Answer:
The linear density of the wire is 0.314 g/m.
Explanation:
It is given that,
Acceleration, [tex]a=5.52\ m/s^2[/tex]
Mass of the ball, m = 112 gm
Speed of the transverse wave, v = 44.4 m/s
The speed of the transverse wave is given by :
[tex]v=\sqrt{\dfrac{T}{\mu}}[/tex]
Where
T = tension in the wire
[tex]\mu[/tex] = mass per unit length
[tex]\mu=\dfrac{T}{v^2}[/tex]
[tex]\mu=\dfrac{ma}{v^2}[/tex]
[tex]\mu=\dfrac{112\ g\times 5.52\ m/s^2}{(44.4\ m/s)^2}[/tex]
[tex]\mu=0.3136\ g/m[/tex]
or
[tex]\mu=0.314\ g/m[/tex]
So, the linear density of the wire is 0.314 g/m. Hence, this is the required solution.
An 92-kg football player traveling 5.0m/s in stopped in 10s by a tackler. What is the original kinetic energy of the player? Express your answer to two significant figures and include the appropriate units. What average power is required to stop him? Express your answer to two significant figures and include the appropriate units.
Explanation:
It is given that,
Mass of the football player, m = 92 kg
Velocity of player, v = 5 m/s
Time taken, t = 10 s
(1) We need to find the original kinetic energy of the player. It is given by :
[tex]k=\dfrac{1}{2}mv^2[/tex]
[tex]k=\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2[/tex]
k = 1150 J
In two significant figure, [tex]k=1.2\times 10^3\ J[/tex]
(2) We know that work done is equal to the change in kinetic energy. Work done per unit time is called power of the player. We need to find the average power required to stop him. So, his final velocity v = 0
i.e. [tex]P=\dfrac{W}{t}=\dfrac{\Delta K}{t}[/tex]
[tex]P=\dfrac{\dfrac{1}{2}\times (92\ kg)\times (5\ m/s)^2}{10\ s}[/tex]
P = 115 watts
In two significant figures, [tex]P=1.2\times 10^2\ Watts[/tex]
Hence, this is the required solution.
Hydro-Quebec transmits power from hydroelectric dams in the far north of Quebec to the city of Montreal at 735kV. The lines are 935 km long and are 3.50 cm in diameter. Given the resistivity of copper is 1.68 x 10^-8 Ω.m. a) find the resistance of one of the lines, and b) the current carried by the wire.
Answer:
a)
16.33 Ω
b)
45009.18 A
Explanation:
a)
L = length of the line = 935 km = 935000 m
d = diameter of the line = 3.50 cm = 0.035 m
ρ = resistivity of the line = 1.68 x 10⁻⁸ Ω.m
Area of cross-section of the line is given as
A = (0.25) πd²
A = (0.25) (3.14) (0.035)²
A = 0.000961625 m²
Resistance of the line is given as
[tex]R=\frac{\rho L}{A}[/tex]
inserting the values
R = (1.68 x 10⁻⁸) (935000)/(0.000961625)
R = 16.33 Ω
b)
V = potential difference across the line = 735 kv = 735000 Volts
i = current carried by the wire
Using ohm's law, current carried by the wire is given as
[tex]i=\frac{V}{R}[/tex]
i = 735000/16.33
i = 45009.18 A
The bending of light as it moves from one medium to another with differing indices of refraction is due to a change in what property of the light? A) amplitude B) period C) frequency D speed E) Color
Answer:
D] speed
Explanation:
An elbow has a loss coefficient K of 0.4. Considering a pipe with friction factor of 0.025 and diameter of 20 mm, what is the equivalent length (Le) for the elbow in mm?
Answer:
320 mm
Explanation:
The equivalent length for a minor loss is:
Le = K D / f
where K is the loss coefficient, D is the diameter, and f is the friction factor.
Le = (0.4) (20 mm) / 0.025
Le = 320 mm
The equivalent length (Le) for the elbow with a loss coefficient (K) of 0.4, a pipe with a friction factor of 0.025, and a diameter of 20 mm, is 320 mm.
To calculate the equivalent length (Le) for the elbow in terms of pipe diameter, we use the relationship between the loss coefficient (K), the friction factor (f), and the diameter (D) of the pipe, given by the equation Le = (K * D) / f. In this case, the loss coefficient for the elbow is 0.4, the friction factor is 0.025, and the diameter of the pipe is 20 mm. Plugging these values into the equation, we get:
Le = (0.4 * 20 mm) / 0.025 = 320 mm.
Therefore, the equivalent length for the elbow is 320 mm.
