantaneous rate of 68.8 cm/s, dr/dt = 0.688 m/s
We need to find the emf induced in the loop at that instant.
[tex]\epsilon=\dfrac{-d\phi}{dt}\\\\=\dfrac{d}{dt}(BA)\\\\=\dfrac{d}{dt}(\pi r^2 B)\\\\=\pi B\dfrac{d}{dt}(r^2)\\\\=2\pi B r\dfrac{dr}{dt}\\\\=2\pi \times 0.88\times 0.112\times 0.688\\\\=0.426\ V[/tex]
So, the magnitude of induced emf is 0.426 volts.
PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP
The first person with the right answer gets to be a brainlest
In the attachment there is a density column where there is colour
Question: tell me why is the red at the bottom of the density column if it is the least dense
Answers
Answer: It is the color of the container
Explanation:
The red at the bottom is the color of the bottom of the container. It is not part of the experiment and is not liquid.
The red liquid that is part of the experiment does indeed have the lowest density of the liquids which is why it is floating at the top of al the liquids with the only thing floating on top of it being the blue cube which has a lower density than it.
Answer: That is not meant to be red, it‘s the bottom of the beaker. The star is at the very bottom of the beaker. it’s just the base of the beaker.
why are curved roads banked
Answers
Answer:
So that vehicles do not topple or skid off the road.
Explanation:
As a vehicle is negotiating a curve, two equal and opposite forces act on the vehicle; centripetal force which keeps it moving and centrifugal force which tends to throw it out of the path.
In order to avoid skidding off the road curved roads are banked. Banking a curved surface provides the centripetal force that points towards the center of the road hence the vehicle or car does not skid off or topple.
Describe how the force between two charged objects changes as the objects move closer to each other
Answers
Answer:
the electrical force between two charged objects is inversely related to the distance of separation between the two objects. Increasing the separation distance between objects decreases the force of attraction or repulsion between the objects. And decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. Electrical forces are extremely sensitive to distance.
Which of the following objects will have more kinetic energy?
Kinetic Energy =
(Joules)
12 x mass x (velocity)?
(kg) (m/s)
KE = 12 mv
O A 6 kg ball thrown a 8 m/s.
O A 2 kg ball thrown at 15 m/s.
O A 4 kg ball thrown at 10 m/s.
Answers
A bag of shells weighs 1.5 N. What is its mass? approximately 1.5 kilograms approximately 1.5 pounds approximately 150 grams approximately 1.5 grams
Answers
Answer:
approximately 150 grams
Explanation:
The weight of an object is a product of its mass and acceleration of gravity acting upon it. Since weight (W) is force, it is measured in Newtons (N) or Kgms-²
W = mg
Where;
W = weight (N)
m = mass (kg)
g = acceleration due to gravity
g is a constant, which is approximately 10m/s²
Therefore, according to this question, the mass of a bag of shells that weighs 1.5N can be calculated thus:
m = W/g
m = 1.5/10
m = 0.15 kg
Converting 0.15kg to grams, we multiply by 1000 i.e 0.15 × 1000 = 150 grams.
Hence, the mass of the object is approximately 150grams.
20 POINTS
An object released from rest at time t = 0 slides down a frictionless
incline a distance of 2 m during the time interval from t=0 s to t = 1 s.
The distance traveled by the object during the time interval from t = 2s
to t = 3s is: (A) 15 m (B) 10 m (C) 5 m (D) 2 m (E) 1m
Answers
Answer:
10m
Explanation:
let's take the acceleration as a constant throughout the complete motion...
therefore first let's find the acceleration
[tex]s = ut + \frac{1}{2} a {t}^{2} \\ 2 = \frac{1}{2} a {1}^{2} \\ a = 4m {s}^{ - 2} [/tex]
then we have to find v1
apply V = u + at
v = 4×1
= 4ms^-1
lets find the distance travel by the object DURING THE TIME INTERVAL 1-2
[tex]s = ut + \frac{1}{2} a {t}^{2} \\ s = 4 \times 1 + \frac{1}{2} \times 4 \times {1}^{2} \\ s= 6m [/tex]
then let's find the V2
[tex] {v}^{2} = {u}^{2} + 2as \\ {v}^{2} = {4 }^{2} + 2 \times 4 \times 6 \\ {v}^{2} = 64 \\ v = \sqrt{64} = 8m {s}^{ - 1} [/tex]
then let's find the distance travel by the object during time interval 2s to 3s[tex]s = ut + \frac{1}{2} a {t}^{2} \\ s = 8 \times 1 + \frac{1}{2} \times 4 \times {1}^{2} \\ s = 10m [/tex]
The distance traveled by the object during the time interval from t = 2 seconds to t = 3 seconds would be 10 meters, therefore the correct answer is option B.
