Answer:
Electromagnetic waves
Electromagnetic waves bring energy into a system by virtue of their electric and magnetic fields. These fields can exert forces and move charges in the system and, thus, do work on them. However, there is energy in an electromagnetic wave itself, whether it is absorbed or not.
So the answer is B electrical energy
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Tires of a Bigfoot truck has a diameter of 2.2 m. If it rotates 60 revolutions find distance travel on the road.
Answer:
[tex]s = 414.7 m\\[/tex]
Explanation:
The relationship between the linear distance covered by an object and its angular displacement is given by the following formula:
s = rθ
where,
s = distance traveled on road = ?
r radius of tires = diameter/2 = 2.2 m/2 = 1.1 m
θ = angular displacement = (60 rev)(2π rad/1 rev) = 377 rad
Therefore,
[tex]s = (1.1 m)(377 rad)\\s = 414.7 m[/tex]
Arrange an 8-, 12-, and 16-Ω resistor in a combination that has a total resistance of 8.89 Ω pls with de work
Earth receives light and heat energy from the sun. Every now and then Ohio has
a
blue, sunny day. When the sun is shining, Ms. Welch can feel the heat of the sun on
her face.
How does light and heat energy from the sun reach Earth?
By mechanical waves that transfer heat and light energy
By radio waves that transfer heat and light energy
By electromagnetic waves that transfer heat and light energy
By sound waves that transfer heat and light energy
his
Answer: By electromagnetic waves that transfer heat and light energy
Explanation: electromagnetic waves travels through space it cant be radio waves use electromagnetic waves so they can travel through spac too but with the help of electromagnetic waves.
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Light and heat energy from sun reach earth by electromagnetic waves that transfer heat and light energy.
What are electromagnetic waves?
The electromagnetic radiation consists of waves made up of electromagnetic field which are capable of propogating through space and carry the radiant electromagnetic energy.
The radiation are composed of electromagnetic waves which are synchronized oscillations of electric and magnetic fields . They are created due to change which is periodic in electric as well as magnetic fields.
In vacuum ,all the electromagnetic waves travel at the same speed that is with the speed of air.The position of an electromagnetic wave in an electromagnetic spectrum is characterized by it's frequency or wavelength.They are emitted by electrically charged particles which undergo acceleration and subsequently interact with other charged particles.
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Which expression can be used to calculate centripetal acceleration?
4p2r/T2
4p2r/T
2pr/T
(2pr)2/T2
Answer:
What is the magnitude of the centripetal acceleration of a car following a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h)? Compare the acceleration with that due to gravity for this fairly gentle curve taken at highway speed. See Figure 2a.
Strategy
Because v and r are given, the first expression in
a
c
=
v
2
r
;
a
c
=
r
ω
2
a
c
=
v
2
r
;
a
c
=
r
ω
2
is the most convenient to use.
Solution
Entering the given values of v = 25.0 m/s and r=500 m into the first expression for ac gives
a
c
=
v
2
r
=
(
25.0
m/s
)
2
500
m
=
1.25
m/s
2
a
c
=
v
2
r
=
(
25.0
m/s
)
2
500
m
=
1.25
m/s
2
.
Discussion
To compare this with the acceleration due to gravity (g = 9.80 m/s2), we take the ratio of
a
c
g
=
(
1.25
m/s
2
)
(
9.80
m/s
2
)
=
0.128
a
c
g
=
(
1.25
m/s
2
)
(
9.80
m/s
2
)
=
0.128
. Thus, ac=0.128 g and is noticeable especially if you were not wearing a seat belt.
Explanation:
QUESTION If the angular acceleration were doubled for the same duration, by what factor would the angular displacement change
Answer:
By a factor of 2
Explanation:
The angular displacement would change by a factor of 2.
The angular acceleration is represented by the formula
α = Δw / Δt, where
α = angular acceleration
Δw = change in velocity.
