Answer:
(A) - 0.273 kg m /s
(B) - 0.546 kg m /s
Explanation:
mass of paintball, m = 0.0032 kg
initial velocity, u = 85.3 m/s
(A) Momentum is defined as the product of mass of body and the velocity of body.
initial momentum, pi = mass x initial velocity
pi = 0.0032 x 85.3 = 0.273 kg m /s
Finally the ball stops, so final velocity, v = 0 m/s
final momentum, pf = mass x final velocity = 0.0032 x 0 = 0 m/s
change in momentum = pf - pi = 0 - 0.273 = - 0.273 kg m /s
(B) initial velocity, u = + 85.3 m/s
final velocity, v = - 85.3 m/s
initial momentum, pi = mass x initial velocity
pi = 0.0032 x 85.3 = 0.273 kg m /s
final momentum, pf = mass x final velocity = - 0.0032 x 85.3 = - 0.273 kgm/s
change in momentum = pf - pi = - 0.273 - 0.273 = - 0.546 kg m /s
(C) According to the Newton's second law, the rate of change of momentum is directly proportional to the force exerted.
As the change in momentum in case (B) is more than the change in momentum in case (A), so the force exerted on the skin is more in case (B).
From a crouched position an excellent human jumper launches herself upwards. The jumpers full height is 168 cm, but in the crouched position her height is half of that value. The jumper accelerates from the crouched position until she reaches her full height. Her motion then carries her upward until her center of mass is 90 cm above its position when she is standing. At what velocity must the jumper leave the ground to reach this height? What constant acceleration must be maintained to reach this initial velocity? Assuming a constant acceleration, how long in seconds does it take to accelerate from rest to this initial velocity?
Answer:
a) 4.2 m/s
b) 13.6 m/s^2
Explanation:
She is jumping, and when her feet no longer touch the ground she is in free fall, only affected by the acceleration of gravity.
The equation for position under constant acceleration is:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
We set up a reference system that has its origin at the point her center of mass is when she is standing and the positive Y axis points upwards, then:
Y0 = 0 m
a = -9.81 m/s^2
The equation for speed under constant acceleration is:
V(t) = Vy0 + a * t
We know that when she reaches her highes point her vertical speed will be zero because that is wehn her movement changes direction. We'll call this moment t1.
0 = Vy0 + a * t1
a * t1 = -Vy0
t1 = -Vy0/a
If we replace this value on the position equation we can find her initial speed:
Y(t1) = Y0 - Vy0 * Vy0/a + 1/2 * a * (-Vy0/a)^2
Y(t1) = - Vy0^2/a + 1/2 * Vy0^2/a
Y(t1) = -1/2 * Vy0^2 / a
Vy0^2 = -2 * a * Y(t1)
[tex]Vy0 = \sqrt{-2 * a * Y(t1)}[/tex]
[tex]Vy0 = \sqrt{-2 * (-9.81) * 0.9} = 4.2 m/s[/tex]
I assume her center of mass is at half her height, so when she is standing it would be at 93 cm of the grouind, and when she is crouching at 46.5 cm.
Therefore when she jumps her centr of mass moves 0.465 m before leaving the ground.
During that trajectory she moves with acceleration.
Y(t) = Y0 + Vy0 * t + 1/2 * a *t^2
In this case her initial position is
Y0 = -0.465
Her initial speed is
Vy0 = 0
At t=t0 her position will be zero
The equation for speed under constan acceleration is
Vy(t) = Vy0 + a * t
Her speed at t0 will be 4.2 m/s
4.2 = a * t0
t0 = 4.2 / a
0 = -0.465 - 1/2 * 9.81 * (4.2 / a)^2
0.465 = 4.9 * 17.6 / a^2
a^2 = 86.2 / 0.465
[tex]a = \sqrt{185.4} = 13.6 m/s^2[/tex]
If your front lawn is 16.016.0 feet wide and 20.020.0 feet long, and each square foot of lawn accumulates 13501350 new snowflakes every minute, how much snow, in kilograms, accumulates on your lawn per hour? Assume an average snowflake has a mass of 2.20 mg.
Answer:
The mass of snow accumulated in the lawn in 1 hour equals 57.024 kilograms
Explanation:
Given
Width of lawn = 16 feet
Length of lawn = 20 feet
Thus total area of lawn equals [tex]16\times 20=320sq.feet[/tex]\
Now it is given that 1 square foot accumulates 1350 new snowflakes each minute thus number of snowflakes accumulated by 320 square feet in 1 minute equals
[tex]320\times 1350=432000[/tex]
Now it is given that average mass of each snowflake is [tex]2.20mg=2.20\times 10^{-3}g=2.20\times 10^{-6}kg[/tex]
Hence the mass accumulated per minute equals [tex]432000\times 2.20\times 10^{-6}=0.9504kg[/tex]
Now since there are 60 minutes in 1 hour thus the mass accumulated in 1 hour equals [tex]0.9504kg\times 60=57.024kg[/tex]
The time base on the oscilloscope is set for 2 ms/cm and the vertical input has a frequency of 3000 Hz. How many wave cycles will appear within the 10 cm width of the screen?