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
As given in the problem, an object released from rest at time t = 0 slides down a frictionless incline a distance of 2 m during the time interval from t=0 seconds to t = 1 seconds.
2 = ut + 1/2*a*t²
2 = 0 + 0.5×a×1²
a = 2 / 0.5
a = 4 meters / second²
Now to find the velocity after the 2 seconds,
v = u + at
v = 0 + 4×2
v = 8 m/s
Now by using the second equation of motion,
S = ut + 1/2×a×t²
S = 8×1 + 0.5×4×1²
S = 8 + 2
S = 10 meters
Thus, The distance traveled by the object during the time interval from t = 2 seconds to t = 3 seconds would be 10 meters, therefore the correct answer is option B.
To learn more about equations of motion here,
brainly.com/question/5955789
#SPJ5
Two prominent research groups came to the same surprising conclusion after taking measurements of the luminosity of Type Ia supernovae at great distances, this being that the universe is accelerating while it expands.
A. True
B. False
Answers
Answer:
A. True
Explanation:
Type la supernova can be described as a phenomenon whereby two stars are in orbit with one another .
Recently , two prominent research groups came to the same surprising conclusion after taking measurements of the luminosity of Type Ia supernovae at great distances that the universe is accelerating while it expands.
An astronaut weighing 190 lbs on Earth is on a mission to the Moon and Mars.
Required:
a. What would he weigh in newtons when he is on the Moon?
b. How much would he weigh in newtons when he is on Mars, where the acceleration due to gravity is 0.38 times that on Earth?
Answers
Answer:
The weight is defined as:
W = m*g
where:
m = mass
g = gravitational acceleration.
We know that in Earth the astronaut weights 190 lb-f (this is force, not mass, the correct unit here is 190 lb*m/s^2)
then:
190 lb*m/s^2 = m*9.8m/s^2
(190 lb*m/s^2)/(9.8m/s^2) = 19.39 lb
Now we know the mass of the astronaut.
a) wieght on the moon in Newtons.
Newtons uses kilograms as the units of mass, then we need to rewrite the mass of the astronaut in kg.
we know that 1lb = 0.454 kg
Then 19.39 lb is equal to: 19.39*0.454 kg = 8.8 kg
We know that the acceleration due to gravity on the Moon is one-sixth that on Earth.
then: g = (9.8m/s^2)/6
And the weight of the astronaut in the moon will be:
W = 8.8 kg*(9.8m/s^2)/6 = 14.37 N
b) The weight on mars, where the acceleration due to gravity is 0.38 times that on Earth, we have:
g = (9.8m/s^2)*0.38
then the weight will be:
W = 8.8kg*(9.8m/s^2)*0.38 = 32.77 N
Is the range of the projectile dependent or independent of the projectile's mass? Explain.
Answers
For objects near the surface of Earth, the magnitude of gravitational pull is
F = mg
where m is the mass of the object and g is a constant the value of which varies from place to place. Its average value on the surface of the earth is 9.81 metres per second. And this value decreases on moving away from the surface in either directions, inwards or outwards.
Though the magnitude of earth's gravitational pull varies from object to object(depending on its mass), the acceleration produced due to this gravitational pull is constant (of course near earth's surface). So when you drop two objects of different masses from the same height, they reach the ground at the same time (neglect air resistance).
Any projectile motion is just a combination of vertical and horizontal motion. The vertical motion is controlled by the gravity because it acts in vertical direction only, no matter in what direction the projectile is moving. And, the acceleration produced due to gravity is independent of the object's mass. So, the vertical motion is independent of the mass.
The horizontal motion of a projectile is very simple. No horizontal force acts on a projectile. And so its horizontal velocity remains constant. And the range is dependent on time of flight and horizontal velocity. Time of flight being dependent on the vertical motion.
So.. We now know that mass doesn't matter for time of flight or range. And, whenever you come across a problem related to projectile motion, just break the projectile motion into horizontal and vertical motions. And deal with it like you do with motion in one direction. Makes things easier, much easier.
Hope I can be of any assistance happy new year
What acceleration will you give a 22.4 kg box if you push it with a force of 83.1N
Answers
Answer:
mass =22.4kg
force=83.1N
a=?
f=ma
a=f/m
a=83.1/22.4
a=3.70m/s^2
a bird of mass 2350g is flying at a height of 20.0m above the ground with a speed of 10m/s2 calculate it's potential energy
Answers
Explanation:
p.e =mgh
given: m=2350g=2.35kg h=20 g=9.8m/s
p.e=mgh
=2.35kg×20.0m×9.8
=460.6j
I am not sure
The late news reports the story of a shooting in the city. Investigators think that they have recovered the weapon and they run ballistics tests on the pistol at the firing range. If a 0.050-kg bullet were fired from the handgun with a speed of 400 m/s and it traveled 0.080 m into the target before coming to rest, what force did the bullet exert on the target?