Δt = change in time taken
The angular displacement is given by the relation
s = rθ
s = arc length
r = distance
θ = angular displacement
We all know that velocity is the ratio of displacement and time, thus,
v = s/t
Angular acceleration on the other hand says that
α = w/t, substitute w for v, we have
α = s/t ÷ t
α = s/t * 1/t
α = s/t²
We see here that multiplying the acceleration by 2 will only be balanced by increasing the displacement by the same number, 2 in this case
The volume of a cube is given by the formula V = s3, where s is the length of one side. What is the volume of the cube including the planet? First, justify your answer using the Product of Powers Property. Then justify your answer using the Power of a Product Property.
Answer:
1000 cm³
Explanation:
The product of powers property is a rule that helps in simplifying the difficulties that comes with multiplying powers of numbers. It states that when multiplying two powers having the same base, we take one of the base, and then just add the exponents.
This is illustrated in the example.
S¹ * S². The product of powers asks us to pick one of the bases, S, and then add up the powers, thus
S¹ * S² = S^(¹+²)
S¹ * S² = S³
Now, using this in our question, volume of the cube is S³. Since we aren't given a length, I'm going to assume a length.
L = 10 cm. We all know that all sides are equal in a cube,
So, S = 10 cm.
S³ = volume of the cube
V = 10³
V = 1000 cm³
Now, to test the property of products, we say
S = 10, even though it doesn't show any visible exponent, we know that it's raised to the power of 1, and thus
S = 10¹
V = S * S * S
V = 10¹ * 10¹ * 10¹
V = 10^(¹+¹+¹)
V = 10³
V = 1000 cm³
Justified.......
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Use the table below to answer the questions concerning absolute and comparative advantages.
Use the table below to answer the questions concerning absolute and comparative advantages.
Lumber
Automobiles
United States
8
8
Canada
4
2
Which country has an absolute advantage in producing automobiles?
Lumber
Automobiles
United States
8
8
Canada
4
2
2. Which country has an absolute advantage in producing automobiles?
Answer:
United States
Explanation:
The United States and South Korean soccer teams are playing in the first round of the World Cup. An American kicks the ball, giving it an initial velocity of 4.3 m/s. The ball rolls a distance of 5.0 m and is then intercepted by a South Korean player. If the ball accelerates at −0.50 m/s2 while rolling along the grass, find its velocity at the time of interception.
Answer:
Vf = 3.67 [m/s]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f} ^{2} =v_{i} ^{2} -(2*a*x)[/tex]
where:
Vf = final velocity [m/s]
Vi = initial velocity = 4.3 [m/s]
a = acceleration or desacceleration = 0.5 [m/s²]
x = distance = 5 [m]
Note: The negative sign in the above equation means that the velocity of the ball is decreasing (desacceleration).
Now replacing:
Vf² = (4.3)² - (2*0.5*5)
Vf² = 18.49 - 5
Vf² = 13.49
using the square root, we have.
Vf = 3.67 [m/s]
Tell what is an element in your words ?
Please help
Answer: an element is a substance that cannot be broken down. it can’t be chemically decomposed into a different substance. elements are made up of just one atom
Explanation:
A grocery store sells 5-lb bags of mangoes. You purchase four bags over the course
of a month and weigh the mangoes cach time. You obtain the following
measurements:
Week 1 weight: 4.8 lb
Week 2 weight: 5.3 lb
Week 3 weight: 4.9 lb
Week 4 weight: 5.4 lb
is the measurement accurate or precise?
Answer:
Accurate
Explanation:
Accuracy deals with the nearness of the measured values to the true value.
Precision is the ability to reproduce a certain result in a repeated fashion.
From the given masses of the 5-lb bags, we see that the mean value of the measurement is 5.1-lbThis value is very close to the original mass of the bag which is 5-lb bags. We can say the reading is accurateThe reading is not precise because, the same weight is not reproduced. Different values of weight was reported from each of the measurement process.
a baseball was thrown with an initial velocity of 14 m/s at an angle above the horizontal . It remained in the air for 2 s. Which of the following quantities was constant while the baseball was in the air?
Answer:
Explanation:acceleration
If a baseball was thrown with an initial velocity of 14 m/s at an angle above the horizontal. It remained in the air for 2 seconds, then the acceleration of the object would remain constant while the baseball was in the air.