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.70 mm. A 21.0-V potential difference is applied to these plates. (a) Calculate the electric field between the plates.
Answer:
12353 V m⁻¹ = 12.4 kV m⁻¹
Explanation:
Electric field between the plates of the parallel plate capacitor depends on the potential difference across the plates and their distance of separation.Potential difference across the plates V over the distance between the plates gives the electric field between the plates. Potential difference is the amount of work done per unit charge and is given here as 21 V. Electric field is the voltage over distance.
E = V ÷ d = 21 ÷ 0.0017 = 12353 V m⁻¹
A 6.0-μF air-filled capacitor is connected across a 100-V voltage source. After the source fully charges the capacitor, the capacitor is immersed in transformer oil (of dielectric constant 4.5). How much ADDITIONAL charge flows from the voltage source, which remained connected during the process?
Answer:
[tex]2.1\cdot 10^{-3} C[/tex]
Explanation:
The initial charge stored on the capacitor is given by
[tex]Q_0 =C_0 V[/tex]
where
[tex]C_0 = 6.0 \mu F = 6.0 \cdot 10^{-6}F[/tex] is the initial capacitance
V = 100 V is the potential difference across the capacitor
Solving the equation,
[tex]Q_0 = (6.0 \cdot 10^{-6}F)(100 V)=6.0 \cdot 10^{-4}C[/tex]
The charge stored in the capacitor when inserting the dielectric is
[tex]Q = k Q_0[/tex]
where
k = 4.5 is the dielectric constant
Substituting,
[tex]Q=(4.5)(6.0 \cdot 10^{-4}C)=2.7\cdot 10^{-3}C[/tex]
So the additional charge is
[tex]\Delta Q=Q-Q_0 = 2.7 \cdot 10^{-3}C - 6.0 \cdot 10^{-4}C=2.1\cdot 10^{-3} C[/tex]
The additional charge that flows into the capacitor once it is immersed in transformer oil (of dielectric constant 4.5) and kept connected to the source is 2100 μC.
Explanation:The question involves a capacitor that is connected across a 100-V voltage source and then submerged in transformer oil. The capacitor initially charges to capacity while in the air. When a capacitor is then immersed in a material with a dielectric constant, its ability to store charge improves. Here, the dielectric constant of the transformer oil is given as 4.5, suggesting that the capacitor's capacity to store charge will increase 4.5 times as compared to when it was in air.
The additional charge that flows into the capacitor can be calculated using the formula for the charge in a capacitor, Q = CV. The initial charge (Q1) on the capacitor when it was in air would be Q1 = CV1 = 6.0 μF * 100 V = 600 μC. After immersing in transformer oil, the capacitance would increase by a factor of 4.5, giving a new capacitance C2 = 4.5 * 6.0 μF = 27.0 μF. The new charge (Q2) would be Q2 = CV2 = 27.0 μF * 100 V = 2700 μC. Hence the additional charge that's flown from the source would be Q2 - Q1 = 2700 μC - 600 μC = 2100 μC.
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A block of mass 0.240 kg is placed on top of a light, vertical spring of force constant 5 200 N/m and pushed downward so that the spring is compressed by 0.096 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise? (Round your answer to two decimal places.)
Answer:
10.19 m
Explanation:
Energy is conserved, so elastic energy stored in spring = gravitational energy of block.
1/2 kx² = mgh
h = kx² / (2mg)
h = (5200 N/m) (0.096 m)² / (2 × 0.240 kg × 9.8 m/s²)
h = 10.19 m
What current is produced if 1473 sodium ions flow across a cell membrane every 3.4ju8? 0 5.81 pA O 694 pA O 7.76 pA 5.99 pA
Answer:
69.4 pA
Explanation:
n = number of sodium ions = 1473
e = magnitude of charge on each sodium ion = 1.6 x 10⁻¹⁹ C
t = time taken to flow across the membrane = 3.4 x 10⁻⁶ sec
Total Charge on sodium ions is given as
q = n e eq-1
Current produced is given as
[tex]i = \frac{q}{t}[/tex]
Using eq-1
[tex]i = \frac{ne}{t}[/tex]
Inserting the values
[tex]i = \frac{(1473)(1.6\times 10^{-19})}{3.4\times 10^{-6}}[/tex]
i = 69.4 x 10⁻¹² A
i = 69.4 pA
The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.3 hours. If 1 gram of this isotope is present initially, how long will it take for 80% of the lead to decay? (Round your answer to two decimal places.)
Final answer:
The question inquires about the time needed for 80% of Pb-209 to decay, knowing its half-life is 3.3 hours. By applying the exponential decay formula, we calculate that approximately 7.39 hours are required for 80% of Pb-209 to decay.