Answer:
60 cycles
Explanation:
The first thing we must do to solve the problem is to find how many cycles are presented in 1cm by multiplying the frequency by the base time of the
K=time base=2ms/cm=2x10-3s/cm
f=frecuency=3000s^-1
N=fk
N=(3000)(2x10^-3)=6cycles/cm
Ntot=6x10=60cycles
Answer:
60 wave cycles
Explanation:
As the horizontal axis in a oscilloscope represents time, the time base is simply the scale, in other words, the amount of time that each division of oscilloscope represents. Therefore, multiplying the width of the screen times the time base will give us the total amount of time graphed on the screen.
The frequency is the amount of oscillations or waves cycles per second. So, in order to find the total amount of oscillations:
[tex]10cm * \frac{0.002 s}{cm} *\frac{3000cycles}{s} = 60 cycles[/tex]
An airplane pilot wishes to fly due west. A wind of 70.0 km/h is blowing toward the south. And I need to find out the speed of the plain over the ground. It is given that the speed is in still air and the airspeed is 435.0 km/h.
Answer:
Speed of the plane over the ground is 429.33 km/hr due West and 70 km/hr towards North
Solution:
According to the question:
Wind flowing in South direction, [tex]v_{s} = 70.0 km/h[/tex]
Air speed, [tex]v_{a} = 435.0 km/h[/tex]
Now,
The velocity vector is required 70 km/h towards north in order to cancel the wind speed towards south.
Therefore,
The ground speed of the plane is given w.r.t fig 1:
[tex]v_{pg} = \sqrt{v_{a^{2}} - v_{s}^{2}}[/tex]
[tex]v_{pg} = \sqrt{435.0^{2} - 70.0^{2}} = 429.33 km/h[/tex]
Final answer:
The speed of the plane over the ground, taking into consideration a wind blowing south at 70.0 km/h and a plane airspeed of 435.0 km/h, is calculated to be approximately 438.3 km/h due west.
Explanation:
An airplane pilot wishes to fly due west, with a wind of 70.0 km/h blowing toward the south, and the plane's airspeed is 435.0 km/h. To determine the speed of the plane over the ground, we must account for both the plane's airspeed and the wind's effect.
The plane's airspeed vector (435.0 km/h) is combined with the wind's vector (70.0 km/h south) to calculate the resultant vector, which represents the plane's actual velocity relative to the ground. This calculation is a vector addition problem that requires the use of the Pythagoras theorem or vector components.
The speed of the plane over the ground (groundspeed) can be found by calculating the magnitude of the resultant vector: Groundspeed = √(air speed² + wind speed²) = √(435² + 70²) km/h, which simplifies to approximately 438.3 km/h to the west. This calculation shows the combined effect of the plane's airspeed and the wind, resulting in the plane's groundspeed.
A squirrel is trying to locate some nuts he buried for the winter. He moves 4.3 m to the right of a stone and dogs unsuccessfully. Then he moves 1.1 m to the left of his hole, changes his mind, and moves 6.3 m to the right of that position and digs a second hole. No luck. Then he moves 8.0 m to the left and digs again. He finds a nut at last. What is the squirrel's total displacement from his starting point
Answer:
The total displacement from the starting point is 1.5 m.
Explanation:
You need to sum and substract, depending on the movement (to the right, sum; to the left, substract).
First, it moves 4.3 m right and return 1.1 m. So the new distance from the starting point is 3.2 m.
Second, it moves 6.3 m right, so the new distance is 9.5 m.
Finally it moves 8 m to the left, so 9.5 m - 8 m= 1.5 m.
Summarizing, at the end the squirrel is 1.5 m from its starting point.
For a growing quantity to reach a value 32 times its initial value, how many doubling times are required? A) 4
B) 5
C) 16
D) 8
Answer:
B) n=5
Explanation:
We call the initial value Xo. We start to double this initial value
X1=2*Xo n=1
We double again:
X2=2*X1=2*(2Xo) n=2
X2=2*X1=2^2*Xo n=2
In general:
Xn=(2^n)*Xo
If we want to reach a value 32 times its initial value:
2^n=32
then: n=5
2^5=2*2*2*2*2=32
At a certain elevation, the pilot of a balloon has a mass of 125 lb and a weight of 119 lbf. What is the local acceleration of gravity, in ft/s2, at that elevation? If the balloon drifts to another elevation where g = 32.05 ft/s2, what is her weight, in lbf, and mass, in lb?
Final answer:
The local acceleration of gravity at the pilot's initial elevation is 0.952 ft/s². At a different elevation with gravity at 32.05 ft/s², her weight would be 4006.25 lbf, but her mass remains the same at 125 lb.
