Answer: The force does not change.
Explanation:
The force between two charges q₁ and q₂ is:
F = k*(q₁*q₂)/r^2
where:
k is a constant.
r is the distance between the charges.
Now, if we increase the charge of each particle two times, then the new charges will be: 2*q₁ and 2*q₂.
If we also increase the distance between the charges two times, the new distance will be 2*r
Then the new force between them is:
F = k*(2*q₁*2*q₂)/(2*r)^2 = k*(4*q₁*q₂)/(4*r^2) = (4/4)*k*(q₁*q₂)/r^2 = k*(q₁*q₂)/r^2
This is exactly the same as we had at the beginning, then we can conclude that if we increase each of the charges two times and the distance between the charges two times, the force between the charges does not change.
Padded dashboards in cars are safer in an accident than nonpadded ones because an occupant hitting the dash has:________.
(a) increased time of impact.
(b) decreased impulse.
(c) decreased impact force.
(d) a and b above.
(e) a and c above.
Answer:
(e) a and c above.
Explanation:
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
[tex] Momentum = Mass * Velocity [/tex]
The law of conservation of momentum states that the total linear momentum of any closed system would always remain constant with respect to time.
Padded dashboards in cars are safer in an accident than non-padded ones because an occupant hitting the dashboard of an automobile car has an increased time of impact and a decreased impact force because the force or shock experienced is high and happens rapidly over a short period of time, thus, the occupant has less time and velocity while absorbing the momentum of the car in the course of the collision.
which hand has a negatively charged?
Answer:
The dryer sheet is negatively charged and your hand is positively charged
Explanation:
PLEASE ANSWER-Why are loose electrons needed for heat conduction?
Answer:
Why do metals conduct heat so well? The electrons in metal are delocalised electrons and are free moving electrons so when they gain energy (heat) they vibrate more quickly and can move around, this means that they can pass on the energy more quickly.
Explain how air pollution impacts each of Earth's systems.
1) Atmosphere
2) Lithosphere
3) Biosphere
4) Hydrosphere
Answer:
1) Atmosphere
Explanation: because if you think about it the only real explaination is atmosphere.
Logs sometimes float vertically in a lake because one end has become water-logged and denser than the other. What is the average density (in kg/m3) of a uniform-diameter log that floats with 60.0% of its length above water
Answer: the average density (in kg/m3) of the log is 400 kg/m³
Explanation:
Given that;
a log with uniform-diameter floats with 60.0% of its length above water.
from the Archimedes principle formula;
friction submerged = Average density of object / density of fluid. --- equ 1
We know that; density of water = 1000 kg/m³
now the log floats with 60.0% of its length above water
so submerged fraction is 100% - 60% = 40% = 0.40
from our equation;
friction submerged = Average density of object / density of fluid
we substitute
0.40 = Average density of object / 1000 kg/m³
Average density of object = 0.40 × 1000
Average density of object = 400 kg/m³
Therefore; the average density (in kg/m3) of the log is 400 kg/m³
You have to deliver some 5.0-kg packages from your home to two locations. You drive for 2.0 h at 30 mi/h due east (call this segment 1 of your trip), then turn around and drive due west for 30 min at 28 mi/h (segment 2). Use a coordinate system with the positive x axis aimed toward the east and the origin at your home.
A. What is your position vector at the instant you reach the end of segment 1?
B. What is your position vector at the instant you reach the end of segment 2?
C. Calculate your displacement during segment 2.
Answer:
A. 60 mi.
B. 46 mi.
C. -14 mi.
Explanation:
A)
Assuming that we use a coordinate system with the positive x-axis aimed toward the east, and the origin at the start point, we can find the position vector at the end of the segment 1, applying the definition of average velocity, as follows:[tex]v_{avg1} = \frac{x_{1f} -x_{1o}}{t} (1)[/tex]
where x₁f = final position at the end of segment 1, x₁₀ = initial position at the start of the segment 1 = 0, t = time traveled during segment 1 = 2.0 h, and v₁, average velocity during segment 1 = 30 mi/h due east.Replacing by the givens, and solving for x₁f, we get:[tex]x_{1f} = v_{1avg} * t =30 mi/h * 2.0 h = 60 mi due east. (2)[/tex]
B)
In order to find the position at the end of the segment 2, we can use the same equation (1), but taking into account that the initial position will not be zero, but the final position at the end of the segment 1, i.e., 60 mi due east.Replacing by the givens, and solving for x₂f, we get:[tex]x_{2f} = x_{1f} + v_{2avg} * t = (-28 mi/h) * 0.5 h = 60 mi - 14 mi = 46 mi (3)[/tex]
C)
The displacement during the segment two, is simply the difference between the final and initial positions for this segment.Since x₂₀= x₁f = 60 mi, and x₂f = 46 mi, we find that the displacement is as follows:Δx = x₂f - x₂₀ = 46 mi - 60 mi = - 14 mi.In an experiment, a cement block of mass 14 kg was placed on top of 8 (carefully arranged) eggs without breaking the eggs.
1) Calculate the force per egg caused by the weight of the block.
2) If the pressure on each egg is 25 N/mm 2 , calculate the contact area, in mm 2 , between the block and one egg.
Answer:
1) The force per egg caused by the weight of the block is 17.15 N
2) [tex]A= 0.686\ mm^2[/tex]
Explanation:
Pressure
The pressure is computed as the force acting upon a surface divided by the area of the surface :
[tex]P=\frac{F}{A}[/tex]
The weight of an object of mass m is:
W = mg
Where g is the gravitational acceleration of [tex]9.8\ m/s^2[/tex]
The cement block has a mass of m=14 kg, thus its weight is:
W = 14*9.8
W = 137.2 N
1) This weight is evenly distributed over the 8 eggs, thus the force per egg is:
F = 137.2 N/8= 17.15 N
The force per egg caused by the weight of the block is 17.15 N
2) It's given the pressure on each egg as [tex]P= 25 \ N/mm^2[/tex] and we know the force acting on each one is F=17.15 N. To find the contact area we solve the formula of the pressure for A:
[tex]\displaystyle A=\frac{F}{P}[/tex]
[tex]\displaystyle A=\frac{17.15}{25}[/tex]
[tex]\mathbf{A= 0.686\ mm^2}[/tex]
Two charges q1 and q2 are separated by a distance r and apply a force F to each other. If both charges are doubled, and the distance between them is halved, the new force between them is
Answer:
The new force between them is increased by a factor of 16.
Explanation:
According to Coulombs law, the force of attraction between two (2) charges is given by the formula;
F = Kq1q2/r²
Given the following data;
q1 = 2q1
q2 = 2q2
r = r/2
Substituting into the equation, we have;
F = 2q1*2q2/(r/2)²
F = 4q1q2/r²/4
F = 4q1q2 * 4/r²
F = 16q1q2/r²
Therefore, the new force between them is increased by a factor of 16.
A projectile is launched with an initial velocity of
200 meters per second at an angle of 30° above the
horizontal. What is the magnitude of the vertical
component of the projectile's initial velocity by?
(1) 200 m/s x cos 30°
(2) 200 m/s X sin 30°
(3) (200 m/s)/(cos 30 °)
(4) (200 m/s)/(sin 30 °)
The magnitude of the vertical component of the projectile's initial velocity is 200 m/s × sin 30°.
The diagrammatic representation of the velocity of the projectile can be seen in the attached image below.
From the diagram, let consider the ΔOAP where Vector OP makes an ∠θ = 30° to the horizontal x-axis.
where;
|OP| = magnitude of projectile velocity|OA| = magnitude of the horizontal component|OB|/|AP| = vertical component of the projectile∴
Using trigonometric approach for ΔOAP;
[tex]\mathbf{sin\theta = \dfrac{AP}{OP}}[/tex]
[tex]\mathbf{AP =OP\times sin \theta}}[/tex]
AP = 200 × sin 30°
Learn more about projectile here:
https://brainly.com/question/20689870?referrer=searchResults
In which three ways does the pattern seen on the screen during a double-slit
experiment support the wave model of light?
O A. All of the light that enters the slits passes through them.
B. The light bends around the edges of the slit.
C. There are bands of dimmer light.
D. There are bands of brighter light.
Answer:
b,c,d
Explanation:
the light bends around the edges of the slit
there are bands of brigter light
there are bands of dimmer light
( just took the quiz )
Option B,C and D shows the three ways does the pattern seen on the screen during a double-slit experiment.
