Solution :
a). From Newtons second law,
F = ma
The total tension force is 2T.
∴ 2T - (m + M)g = (m+ M)a
Then
[tex]$a=\frac{2T-(m+M)g}{m+M}$[/tex]
[tex]$a=\frac{2\times 600-(52+63)9.8}{52+63}$[/tex]
[tex]$=0.63 \ m/s^2$[/tex]
b). From the person,
F = ma
T - Mg + N = Ma
or N = Ma + Mg - T
= (63 x 9.8) + (52 x 9.8) - 600
= 617.4 + 509.6 - 600
= 527 N
If a car is moving to the left with constant velocity, one can conclude that :
a. there must be no forces exerted on the car.
b. the net force exerted on the car is directed to the left.
c. the net force exerted on the car is zero.
d. there is exactly one force exerted on the car.
Answer:
b. the net force exerted on the car is directed to the left.
Explanation:
Applying Newton's second law of motion, the car will move in the direction of the applied force. If the applied forces are on different directions, the car will move in the direction of the greater force (net force).
Therefore, if a car is moving to the left with constant velocity, one can conclude that the net force exerted on the car is directed to the left.
The correct option is "b. the net force exerted on the car is directed to the left"
If the diameter of a moose eye is 40 mm, what is the total refractive power of the anterior portion of the eye?
Answer:
-the ratio of the speed of light
in air to the speed of light in the substance.
-speed of light in air 300,000 km/sec, which decreases when it passes through a transparent substance.
-e.g.. speed of light in substance = 200,000 km/sec, R.I. = 300,000/200,000 = 1.5
Explanation:
Identical balls are dropped from the same initial height and bounce back to half the initial height. In Case 1, the ball bounces off a cement floor, and in Case 2, the ball bounces off a piece of stretchy rubber. Two cases of a ball dropping and bouncing off of a surface. In case one, the ball approaches a cement floor, comes in contact, momentarily comes to rest, and then bounces off with a velocity smaller than the approach velocity. In case 2, the ball approaches a piece of stretchy rubber. When the ball comes in contact with the rubber, it deforms the rubber and comes momentarily to rest. The ball bounces off again with a velocity smaller than the approach velocity, and the rubber regains its original shape. In both cases, the balls are dropped from the same initial height and reach the same final height. In which case is the magnitude of the ball's change in momentum the greatest
Answer:
the impulse must be the same in these two cases F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])
Explanation:
For this exercise we use the relationship between momentum and momentum
I = Δp
F t = m v_f - m v₀
To know the speed we use the conservation of energy
starting point. Highest point
Em₀ = U = m g h
fincla point. Just before the crash
Em_f = K = ½ m v²
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v = [tex]\sqrt{2gh}[/tex]
we substitute in the impulse relation
F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])
therefore we can see that as in case the initial and final heights are equal, the impulse must be the same in these two cases
.You have always been impressed by the speed of the elevators in your apartment building. You wonder about the maximum acceleration for these elevators during normal operation, so you decide to measure it by using your bathroom scale. While the elevator is at rest on the ground floor,you get in, put down your scale, and stand on it. The scale reads 50 kg. You continue standing on the scale when the elevator goes up, carefully watching the reading. During the trip to the 10th floor, the greatest scale reading was
Answer:
5.51 m/s^2
Explanation:
Initial scale reading = 50 kg
assume the greatest scale reading = 78.09 kg
Determine the maximum acceleration for these elevators
At rest the weight is = 50 kg
Weight ( F ) = mg = 50 * 9.81 = 490.5 N
At the 10th floor weight = 78.09 kg
Weight at 10th floor ( F ) = 78.09 * 9.81 = 766.11 N
F = change in weight
Change in weight( F ) = ma = 766.11 - 490.5 (we will take the mass as the starting mass as that mass is calculated when the body is at rest)
50 * a = 275.61
Hence the maximum acceleration ( a ) = 275.61 / 50 = 5.51 m/s^2
A freight train has a mass of [02] kg. The wheels of the locomotive push back on the tracks with a constant net force of 7.50 × 105 N, so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 km/h?