Explanation:
To find the local acceleration of gravity, we use the formula weight = mass × gravity. The pilot's weight is 119 lbf, and her mass is 125 lb. We rearrange the formula to find gravity: gravity = weight / mass, which gives us 119 lbf / 125 lb.
The local acceleration of gravity at the pilot's elevation is therefore 0.952 ft/s². Now, if the pilot drifts to another elevation where gravity is 32.05 ft/s², her weight in pounds-force would be her mass times the new acceleration due to gravity, which is 125 lb × 32.05 ft/s². Hence, her new weight would be 4006.25 lbf. Her mass remains unchanged as mass is not dependent on gravity.
Final answer:
The local acceleration of gravity at the given elevation is 0.952 ft/s². When the balloon drifts to another elevation with an acceleration of gravity of 32.05 ft/s², the pilot's weight is 4006.25 lbf and the mass is 125 lb.
Explanation:
At a certain elevation, the pilot's weight is less than the mass due to the reduction in the acceleration of gravity. To find the local acceleration of gravity, we need to use the equation:
weight = mass * acceleration of gravity
For the given values, the pilot's weight is 119 lbf, and the mass is 125 lb. Rearranging the equation, we have:
acceleration of gravity = weight / mass
Substituting the values, we get:
acceleration of gravity = 119 lbf / 125 lb = 0.952 ft/s²
When the balloon drifts to another elevation where the local acceleration of gravity is 32.05 ft/s², we can use the same equation to find the new weight and mass. Rearranging the equation, we have:
weight = mass * acceleration of gravity
Substituting the new acceleration of gravity and the previous mass, we get:
weight = 125 lb * 32.05 ft/s² = 4006.25 lbf
Therefore, at the new elevation, the pilot's weight is 4006.25 lbf and the mass is 125 lb.
Paintball guns were originally developed to mark trees for logging. A forester aims his gun directly at a knothole in a tree that is 4.0 m above the gun. The base of the tree is 15 m away. The speed of the paintball as it leaves the gun is 50 m/s. How far below the knothole does the paintball strike the tree? Express your answer with the appropriate units.
Answer:
The distance between knothole and the paint ball is 0.483 m.
Explanation:
Given that,
Height = 4.0 m
Distance = 15 m
Speed = 50 m/s
The angle at which the forester aims his gun are,
[tex]\tan\theta=\dfrac{4}{15}[/tex]
[tex]\tan\theta=0.266[/tex]
[tex]\cos\theta=\dfrac{15}{\sqrt{15^2+4^2}}[/tex]
[tex]\cos\theta=0.966[/tex]
Using the equation of motion of the trajectory
The horizontal displacement of the paint ball is
[tex]x=(u\cos\theta)t[/tex]
[tex]t=\dfrac{x}{u\cos\theta}[/tex]
Using the equation of motion of the trajectory
The vertical displacement of the paint ball is
[tex]y=u\sin\theta(t)-\dfrac{1}{2}gt^2[/tex]
[tex]y=u\sin\theta(\dfrac{x}{u\cos\theta})-\dfrac{1}{2}g(\dfrac{x}{u\cos\theta})^2[/tex]
[tex]y=x\tan\theta-\dfrac{gx^3}{2u^2(\cos\theta)^2}[/tex]
Put the value into the formula
[tex]y=(15\times0.266)-(\dfrac{9.8\times(15)^2}{2\times(50)^2\times(0.966)^2})[/tex]
[tex]y=3.517\ m[/tex]
We need to calculate the distance between knothole and the paint ball
[tex]d=h-y[/tex]
[tex]d=4-3.517[/tex]
[tex]d=0.483\ m[/tex]
Hence, The distance between knothole and the paint ball is 0.483 m.
Consider projectile thrown horizontally at 50 m/s from height of 19.6 meters. The projectile will take ______________ time to hit the ground as it would if it were dropped from the same height. a) More
b) Less
C)The Same
Answer:
C)The Same
Explanation:
Kinematics equation:
[tex]y=v_{oy}*t+1/2*g*t^2[/tex]
for both cases the initial velocity in the axis Y is the same, equal a zero.
So the relation between the height ant temps is the same for both cases (the horizontal velocity does not play a role)
C)The Same
The average velocity for a trip has a positive value. Isit
possible for the instantaneous velocity at any point during thetrip
to have a negative value? Justify the answer.
Answer:
Yes, it's possible.
Explanation:
The average velocity is a mean value:
[tex]Vavg=\frac{displacement}{time taken}[/tex].
during that displacement, it may occur that the acceleration would negative at any time so at that moment if the velocity goes in the same direction with the acceleration, the velocity will be negative, it may take just a few moments and then go positive again. The velocity can also take negative values if for a moment the object was going backward (opposite direction). so the average velocity only means that the major of the velocity was positive.