What is double-slit experiment?The double-slit experiment shows that light and matter may have properties that are both conventionally defined waves and particles, as well as the essentially probabilistic nature of quantum mechanical processes.
The three ways does the pattern seen on the screen during a double-slit experiment support the wave model of light;
B. The light bends around the edges of the slit.
C. There are bands of dimmer light.
D. There are bands of brighter light.
Hence, B,C and D are the three ways does the pattern seen on the screen during a double-slit experiment.
To learn more about the double-slit experiment, refer to the link;
https://brainly.com/question/17167388
#SPJ2
A dwari planet is a:
small rocky object that orbits the Sun and is usualy found in a bet beiween me orbits of Mars and Jupiter.
small object made of ice and dust thai orbits the Sun and forms a coma as i approaches the Sun.
round object that orbits the Sun but lacks the abity to clear the neghborhood around is orbit.
meteoroid thai bums up in Earth's amosphere, producing a streak of light.
Answer:
round object that orbits the Sun but lacks the ability to clear the neighborhood around is orbit.
Explanation:
It is true that a dwarf planet is a round object that orbits the Sun but lacks the ability to clear the neighborhood around its orbit.
Due to the size of a dwarf planet, it does not possess enough gravitational attraction or force to clear the orbit around it or other bodies.
One of the notable dwarf planets is Pluto. It was formerly thought to be planet but it has since been delisted.
Pluto's orbit lies beyond that of Neptune.
2. A girl of weight 400 N has feet of area 100 cm2. What
pressure does she put on the ground ?
Answer:
Explanation:
Given:
Weight = 400N
Area = 100cm^2
If 1cm^2 = 1×10^-4m^2
Hence 100 cm^2 = 0.01m^2
Pressure = force/area .........1
Force = ma = mg which is also the weight of the girl. Hence substitute the values into 1
P = 400/0.01
P = 40000 N/m^2
The pressure she puts on the ground is 40000 N/m^2
a car traveled at 100 km h for 3 hours how far did it travel
100 km
300 km/h
300 km
3 km/h
Answer:
300km
Explanation:
Given parameters:
Speed of the car = 100km/h
Time taken for the travel = 3hrs
Unknown:
How far did it travel = ?
Solution:
To solve this problem, we must understand that;
Speed = [tex]\frac{distance}{time}[/tex]
Distance = speed x time
Distance = 100km/hr x 3hr = 300km
A 5kg box is sliding down a ramp with a rough surface as seen below. The height of the ramp is 20m and the distance the box travels down the ramp (from A to B) is 15m. At point A the velocity of the box is 8 m/s. If the velocity at point B is 3m/s, what was the impulse caused by friction? If the force of friction is 5N, how long did it take the box to slide the 15 m?
Answer:
A Impulse = – 25 Ns
B. Time = 5 s
Explanation:
From the question given above, the following data were obtained:
Mass (M) = 5 Kg
Initial velocity (u) = 8 m/s
Final velocity (v) = 3 m/s
Impulse (I) =?
Time (t) =?
A. Determination of the Impulse.
Mass (M) = 5 Kg
Initial velocity (u) = 8 m/s
Final velocity (v) = 3 m/s
Impulse (I) =?
I = Ft = M(v – u)
I = M(v – u)
I = 5 (3 – 8)
I = 5 × – 5
I = – 25 Ns
NOTE: the negative sign indicates that the net force is acting in the negative direction.
B. Determination of the time.
Impulse (I) = 25 Ns
Force (F) = 5 N
Time (t) =?
I = Ft
25 = 5 × t
Divide both side by 5
t = 25 / 5
t = 5 s
Thus, it will take 5 s for the box to slide through the 15 m long ramp.
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 39 m in front of you. Your reaction time before stepping on the brakes is 0.50 s , and the maximum deceleration of your car is 11 m/s2.
a. How much distance is between you and the deer when you come to a stop?
b. What is the maximum speed you could have and still not hit the deer?