What is a projectile motion?It can be defined as the motion of any object or body when thrown from the earth's surface and follows any curved path under the effect of the gravitational force of the earth.
As given in the problem a baseball was thrown with an initial velocity of 14 m/s at an angle above the horizontal. It remained in the air for 2 seconds.
Thus, If a baseball was thrown with an initial velocity of 14 m/s at an angle above the horizontal. It remained in the air for 2 seconds, then the acceleration of the object would remain constant while the baseball was in the air.
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A bike traveling initially at a speed of 32 m/s accelerates
uniformly at the rate of 3 m/s2 for a distance of 40 meters. The
bike's velocity after covering this distance is _m/s.
We are given:
Initial velocity (u) = 32 m/s
Acceleration (a) = 3 m/s²
Displacement (s) = 40 m
Final Velocity (v) = v m/s
Solving for the Final Velocity:
from the third equation of motion:
v² - u² = 2as
replacing the variables
v² - (32)² = 2(3)(40)
v² = 240 + 1024
v² = 1264
v = √1264
v = 35.5 m/s
Therefore, the velocity of the bike after travelling 40 m is 35.5 m/s
The bike's velocity after covering 40 m distance is 35.55 m/s.
Explanation:
Given:
The initial velocity of the bike = 32 m/s
The rate of acceleration of the bike = 3 ms^2
Distance covered by the bike = 40 m
To find:
The final velocity of the bike.
Solution:
The initial velocity of the bike = u = 32 m/s
The rate of acceleration of the bike = a = 3 ms^2
Distance covered by the bike = s = 40 m
The final velocity of the bike = v
Using the third equation of motion, which is written as :
[tex]v^2-u^2=2as\\v^2-(32m/s)^2=2\times 3m/s^2\times 40m\\v^2-1024 m^2/s^2=240 m^2/s^2\\v^2=240 m^2/s^2+1024 m^2/s^2\\v^2=1264 m^2/s^2\\v=35.55 m/s[/tex]
The bike's velocity after covering 40 m distance is 35.55 m/s.
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If the change in kinetic energy of a tennis ball hit by the racket
is 29), and the average force that the racket exerts on the ball
is 80. N, what is the distance that the force is exerted over?
Answer: .36 m
Explanation:
The distance that the force is exerted over is equal to 0.3625 meter.
Given the following data:
Change in kinetic energy = 29 Joules.Average force = 80 NewtonTo calculate the distance that the force is exerted over, we would apply the law of conservation of energy.
The law of conservation of energy.According to the law of conservation of energy, the work done by an external force equals the change in kinetic energy for the motion of the tennis ball hit by the racket.
Mathematically, this is given by this expression:
[tex]Work\;done = \Delta K.E = Fd[/tex]
Where:
F is the force.d is the distance.Substituting the given parameters into the formula, we have;
[tex]29 = 80 \times d\\\\d=\frac{29}{80}[/tex]
Distance, d = 0.3625 meter.
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A piston above a liquid in a closed container has an area of 0.75m^2, and the piston carries a load of 200kg. What will be the external pressure on the upper surface of the liquid?
Answer:
2613.3 pa
Explanation:
p=F/A
p=ma/A
p=200×9.8/0.75
p=2613.3
A ball is projected horizontally from the top of a 92.0-meter high cliff with an initial speed of 19.8m/s. Determine the horizontal displacement.