Explanation:
The question involves the concept of radioactive decay and specifically asks how long it will take for 80% of Pb-209 to decay, given that its half-life is 3.3 hours. To find the time required for 80% of the lead to decay, we use the half-life formula and the property that radioactive decay is an exponential process. Since 80% decay means 20% remains, we set up the equation based on the exponential decay formula: N = N0(1/2)(t/T), where N is the remaining amount of substance, N0 is the initial amount, t is the time elapsed, and T is the half-life of the substance.
Substituting the given values and solving for t, we find:
N0 = 1 gram (100% initially)
N = 0.2 grams (20% remains)
T = 3.3 hours
Thus, the equation becomes 0.2 = 1(1/2)(t/3.3). Solving for t gives us the time required for 80% decay.
After calculations, the result is that it takes approximately 7.39 hours for 80% of the Pb-209 to decay. This showcases the practical application of exponential decay and half-life in determining the amount of a radioactive substance that remains after a given period.
106 m/s in a uniform 1.9 x 105 N/C electric field. The field accelerates the Problem 6: An electron has an initial velocity of 5.25 electron in the direction opposite to its initial velocity. Part (a) What is the direction of the electric field? MultipleChoice 1) The field is in the direction of the electron's initial velocity 2) The field is in the direction to the right of the clectron's initial velocity 3) The ficld is in the opposite dircction of the elcctron's initial velocity 4) The field is in another direction not listed here Part (b) How far does the electron travel before coming to rest in m? Numeric : A numeric value is expected and not an expression Part (c) How long does it take the clectron to come to rest in s? Numeric A numeric value is expected and not an expression. Part (d) What is the magnitude of the electron's velocity (in m/s) when it returns to its starting point in the opposite direction of its initial velocity? Numeric : A numeric value is expected and not an expression
(a) 1) The field is in the direction of the electron's initial velocity
The electric field is in a direction opposite to the initial velocity of the electron.
Let's remind that, when an electric charge is immersed in an electric field:
- if the charge is positive, the charge experiences a force in the same direction as the electric field direction
- if the charge is negative, the charge experiences a force in the opposite direction to the electric field direction
In this case, we have an electron: so the electric force exerted on the electron will be in a direction opposite to the direction of the electric field. Since the electron is accelerated in a direction opposite to the electron's initial velocity, this means that the electric force is in a direction opposite to the initial velocity, and so the electric field must be in the same direction as the electron's initial velocity.
(b) [tex]4.13\cdot 10^{-4} m[/tex]
We have:
Electron's initial velocity: [tex]u=5.25\cdot 10^6 m/s[/tex]
Electric field magnitude: [tex]E=1.9 \cdot 10^5 N/C[/tex]
Electron charge: [tex]q=-1.6\cdot 10^{-19} C[/tex]
Mass of the electron: [tex]m=9.11\cdot 10^{-31}kg[/tex]
The electric force exerted on the electron is:
[tex]F=qE=(-1.6\cdot 10^{-19} C)(1.9\cdot 10^5 N/C)=-3.04\cdot 10^{-14}N[/tex] (the negative sign means the direction of the force is opposite to its initial velocity)
The electron's acceleration is given by:
[tex]a=\frac{F}{m}=\frac{3.04\cdot 10^{-14} N}{9.11\cdot 10^{-31} kg}=-3.34\cdot 10^{16} m/s^2[/tex]
Now we can use the SUVAT equation:
[tex]v^2 - u^2 = 2ad[/tex]
where
v = 0 is the final speed (the electron comes to rest)
d is the total distance travelled by the electron
Solving for d,
[tex]d=\frac{v^2-u^2}{2a}=\frac{0-(5.25\cdot 10^6 m/s)^2}{2(-3.34\cdot 10^{16} m/s^2)}=4.13\cdot 10^{-4} m[/tex]
(c) [tex]1.57\cdot 10^{-10}s[/tex]
We can use the following equation:
[tex]a=\frac{v-u}{t}[/tex]
where we have
[tex]a=-3.34\cdot 10^{16}m/s^2[/tex] is the electron's acceleration
v = 0 is its final speed
[tex]u=5.25\cdot 10^6 m/s[/tex] is the initial speed
t is the time it takes for the electron to come at rest
Solving for t,
[tex]t=\frac{v-u}{a}=\frac{0-(5.25\cdot 10^6 m/s)}{-3.34\cdot 10^{16} m/s^2}=1.57\cdot 10^{-10}s[/tex]
(d) [tex]5.25\cdot 10^6 m/s[/tex]
This part of the problem is symmetrical to the previous part. In fact, the force exerted on the electron is the same as before (in magnitude), but in the opposite direction. This also means that the acceleration is the same (in magnitude), but in the opposite direction.