Answer:
t = 300.3 seconds
Explanation:
Given that,
The mass of a freight train, [tex]m=1.01\times 10^7\ kg[/tex]
Force applied on the tracks, [tex]F=7.5\times 10^5\ N[/tex]
Initial speed, u = 0
Final speed, v = 80 km/h = 22.3 m/s
We need to find the time taken by it to increase the speed of the train from rest.
The force acting on it is given by :
F = ma
or
[tex]F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.01\times 10^7\times (22.3-0)}{7.5\times 10^5}\\\\t=300.3\ s[/tex]
So, the required time is 300.3 seconds.
What is the Radiation left over from the big bang called?
Answer:
The Big Bang theory predicts that the early universe was a very hot place and that as it expands, the gas within it cools. Thus the universe should be filled with radiation that is literally the remnant heat left over from the Big Bang, called the “cosmic microwave background", or CMB.
Explanation:
The Big Bang theory suggest that the universe in early stage was at very hot place and which can be expanded, the gas within it cools. It is in an infinite universe and it has no edge.
What is big bang theory ?The Big Bang theory is a cosmological model which explain the existence of the observable universe from the earliest periods to the large-scale evolution.
The model describes the mechanism behind the universe expansion from an initial state of high density and temperature, it is very important concept as a lot of research is going on in this field to find out exactly how the universe began billions of years ago.
The universe began to cool down in order to allow the formation of particles become atoms after its initial phase of expansion, Primordial elements such as Hydrogen, Helium, and Lithium are condensed through gravity are formed early stars and galaxies.
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A fan has four identical, symmetrically placed blades. The blades are rotating clockwise at twenty revolutions per second.
A) What is the smallest time interval between stroboscope flashes that will make the fan blades appear motionless?
B) What is the highest frequency (in flashes per second) at which a stroboscope will make the
fan blades appear to stand still? Show your calculation.
C) The same questions as (a) and (b), but someone has put a yellow dot on one blade, and now you want the yellow dot to appear to be standing still. Explain, and show your calculation.
D) Now the stroboscope is set for nineteen flashes per second, and the yellow dot appears to be slowly rotating. Which direction does it appear to rotate, clockwise or counterclockwise? Explain, and show your calculation.
E) The same as (d), but the stroboscope is set for twenty-one flashes per second. Explain, and show your calculation.
Answer:
A) t = 1.249 10⁻² s, B) f = 80 Hz, C) f = 20 Hz,
D) slowly advancing an angle of approximately Δθ = 0.05 rad each flash
E) In each flash it seems to go backward an angle of Δθ = -0.05 rad
Explanation:
A) To make it appear that the blades are immobile, it implies that every time the light turns on, a blade should be in the same position, therefore, as we have 4 blades, they must rotate an angle of 2π/4,
θ = π / 2
θ = 1.57 rad
taking the angle let's use the endowment kinematics relations
θ = w₀ t + ½ α t²
in general the fans rotate at constant speed α= 0
θ = w₀ t
t = θ / w₀
let's reduce the magnitudes to the SI system
w₀ = 20 rev / s (2π rad / 1rev) = 125.66 rad / s
let's calculate
t = 1.57 / 125.66
t = 1.249 10⁻² s
B) the fastest speed for the blades to rotate is when one blade of a complete turn , we use the relationship between the fecuance and the period
f = 1 / T
f = 1 / 1.25 10⁻²
f = 80 Hz
C) we have two possibilities:
* a yellow dot is placed on each sheet
In this case the angular velocity of the blade is the same at all points, therefore the results obtained should not change
* a yellow dot is placed on a single sheet.
Here for the point to remain fixed the angle of rotation must be
θ= 2π rad
the time is
t = 2π / 125.66
t = 5 10⁻² s
the maximum frequency is
f = 1/5 10⁻²
f = 20 Hz
D) The copy strobe rotates at f = 19 Hz, the time between each flash is
t = 1/19
t = 5.26 10⁻² s
this time is higher, so the angle turned is large
θ = w t
θ = 125.66 5.26 10⁻²
θ = 6.61 rad
the relationship between this angle and the angle of a circle is
θ = 1,052
We can see that it is this time the blade rotates 1 complete turns, for this the position of the blade changes us, for the other 0.052 rad the blade rotates a little more than the circumference therefore it seems that it is slowly advancing an angle of approximately
Δθ = 0.05 rad each flash
E) in this case changes the flash speed
t = 1/21
t = 4.76 10⁻² s
the angle rotated is
θ = 125.66 4.76 10⁻²
θ = 5.984 rad
θ / 2π = 0.95
in that case, the blade did not complete the turn, therefore in each flash it seems to go backward an angle
Δθ = -0.05 rad
In the Styrofoam ball investigation, it is likely that the charges on the ball and rod are
?