Answer:
85.14 m
__________________________________________________________
(y) denotes "in the vertical direction"
(x) denotes "in the horizontal direction"
We are given:
Initial Horizontal velocity of the Ball (u(x)) = 19.8 m/s
Initial height of the ball (s(y)) = 92 m
Initial Vertical velocity of the Ball (u(y)) = 0 m/s
Time taken to reach the ground:
taking downwards direction as positive
Since the horizontal velocity is not opposed by any force, it will be the same until the ball reaches the ground
The vertical velocity will be increasing at a rate of (10 m/s)/s until the ball hits the ground
ay = 10 m/s²
So, while calculating the time. we can just ignore the horizontal velocity
Solving for the time taken:
s(y) = u(y)t + 1/2a(y)t² [second equation of motion]
92 = (0)(t) + 1/2(10)(t)² [replacing the variables]
92 = 5t²
t² = 92/5 [dividing both sides by 5]
t = √18.4 [taking the square root of both sides]
t = 4.3 seconds
So, it took the ball 4.3 seconds to reach the ground
Horizontal Distance travelled by the ball:
We know that the ball will reach the ground in 4.3 seconds
Since the horizontal velocity will not change, the ball will move with a constant velocity of 19.8 m/s in the horizontal direction
Horizontal distance travelled:
s(x) = u(x)t + 1/2a(x)t² [second equation of motion]
s(x) = (19.8)(4.3) + 1/2(0)(t)² [replacing the variables]
s(x) = 85.14 m
Hence, the ball travels 85.14 m horizontally
At what speed do a bicycle and its rider, with a combined mass of 110 kgkg , have the same momentum as a 1400 kgkg car traveling at 6.0 m/sm/s
Answer:
76.36m/sExplanation:
Step one:
given data
let the mass of the rider be m,=110kg
let the velocity be v=?
let the mass of the car be M,=110kg
let velocity of the car V=6m/s
Step two:
the momentum of an object is given by the expression
P=mv
Step two:
since the two the bicycle rider must have the same momentum as the car,
then we need to equate their momentum
mv=MV
110*v=1400*6
110v=8400
divide both sides by 110 to find v
v=8400/110
v=76.36m/s
The rider is expected to have a speed of 76.36m/s
Which chemical equation is balanced?
O Na + O2--> Na20
O 2Na + 202 --> 2Na20
O 2Na2 + 02 --> 2Na20
4Na + 02 --> 2Na20
G
A wagon wheel consists of 8 spokes of uniform diamter, each of mass m, and length L. The outer ring has a mass m rin. What is the moment of inertia of the wheel through an axis through the center and perpendicular to the plane of the ring? Assume that each spoke extends from the center to the outer ring is of negligible thickness.
Answer:
[tex]L^2(\dfrac{8m}{3}+m_r)[/tex]
Explanation:
m = Mass of each rod
L = Length of rod = Radius of ring
[tex]m_r[/tex] = Mass of ring
Moment of inertia of a spoke
[tex]\dfrac{mL^2}{3}[/tex]
For 8 spokes
[tex]8\dfrac{mL^2}{3}[/tex]
Moment of inertia of ring
[tex]m_rL^2[/tex]
Total moment of inertia
[tex]8\dfrac{mL^2}{3}+m_rL^2\\\Rightarrow L^2(\dfrac{8m}{3}+m_r)[/tex]
The moment of inertia of the wheel through an axis through the center and perpendicular to the plane of the ring is [tex]L^2(\dfrac{8m}{3}+m_r)[/tex].
A 310 kg car is parked in a parking garage. If the car has 35,944 J of potential energy, how many
meters above ground is the car?
Gravitational potential energy is the energy stored in an object as the result of its vertical position or height. It can be calculated as its weight times the height. This is:
[tex]\boxed{U_g=W\cdot h=mgh}[/tex]
Where:
W is the weight of the objectm is the massg is gravitational accelerationh is the heightSolving for the h:
[tex]h=\dfrac{U_g}{mg} = \dfrac{35944\;J}{310\;kg\cdot 9.8\;m/s^2} = 11.8\;m[/tex]
R/ The car is parked 11.8 m above ground
Difference between relating velocity and angular velocity?
Answer:
Linear velocity is speed in a straight line (measured in m/s) whileangular velocity is the change in angle over time (measured in rad/s, which can be converted into degrees as well).
Explanation:
hope this helps :)
How many electrons and how many protons
are in an atom of Fluorine?
19
10
9
Explanation:
a fluorine atom has nine protons and electrons so it is electrically neutral
Answer:
no bloody idea
Explanation:
???