So we have:
u = 0 is the initial speed of the electron
[tex]a=3.34\cdot 10^{16}m/s^2[/tex]
[tex]d=4.13\cdot 10^{-4} m[/tex] is the distance covered to go back
So we can use the following equation:
[tex]v^2 - u^2 = 2ad[/tex]
to find v, the new final speed:
[tex]v=\sqrt{u^2 +2ad}=\sqrt{0^2 + 2(3.34\cdot 10^{16} m/s^2)(4.13\cdot 10^{-4} m)}=5.25\cdot 10^6 m/s[/tex]
A projectile is shot from the edge of a cliff 140 m above ground with an initial speed of 120 m/s at an angle of 38 degrees above the horizontal. What is the time taken by the projectile to hit the ground 140 m below the cliff?(g = 9.8 m/s²)
Answer:
17 seconds
Explanation:
In the y direction:
y = y₀ + v₀ᵧ t + ½ gt²
0 = 140 + (120 sin 38) t + ½ (-9.8) t²
4.9 t² - 73.9 t - 140 = 0
Solve with quadratic formula:
t = [ -b ± √(b² - 4ac) ] / 2a
t = [ 73.9 ± √((-73.9)² - 4(4.9)(-140)) ] / 9.8
t = -1.7, 16.8
Since t can't be negative, t = 16.8. Rounding to 2 sig-figs, the projectile lands after 17 seconds.
A 6 N force and a 15 N force act on an object. The moment arm of the 6 N force is 0.4 m. If the 15 N 20. force provides 5 times the torque of the 6 N force, what is its moment arm?
Answer:
0.8 m
Explanation:
For "6 N" force :
F = magnitude of the force = 6 N
r = moment arm = 0.4 m
Torque due to "6 N" force is given as
τ = r F
τ = (0.4) (6)
τ = 2.4 Nm
For " 15 N" force :
F' = magnitude of the force = 15 N
r' = moment arm = ?
τ' = Torque = 5 τ = 5 x 2.4 = 12 Nm
Torque due to "15 N" force is given as
τ' = r' F'
12 = r' (15)
r' = 0.8 m
So the moment arm for "15 N" force is 0.8 m
Two identical circular plates have a diameter of 8.2 cm. They are placed parallel to each other separated by a distance of 2.1 mm. What is the capacitance if paper, with a k-3.7, is placed between the plates? O 92 pF O 82 p O 62 pF ? 13oF
Answer:
82 pF
Explanation:
diameter = 8.2 cm, distance, d = 2.1 mm = 0.0021 m,
dielectric constant, k = 3.7
radius = 1/2 x diameter = 8.2 / 2 = 4.1 cm = 0.041 m
The formula for the capacitance of parallel plate capacitor is given by
[tex]C = \frac{k \varepsilon _{0}\times A}{d}[/tex]
[tex]C = \frac{3.7 \times 8.854\times 10^{-12} \times \pi \times 0.041 \times 0.041}{0.0021}[/tex]
C = 8.23 x 10 ^-11 F
C = 82 pF
Tom kicks a soccer ball on a flat, level field giving it an initial speed of 20 m/s at an angle of 35 degrees above the horizontal. a) How long will the ball be in the air? b) What maximum height will the ball attain? c) How far away from Tom will the ball land? d) What speed will the ball have in the instant just before it lands?
Answer:
(a) 2.34 s
(b) 6.71 m
(c) 38.35 m
(d) 20 m/s
Explanation:
u = 20 m/s, theta = 35 degree
(a) The formula for the time of flight is given by
[tex]T = \frac{2 u Sin\theta }{g}[/tex]
[tex]T = \frac{2 \times 20 \times Sin35 }{9.8}[/tex]
T = 2.34 second
(b) The formula for the maximum height is given by
[tex]H = \frac{u^{2} \times Sin^{2}\theta }{2g}[/tex]
[tex]H = \frac{20^{2} \times Sin^{2}35 }{2 \times 9.8}[/tex]
H = 6.71 m
(c) The formula for the range is given by
[tex]R = \frac{u^{2} \times Sin 2\theta }{g}[/tex]
[tex]R = \frac{20^{2} \times Sin 2 \times 35}{9.8}[/tex]
R = 38.35 m
(d) It hits with the same speed at the initial speed.