A. the same
B. opposite
C. constantly changing.
Answer:
answer = opposite
they are not changing being in a place as being they-self
the same would not be true because its different then the Styrofoam and it could only be different and its make the most sense aswell.
Both ball and rod has opposite charges on their bodies because of presence of different charges on it.
What are the charges on ball and rod?In the Styrofoam ball investigation, it is likely that the charges on the ball and rod are opposite because Styrofoam ball is negatively charged due to the presence of electrons while on the other hand, the rod is positively charged because of lining of positive charges on the rod.
So we can conclude that both ball and rod has opposite charges on their bodies because of presence of different charges on it.
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A 450.0 kg roller coaster is traveling in a circle with radius 15.0m. Its speed at point A is 28.0m/s and its speed at point B is 14.0 m/s. At point A the cart is already moving with circular motion. a) Draw free bodydiagramsfor the cartatpointsAand B(two separate free body diagrams). b) Calculate the acceleration of the cartat pointsAandB(magnitude and direction). c) Calculate the magnitude of the normal force exerted by the trackson the cartat point A. d) Calculate the magnitude of the normal force exerted by the tracks on the cart at point B.
Answer:
b) a = 52.26 m / s², a ’= 13.06 m / s², c) N = 2.79 10⁴ N, d) N = 1.89 10³ N
Explanation:
a) In the attached we can see the free body diagrams for the two positions, position A in the lower part of the circle and position B in the upper part of the circle
b) Let's start at point A
Let's use that the acceleration is centripetal
a = v² / r
let's calculate
a = 28² / 15.0
a = 52.26 m / s²
as they relate it is centripetal it is directed towards the center of the circle, therefore for this point it is directed vertically upwards
Point B
a ’= 142/15
a ’= 13.06 m / s²
in this case the acceleration is vertical downwards
c) The values of the normal force
point A
let's use Newton's second law
∑ F = m a
N- W = m a
N = mg + ma
N = m (g + a)
N = 450.0 (9.8 + 52.25)
N = 2.79 10⁴ N
d) Point B
-N -W = m (-a)
N = ma -m g
N = m (a-g)
N = 450.0 (14.0 - 9.8)
N = 1.89 10³ N
Cameron is standing on the edge of a 60 m high cliff. He horizontally throws a football
with an initial velocity of 10 m/s. How long does it take for the football to hit the
ground?
Answer:34.6 m/s
Explanation: It is asking how long meaning the answer is in time
Two objects are electrically charged. The net charge on one object is doubled.
Therefore, the electric force _____.
reverses
doubles
quadruples
divides
Which phrase describes velocity?
u
A. A quantity with direction only
B. A quantity with magnitude only
C. A quantity with no units
D. A quantity with magnitude and direction
SUBMI
QUCIK!! SOMEONE PLEASE HELP! I’LL MARK BRAINLIEST!!
Answer:
A. v = √2gh
B. No! The final velocity does not depend on the mass of the car.
C. Yes! the final velocity depends on the steepness of the hill
D. 3.28 m/s
Explanation:
A. Determination of the final velocity.
½mv² = mgh
Cancel out m
½v² = gh
Cross multiply
v² = 2gh
Take the square root of both side
v = √2gh
B. Considering the formula obtained for the final velocity i.e
v = √2gh
We can see that there is no mass (m) in the formula.
Thus, the final velocity does not depend on the mass of the car.
C. Considering the formula obtained for the final velocity i.e
v = √2gh
We can see that there is height (h) in the formula.