Answer:
a) 10.8 m
b) 24.3 m/s
Explanation:
a)
In order to get the total distance traveled since you see the deer till the car comes to an stop, we need to take into account that this distance will be composed of two parts.The first one, is the distance traveled at a constant speed, before stepping on the brakes, which lasted the same time that the reaction time, i.e., 0.5 sec.We can find this distance simply applying the definition of average velocity, as follows:[tex]\Delta x_{1} = v_{1o} * t_{react} = 20 m/s * 0.5 s = 10 m (1)[/tex]
The second part, is the distance traveled while decelerating at -11 m/s2, from 20 m/s to 0.We can find this part using the following kinematic equation (assuming that the deceleration keeps the same all time):[tex]v_{1f} ^{2} - v_{1o} ^{2} = 2* a* \Delta x (2)[/tex]
where v₁f = 0, v₁₀ = 20 m/s, a = -11 m/s².Solving for Δx, we get:[tex]\Delta x_{2} = \frac{-(20m/s)^{2}}{2*(-11m/s)} = 18.2 m (3)[/tex]
So, the total distance traveled was the sum of (1) and (3):Δx = Δx₁ + Δx₂ = 10 m + 18.2 m = 28.2 m (4)Since the initial distance between the car and the deer was 39 m, after travelling 28.2 m, the car was at 10.8 m from the deer when it came to a complete stop.b)
We need to find the maximum speed, taking into account, that in the same way that in a) we will have some distance traveled at a constant speed, and another distance traveled while decelerating.The difference, in this case, is that the total distance must be the same initial distance between the car and the deer, 39 m.⇒Δx = Δx₁ + Δx₂ = 39 m. (5)Δx₁, is the distance traveled at a constant speed during the reaction time, so we can express it as follows:[tex]\Delta x_{1} = v_{omax} * t_{react} = 0.5* v_{omax} (6)[/tex]
Δx₂, is the distance traveled while decelerating, and can be obtained using (2):[tex]v_{omax} ^{2} = 2* a* \Delta x_{2} (7)[/tex]
Solving for Δx₂, we get:[tex]\Delta x_{2} = \frac{-v_{omax} ^{2} x}{2*a} = \frac{-v_{omax} ^{2}}{(-22m/s2)} (8)[/tex]
Replacing (6) and (8) in (5), we get a quadratic equation with v₀max as the unknown.Taking the positive root in the quadratic formula, we get the following value for vomax:v₀max = 24.3 m/s.OFFERING 60 POINTS IF YOU CAN SHOW THE WORK!!!!
A 1000 kg roller coaster begins on a 10 m tall hill with an initial velocity of 6m/s and travels down before traveling up a second hill. As the coaster moves from its initial height to its lowest position, 1700J of energy is transformed to thermal energy by friction.
Answer; 10.6 i think
Explanation:
(a) At the top of the hill, the coaster has total energy (potential and kinetic)
E = (1000 kg) g (10 m) + 1/2 (1000 kg) (6 m/s)² = 116,000 J
As it reaches its lowest position, its potential energy is converted to kinetic energy, and some is lost to friction, making its speed v such that
1/2 (1000 kg) v ² = 116,000 J - 1700 J = 114,300 J
===> v ≈ 15.2 m/s
If no energy is lost to friction as the coaster makes its way up the second hill, all of its kinetic energy would be converted to potential energy at the maximum possible height H.
1/2 (1000 kg) (15.2 m/s)² = (1000 kg) g H
===> H ≈ 11.7 m
(b) At the top of the second hill with minimum height h, and with maximum speed 4.6 m/s, the coaster has energy
E = P + K = (1000 kg) g h + 1/2 (1000 kg) (4.6 m/s)²
Assuming friction isn't a factor again, the energy here should match the energy at the lowest point in part (a), 114,300 J.