A small metal bead, labeled A has a charge of 25 nC. It is touched to metal bead B, initially neutral, so that the two beads share the 25 nC charge, but not necessarily equally. When the two beads are then placed 5.0 cm apart, the force between them is 5.4 x 10e-4 N. What are the charges qa and qb on the beads?
Answer:
15nC & 10 nC
Explanation:
We will use the formula:
[tex]q_{A}q_{B} = \frac{Fr^{2}}{k}[/tex]
Plugging values for F,r, and k, we get:
[tex]\frac{(5.4*10^{-4} N)(0.050m)^{2}}{9.0 * 10x^{9}N*m^{2}/C^{2}}=1.5 * 10 ^{-16} C^{2}[/tex]
Now, use the equation qB = 25nC - qA: (we know this from the problem)
[tex]q_{A}(25 nC - q_{A})=1.5 *10^{-16} C^{2}[/tex]
This is a quadratic equation that is solved to yield
qA = 10nC or qA = 15nC.
qB is of course the one that qA is not, but we do not know which is which, however that is irrelevant for the problem.
The charges have the same magnitude after sharing but are not necessarily equal, the charges on bead A and bead B are 0.0012 C each.
Given:
Charge on bead A, [tex]qa = 25\ nC = 25 \times 10^{-9}\ C[/tex]
Electric force between the beads, [tex]F = 5.4 \times 10^{-4}\ N[/tex]
Distance between the beads, [tex]r = 5.0\ cm = 0.05\ m[/tex]
Coulomb's constant, [tex]k = 8.99 \times 10^9\ N m^2/C^2[/tex]
Using Coulomb's law, we have on substituting the value:
[tex]F = k \times |q_a \times q_b| / r^2[/tex]
Substitute the values:
[tex]5.4 \times 10^{-4} = (8.99 \times 10^9) \times |(25 \times 10^{-9}) \times q_b| / (0.05)^2[/tex]
Now solve for qb:
[tex]|q_b| = (5.4 \times 10^{-4} \times (0.05)^2) / (8.99 \times 10^9 \times 25 \times 10^{-9})\\|q_b| = 0.0012\ C[/tex]
Since the charges have the same magnitude after sharing, but are not necessarily equal, the charges on bead A and bead B are 0.0012 C each.
To summarize:
Charge on bead A (qa) = 0.0012 C
Charge on bead B (qb) = 0.0012 C
To know more about the charges:
https://brainly.com/question/28721069
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Joseph DeLoach of the United States set an Olympic record in 1988 for the 200-meter dash with a time of 19.75 seconds. What was his average speed? Give your answer in meters per second and miles per hour.
Answer:
The average speed was 10.12 m/s or 22.96 mi/h
Explanation:
The average speed is defined as:
[tex]v = \frac{d}{t}[/tex] (1)
Where d is the total distance traveled and t is the passed time.
For the special case of Joseph DeLoach, he traveled a total distance of 200 meters in 19.75 seconds. Those values can be introduced in equation 1:
[tex]v = \frac{200 m}{19.75 s}[/tex]
[tex]v = 10.12 m/s[/tex]
That means that Joseph DeLoach traveled a distance of 10.12 meters per second.
To represent the result in miles per hour, it is necessary to know that 1 mile is equivalent to 1609 meters.
[tex]200 m x \frac{1 mi}{1609 m}[/tex] ⇒ 0.124 mi
it is needed to express the given time in units of hour. 1 hour is equivalent to 3600 seconds.
[tex]19.75 s x \frac{1 h}{3600 s}[/tex] ⇒ 0.0054 h
Then, equation 1 is used with the new representation of the values.
[tex]v = \frac{0.124 mi}{0.0054 h}[/tex]
[tex]v = 22.96 mi/h[/tex]
A small silver (10.5 g/cm^3) cylinder of diameter 1.4 cm and a cylinder of lead (11.3 g/cm^3) balance each other when placed on a triple beam balance. If they have the same length, what must be the diameter of the lead cylinder?
Answer:
The diameter of the lead cylinder is 1.35 cm.
Explanation:
Given that,
Density of silver = 10.5 g/cm³
Density of lead = 11.3 g/cm³
Diameter = 1.4 cm
As mass of both is equal.
Let diameter of lead [tex]d_{l}[/tex]
We need to calculate the the diameter of the lead cylinder
Using balance equation of density
[tex]V\times \rho_{s}=V\times \rho_{l}[/tex]
[tex]\dfrac{\pi\times d_{s}^2\times h}{4}\times\rho_{s}=\dfrac{\pi\times d_{l}^2\times h}{4}\times\rho_{s}[/tex]
[tex]d_{l}^2=\dfrac{d_{s}^{2}\times\rho_{s}}{\rho_{l}}[/tex]
put the value into the formula
[tex]d_{l}^2=\dfrac{(1.4\times10^{-2})^2\times10.5}{11.3}[/tex]
[tex]d_{l}=\sqrt{0.00018212}[/tex]
[tex]d_{l}=0.0135\ m[/tex]
[tex]d_{l}=1.35\ cm[/tex]
Hence, The diameter of the lead cylinder is 1.35 cm.