An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period of 0.640 s. What is the spring constant of the spring
A box is pressed against a vertical wall by a force F which is directed horizontally. If the magnitude of F is the minimum required to hold the box at rest, what is its value
Answer:
F = mg /μ
Explanation:
For this exercise we define a coordinate system with the horizontal x axis and the vertical y axis, let's write the equilibrium equations for each axis
X axis
F -N = 0
F = N
Y axisy
fr - W = 0
fr = W (2)
the force and touch have the expression
fr = μ N
we substitute
fr = μ F
we substitute in 2
μ F = m g
F = mg /μ
Atoms of elements in the same group of the periodic table share the same
Answer:
Atoms of elements in the same group of the periodic table share the same ELECTRONS
Explanation:
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Aman tosses a dart upward with a velocity of 14.1 m/s at 60° angle.
How much time is it aloft?
What is its max height and range?
Angle of projection of the dart = 60°
initial velocity of the dart = 14.1 m/s
Horizontal and vertical component of the Dart:Using basic trigonometry, we can see that:
Vertical component of the Dart = 14.1*Sin 60° = 12.2 m/s
Horizontal component of the Dart = 14.1*Cos 60° = 7.05 m/s
Time the Dart is Aloft:
Explaining the concept:
Once the dart is in mid-air, there will be no force that will reduce its horizontal velocity
So, the dart will be moving at a constant horizontal velocity of 7.05 m/s
but the force of gravity will be applied on the dart while mid-air and will pull it downwards with an acceleration of 9.8 m/s²
Hence, the time the dart will be aloft is the same as the time taken by the dart to reach the ground if we threw it vertically upwards at a velocity of 12.2 m/s
Solving for the time taken for the dart to reach the ground:
initial velocity = 12.2 m/s
velocity at max height = 0 m/s
acceleration = -9.8 m/s
time taken to reach max height:
v = u + at using the first equation of motion
replacing the variables
0 = 12.2 + (-9.8)t
t = -12.2 / -9.8
t = 1.24 seconds
Time taken to reach the ground:
time taken to reach the ground = 2 * time taken to reach max height
(Since the dart will take the same time to come back down)
Time taken to reach the ground = 2 * 1.24
Time taken to reach the ground = 2.48 seconds
Therefore, the dart will hit the ground and stop 2.48 seconds after throwing
Max height of the Dart:
Explaining the concept:
From newton's second equation of motion
s = ut + 1/2 at²
Since we need the max vertical height, we need the vertical component of the values,
so this equation can be rewritten for vertical height as :
s(vertical) = u(vertical)* t + 1/2 * a(vertical) * t²
from the last section, we know that the dart reaches its maximum height at t = 1.24 seconds
replacing the values in this equation:
s(vertical) = (12.2 * 1.24) + 1/2 * (-9.8) * (1.24)²
s(vertical) = 15.128 + (-7.5)
s(vertical) = 7.63 m
Therefore, the maximum height of the dart is 7.63 m
The range of the Dart:
Explaining the concept:
The range of the dart is the horizontal distance covered by the dart
from the second section, we know that the dart travels horizontally with a constant velocity of 7.05 m/s
from the third section, we know that the total time taken by the dart to hit the ground is 2.48 seconds
Since the horizontal velocity of the dart is constant, we can say that it moved horizontally for 2.48 seconds at a constant velocity of 7.05 m/s
Solving for the range:
s = ut + 1/2 at² From the second equation of motion
this equation can be rewritten for horizontal distance as:
s(horizontal) = u(horizontal)* t + 1/2 (a)(t)²
The dart is moving at a constant velocity, that means that its acceleration is 0
s(horizontal) = u(horizontal)* t + 1/2 (0)(t)²
s(horizontal) = u(horizontal)* t
replacing the variables
s(horizontal) = 7.05 * 2.48 [here, 2.48 is the time taken by the dart to reach the ground and stop]
s(horizontal) = 17.484 m
The horizontal distance covered by the dart is 17.484 m
Therefore, the range of the dart is 17.484 m
Answer:
time = 2.49 s
max height = 7.61 m
range = 17.57 m
See below for exact values.
Explanation:
Vertical/Horizontal Components:Since we are given the angle at which the object is tossed, this object is in projectile motion.
We are given the initial velocity of the dart, and we can use the angle to solve for its horizontal and vertical components.