There are competitions in which pilots fly small planes low over the ground and drop weights, trying to hit a target. A pilot flying low and slow drops a weight; it takes 2.0 s to hit the ground, during which it travels a horizontal distance of 100 m. Now the pilot does a run at the same height but twice the speed. How much time does it take the weight to hit the ground? How far does it travel before it lands?
Answer:
2.0 s, 200 m
Explanation:
Time to hit the ground depends only on height. Since the plane is at the same height, the weight lands at the same time as before, 2.0 s.
Since the plane is going twice as fast, the weight will travel twice as far (ignoring air resistance). So it will travel a horizontal distance of 200 m.
Answer:
1) 2 seconds
2) 200 m
Explanation:
1) Fall time at initial speed [tex]s_{1}[/tex] = [tex]t_{1}[/tex]
Fall time at final speed [tex]s_{2}[/tex] = [tex]t_{2}[/tex]
Initial fall height [tex]h_{1}[/tex] at initial speed = Final fall height [tex]h_{2}[/tex] at final speed i.e [tex]h_{1}[/tex] = [tex]h_{2}[/tex]
s = speed
t = time
h = height
Therefore, since fall time depends on fall height where acceleration due to gravity (g) is constant,
Fall time at [tex]s_{1}[/tex] = Fall time at [tex]s_{2}[/tex]
i.e [tex]t_{1}[/tex] = [tex]t_{2}[/tex] = 2.0 s
Time taken to land = 2.0 s
2) Initial distance traveled ([tex]S_{1}[/tex]) at initial speed [tex]s_{1}[/tex] = 100 m
Final speed [tex]s_{2}[/tex] is double initial speed i.e [tex]s_{2}[/tex] = [tex]2s_{1}[/tex]
Therefore, since distance traveled is directly proportional to speed,
Final distance traveled [tex]S_{2}[/tex] at final speed [tex]s_{2}[/tex] is double initial distance [tex]S_{1}[/tex]
i.e [tex]S_{2}[/tex] = [tex]2S_{1}[/tex]
[tex]2S_{1}[/tex] = 2 x 100 m = 200 m
Distance traveled = 200 m
How many electrons leave a 9.0V battery every minute if it is connected to a resistance of 1.4?? O 80x 1020 O 6.7x 1020 O 5.1 x 1021 O 24x 1021
Answer:
2.4 x 10^21
Explanation:
V = 9 V, R = 1.4 ohm, t = 1 minute = 60 second
Use Ohm's law
V = I R
I = V / R
I = 9 / 1.4
I = 6.43 A
Now use Q = I t
Q = 6.43 x 60 = 385.7 C
Number of electrons passing in 1 minute , n
= total charge in one minute / charge of one electron
n = 385.7 / (1.6 x 10^-19) = 2.4 x 10^21
An object moving in a straight line changes its velocity uniformly from 2m/s to 4 m/s over a distance of 12 m. What was its acceleration? (A) 0.5 m/s^2 (B) 1 m/s^2 (C) 2 m/s^2 (D) 3 m/s^2
Answer:A
Explanation:
Initial velocity, u = 2m/s
Final velocity, v = 4m/s
Distance covered, s = 12m
Acceleration, a = ?
Using
v² = u² + 2as
2as = v² - u²
a = v²-u²/2s
a = 4²-2²/2 x 12
a = 16-4/24
a = 12/24
a = 0.5m/s²
A tiny object carrying a charge of +35 μC and a second tiny charged object are initially very far apart. If it takes 32 J of work to bring them to a final configuration in which the +35 μC object i is at x = 1.00 mm, y = 1.00 mm, and the other charged object is at x = 1.00 mm, y = 3.00 mm (Cartesian coordinate system), find the magnitude of the charge on the second object. (k = 1/4πε 0 = 8.99 × 109 N · m2/C2)
The magnitude of the charge on the second object is 0.025 μC and its sign is negative because it is required work to bring the two charges together, suggesting these are opposite charges and repel each other.
Explanation:The problem can be solved using the formula for the work done on a charge moving in an electric field, which is determined by the formula W = k * q1 * q2 / r, where k is the Coulomb's constant (8.99 × 10⁹ N · m²/C²), q1 and q2 are the charges, and r is the distance between them. From the problem, we know W = 32J, q1 = +35 μC, and r = 2.00 mm (the difference in the y-coordinates). Solving for q2 gives q2 = W * r / (k * q1) = 32J * 2.00 x 10⁻³m / (8.99 × 10⁹ N · m²/C² * 35 x 10⁻⁶C) = approximately -0.025 μC. Therefore, the magnitude of the charge on the second object is 0.025 μC, and its sign is negative because it takes work to bring the two charges together.