I know that my heat recovery ventilator consumes 83.2 Watts of power to run its fans when supplying balanced fresh air into my house. If it runs 24 hrs a day, how many kWh's of energy would it consume in a year?
Answer: 730 kWh
Explanation: To solve this problem we have to use the following considerations:
If teh heat recovery ventilator consumes 83.2 W running 24 hs we have a consumed energy of:
Energy consumed= 83.2* 24= 2 kWh
so in a year we must multiply by 365 then
Energy consumed in a year= 2kWh*365= 730 kWh
An object (with charge -2.4 µC and mass 0.026 kg) hovers at rest above the ground, its weight being held up by a uniform electric field. (a) Find the size and direction of this electric field.
(b) If the electric charge on the object is quadrupled while its mass remains the same, find the size and direction of its acceleration.
Answer:
a ) 10.61 X 10⁴ N/C
b )29.37 ms⁻² .
Explanation:
Since the weight of the object is balanced by force by electric field , the force must be acting in upward direction . In an electric field , an electron experiences a force opposite to the direction of field , therefore ,direction of electric field must be in downward direction .
Let E be the electric field . Force on the charge
= electric field x charge = E x 2.4 x 10⁻⁶
Weight acting downwards = .026 x 9.8
For balancing
.026 x 9.8 = E x 2.4 x 10⁻⁶
E = .026 x 9.8 / 2.4 x 10⁻⁶
= 10.61 X 10⁴ N/C
b )
If the charge is quadrupled , charge becomes 4 x 2.4 x 10⁻⁶
Upward force = 4 x 2.4 x 10⁻⁶ x 10.61 X 10⁴ = 101.85 x 10⁻²
Down ward force =.026 x 9.8
Net force = 101.85 x 10⁻² - .026 x 9.8
= .7637 N
acceleration = .7637 / .026 = 29.37 ms⁻² .
Zeta Anderson, futuristic super-spy for the Terran Confederation, has completed her objective of stealing intelligence from the Zorn collective. Stealthily, she slips into her space suit (with jet-pack), and slips from an airlock, headed for her stealth ship. Her jet-pack can supply her with a constant acceleration, and gravity can be neglected. When she turns on her jet-pack, how does her velocity change? Since she does not want to be going too fast (and either overshoot, or collide with her ship), how does her velocity change when she turns the jet-pack off?
Explanation:
First lets understand the 2nd law of motion by Sir Isaac Newton. According to this law,
[tex]Force = mass\times acceleration[/tex]
[tex]Velocity = acceleration\times time[/tex]
Since the spy is in space there is no medium and hence no friction to restrict the motion. Thus, when the spy turns on the jet pack, she will be accelerated and her velocity will increase. As there is no medium so no friction. So even when she turns her jet-pack off the velocity will not change. Although the acceleration will be zero but she will be moving with a constant velocity until an opposite force is applied. That can be done using reverse thrust.
A jet lands with a speed of 100 m/s and can accelerate uniformly at a rate of -5.0 m/s^2 as it comes to rest. What is the minimum length runway to land this plane? A. 1000 m
B. 20 m
C. 675 m
D. 500 m
Answer:
A ) 1000 m.
Explanation:
Here initial velocity u = 100 m /s
Final velocity v = 0
Acceleration a = -5 ms⁻²
Distance travelled = S
v² = u² + 2aS
0 = (100)² -2 x 5 S
S = 10000/ 10
=1000 m.
Red and blue light enter together into a glass plate of 10 cm. What is the distance between red and blue light whenthe light emerges from the plate. n(blue) = 1.6, n(red) = 1.3
Answer:
The shift in the color's depends on the angle of incidence, for a special case when the angle of incidence is along the normal to the surface no shift will be observed.
Explanation:
When a ray of light is incident on a medium perpendicular to it it does not undergo any refraction thus no shift will be seen.
Answer:
The distance between the emergent red and blue light is 3 cm
Solution:
As per the question:
Thickness of the glass plate, s = 10 cm = 0.1 m
Refractive index of blue light, [tex]n_{blue} = 1.6[/tex]
Refractive index of blue light, [tex]n_{red} = 1.3[/tex]
Now, to calculate the distance between red and blue light as it emerges from the plate:
We know that refractive index is given as the ratio of speed of light in vacuum, c or air to that in medium, [tex]v_{m}[/tex].
[tex]n = \frac{c}{v_{m}}[/tex]
[tex]v_{m} = \frac{c}{n}[/tex] (1)
Since, c is constant, thus
n ∝ [tex]\frac{1}{v_{m}}[/tex]
Now, the refractive index of blue light is more than that of red light thus its speed in medium is lesser than red light.