Thus, the final velocity depends on the steepness of the hill
D. Determination of the final velocity.
Height (h) = 0.55 m
Acceleration due to gravity (g) = 9.8 m/s²
Velocity (v) =?
v = √2gh
v = √(2 × 9.8 × 0.55)
v = √10.78
v = 3.28 m/s
A liquid fueled rocket is red on a test stand. The rocket nozzle has an exit diameter of 30 cm and the combustion gases leave the nozzle at a velocity of 3800 m/s and a pressure of 100 kPa, which is the same as the ambient pressure. The temperature of the gases in the combustion area is 2400 C. Find (a) the temperature of the gases at the nozzle exit plane, (b) the pressure in the combustion area, and (c) the thrust developed. Assume that the gases have a speci c heat ratio of 1.3, and a molar mass of 9. Assume that the ow in the nozzle is isentropic.
Answer:
1. Temperature= 869.35 K
2. Pressure of combustion = 12994.043 kpa
3. Thrust = 127x10⁶N
Explanation:
this problem has been fully explained in the attachment. please use it to get a clearer explanation of the answer.
1.
The temperature = (273+2400k) - (3800)²/2(4003)
= 2673 - 14440000/8006
= 2673 - 1803.65
= 869.35 K
Approximately 869.4K
2. We first get mach number
= 3800/√1.3(923.8)(869.35)
= 3800/1021.78
= 3.719
Pressure = 100kpa[1+2.07464415]^1.3/0.3
= 12995.043kpa
C. Thrust
Pi/4(3800)²(0.3)²(100x10³)/(923.8)(869.4)
= 12678.621
= 126.781 kN
Thrust is approximately 127kN = 127x10⁶N
The sides of a right triangle that has any given vector for the hypotenuse are called _____
A. Scalar
B. Component
C. Resultant
D. Vector
Answer:
They are the resultant vector.
Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25 cm and the point-like object with charge q2 = −2.14 µC is located at x2 = −1.80 cm.
A) Determine the total electric potential (in V) at the origin.
B) Determine the total electric potential (in V) at the point with coordinates (0, 1.50 cm).
Answer:
a) the total electric potential is 2282000 V
b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
Explanation:
Given the data in the question and as illustrated in the image below;
a) Determine the total electric potential (in V) at the origin.
We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges
so
Electric potential at p in the diagram 1 below is;
Vp = V1 + V2
Vp = kq1/r1 + kq2/r2
we know that; Coulomb constant, k = 9 × 10⁹ C
q1 = 4.60 uC = 4.60 × 10⁻⁶ C
r1 = 1.25 cm = 0.0125 m
q2 = -2.06 uC = -2.06 × 10⁻⁶ C
location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m
so we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )
Vp = (3312000) + ( -1030000 )
Vp = 3312000 -1030000
Vp = 2282000 V
Therefore, the total electric potential is 2282000 V
b)
the total electric potential (in V) at the point with coordinates (0, 1.50 cm).
As illustrated in the second image;
r1² = 0.015² + 0.0125²
r1 = √[ 0.015² + 0.0125² ]
r1 = √0.00038125
r1 = 0.0195
Also
r2² = 0.015² + 0.018²
r2 = √[ 0.015² + 0.018² ]
r2 = √0.000549
r2 = 0.0234
Now, Electric Potential at P in the second image below will be;
Vp = V1 + V2
Vp = kq1/r1 + kq2/r2
we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )
Vp = 2123076.923 + ( -762962.962 )
Vp = 2123076.923 -792307.692
Vp = 1330769.23 V
Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
a) The total electric potential is 2282000 V
b) The total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
What is electric potential?The electric potential is defined as the amount of work energy needed to move a unit of electric charge from a reference point to a specific point in an electric field.
Given the data in the question and as illustrated in the image below;
a) Determine the total electric potential (in V) at the origin.
We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges
Electric potential at p in diagram 1 below is;
[tex]V_P=V_1+V_2[/tex]
[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]
we know that; the Coulomb constant, k = 9 × 10⁹ C
q1 = 4.60 uC = 4.60 × 10⁻⁶ C
r1 = 1.25 cm = 0.0125 m
q2 = -2.06 uC = -2.06 × 10⁻⁶ C
location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m
so we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )
Vp = (3312000) + ( -1030000 )
Vp = 3312000 -1030000
Vp = 2282000 V
Therefore, the total electric potential is 2282000 V
b)The total electric potential (in V) at the point with coordinates (0, 1.50 cm).