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Select the impulse-momentum bar charts for the next problems. A 1.0-kg block moving to the right at speed 3.0 m/s collides with an identical block also moving to the right at a speed 1.0 m/s. Both blocks stick together and move to the right. What is their speed after collision?
Answer with Explanation:
since the two blocks move (stick) together, the collision is inelastic, which does not conserve kinetic energy. So do not use kinetic energy consideration.
Fortunately, in such a situation, momentum is still conserved.
Momentum of 1.0 kg block
= 1.0 * 3.0 = 3.0 kg-m/s
Momentum of second block
= 1.0 * 1.0 = 1.0 kg-m/s
Total mass after collision = 1.0+1.0 = 2.0 kg
Common velocity after collision
= total momentum / total mass
= (3.0+1.0)/2.0 = 2.0 m/s
A bearing is designed to ____ A reduce friction B. support a load C.guide moving parts such as wheels, shafts and pivots D. all of the above
Answer:
Option (A)
Explanation:
A ball bearing is a device which is use to reduce the friction.
The outer rim of the bearing is fixed with the part of machine and inner rim is fitted into shafts. Now the shafts rotates and only the small spheres in the bearing will rotate. The friction can be further reduced by apply the oil or grease to the bearing.
A set of crash tests consists of running a test car moving at a speed of 11.4 m/s (25.08 m/h) into a solid wall. Strapped securely in an advanced seat belt system, a 55 kg (121 lbs) dummy is found to move a distance of 0.78 m from the moment the car touches the wall to the time the car is stopped. Calculate the size of the average force which acts on the dummy during that time.
Answer:
4582 N
Explanation:
The initial speed of the test car is
u = 11.4 m/s
While the final speed is
v = 0
The displacement of the test car during the collision is
d = 0.78 m
So we can find the acceleration of the car by using the following SUVAT equation:
[tex]v^2 - u^2 = 2ad\\a=\frac{v^2-u^2}{2d}=\frac{0-(11.4)^2}{2(0.78)}=-83.3 m/s^2[/tex]
Now we can find the average force acting on the dummy by using Newton's second law:
F = ma
Where m = 55 kg is the mass. Substituting,
[tex]F=(55 kg)(-83.3 m/s^2)=-4582 N[/tex]
So the size of the average force is 4582 N.
A 4350 kg truck, driving 7.39 m/s, runs into the back of a stationary car. After the collision, the truck moves 4.55 m/s and the car moves 11.5 m/s, both in the same direction. What is the mass of the car?
Answer:
Mass of car = 1098 kg
Explanation:
Here law of conservation of momentum is applied.
Let mass of car be m.
Initial momentum = Final momentum.
Initial momentum = 4350 x 7.39 + m x 0 = 32416.5 kgm/s
Final momentum = 4350 x 4.55 + m x 11.5 = 19792.5+11.5m
We have
19792.5+11.5m = 32416.5
m = 1097.97 kg
Mass of car = 1098 kg
Answer:1074.26
Explanation:just got it right on my accelus
A marble is dropped from a toy drone that is 25 m above the ground, and slowly rising with a vertical velocity of 0.8 m/s. How long does it take the marble to reach the ground?
Answer:
2.18 s
Explanation:
H = - 25 m ( downwards)
U = - 0.8 m/s
g = - 9.8 m/s^2
Let time taken is t.
Use second equation of motion
H = u t + 1/2 g t^2
- 25 = - 0.8 t - 1/2 × 9.8 × t^2
4.9 t^2 + 0.8 t - 25 = 0
By solving we get
t = 2.18 s
A particle travels in a circular orbit of radius 21 m. Its speed is changing at a rate of 23.1 m/s2 at an instant when its speed is 37.2 m/s. What is the magnitude of the acceleration (in m/s?) of the particle?