Now, time taken, t by red and blue light to emerge out of the glass slab:
[tex]s = v_{m}\times t[/tex]
[tex]t = \frac{s}{v_{blue}} = \frac{sn_{blue}}{c}[/tex]
In the same time, red light also traveled through the glass covering some distance in air say x
[tex]t' = \frac{s}{v_{red}} = \frac{sn_{red}}{c}[/tex] (2)
Time taken by red light to cover 'x' distance in vacuum is t'':
[tex]t" = \frac{x}{c}[/tex]
Now,
t = t' + t" (3)
From eqn (1), (2) and (3):
[tex]\frac{sn_{blue}}{c} = \frac{sn_{red}}{c} + \frac{x}{c}[/tex]
Now, putting appropriate values in the above eqn:
[tex]\frac{0.1\times 1.6}{c} = \frac{0.1\times 1.3}{c} + \frac{x}{c}[/tex]
[tex]\frac{0.16}{c} - \frac{0.13}{c} = \frac{x}{c}[/tex]
x = 0.03 m = 3 cm
A space vehicle is traveling at 4150 km/h relative to the Earth when the exhausted rocket motor (mass 4m) is disengaged and sent backward with a speed of 79 km/h relative to the command module (mass m). What is the speed of the command module relative to Earth just after the separation? km/h
Answer:
4213.2 Km/h
Explanation:
Given:
Initial Speed of space vehicle relative to Earth, u = 4150 km/h
Mass of rocket motor = 4m
speed of the rocket motor relative to command module , v' = 79 Km/h
Mass of the command module = m
Now,
let the speed of command module relative to earth be 'v'
From the conservation of momentum, we have
( 4m + m ) × u = m × v + (4m × (v - v'))
or
5m × 4150 = mv + 4mv - 4mv'
or
20750 = 5v - ( 4 × 79 )
or
20750 = 5v - 316
or
v = 4213.2 Km/h
A bird, accelerating from rest at a constant rate,
experiencesa displacement of 28 m in 11 s. What is the final
velocity after 11s?
Answer:
the speed of the bird is 2.8 m/s in after 11 seconds.
Explanation:
given,
displacement of bird = 28 m
time taken of the displacement of 28 m = 11 s
distance = speed × time
velocity = [tex]\dfrac{displacement}{time}[/tex]
=[tex]\dfrac{28}{10}[/tex]
= 2.8 m/s
hence, the speed of the bird is 2.8 m/s in after 11 seconds.
What is the state of an object being acted upon by an unbalanced force? A. at rest B. zero speed C. in motion with a constant velocity D. accelerated E. any of these
Answer:
D. accelerated
Explanation:
According to Newton's first law of motion, an object always remains in its state of rest or in uniform motion until an external unbalanced force is acted upon the object. This means if an external unbalanced force is acted on the object, it will come to a non-uniform motion i.e., an accelerated motion.
Let me first get you clear that uniform motion is determined by a constant velocity and a state of rest is considered by a zero velocity or speed. But, here we have an unbalanced force acting on the object. This means the object will change its velocity and hence, it has an accelerated motion.
Final answer:
The state of an object being acted upon by an unbalanced force is accelerated. This is because an unbalanced force results in a change in the velocity of the object, causing it to accelerate as per Newton's First Law of Motion.
Explanation:
The state of an object being acted upon by an unbalanced force is accelerated. According to Newton's First Law of Motion, an object at rest stays at rest, and an object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced external force. This implies that the presence of an unbalanced force (net force not equal to zero) changes the velocity of an object, causing it to accelerate.
In other words, option D, accelerated, is the correct answer. When an unbalanced force acts on an object, it results in the object's acceleration, which is a change in its velocity over time. This scenario directly contradicts situations where an object is at rest, at zero speed, or moving with a constant velocity, where no net force or balanced forces are acting on the object.
You are jogging for one hour at a speed of 5 m/s and for the final 200 m you decide to increase your velocity to 11 m/s. Assuming you were able to do so instantaneously. Once you reach that speed how long it will take to cover 200m?
Answer:
[tex]t=18.18s[/tex]
Explanation:
From the exercise we know that the person change its initial velocity from 5m/s to 11 m/s for 200m
According to the formula:
[tex]v=\frac{x}{t}[/tex]
If we want to know how much time does it take to cover 200 m at 11 m/s we need to calculate the following formula:
[tex]t=\frac{x}{v}=\frac{200m}{11m/s}=18.18s[/tex]
A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.727 g, q = 2.73 µC is located on the x axis at x = 19.3 cm, moving with a speed of 49.2 m/s in the positive y direction. For what value of Q will the moving particle execute circular motion? (Neglect the gravitational force on the particle.)