(1000 kg) g h + 1/2 (1000 kg) (4.6 m/s)² = 114,300 J
===> h ≈ 10.6 m
A child is twirling a 0.0113-kg ball on a string in a horizontal circle whose radius is 0.147 m. The ball travels once around the circle in 0.388 s. (a) Determine the centripetal force acting on the ball. (b) If the speed is doubled, by what factor does the centripetal force increase
Answer:
0.435 N
The centripetal force increases by a factor of 4
Explanation:
Let the centripetal force be F
F =mv^2/r
m= mass of the object
v = linear velocity of the object
r = radius of the circular path
but v = 2πr/T = 2 * 3.142 * 0.147 / 0.388
v = 2.38 m/s
F = 0.0113 * (2.38)^2/0.147
F= 0.435 N
if v= 2v
Then;
F = m(2v)^2/r
F = m4v^2/r
F= 4mv^2/r
The centripetal force increases by a factor of 4
You want to build a snowman, so you accelerate a 2kg snowball across your yard at a rate of 0.5m/s2. Calculate the amount of force you applied to your friend.
Answer:
4
Units:
Newtons
A car with mass 1141 kg, moving at 16 m/s, strikes a(n) 2916 kg car at rest. If the two cars stick together, with what speed do they move
Answer:
V = 4.49 m/s
Explanation:
Given that,
Mass of the car 1, m₁ = 1141 kg
Initial speed of car 1, u₁ = 16 m/s
Mass of car 2, m₂ = 2916 kg
Initial speed of car 2, u₂ = 0
We need to find the speed of the cars if they stick together. Let the speed be V. The momentum will remain conserved in the process. Using the conservation of momentum to find it.
m₁u₁ + m₂u₂ = (m₁+m₂)V
[tex]V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\=\dfrac{1141\times 16+2916 \times 0}{(1141 +2916 )}\\\\=4.49\ m/s[/tex]
So, the required speed of the two cars is 4.49 m/s.
Two charged particles attract each other with a force of magnitude F. If the distance between the charges is made 3.5 times as large and the charge on one of the particles is made 4 times as big, what is the ratio of the new F to the old one
Answer:
The ratio between the forces is:
[tex]\frac{F_{new}}{F_{old}}=0.33[/tex]
Explanation:
The electrostatic force equation is:
[tex]F_{old}=k\frac{q_{1}q_{2}}{d^{2}}[/tex]
Where:
q1 and q2 are the electric charges
d is the distance between them
k is the electrostatic constant
Now, the distance is 3.5 times as large and q1 is 4 times as big, then the new force will be:
[tex]F_{new}=k\frac{4q_{1}q_{2}}{(3.5d)^{2}}[/tex]
[tex]F_{new}=\frac{4}{3.5^{2}}k\frac{q_{1}q_{2}}{d^{2}}[/tex]
We can rewrite this equation in terms of F(old)
[tex]F_{new}=\frac{4}{3.5^{2}}F_{old}[/tex]
Therefore, the ratio between the forces is:
[tex]\frac{F_{new}}{F_{old}}=\frac{4}{3.5^{2}}[/tex]
[tex]\frac{F_{new}}{F_{old}}=0.33[/tex]
I hope it helps you!
A plane drops a package for delivery. The plane is flying horizontally at a speed of 120\,\dfrac{\text m}{\text s}120 s m 120, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction, and the package travels 255\,\text m255m255, start text, m, end text horizontally during the drop. We can ignore air resistance.
Answer:
-22.1
Explanation:
1 / 4
Step 1. List horizontal (xxx) and vertical (yyy) variables
xxx-direction yyy-direction
t=\text?t=?t, equals, start text, question mark, end text t=\text?t=?t, equals, start text, question mark, end text
a_x=0a
x
=0a, start subscript, x, end subscript, equals, 0 a_y=-9.8\,\dfrac{\text m}{\text s^2}a
y
=−9.8
s
2
m
a, start subscript, y, end subscript, equals, minus, 9, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction
\Delta x=255\,\text mΔx=255mdelta, x, equals, 255, start text, m, end text \Delta y=\text ?Δy=?delta, y, equals, start text, question mark, end text
v_x=v_{0x}v
x
=v
0x
v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript v_y=?v
y
=?v, start subscript, y, end subscript, equals, question mark
v_{0x}=120\,\dfrac{\text m}{\text s}v
0x
=120
s
m
v, start subscript, 0, x, end subscript, equals, 120, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction v_{0y}=0v
0y
=0v, start subscript, 0, y, end subscript, equals, 0
Note that there is no horizontal acceleration, so v_x=v_{0x}v
x
=v
0x
v, start subscript, x, end subscript, equals, v, start subscript, 0, x, end subscript. The time is the same for the xxx and yyy directions.