The particle has an acceleration vector with one component directed toward the center of its orbit, and the other directing tangentially to its orbit. Call these components [tex]\vec a_c[/tex] ([tex]c[/tex] for center) and [tex]\vec a_t[/tex] ([tex]t[/tex] for tangent). Then its acceleration vector has magnitude
[tex]|\vec a|=\sqrt{\|\vec a_c\|^2+\|\vec a_t\|^2}[/tex]
We have
[tex]\|\vec a_c\|=\dfrac{\|\vec v\|^2}r[/tex]
where [tex]\|\vec v\|[/tex] is the particle's speed and [tex]r[/tex] is the radius of orbit, so
[tex]\|\vec a_c\|=\dfrac{\left(37.2\frac{\rm m}{\rm s}\right)^2}{21\,\rm m}=65.9\dfrac{\rm m}{\mathrm s^2}[/tex]
We're given that the particle's speed changes at a rate of 23.1 m/s^2. Its velocity vector points in the same direction as [tex]\vec a_t[/tex], i.e. perpendicular to [tex]\vec a_c[/tex], so
[tex]\|\vec a_t\|=23.1\dfrac{\rm m}{\mathrm s^2}[/tex]
Then the magnitude of the particle's acceleration is
[tex]\|\vec a\|=\sqrt{\left(65.9\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(23.1\dfrac{\rm m}{\mathrm s^2}\right)^2}=\boxed{69.8\dfrac{\rm m}{\mathrm s^2}}[/tex]
The magnitude of the acceleration of the particle is approximately 70.55 m/s^2, calculated by using the formulas for combined radial and tangential acceleration in circular motion.
Explanation:In this physics problem, the particle not only moves around in a circle but is also experiencing an increase in speed which is a case of combined radial and tangential acceleration. Radial acceleration, known as centripetal acceleration (ar), is the result of the change in direction of the velocity vector, while tangential acceleration (at) comes from changes in speed.
The total acceleration of an object in circular motion is given by:
a = sqrt((ar^2) + (at^2))
Centripetal acceleration can be calculated using the formula ar = v^2 / r, where: v = speed (37.2 m/s), r = radius of the circle (21 m). This gives us ar = (37.2^2) / 21, which approximately equals 66.62 m/s^2.
The tangential acceleration is given in the problem: at = 23.1 m/s^2.
We therefore calculate the total acceleration using the formula above which gives us:
a = sqrt((66.62^2) + (23.1^2)) which approximately equals 70.55 m/s^2.
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A planet of mass m 6.75 x 1024 kg is orbiting in a circular path a star of mass M 2.75 x 1029 kg. The radius of the orbit is R 8.05 x107 km. What is the orbital period (in Earth days) of the planet Tplanet?
The planet's orbital period is about 388 days
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Further explanationCentripetal Acceleration can be formulated as follows:
[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]
a = Centripetal Acceleration ( m/s² )
v = Tangential Speed of Particle ( m/s )
R = Radius of Circular Motion ( m )
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Centripetal Force can be formulated as follows:
[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]
F = Centripetal Force ( m/s² )
m = mass of Particle ( kg )
v = Tangential Speed of Particle ( m/s )
R = Radius of Circular Motion ( m )
Let us now tackle the problem !
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Given:
mass of the planet = m = 6.75 × 10²⁴ kg
mass of the star = M = 2.75 × 10²⁹ kg
radius of the orbit = R = 8.05 × 10⁷ km = 8.05 × 10¹⁰ m
Unknown:
Orbital Period of planet = T = ?
Solution:
Firstly , we will use this following formula to find the orbital period:
[tex]F = ma[/tex]
[tex]G \frac{ Mm}{R^2}=m \omega^2 R[/tex]
[tex]G M = \omega^2 R^3[/tex]
[tex]\frac{GM}{R^3} = \omega^2[/tex]
[tex]\omega = \sqrt{ \frac{GM}{R^3}}[/tex]
[tex]\frac{2\pi}{T} = \sqrt{ \frac{GM}{R^3}}[/tex]
[tex]T = 2\pi \sqrt {\frac{R^3}{GM}}[/tex]
[tex]T = 2 \pi \sqrt {\frac{(8.05 \times 10^{10})^3}{6.67 \times 10^{-11} \times 2.75 \times 10^{29}}}[/tex]
[tex]T \approx 3.35 \times 10^7 \texttt{ seconds}[/tex]
[tex]T \approx 388 \texttt{ days}[/tex]
[tex]\texttt{ }[/tex]
Learn moreImpacts of Gravity : https://brainly.com/question/5330244Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454The Acceleration Due To Gravity : https://brainly.com/question/4189441[tex]\texttt{ }[/tex]
Answer detailsGrade: High School
Subject: Physics
Chapter: Circular Motion
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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant
To calculate the orbital period of the planet, convert the radius to meters, substitute the mass of the star and the radius into Kepler's third law, solve for the orbital period squared, and convert to Earth days.
Explanation:The question involves applying Kepler's third law of planetary motion to calculate the orbital period of a planet revolving around a star. The law relates the orbital period (T) to the radius (r) of the orbit and the mass (M) of the star around which the planet orbits. Kepler's third law can be expressed as:
T² = (4π²/GM) · r³,
where G is the gravitational constant (6.67430 × 10⁻¹¹ m³/kg · s²).
