Answer: 2.80 N/C
Explanation: In order to calculate the electric firld inside the solid cylinder
non conductor we have to use the Gaussian law,
∫E.ds=Q inside/ε0
E*2πrL=ρ Volume of the Gaussian surface/ε0
E*2πrL= a*r^2 π* r^2* L/ε0
E=a*r^3/(2*ε0)
E=6.2 * (0.002)^3/ (2*8.85*10^-12)= 2.80 N/C
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]
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
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.
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.
You drive 3.4 km in a straight line in a direction 6.9º east of north. If an alternate route to this same destination takes you straight east and then turns directly north to arrive at the same point, find the distance you would have to drive north.
Answer:
You would have to drive in north direction 0.41km
Explanation:
We can solve this with trigonometry, We know that:
[tex]sin(\alpha )=\frac{opposite}{hypotenuse}\\where: \\opposite=north\\hypotenuse=distance[/tex]
[tex]North=sin(6.9^o)*3.4km=0.41km[/tex]
So the distance I will have to drive in north direction is 0.41km
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 daring stunt woman sitting on a tree limb wishes to drop vertically onto a horse gallop ing under the tree. The constant speed of the horse is 12.8 m/s, and the woman is initially 2.23 m above the level of the saddle. How long is she in the air? The acceleration of gravity is 9.8 m/s2 Answer in units of s What must be the horizontal distance between the saddle and limb when the woman makes her move? Answer in units of m.
Answer:
0.67 seconds
8.576 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.8 m/s²
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 2.23=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2.23\times 2}{9.8}}\\\Rightarrow t=0.67\ s[/tex]
Time taken by the stunt woman to drop to the saddle is 0.67 seconds which is the time she will stay in the air.
Speed of the horse = 12.8 m/s
Distance = Speed × Time
⇒Distance = 12.8×0.67
⇒Distance = 8.576 m
Hence, the distance between the horse and stunt woman should be 8.576 m when she jumps.
The stuntwoman will be in the air for approximately 0.677 seconds, and the horizontal distance or freefall between the saddle and the limb when the woman makes her move must be approximately 8.668 meters.
Explanation:To find out how long the stuntwoman is in the air, we need to apply the physics concept of free fall. The equation for the time of fall under gravity is sqrt(2h/g), where h is the height and g is the acceleration due to gravity. If we substitute the given values, we get √((2 × 2.23m)/9.8m/s²) = 0.677s.
Next, we need to find horizontal distance between the saddle and the limb when the woman makes her move. Since horizontal distance is simply speed × time, and the speed of the horse is constant, we find that 0.677s × 12.8m/s = 8.668m. So, the branch from which the stunt woman drops must be 8.668m in front of the horse when she drops.
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What properties does a loud, shrill whistle have? a.) high amplitude, high frequency
b.) high amplitude, low frequency
c.) low amplitude, high frequency
d.) low amplitude, low frequency
Answer:
a.) high amplitude, high frequency
Explanation:
Frequency and amplitude are properties of sound. Varying these properties changes how people perceive sound.
While hearing sound of a particular frequency we call it pitch i.e., the perception of a frequency of sound.
High pitch means high frequency and high frequency is perceived to have a shrill sound.
The loudness of a sound is measured by the intensity of sound i.e., the energy the sound possesses per unit area. As the amplitude increases the intensity increases. So, a loud sound will have higher density.
Hence, the loud shrill whistle will have high frequency and high amplitude.
Answer:
B
Explanation:
I think this is right.
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.
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
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]
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
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
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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]
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.
Calculate the frequency of each of the following wave lengths of electromagnetic radiation. Part A 488.0 nm (wavelength of argon laser) Express your answer using four significant figures. ν1 ν 1 = nothing s−1 Request Answer Part B 503.0 nm (wavelength of maximum solar radiation) Express your answer using four significant figures.
The frequency of electromagnetic radiation are:
Part A: 6.1432 × 10¹⁴ Hz
Part B: 5.96× 10¹⁴Hz.
The frequency of electromagnetic radiation can be calculated using the speed of light formula:
[tex]v=\frac{c}{\lambda}[/tex]
Where: v is the frequency in hertz (Hz)
c is the speed of light in a vacuum (299,792,458m/s)
λ is the wavelength in meters (m)
Given the wavelengths in nanometers (nm), we need to convert them to meters by dividing by 10⁹ (since 1 nm = 10⁻⁹m).
Part A: Wavelength = 488.0 nm
λ₁ = 488.0/10⁹
=4.88 × 10⁻⁷
v₁ = c/λ₁
=299,792,458/4.88 × 10⁻⁷
= 61432880.7377× 10⁷
=6.1432 × 10¹⁴
Part B:
Wavelength = 503.0 nm
λ₂ = 503.0/10⁹
=5.03 × 10⁻⁷m
v₂=c/λ₂
=299,792,458/5.03 × 10⁻⁷
=59600886.2× 10⁷
=5.96× 10¹⁴
Hence, the frequency of electromagnetic radiation is 5.96× 10¹⁴Hz.
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The frequency of a wave with a wavelength of 488.0 nm (argon laser) is approximately 6.1475 x 10¹⁴ Hz. The frequency of a wave with a wavelength of 503.0 nm (maximum solar radiation) is approximately 5.9682 x 10¹⁴ Hz.
Explanation:The wavelength and frequency of electromagnetic waves are inversely related by the speed of light equation, c = λ ν, where c = 3×10⁸ m/s (speed of light), λ = wavelength, and ν = frequency. To find the frequency of a wave when the wavelength is given, rearrange the equation to ν = c/λ. Applying the equation we get:
Part A: ν₁ = 3×10⁸ m/s ÷ 488.0 x 10⁻⁹ m = 6.1475 x 10¹⁴ Hz. Part B: ν₂ = 3×10⁸ m/s ÷ 503.0 x 10⁻⁹ m = 5.9682 x 10¹⁴ Hz. Learn more about Electromagnetic Waves here:
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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
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
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
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]
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.
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.
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.
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 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.
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
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]
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.
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]
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.