Also, the package has no initial vertical velocity.
Our yyy-direction variable list has too many unknowns to solve for \Delta yΔydelta, y directly. Since both the yyy- and xxx-directions have the same time ttt and horizontal acceleration is zero, we can solve for ttt from the xxx-direction motion by using equation:
\Delta x=v_xtΔx=v
x
tdelta, x, equals, v, start subscript, x, end subscript, t
Once we know ttt, we can solve for \Delta yΔydelta, y using the kinematic equation that does not include the unknown variable v_yv
y
v, start subscript, y, end subscript:
\Delta y=v_{0y}t+\dfrac {1}{2}a_yt^2Δy=v
0y
t+
2
1
a
y
t
2
delta, y, equals, v, start subscript, 0, y, end subscript, t, plus, start fraction, 1, divided by, 2, end fraction, a, start subscript, y, end subscript, t, squared
Hint #22 / 4
Step 2. Find ttt from horizontal variables
\begin{aligned}\Delta x&=v_xt \\\\ t&=\dfrac{\Delta x}{v_{0x}} \\\\ t&=\dfrac{255\,\text m}{120\dfrac{\text m}{\text s}} \\\\ &=2.125\,\text s \end{aligned}
Δx
t
t
=v
x
t
=
v
0x
Δx
=
120
s
m
255m
=2.125s
Hint #33 / 4
Step 3. Find \Delta yΔydelta, y using ttt
Using ttt to solve for \Delta yΔydelta, y gives:
\begin{aligned}\Delta y&=v_{0y}t+\dfrac{1}{2}a_yt^2 \\\\ &=\cancel{ (0 )t}+\dfrac{1}{2}\left (-9.8\dfrac{\text m}{\text s^2}\right )\left(2.125\,\text s\right)^2 \\\\ &=-22.1\,\text m \end{aligned}
Δy
=v
0y
t+
2
1
a
y
t
2
=
(0)t
+
2
1
(−9.8
s
2
m
)(2.125s)
2
=−22.1m
Hint #44 / 4
The correct answer is -22.1\,\text m−22.1mminus, 22, point, 1, start text, m, end text.
Help can’t find the answer no where
i'm stuck on that question also
A dog sees a flowerpot sail up and then back past a window H high. If the total time the pot is in sight is t seconds, find the height above the window that the pot rises. (Let H = 2 m and t = 1.0 s, find the height above the window-- after you have found an algebraic solution.)
Answer:
maximum height = 0.1225 m
Explanation:
given data
H = 2m
t = 1 sec
solution
we consider here velocity of pot at lower side is u
and final velocity is v with acceleration a
time take is t/2
so
height h = u × [tex]\frac{t}{2}[/tex] - 0.5 × g × [tex](\frac{t}{2})^2[/tex] .................1
here
u = [tex]\frac{2}{t} \times ( h + \frac{gt^2}{8} )[/tex]
and
v = u +a t/2 .........................2
v = u + g t/2
v = [tex]\frac{2h}{t} + \frac{gt}{4} - \frac{gt}{2}[/tex]
v = [tex]\frac{2h}{t} - \frac{gt}{4}[/tex]
so that
maximum height is = [tex]\frac{v^2}{2g}[/tex]
maximum height = [tex]\frac{(\frac{2h}{t} - \frac{gt}{4})^2}{2g}[/tex]
put here value of h and t
maximum height = [tex]\frac{(\frac{2(2)}{1} - \frac{g(1)}{4})^2}{2g}[/tex]
maximum height = 0.1225 m
The Earth (geosphere) comprises several major layers, differing in chemical and mineral composition, material strength, and other physical properties. Which of the layers listed below constitutes the largest proportion of Earth's volume?
a) Lithosphere
b) Crust
c) Outer core ·
d) Inner core
e) Mantle
Answer:
Option E:
The mantle
Explanation:
The earth's mantle is the mushy, semi-solid portion of the earth that makes up most of the earth's volume. The mantle extends for a depth of about 2800km downwards into the earth, making it the largest internal portion of the earth. It makes up about 84 percent of the earth's structure, leaving the core and the crust with 15 percent and 1 percent respectively.