Horizontal component:
[tex](v_i)_x =v_i \cdot cos(\theta)[/tex] [tex](v_i)_x=14.1\cdot cos(60)[/tex]Vertical component:
[tex](v_i)_y =v_i \cdot sin(\theta)[/tex] [tex](v_i)_y=14.1 \cdot sin(60)[/tex] Time In The Air:In order to find the time that the dart stays in the air, we can use the constant acceleration equation that does not include displacement. This equation is:
[tex]v_f=v_i+at[/tex]Since we solve for time using the vertical motion of the projectile, we will use this equation in terms of the y-direction.
[tex](v_f)_y = (v_i)_y+a_yt[/tex]We don't know what the final velocity of the dart when it reaches the ground is, but we do know that its final velocity when it reaches its maximum height is 0 m/s.
Therefore, we can solve for half of the full time that it takes the object to reach the ground, and double this value at the end.
Let's set the downwards direction to be negative and the upwards direction to be positive.
Using our knowledge and previous calculations, we know that:
[tex](v_f)_y = 0[/tex] [tex](v_i)_y=14.1\cdot sin(60)[/tex] [tex]a_y=-9.8[/tex]The acceleration of gravity is in the y-direction and facing downwards, so that's why it is -9.8 m/s².
Substitute these values into the constant acceleration equation.
[tex]0=14.1\cdot sin(60) + (-9.8)t[/tex]Subtract 14.1 * sin(60) from both sides of the equation.
[tex]-14.1\cdot sin(60)=-9.8t[/tex]Divide both sides of the equation by -9.8 to solve for t.
[tex]\displaystyle{\frac{-14.1\cdot sin(60)}{-9.8} =t}[/tex] [tex]t=1.246016142[/tex]Now, this is only the time for half of the object's trajectory. Double it to find the total time the dart is aloft:
[tex]2t = 2.492032284[/tex]The dart is in the air for a total of 2.49 seconds.
Maximum Height:We can find the maximum height of the dart by using another constant acceleration equation.
Since we don't have the final velocity of the object, we can use this equation:
[tex]x_f=x_i+v_it+\frac{1}{2}at^2[/tex]Subtract [tex]x_i[/tex] from both sides to get the change in position, or delta x.
[tex]\triangle x=v_it+\frac{1}{2}at^2[/tex]In order to find the maximum height, we need to use this equation in terms of the y-direction.
[tex]\triangle x_y=(v_i)_yt+\frac{1}{2}a_yt^2[/tex]Remember that time is the same regardless of the x- or y- direction.
Now, we can solve for the displacement in the y-direction by plugging in the values that we know.
[tex](v_i)_y=14.1\cdot sin(60)[/tex] [tex]t=1.246016142[/tex] [tex]a_y=-9.8[/tex]Note that we are using half of the time t, since this is where the maximum height occurs.
Plug these known values into the constant acceleration equation:
[tex]\triangle x_y=[14.1 \cdot sin(60)] (1.246016142) + \frac{1}{2}(-9.8)(1.246016142)^2[/tex] [tex]\triangle x_y=(15.21505102) + (-4.9)(1.552556226)[/tex] [tex]\triangle x_y=15.21505102-7.607525508[/tex] [tex]\triangle x_y=7.60752551[/tex]The maximum height of the dart is 7.61 meters.
Range:Finding the range of the object in projectile motion involves the same constant acceleration equation, but this time we are solving for the displacement in the horizontal direction (x-direction). Therefore:
[tex]\triangle x_x=(v_i)_xt+\frac{1}{2}a_xt^2[/tex]We know the horizontal component of the initial velocity vector, which is what we will be using.
The acceleration, of an object in projectile motion, in the x-direction is always 0 m/s².
We will use the full time of the object since we want to find the entire horizontal distance that the object covers. We have:
[tex](v_i)_x=14.1\cdot cos(60)[/tex] [tex]t=2.492032284[/tex] [tex]a_x=0[/tex]Plug these values into the constant acceleration equation.