Answers
Answer:
50000 N
Explanation:
From the question given above, the following data were obtained:
Mass (m) of bullet = 0.050 kg
velocity (v) = 400 m/s
Distance (s) = 0.080 m
Force (F) =?
Next, we shall determine the acceleration of the bullet. This can be obtained as follow:
Initial velocity (u) = 0 m/s
Final velocity (v) = 400 m/s
Distance (s) = 0.080 m
Acceleration (a) =?
v² = u² + 2as
400² = 0 + (2 × a × 0.08)
160000 = 0 + 0.16a
160000 = 0.16a
Divide both side by 0.16
a = 160000 / 0.16
a = 1×10⁶ m/s²
Finally, we shall determine the force exerted by the bullet on the target. This can be obtained as follow:
Mass (m) of bullet = 0.050 kg
Acceleration (a) of bullet = 1×10⁶ m/s²
Force (F) =?
F = ma
F = 0.050 × 1×10⁶
F = 50000 N
Thus, the bullet exerted a force of 50000 N on the target.
i)x-1 by 4=3
ii)3x+6 by 2=3 by 2
iii)3x-1by5=2x+3by7
Answers
Answer:
i) x = 13
ii) x = -1
iii) x = 2
Explanation:
Given the following expressions, we are to find the corresponding value of x.
1) x-1 by 4=3
x-1/4=3
Cross multiply
x-1 = 4*3
x-1 = 12
x = 12+1
x = 13
2) Given 3x+6 by 2 = 3 by 2
3x+6/2=3/2
Cross multiply
2(3x+6) = 2(3)
2(3x) + 2(6) = 6
6x + 12 = 6
6x = 6-12
6x = -6
x = -6/6
x = -1
3) 3x-1by5=2x+3by7
This can also be written as:
(3x-1)/5 = (2x+3)/7
Cross multiply
7(3x-1) = 5(2x+3)
Open the bracket
7(3x) - 7(1) = 5(2x) + 5(3)
21x - 7 = 10x + 15
Collect like terms
21x - 10x = 15 + 7
11x = 22
x = 22/11
x = 2
The Earth has a diameter of 12,800 km and a globe has a diameter of 35.5 cm. What is the scale of the globe
Answers
Answer:
1:36,056,338
Explanation:
First thing we do is convert the diameter of the earth from kilometers into centimeters. Thus, we have
12800 km = 12800 * 100000 cm
12800 km = 1280000000 cm
Then we have diameter of the globe to be 35.5 cm.
To get the scale, what we do is divide the diameter of the earth by that of the globe, and as such we have
1280000000 / 35.5 =
36056338.
Therefore, the scale of the globe is 1:36,056,338
For every 1 cm on the globe, it is 36,056,338 cm or 360.6 km on earth
The scale of this globe is equal to 1 : 36,056,338.
Given the following data:
Earth diameter = 12,800 kmGlobe diameter = 35.5 cmIn Science, a scale can be defined as a ratio of the distance on a map to the actual (corresponding) distance on planet Earth.
A globe refers to a scale model of planet Earth that accurately depicts various geographic information such as distance, circumference, area, etc.
To determine the scale of this globe:
First of all, we would convert the value of Earth diameter in kilometers to centimeters as follow:
Conversion:
1 kilometer = 100,000 centimeter
12,800 kilometer = [tex]12800 \times 10^5 = 128 \times 10^7\;centimeters[/tex]
Now, we can calculate the scale of the globe by using this formula:
[tex]Distance = \frac{Earth\;diameter}{Globe\;diameter}[/tex]
Substituting the given parameters into the formula, we have;
[tex]Distance = \frac{128 \times 10^7}{35.5}[/tex]
Distance = 36,056,338.03 ≈ 36,056,338 centimeters.
Scale = 1 : 36,056,338.
Read more on globe here: https://brainly.com/question/5659485
Cork has a density of about 0.60 g/cm3 . Cork will partially float in water which has a density of 1.0 g/cm3 . If the piece of cork has a total volume of 5.0 cm3 what volume of cork is below the water
Answers
Answer:
3 cm³
Explanation:
Density of the cork = 0.6 g/cm³
Density of water = 1 g/cm³
Volume of cork = 5 cm³
We all know that the formula for density is given as
Density = mass/volume,
The mass of the cork is
Mass = density * volume
Mass = 0.6 * 5
Mass = 3 gram
Given that the density is 0.6, and it is partially floating, then we can say that the volume of the cork below the water is
5 * 0.6 = 3 cm³
The volume of cork is below the water 3 cm³.