Due its nature, convection currents are set up predominantly in the mantle of the earth, which leads to movements in the upper layers of the earth (the crust).
The mantle is large enough for the lighter crust to float on its surface.
While driving his sports car at 20.00 m/s down a four lane highway, a
man comes up behind a very slow school bus. The man is in a hurry so he
decides to pass the bus. If the man's car can accelerate at 6.00 m/s^2,
how long will it take for him to reach a speed of 35.00 m/s in order to pass
the bus? Round to the nearest hundredth. (2 decimal places) Type the
number only.
Answer:
2.50 s
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 20 m/s
Acceleration (a) = 6 m/s²
Final velocity (v) = 35 m/s
Time (t) =?
Thus, we can obtain the time taken for the man to reach a speed of 35 m/s as follow:
v = u + at
35 = 20 + 6t
Collect like terms
35 – 20 = 6t
15 = 6t
Divide both side by 6
t = 15/6
t = 2.50 s
Thus, it will take the man 2.50 s to reach a speed of 35 m/s.
g A box of mass 10 kg attached to a spring is pulled to a maximum of 44 cm. The box is released. What is the speed when the box reaches a point 10 cm from the equilibrium position
Answer:
The speed when the box reaches a point 10 cm from the equilibrium position is 2.02 m/s.
Explanation:
Mass attached to the spring, m = 10 kg
maximum displacement of the spring, A = 0.44 m
The spring constant is calculated from Hook's law;
F = kx
mg = kx
k = (mg) / x
k = (10 x 9.8) / 0.44
k = 222.73 N/m
The angular speed of the spring is calculated as;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega =\sqrt{\frac{222.73}{10} } \\\\\omega = 4.72 \ rad/s[/tex]
The speed when the box reaches a point 10 cm from the equilibrium position is calculated as;
[tex]v = \omega \sqrt{A^2-x^2} \\\\v = 4.72\sqrt{0.44^2-0.1^2}\\\\v = 2.02 \ m/s[/tex]
Therefore, the speed when the box reaches a point 10 cm from the equilibrium position is 2.02 m/s.
What is the momentum of a dirt bike with a mass of 112 kg
moving 25 m/s?
a. 4.88 kg m/s
b. 87 kg m/s
c. 2800 kg m/s
A steam catapult launches a jet aircraft from the aircraft carrier John C. Stennis, giving it a speed of 195 mi/h in 2.90 s. (a) Find the average acceleration of the plane.
Answer:
The average acceleration of the plane is 0.0186 [tex]\frac{mi}{s^{2} }[/tex]
Explanation:
The acceleration of an object is a magnitude that indicates how the speed of the object changes in a unit of time. That is, acceleration is a magnitude that relates changes in speed with the time it takes to occur.
The average acceleration of an object is calculated using the following expression:
[tex]acceleration=\frac{change of speed}{time}[/tex]
In this case:
change of speed=[tex]195 \frac{mi}{h} *\frac{1 h}{60 minutes} *\frac{1 minute}{60 s} =0.054 \frac{mi}{s}[/tex]time= 2.90 sReplacing:
[tex]acceleration=\frac{0.054 \frac{mi}{s} }{2.90 s}[/tex]
and solving you get:
acceleration=0.0186 [tex]\frac{mi}{s^{2} }[/tex]
The average acceleration of the plane is 0.0186 [tex]\frac{mi}{s^{2} }[/tex]
The velocity of a Froghopper flea, which is represented by a vector, has a direction and a magnitude. If a coordinate system is oriented where the x-axis is horizontal, and the y-axis is vertical, is it possible to write expressions for the x and y components of the velocity vector in terms of the vector's magnitude and direction.
Answer:
Explanation:
The solution of the question cab e found in attachment below:
Exercise 3 In an equation y mx+c; y and x have dimensions of length and c is constant. What are the dimensions of m'
A. Mass
B. Length
C. Time
D. m is Dimensionless
Answer:
D. m is Dimensionless
Explanation:
The equation of a straight line is given as:
y = mx + c
Dimension of y = l
x = l
c has no dimension
So;
if we do a dimensional analysis:
L = m L + 0
m = 1
So, m has no dimension