As illustrated in the second image;
[tex]r_1^2=0.015^2+0.0125^2[/tex]
[tex]r_1 = \sqrt{[ 0.015^2 + 0.0125^2 ][/tex]
[tex]r_1 = \sqrt{0.00038125}[/tex]
[tex]r_1 = 0.0195[/tex]
Also
[tex]r_2^2 = 0.015^2 + 0.018^2[/tex]
[tex]r_2 = \sqrt{0.015^2 + 0.018^2}[/tex]
[tex]r_2 = \sqrt{0.000549[/tex]
[tex]r_2 = 0.0234[/tex]
Now, Electric Potential at P in the second image below will be;
Vp = V1 + V2
[tex]Vp = \dfrac{kq_1}{r_1} + \dfrac{kq_2}{r_2}[/tex]
we substitute
Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )
Vp = 2123076.923 + ( -762962.962 )
Vp = 2123076.923 -792307.692
Vp = 1330769.23 V
Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V
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Please help 25 points!
Three waves with frequencies of 1 Hertz (Hz), 3 Hz, and 9Hz travel at the same speed. Which of the following statements is correct?
A. The 1 Hz wave contains the most energy.
B. The crests of all three waves are of equal height.
C. The wavelength of the 9Hz wave is three times that of the 3 Hz wave.
D. The 1 Hz wave has the longest wavelength.
Answer:
B
Explanation:
The crest of all three waves are of equal height
A particle move in the xy plane so that its position vector r=bcosQi +bsinQj+ ctk, where b, Q and c are constants. show that the partial move with constant speed.
Answer:
The speed of this particle is constantly [tex]c[/tex].
Explanation:
Position vector of this particle at time [tex]t[/tex]:
[tex]\displaystyle \mathbf{r}(t) = b\, \cos(Q)\, \mathbf{i} + b\, \sin(Q) \, \mathbf{j} + c\, t\, \mathbf{k}[/tex].
Write [tex]\mathbf{r}(t)[/tex] as a column vector to distinguish between the components:
[tex]\mathbf{r}(t) = \begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}[/tex].
Both [tex]b[/tex] and [tex]Q[/tex] are constants. Therefore, [tex]b\, \cos(Q)[/tex] and [tex]b \sin (Q)[/tex] would also be constants with respect to [tex]t[/tex]. Hence, [tex]\displaystyle \frac{d}{dt}[b\, \cos(Q)] = 0[/tex] and [tex]\displaystyle \frac{d}{dt}[b\, \sin(Q)] = 0[/tex].
Differentiate [tex]\mathbf{r}(t)[/tex] (component-wise) with respect to time [tex]t[/tex] to find the velocity vector of this particle at time [tex]t\![/tex]:
[tex]\begin{aligned}\mathbf{v}(t) &= \frac{\rm d}{{\rm d} t} [\mathbf{r}(t)] \\ &=\frac{\rm d}{{\rm d} t} \left(\begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}\right) \\ &= \begin{bmatrix}\displaystyle \frac{d}{dt}[b\, \cos(Q)] \\[0.5em] \displaystyle \frac{d}{dt}[b\, \sin(Q)]\\[0.5em]\displaystyle \frac{d}{dt}[c \cdot t]\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\end{aligned}[/tex].
The speed [tex]v[/tex] (a scalar) of a particle is the magnitude of its velocity :
[tex]\begin{aligned}v(t) &= \| \mathbf{v}(t) \| \\ &= \left\|\begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\right\| \\ &= \sqrt{0^2 + 0^2 + c^2} = c\end{aligned}[/tex].
Therefore, the speed of this particle is constantly [tex]c[/tex] (a constant.)
Albert Bandura emphasized the idea of __________, which is the belief one has in one’s own ability to succeed. A. operant conditioning B. determinism C. self-efficacy D. self-worth
Answer:
C
Explanation:
Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.
What is Self efficacy?
A person's self-efficacy relates to their confidence in their ability to carry out the behaviors required to achieve particular performance goals (Bandura, 1977, 1986, 1997).
The belief in one's capacity to exercise control over one's own motivation, behavior, and social environment is known as self-efficacy. The goals for which people strive, the amount of effort put out to obtain goals, and the possibility of achieving particular levels of behavioral performance are all influenced by these cognitive self-evaluations.
Self-efficacy beliefs, unlike conventional psychological notions, are anticipated to change according to the operating domain and the environment in which an action occurs.