Given data:
The density of Cork is, [tex]\rho = 0.60 \;\rm cm^{3}[/tex].
The density of water is, [tex]\rho_{w} = 1.0 \;\rm g/cm^{3}[/tex].
The total volume of piece of cork is, [tex]V = 5.0 \;\rm cm^{3}[/tex].
In the given problem, we can use the concept of density. The mass of substance occupied per unit volume is known as Density of substance. The expression for the density is given as,
Density = mass/volume,
And, the mass of the cork is
Mass = density * volume
Mass = 0.6 * 5
Mass = 3 gram
Given that the density is 0.6, and it is partially floating, then we can say that the volume of the cork below the water is,
= 5 * 0.6 = 3 cm³
Thus, we can conclude that the volume of cork is below the water 3 cm³.
learn more about the concept of density here:
https://brainly.com/question/21667661
Is Florida an example of an isthmus or a peninsula? Both Peninsula Neither Isthmus
Answers
Answer:
Peninsula
Explanation:
Just did the Ed puzzle and got it right
If a nucleus was as big as a nonpareil, an atom would be ____
Answers
Answer:
small marble
Explanation:
A nonpareil are confectionery tiny ball items that are made up of starch and sugar. It is very small of the size of sugar crystal or sand grains. They were the miniature version of the comfits. They are generally opaque white but bow available in all colors.
In the context, if the nucleus is compared to the size of a nonpareil then its atom would be of the size of small size marble. An atom is bigger in size than that of nucleus as the nucleus is located inside the atoms.
The displacement of a car is a function of time as follows: x(t)=25+3.0t², with x is in meters. Find the average velocity between t1 = 1.0s and t2 = 4.0s.
Answers
Answer: 15m/s
Explanation: Average Velocity is vector describing the total displacement of an object and the time taken to change its position. It is represented as:
[tex]v=\frac{\Delta x}{\Delta t}[/tex]
At t₁ = 1.0s, displacement x₁ is:
[tex]x(1)=25+3(1)^{2}[/tex]
x(1) = 28
At t₂ = 4.0s:
[tex]x(4)=25+3(4)^{2}[/tex]
x(4) = 73
Then, average speed is
[tex]v=\frac{73-28}{4-1}[/tex]
v = 15
The average velocity of a car between t₁ = 1s and t₂ = 4s is 15m/s
Determination of the energy imparted to matter by radiation is the subject of ______. A. photometry B. magnification C. dosimetry D. optometry
Answers
Answer:
dosimetry
Explanation:
dosimetry is the determination of energy imparted to matter by radiation
g a 100 m3 container with 1/3 water with pressure of 100 MPa drops to 90 Mpa how much heat transfer is required to bring back to the initial condition
Answers
Answer:
Q = 3.33 108 J
Explanation:
This is an exercise in thermodynamics, specifically isobaric work
W = V ([tex]P_{f}[/tex]- P₀)
They tell us that we have ⅓ of the volume of the container filled with water
V = ⅓ 100
V = 33.3 m³
let's calculate
W = 33.3 (90-100) 10⁶
W = - 3.33 10⁸ J
To bring the system to its initial condition if we use the first law of thermodynamics
ΔE = Q + W
as we return to the initial condition the change of internal energy (ΔE) is zero
W = -Q
therefore the required heat is
Q = 3.33 108 J
A Light spiral spring is loaded with
a mass of 50g and it extends by
10cm.
(is calculate the period of
small vertical oscillations.
Answers
Answer:
0.63 s
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 50 g
Extention (e) = 10 cm
Period (T) =?
Next, we obtained 50 g to Kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
50 g = 50 g × 1 Kg / 1000 g
50 g = 0.05 kg
Next, we shall convert 10 cm to m. This is illustrated below:
100 cm = 1 m
Therefore,
10 cm = 10 cm × 1 m / 100 cm
10 cm = 0.1 m
Next, we shall determine the force exerted on the spring. This can be obtained as follow:
Mass = 0.05 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = mg
F = 0.05 × 9.8
F = 0.49 N
Next, we shall determine the spring constant of the spring.
Extention (e) = 0.1 m
Force (F) = 0.49 N
Spring constant (K) =?
F = Ke
0.49 = K × 0.1
Divide both side by 0.1
K = 0.49 /0.1
K = 4.9 N/m
Finally, we shall determine the period as follow:
Mass = 0.05 Kg
Spring constant (K) = 4.9 N/m
Pi (π) = 3.14
Period (T) =?
T = 2π√(m/k)
T = 2 × 3.14 × √(0.05 / 4.9)
T = 6.28 × √(0.05 / 4.9)
T = 0.63 s
Thus, the period of oscillation is 0.63 s