Answer:
Q = 12.466μC
Explanation:
For the particle to execute a circular motion, the electrostatic force must be equal to the centripetal force:
[tex]Fe = \frac{K*q*Q}{r^{2}} = \frac{m*V^{2}}{r}[/tex]
Solving for Q:
[tex]Q = \frac{m*V^{2}*r}{K*q}[/tex]
Taking special care of all units, we can calculate the value of the charge:
Q = 12.466μC
Two foreces act on a block of mass 4.5 kg resting on
africtionless, horizontal surface, as shown. The horizontal force
is3.7 N; The other force of 5.9 N acts at an angle of 43 degrees
fromthe horizontal. what is the magnitude of the acceleration of
theblock?
Answer:[tex]=\frac{8.014}{4.5}=1.78 m/s^2[/tex]
Explanation:
Given
mass of block(m)=4.5 kg
Horizontal force([tex] F_h[/tex])=3.7 N
another force F at angle of [tex]43^{\circ}[/tex]
if F is pulling Block then
Net Normal reaction=mg-Fsin43=40.12 N
Net Force in Horizontal direction =3.7+Fcos43
=3.7+4.31=8.014 N
thus Net acceleration is a[tex]=\frac{8.014}{4.5}=1.78 m/s^2[/tex]
Explain the meaning of the term "potential divider" as applied to the circuit?
Answer and Explanation:
A potential divider, as the name itself gives a clear description of a simplified circuit which divides the potential across different circuit elements.
The potential divider circuit divides the potential depending upon the respective value of the elements of the circuit,
Given circuit diagram shows a potential divider circuit.
The working of this circuit is as :
[tex]V' = \frac{R'}{R + R'}V[/tex]
where
V = Input voltage
V' = output voltage
A 50kg boy runs at a speed of 10.0m/s and jumpsonto a cart
originally at rest. the cart, with the boy on it, thentakes off in
the same direction in which the boy was running. ifthe cart with
the boy has a velocity of 2.5m/s, what is the mass ofthe
cart?
Answer:
the mass of the cart is 150 kg
Explanation:
given,
mass of boy(m) = 50 kg
speed of boy (v)= 10 m/s
initial velocity of cart (u) = 0
final velocity of cart(V) = 2.5 m/s
mass of the cart(M) = ?
m v + M u = (m + M ) V......................(1)
50× 10 + 0 = (50 + M ) 2.5
M =[tex]\dfrac{500}{2.5} - 50[/tex]
M = 150 Kg
hence, the mass of the cart is 150 kg
Answer:
Mass of the cart is 750 kg
Given:
Mass of the boy, m = 50 kg
Speed of the boy, v = 10.0 m/s
Final speed of the boy with the cart, v' = 2.5 m/s
Solution:
Initially the cart is at rest and since its on the ground, height, h = 0
Now, by the conservation of energy, mechanical energy before and after will remain conserved:
KE + PE = KE' + PE' (1)
where
KE = Initial Kinetic energy
KE' = Final Kinetic Energy
PE = Initial Potential Energy
PE' = Final Potential Energy
We know that:
Kinetic enrgy = [tex]\frac{1}{2}mv^{2}[/tex]
Potential energy = mgh
Since, potential energy will remain zero, thus we apply the conservation of Kinetic Energy only.
Let the mass of cart be M, thus the mass of the system, m' = 50 + M
Using eqn (1):
[tex]\frac{1}{2}mv^{2} = \frac{1}{2}m'v^{2}[/tex]
[tex]\frac{1}{2}\times 50\times 10^{2} = \frac{1}{2}(50 + M)\times 2.5^{2}[/tex]
[tex]5000 = 6.25(50 + M)[/tex]
M = 750 kg
The gravitational acceleration on the surface of the moon is 1/6 what it is on earth. If a man could jump straight up 0.7 m (about 2 feet) on the earth, how high could he jump on the moon?
Answer:
on moon he can jump 4.2 m high
Explanation:
given data
gravitational acceleration on moon a(m) = 1/6
jump = 0.7 m
to find out
how high could he jump on the moon
solution
we know gravitational acceleration on earth a = g = 9.8 m/s²
so on moon am = [tex]9.8 * \frac{1}{6}[/tex] = 1.633 m/s²
so if he jump on earth his speed will be for height 0.7 m i s
speed v = [tex]\sqrt{2gh}[/tex]
v = [tex]\sqrt{2(9.8)0.7}[/tex]
v = 3.7 m/s
so if he hump on moon
height will be
height = [tex]\frac{v^2}{2*a(m)}[/tex]
put here value
height = [tex]\frac{3.7^2}{2*1.633)}[/tex]
height = 4.2 m
so on moon he can jump 4.2 m high
A blacksmith drops a 550 °F piece of iron into a vat of 75 °F water in order to cool it to 100 °F. How many kilograms of water are needed per kilogram of iron? Assume all the thermal energy from the iron is transferred to the water and none of the water evaporates. The specific heats of water and iron are 4186 J/kg×°C and 448 J/kg×°C, respectively.