Therefore, Albert Bandura emphasized the idea of Self efficacy which is the belief one has in one’s own ability to succeed.
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If you have a 0.125 kg lead piece at
20.0°C, how much heat must you
add to melt it? (Remember, you
must warm it to its melting point
first.)
Material
Lead
Melt Pt (°C)
327
L (1/kg)
2.32.104
Boil Pt (°C) Lv (1/kg)
1750 8.59.105
c (1/(kg*c)
128
(Unit = J)
Answer:
7,812 J
Explanation:
Using the relation:
Q = mcΔθ
Q = quantity of heat
C = specific heat capacity of lead
Δθ = temperature change (T2 - T1)
M = mass of substance
Q = mass * specific heat * Δθ
Q = 0.125kg * 128 * (327 – 20)
Q = 0.125 * 128 * 307
Q = 4912 J
For melting:
Q = mass * Hf
0.125 * (2.32 * 10^4)
= 2,900 J
Total = 4,912 J + 2,900 J = 7,812 J
Imagin you have mixed together some sand and salt Based on the venn diagram this mixture would be placed where
Answer:
a
Explanation:
ajjkiikkkkkhglutfgkitfgghiiij
Question 3 (5 points)
Yissel was going to be late to Mr. Scharff's science class. Just as the bell was about to ring. Vissel ran the last little bit of the hallway at 2.47
meters/second for 8 seconds to beat the bell. How far away was Yissel from Mr. Scharff's classroom when she started to run?
A rocket, with a mass of 5100 kg, has an engine that provides a net upward force of 8.0 x 10^5 N. It starts from rest and reaches a maximum speed of 900 m/s. How long does it take to reach that maximum velocity?
Answer:
5.7375 seconds
Explanation:
The computation of the time required to reach that maximum velocity is shown below:
Given that
Mass = m = 5100 kg
Net upward force F = 8 × times 10^5 N
Initial speed = V_i = 0
Maximum speed = V = 900 m.s
Based on the above information
Impluse J = m(V - V_i)
= 5100 (900 - 0)
= 459 × 10^4 kg m.s
As we know that
J = FT
So
T = J ÷ F
= (459 × 10^4) ÷ (8 × 10^5)
= 5.7375 seconds
A spring with a constant of 76 N/m is extended by 0.9 m. How much energy is stored in the extended spring?
Answer:
[tex]E=30.78\ J[/tex]
Explanation:
The force constant of the spring, k = 76 N/m
The extension in the spring, x = 0.9 m
We need to find the energy is stored in the extended spring. The energy stored in the spring is given by :
[tex]E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 76\times (0.9)^2\\\\E=30.78\ J[/tex]
So, 30.78 J of energy is stored in the spring.
if a current of 5A flows for 2minutes, find the quantity of electricity transfered
because there the sperm and eggs are combining together to produce
so thats why they look alike
Answer:
yes. that is how a baby is conceived.
what is the effect of divorce on females?
Answer:
Numerous studies have shown that the economic costs of divorce fall more heavily on women. After separation, women experience a sharper decline in household income and a greater poverty risk (Smock 1994; Smock and Manning
Answer:
sadness and stress...................
39. What is the change in momentum for a 5,000 kg ship in
outer space that experiences no net force over a 1 hr
period?
Answer:
Change in momentum is zero.
Explanation:
The following data were obtained from the question:
Mass (m) = 5000 kg
Time (t) = 1 h
Net force (F) = 0
Change in momentum =?
Force = Rate of change of momentum
0 = change in momentum
Change in momentum = 0
We can see from the above illustration that the net force is zero. Thus, the change in momentum is also zero.
Which statement best compares potential and kinetic energy?
O Objects always have more potentiał energy than kinetic energy.
O Kinetic energy increases and potential energy decreases when the velocity of an object increases
O Only potential energy decreases when an object's height increases.
O Objects always have more kinetic energy than potential energy.
Answer:
Kinetic energy increases and potential energy decrease when velocity of an object increase.
A soccer ball was kicked over the edge of a wall and traveled 35 m horizontally at a speed of 5.6m/s. Calculate the vertical height of the wall.
Answer:
Are you sure it was soccer ball? Or meine hearts
Explanation: