Answer:
Explanation:
The ball is going down with velocity. It must have momentum . It is stopped by
catcher so that its momentum becomes zero . There is change in momentum . So force is applied on the ball by the gloves .
The rate of change of momentum gives the magnitude of force. This force must be in upward direction to stop the ball. So force is in positive direction .
Let us measure the force applied on the ball .
Final velocity after the fall by 66 m
V = √ 2gh
= √ 2x9.8 x 66
35.97 m /s
Momentum = m v
0.145 x 35.97
= 5.2156 kgms⁻¹
Change in momentum
= 5.2156 - 0
= 5.2156
Rate of change of momentum
= Change of momentum / time = 5.2156 / .0118
= 442 N
if R be the force exerted by gloves to stop the ball
R - mg represents the net force which stops the ball so
R - mg = 442
R = 442 + mg
= 442 + .145 x 908
443.43 N
According to Newton's second law of motion rate of change of momentum is equal to the applied force. The force exerted by the glove of the player on the ball will be 497.4 N.
What is Newton's second law of motion?According to Newton's second law of motion rate of change of momentum is equal to the applied force.
Momentum is given by the product of mass and velocity. it is denoted by P.it leads to the impulsive force.
P = mv
ΔP = mΔv
[tex] \rm{ F = \frac{\delP}{\delt} } [/tex]
[tex]V = \sqrt{2gh} [/tex]
[tex]\rm{V = \sqrt{2\times9.81\times60} [/tex]
v = 34.31 m/sec
ΔV = v -u
ΔV = 34.31-0
ΔV = 34.31 m /sec.
Δt = 0.0118 sec
[tex]\rm{\frac{\delv}{\delt} } [/tex]= 34.31
F = [tex]m\rm{\frac{\delv}{\delt} }[/tex]
F = [tex] 0.145\times3430 [/tex]
F = 497.4 N
Thus the force exerted by the glove of the player on the ball will be 497.4 N.
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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.
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.
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.
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 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.
An archer fires an arrow at a castle 230 m away. The arrow is in flight for 6 seconds, then hits the wall of the castle and sticks 16 m above the ground. You may ignore air resistance. a) What was the initial speed of the arrow? b) At what angle above horizontal was the arrow shot?
Answer:
(a). The initial speed of the arrow is 49.96 m/s.
(b). The angle is 39.90°.
Explanation:
Given that,
Horizontal distance = 230 m
Time t = 6 sec
Vertical distance = 16 m
We need to calculate the horizontal component
Using formula of horizontal component
[tex]R =u\cos\theta t [/tex]
Put the value into the formula
[tex]\dfrac{230}{6} = u\cos\theta[/tex]
[tex]u\cos\theta=38.33[/tex].....(I)
We need to calculate the height
Using vertical component
[tex]H=u\sin\theta t-\dfrac{1}{2}gt^2[/tex]
Put the value in the equation
[tex]16 =u\sin\theta\times6-\dfrac{1}{2}\times9.8\times6^2[/tex]
[tex]u\sin\theta=\dfrac{16+9.8\times18}{6}[/tex]
[tex]u\sin\theta=32.06[/tex].....(II)
Dividing equation (II) and (I)
[tex]\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}[/tex]
[tex]\tan\theta=0.8364[/tex]
[tex]\theta=\tan^{-1}0.8364[/tex]
[tex]\theta=39.90^{\circ}[/tex]
(a). We need to calculate the initial speed
Using equation (I)
[tex]u\cos\theta\times t=38.33[/tex]
Put the value into the formula
[tex]u =\dfrac{230}{6\times\cos39.90}[/tex]
[tex]u=49.96\ m/s[/tex]
(b). We have already calculate the angle.
Hence, (a). The initial speed of the arrow is 49.96 m/s.
(b). The angle is 39.90°.
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
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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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
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 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 20-lb force acts to the west while an 80-lb force acts 45° east of north. The magnitu 80 lb 70 lb 067 lb 100 lb
Answer:
option C
Explanation:
given,
force act on west = 20 lb
force act at 45° east of north = 80 lb
magnitude of force = ?
∑ F y = 80 cos 45⁰
F y = 56.57 lb
magnitude of forces in x- direction
∑ F x = -20 + 80 sin 45⁰
= 36.57 lb
net force
F = [tex]\sqrt{F_x^2+F_y^2}[/tex]
F = [tex]\sqrt{56.57^2+36.57^2}[/tex]
F = 67.36 lb≅ 67 lb
hence, the correct answer is option C
In a vacuum, two particles have charges of q1 and q2, where q1 = +3.2 µC. They are separated by a distance of 0.28 m, and particle 1 experiences an attractive force of 2.9 N. What is q2 (magnitude and sign)?
Answer:
q2 = - 8 × [tex]10^{-6}[/tex] C
negative sign because attract together
Explanation:
given data
q1 = +3.2 µC = 3.2 × [tex]10^{-6}[/tex] C
distance r = 0.28 m
force F = 2.9 N
to find out
q2 (magnitude and sign)
solution
we know that here if 2 charge is unlike charge
than there will be electrostatic force of attraction , between them
now we apply coulomb law that is
F = [tex]\frac{q1q2}{4\pi \epsilon *r^{2}}[/tex] ............1
here we know [tex]\frac{1}{4\pi \epsilon}}[/tex] = 9 × [tex]10^{9}[/tex] Nm²/C²
so from equation 1
2.9 = 9 × [tex]10^{9}[/tex] × [tex]\frac{3.2*10^{-6}*q2}{0.28^{2}}[/tex]
q2 = [tex]\frac{2.9*0.28^2}{9*10^9*3.2*10^{-6}}[/tex]
q2 = - 8 × [tex]10^{-6}[/tex] C
Final answer:
Using Coulomb's Law, it's found that the magnitude of charge q2 is 3.2 µC. Since the force experienced by particle 1 is attractive, q2 must have an opposite sign to q1. Therefore, q2 is -3.2 µC.
Explanation:
To determine the magnitude and sign of charge q2, we can use Coulomb's Law, which states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. The equation for Coulomb's Law is F = k * |q1 * q2| / r², where F is the force between the charges, k is Coulomb's constant (8.988 × 10⁹ N·m²/C²), q1 and q2 are the charges, and r is the distance between the charges.
Given that q1 = +3.2 µC (+3.2 × 10⁻⁶ C), r = 0.28 m, and F = 2.9 N, we can rearrange the equation to solve for q2:
|q2| = (F * r²) / (k * |q1|).
Substituting the given values into the equation, we get:
|q2| = (2.9 N * (0.28 m)²) / (8.988 × 10⁹ N·m²/C² * 3.2 × 10⁻⁶ C),
|q2| = 3.2 µC or -3.2 µC. Because particle 1 experiences an attractive force, it implies that q2 must have an opposite sign to q1. Thus, q2 = -3.2 µC.
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 woman is reported to have fallen 144 ft from the 17th floor of a building, landing on a metal ventilator box, which she crushed to a depth of 18.0 in. She suffered only minor injuries. Neglecting air resistance, calculate (a) the speed of the woman just before she collided with the ventilator, (b) her average acceleration while in contact with the box, and (c) the time it took to crush the box.
Answer:
(a) 96 ft/s
(b) - 3072 ft/s^2
(c) 0.03125 s
Explanation:
h = 144 ft
u = 0 ft/s
g = 32 ft/s^2
(a) let she strikes the box with velocity v.
Use third equation of motion
[tex]v^{2}=u^{2}+2gh[/tex]
[tex]v^{2}=0^{2}+2\times 32 \times 144[/tex]
v = 96 ft/s
(b) Let the average acceleration is a.
initial velocity, u = 96 ft/s
final velocity, v = 0
h = 18 in = 1.5 ft
Use third equation of motion
[tex]v^{2}=u^{2}+2ah[/tex]
[tex]0^{2}=96^{2}+2\times a \times 1.5[/tex]
a = - 3072 ft/s^2
(c) Let the time taken is t.
Use first equation of motion
v = u + at
0 = 96 - 3072 x t
t = 0.03125 second
Using basic equations of motion, we calculated the speed of the woman just before impact with the ventilator as 29.3 m/s, her average acceleration during the impact as 930 m/s², and the time it took to crush the box as approximately 0.0315 seconds.
Explanation:The subject of this question is physics, specifically dealing with concepts of kinematics and mechanics. The falling woman problem requires application of the laws of free fall in physics along with some basic algebraic manipulation, and is quite suitable for a high school student.
(a) The speed of the woman just before she collided with the ventilator can be calculated using the formula v=√(2g h) where g is the gravity (9.8 m/s²) and h is the height (144 ft, which is approximately 43.9 m). Plugging in these values gives us v=√(2*9.8*43.9) ≈ 29.3 m/s.
(b) Knowing the depth the box was crushed (converted to meters) we can use the third equation of motion v² = u² + 2a s (s is displacement) we get a = (v² - u²) / 2s = (29.3 m/s)² / 2*0.46m = 930 m/s² as the average acceleration.
(c) Finally, using v = u + at, as initial speed is 0, we get time t = v/a = 29.3m/s / 930m/s² ≈ 0.0315 seconds.
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A rectangular container measuring 20cm x 30cm x 50cm is filled with water. What is the mass of this volume of water in kilograms? A. 30 kg B. 30,000 kg C. 30 milligrams D. 30,000 lbs
Answer:
A. 30 kg
Explanation:
As we know that,
[tex]1 litre=1000cm^{3}[/tex]
And 1 litre is equivalent to 1kg.
Given that, The volume of the rectangular container is,
[tex]V=20cm\times 30cm\times 50cm\\V=30000cm^{3}[/tex]
And this volume will be equal to [tex]V=30000cm^{3}=30litres[/tex]
And this litres in kg will be equal to,
[tex]V=30litres=30 kg[/tex]
Therefore the mass of this volume of water is 30 kg.
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
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:
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A system of 1610 particles, each of which is either an electron or a proton, has a net charge of −5.376×10^−17 C. How many electrons are in this system? What is the mass of this system?
Answer:
Number of electrons in the system = 973.
Total mass of the system = [tex]\rm 1.064\times 10^{-24}\ kg.[/tex]
Explanation:
Assumptions:
[tex]\rm n_e[/tex] = number of electrons in the system.[tex]\rm n_p[/tex] = number of protons in the system.[tex]\rm q_e[/tex] = charge on an electron = [tex]\rm -1.6\times 10^{-19}\ C.[/tex][tex]\rm q_p[/tex] = charge on a proton = [tex]\rm +1.6\times 10^{-19}\ C.[/tex][tex]\rm m_e[/tex] = mass of an electron = [tex]\rm 9.11\times 10^{-31}\ kg.[/tex][tex]\rm m_p[/tex] = mass of a proton = [tex]\rm 1.67\times 10^{-27}\ kg.[/tex]Given:
Total number of particles in the system, N = 1610.Net charge on the system, q = [tex]\rm -5.376\times 10^{-17}\ C.[/tex]Since, the system is comprised of electrons and protons only, therefore,
[tex]\rm N = n_e+n_p\\n_p=N-n_e\ \ \ \ \ \ ................\ (1).[/tex]
The net charge on the system can be written in terms of charges on electrons and protons as
[tex]\rm q=n_eq_e+n_pq_p\ \ \ \ \ ...................\ (2).[/tex]
Putting the value of (2) in (1), we get,
[tex]\rm q=n_eq_e+(N-n_e)q_p\\q=n_eq_e+Nq_p-n_eq_p\\q=n_e(q_e-q_p)+Nq_p\\n_e(q_e-q_p)=q-Nq_p\\n_e=\dfrac{q-Nq_p}{q_e-q_p}\\=\dfrac{-5.3756\times 10^{-17}-1610\times 1.6\times 10^{-19}}{-1.6\times 10^{-19}-1.6\times 10^{-19}}=972.98\\\Rightarrow n_e\approx 973\ electrons.[/tex]
It is the number of electrons in the system.
Therefore, the number of protons is given by
[tex]\rm n_p = N-n_e=1610-973=637.[/tex]
The total mass of the system is given by
[tex]\rm M=n_em_e+n_pm_p\\=(973\times 9.11\times 10^{-31})+(637\times 1.67\times 10^{-27})\\=1.064\times 10^{-24}\ kg.[/tex]
Final answer:
The system consists of 336 electrons and 1274 protons, given the net charge. To calculate the total mass, multiply the number of each particle by its respective mass and then sum the results., which gives 5192.362×10−31 kg as answer.
Explanation:
To find the number of electrons in a system with a net charge, we can use the formula:
Total number of electrons = (Total net charge) / (Charge per electron).Given that the net charge of the system is -5.376×10−17 C and the charge of an electron is approximately -1.6022×10−19 C, we can calculate the total number of electrons using the formula:
Number of electrons = (-5.376×10−17 C) / (-1.6022×10−19 C/electron)
Number of electrons = 3.3555×102
Since we can’t have a fraction of an electron, we round to the nearest whole number, which is 336 electrons.
To determine the mass of the system, first, we need to find the number of protons, which would be 1610 - 336 electrons = 1274 protons. Now we multiply the number of protons by the mass of a proton and the number of electrons by the mass of an electron to get the total mass:
Mass of electrons = 336 × 9.11×10−31 kg
Mass of protons = 1274 × 1.673×10−27 kg
Adding these two gives the total mass of the system, which is 5192.362×10−31 kg.
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 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
You have a rod with a length of 146.4 cm. You prop up one end on a brick which is 3.8 cm thick. Your uncertainty in measuring these distances is ±0.05 cm. What is the angle that the rod makes with the table?
_______ degrees
What is the uncertainty in that angle?
________ degrees
Answer:
[tex]\partial \theta = 0.003[/tex]
Explanation:
we know that
[tex]sin\theta = \frac{3.8}{146.4}[/tex]
[tex]\theta = sin^{-1} \frac{3.8}{146.4}[/tex]
[tex]\theta = 1.484°[/tex]
[tex]\theta = 1.484° *\frac{\pi}{180} = 0.0259 radians[/tex]
as we see that [tex]sin\theta = \theta[/tex]
relative error[tex] \frac{\partial \theta}{\theta} = \frac{\partial X}{X_1} +\frac{\partial X}{X_2}[/tex]
Where X_1 IS HEIGHT OF ROCK
[tex]X_2[/tex] IS THE HEIGHT OF ROAD
[tex]\partial X[/tex] = uncertainity in measuring distance
[tex]\partial X = 0.05[/tex]
Putting all value to get uncertainity in angle
[tex]\frac{\partial \theta}{0.0259} = \frac{0.05}{3.8} +\frac{0.05}{146.4}[/tex]
solving for [tex]\partial \theta[/tex] we get
[tex]\partial \theta = 0.003[/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]
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]
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.
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.
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.
A professional diver steps off of a cliff that is 18 m high. Draw a sketch of the cliff, defining your origin and final position. (Careful with negative and positive signs.) Unlike the WB assignment, assume the diver jumps up first and has an initial vertical velocity is 4 m/s. (Ignore air resistance.) (a) How long does it take the diver to hit the water? (b) What's the diver's velocity on impact with the water? (Careful with negative and positive signs.)
Answer:
19.2 m/s
Explanation:
We set a frame of reference with the origin at the cliff top and the positive X axis pointing down.
Then the initial position is:
X0 = 0
The initial speed is:
V0 = -4 m/s
It is negative because it is speed upwards and the frame of reference is positive downwards.
Since the diver is in free fall, he is affected only by the acceleration of gravity, we can consider him moving under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
He will hit the water at X = 18 m, so:
18 = 0 - 4 * t + 1/2 * 9.81 * t^2
4.9 * t^2 - 4 * t - 18 = 0
Solving this equation electronically:
t = 2.37 s
The diver will hit the water 2.37 s after jumping.
The equation for speed under constant acceleration is:
V(t) = V0 + a * t
V(2.37) = -4 + 9.81 * 2.37 = 19.2 m/s
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
An uncharged metal sphere hangs from a nylon thread. When a positively charged glass rod is brought close to the metalsphere, the sphere is drawntoward the rod. But if the sphere touches the rod, it suddenly flies away from the rod. Explain why the sphere is first attracted and then repelled.
When a charged rod is brought near a neutral metal sphere, the opposite charges are attracted towards the rod, causing an initial attraction. However, when the sphere touches the rod, the charges redistribute, leading to like charges repelling each other with a stronger force than the attraction between opposite charges, resulting in the sphere being repelled.
Explanation:When a charged rod is brought near a neutral substance, the distribution of charge in atoms and molecules is shifted slightly. Opposite charge is attracted nearer the external charged rod, while like charge is repelled. Since the electrostatic force decreases with distance, the repulsion of like charges is weaker than the attraction of unlike charges, and so there is a net attraction.
Thus, when a positively charged glass rod is brought close to the neutral metal sphere, the opposite charges in the metal sphere are attracted towards the rod. However, when the sphere touches the rod, it suddenly flies away from the rod. This is because the charges in the metal sphere redistribute, and now the like charges repel each other with a stronger force compared to the attraction between the opposite charges.
This repulsion causes the metal sphere to be repelled away from the rod, creating a net repulsive force.
The metal sphere is first attracted to the positively charged glass rod due to induction, and then repelled after touching the rod due to the repulsion between like charges.
When the positively charged glass rod is brought close to the uncharged metal sphere, the sphere experiences electrostatic induction. The electrons in the metal sphere are repelled by the positive charges on the rod and move to the far side of the sphere, leaving the side closest to the rod with a net positive charge. This positive side of the sphere is attracted to the negative charges in the rod (or the electrons in the rod are attracted to the positive side of the sphere), causing the sphere to be drawn toward the rod.
As the sphere gets closer to the rod, the attraction increases until they make contact. When the sphere touches the rod, electrons from the rod flow onto the sphere, as the rod has a surplus of electrons due to its negative induction on the side opposite the sphere. This transfer of electrons results in the sphere acquiring a net negative charge, as it gains more electrons.
Once the sphere has the same negative charge as the side of the rod it touched, the electrostatic force of repulsion between the like charges takes over. The negatively charged sphere is now repelled by the negatively charged side of the glass rod. This repulsion is strong enough to overcome the gravitational force and the nylon thread's tension, causing the sphere to suddenly fly away from the rod.