Answer:
1.93 kg water/kg iron
Explanation:
All the thermal energy from the iron is transferred to the water. In equilibrium, the temperature of both the water and the iron will be the same. The heat that an object loses or gains after a change in value in its temperature is equal to:
[tex]Q = mc*(T_f-T_o)[/tex]
Then,
[tex]-Q_{iron} = Q_{water}[/tex]
Before solving the problem, let's convert the values of temperature to Celsius:
(550°F -32)*5/9 = 287.78 °C
(75°F - 32)*5/9 = 23.89 °C
(100°F -32)*5/9 = 37.78°C
Now, we can solve:
[tex]-Q_{iron} = Q_{water}\\m_{iron}*c_{iron}*(T_o_i-T_f_i) = m_{water}*c_{water}*(T_f_w-T_o_w)\\\frac{m_{water}}{m_{iron}} = \frac{c_{iron}(T_o_i-T_f_i)}{c_{water}(T_f_w-T_o_w)} =\frac{448J/kg^oC(287.78^oC - 37.78^oC)}{4186J/kg^oC(37.78^oC-23.89^oC)}= 1.93 kg_{water}/kg_{iron}[/tex]
When leaping at an angle of 47.7° above the horizontal, a froghopper reaches a maximum height of 42.4 cm above the level ground. What was the takeoff speed for such a leap?
What horizontal distance did the froghopper cover for this world-record leap?
Answer:
39 cm /s
77.25 cm approx
Explanation:
Angle of projection θ = 47.7°
Maximum height H = 42.4 cm
Initial velocity = u =?
we know that
maximum height
H = U² x sin²θ / 2g
U² = H x 2g /sin²θ
Putting the values
U² =( 42.4 X 2 X9.8 ) / (sin47.7)²
U = 39 cm /s
Horizontal Range R = U²sin2θ / 2g
= 39 x 39 x (sin95.4) / 2 x 9.8
R = 77.25 cm approx
Final answer:
In this question, we determine the takeoff speed and horizontal distance covered by a froghopper during a leap. Calculations involve kinematic equations and trigonometry to find these values accurately.
Explanation:
Takeoff speed: To find the takeoff speed, we can use the fact that the maximum height reached by the froghopper is related to its initial velocity. By using the kinematic equation for projectile motion, we can calculate the takeoff speed to be approximately 0.75 m/s.
Horizontal distance covered: The horizontal distance covered can be determined by analyzing the horizontal component of the motion. With the takeoff angle and the calculated initial speed, the horizontal distance traveled can be found using trigonometry to be around 0.91 m.
A mail carrier parks her postal truck and delivers packages. To do so, she walks east at a speed of 0.80 m/s for 4.0 min, then north at a speed of 0.50 m/s for 5.5 min, and finally west at a speed of 1.1 m/s for 2.8 min. Define east as +x and north as +y. (a) Write unit-vector velocities for each of the legs of her journey. (b) Find unit-vector displacements for each of the legs of her journey. (c) Find her net displacement from the postal truck after her journey is complete.
Explanation:
Given that,
She walks in east,
Speed = 0.80 m/s
Time = 4.0 min
In north,
Speed = 0.50 m/s
Time = 5.5 min
In west,
Speed = 1.1 m/s
Time = 2.8 min
(a). We need to calculate the unit-vector velocities for each of the legs of her journey.
The velocity of her in east
[tex]\vec{v_{1}}=0.80\ \hat{x}\ m/s[/tex]
[tex]\vec{v_{2}}=0.50\ \hat{y}\ m/s[/tex]
[tex]\vec{v_{3}}=1.1\ \hat{-x}\ m/s[/tex]
(b). We need to calculate the unit-vector displacements for each of the legs of her journey
Using formula of displacement
[tex]\vec{d_{1}}=v_{1}\times t_{1}[/tex]
In east ,
[tex]\vec{d_{1}}=0.80\times4.0\times60[/tex]
[tex]\vec{d_{1}}=192\ \hat{x}\ m[/tex]
In north,
[tex]\vec{d_{2}}=0.50\times5.5\times60[/tex]
[tex]\vec{d_{2}}=165\ \hat{y}\ m[/tex]
In west,
[tex]\vec{d_{3}}=1.1\times2.8\times60[/tex]
[tex]\vec{d_{3}}=184.8\ \hat{-x}\ m[/tex]
(c). We need to calculate the net displacement from the postal truck after her journey is complete
[tex]\vec{d}=\vec{d_{1}}+\vec{d_{2}}+\vec{d_{3}}[/tex]
Put the value in the formula
[tex]\vec{d}=192\hat{x}+165\hat{y}+184.8\hat{-x}[/tex]
[tex]\vec{d}=7.2\hat{x}+165\hat{y}[/tex]
We need to calculate the magnitude of the displacement
[tex]d=\sqrt{(7.2)^2+(165)^2}[/tex]
[tex]d=165.16\ m[/tex]
The magnitude of the displacement is 165.16 m.
Hence, This is the required solution.