[tex]\triangle x=[14.1\cdot cos(60)](2.492032284)+\frac{1}{2} (0)(2.492032284)^2[/tex] [tex]\triangle x=[14.1\cdot cos(60)](2.492032284)[/tex] [tex]\triangle x=17.5688276[/tex]The range of the dart is 17.57 meters.
1. A ball is launched horizontally at 40 m/s while 6.0 m above a level field . How far will the ball move horizontally before striking the ground ?
Answer:
44 m
Explanation:
Given that,
Horizontal velocity of the ball, u = 40 m/s
It is 6 m above the level field.
We need to find the distance covered by the ball when move horizontally before striking the ground. Let it is d.
Firstly, we will find time taken for the ball to hit the ground. Using second equation of motion as follows :
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
Put u = 0 and a = g
[tex]s=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2s}{g}} \\\\t=\sqrt{\dfrac{2\times 6}{9.8}} \\\\t=1.1\ s[/tex]
No finding the horizontal distance as follows :
d = ut
d = 40 m/s × 1.1 s
d = 44 m
So, the ball will move 44 m horizontally before striking the ground.
What will happen to the speed of the Greenhouse effect if all the methane trapped in the permafrost is released into the
earth's atmosphere?
Answer:
The atmosphere would become warmer
Explanation:
methane absorbs the suns heat, therefore warming the atmosphere. if all the methane from the premafrost melted, would would have a huge large scale global warming, the giant increase of temperature could have devistating effects on all life on our planet.
A certain frictionless simple pendulum having a length L and mass M swings with period T. If both L and M are doubled, what is the new period?
a. T
b. 2T
c. √2T
d. T/2
e. T/4
Answer:
c. √2T
Explanation:
The period of a simple pendulum is given by;
[tex]T = 2\pi \sqrt{\frac{L}{g} } \\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g}} \\\\\frac{T^2}{4\pi^2} = \frac{L}{g}\\\\\frac{g}{4\pi^2} = \frac{L}{T^2}\\\\ \frac{L_1}{T_1^2}= \frac{L_2}{T_2^2}\\\\T_2^2 = \frac{L_2T_1^2}{L_1}\\\\L_2 = 2L_1\\\\ T_2^2 = \frac{2L_1T_1^2}{L_1}\\\\ T_2^2 =2T_1^2\\\\T_2 = \sqrt{2T_1^2}\\\\T_2 = T_1\sqrt{2}[/tex]
Thus, the the new period will be √2T
The heating element on an electric iron has a resistance of 24 ohm and draws a current of
5.0 A. How much work is done if the iron is used for 45 minutes?
Answer:
1620000J
Explanation:
Given parameters:
Resistance = 24ohm
Current = 5A
Time = 45min
Unknown:
Work done = ?
Solution:
Electrical work done is given as the product of power with time;
Electrical work done = power x time
Power = I²R
Electrical work done = I²R x time
Convert the given time to seconds;
1min = 60s
45 min = 45 x 60 = 2700s
Now,
Work done = 5² x 24 x 2700 = 1620000J
What is the momentum of a 533 kg blimp moving east at +75 m/s
Answer:
39975kgm/s due east
Explanation:
Given parameters:
Mass of the blimp = 533kg
Velocity = +75m/s due east
Unknown:
Momentum of the body = ?
Solution:
The momentum of a body is the amount of motion it posses.
Momentum is the product of mass and velocity;
Momentum = mass x velocity
Insert the parameters and solve;
Momentum = 533 x 75 = 39975kgm/s
The momentum of the body is 39975kgm/s due east
The momentum of a 533 kg blimp moving east at +75 m/s is 39,975kgm/s.
MOMENTUM:Momentum of a body can be calculated by multiplying the mass of the substance by its velocity. That is;
Momentum (p) = mass (m) × velocity (v)
According to this question, a 533 kg blimp is said to be moving east at +75 m/s. The momentum is calculated thus:
Momentum = 533 × 75
Momentum = 39,975kgm/s.
Therefore, momentum of a 533 kg blimp moving east at +75 m/s is 39,975kgm/s
Learn more about momentum at: https://brainly.com/question/19636349?referrer=searchResults