To find the orbital period of the planet:
Convert the radius of the orbit from kilometers to meters (R = 8.05 x 10⁷ km = 8.05 x 10¹ m).Substitute the given values into the equation (G = 6.67430 × 10⁻¹¹ m³/kg · s², M = 2.75 x 10²¹ kg, r = 8.05 x 10¹ m).Solve for T² and then calculate T by taking the square root of T².Convert T from seconds to Earth days by dividing by the number of seconds in a day (86,400 s).The student will then have the orbital period of the planet in Earth days.
A bullet is shot at an angle of 32° above the horizontal on a level surface. It travels in the air for 6.4 seconds before it strikes the ground 92m from the shooter. What was the maximum height reached by the bullet?
Answer:
H = 4.12 m
Explanation:
As we know that horizontal range is the distance moved in horizontal direction
Since horizontal direction has no acceleration
so here we have
[tex]Range = v_x T[/tex]
here we know that
[tex]v_x = vcos32[/tex]
so from above formula
[tex]92 = (vcos32)(6.4)[/tex]
[tex]v = 16.95 m/s[/tex]
now we have maximum height is given as
[tex]H = \frac{(vsin32)^2}{2g}[/tex]
[tex]H = \frac{(16.95 sin32)^2}{2(9.8)}[/tex]
[tex]H = 4.12 m[/tex]
An object is moving east, and it’s velocity changes from 65m/s to 25m/s in 10 seconds. Which describes the acceleration?
Answer:
4 m/s2 in negative acceleration
Explanation:
If An object is moving east, and it’s velocity changes from 65m/s to 25m/s in 10 seconds, it is 4 m/s2 in negative acceleration.
Hope this helps!
Answer:
[tex]-4\frac{m}{s^2}[/tex]
Explanation:
The object changes its speed over some time, this means that there is an acceleration.
It has a uniformly accelerated movement.
The Formula for finding speed in a uniformly accelerated motion is
[tex]a=\frac{V_{f}-V_{o}}{t}[/tex]
[tex]V_{o}= 65\frac{m}{s}\\V_{f}= 25\frac{m}{s} \\t= 10s[/tex]
Replace
[tex]a=\frac{(25-65)\frac{m}{s} }{10 s}\\ a=\frac{-40\frac{m}{s} }{10s}\\a= -4\frac{m}{s^2}[/tex]
Acceleration gives us a negative value this means that it is slowing.
A 1240-kg car is traveling with a speed of 15.0 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 31.8 m?
Answer:
Force required is 4387 N in the opposite direction of motion.
Explanation:
We have equation of motion v² = u² + 2as
v = 0m/s , u = 15 m/s, s = 31.8 m
Substituting
0² = 15² + 2 x a x 31.8
a = -3.54 m/s²
So, deceleration = 3.54 m/s²
Force = Mass x Acceleration
= 1240 x -3.54 = -4387 N
So force required is 4387 N in the opposite direction of motion.
An electron travels undeflected in a path that is perpendicular to an electric feld of 8.3 x 10 v/m. It is also moving perpendicular to a magnetic field with a magnitude of 7.3 x 103 T. If the electric field is turned off, at what radius would the electron orbit? O 124 x 10*m 889 x 104 m O 9.85 x 104m O 1.06 x 10o m
Answer:
[tex]8.6\cdot 10^{-18} m[/tex]
Explanation:
Initially, the electron is travelling undeflected at constant speed- this means that the electric force and the magnetic force acting on the electron are balanced. So we can write
q E = q v B
where
q is the electron's charge
[tex]E=8.3\cdot 10 V/m[/tex] is the electric field magnitude
v is the electron's speed
[tex]B=7.3\cdot 10^3 T[/tex] is the magnitude of the magnetic field
Solving for v,
[tex]v=\frac{E}{B}=\frac{8.3 \cdot 10 V/m}{7.3\cdot 10^3 T}=0.011 m/s[/tex]
Then the electric field is turned off, so the electron (under the influence of the magnetic field only) will start moving in a circle of radius r. Therefore, the magnetic force will be equal to the centripetal force:
[tex]qvB= m \frac{v^2}{r}[/tex]
where
[tex]q=1.6\cdot 10^{-19} C[/tex] is the electron's charge
[tex]m=9.11\cdot 10^{-31} kg[/tex] is the electron's mass
Solving for r, we find the radius of the electron's orbit:
[tex]r=\frac{mv}{qB}=\frac{(9.11\cdot 10^{-31} kg)(0.011 m/s)}{(1.6\cdot 10^{-19} C)(7.3\cdot 10^3 T)}=8.6\cdot 10^{-18} m[/tex]
A sample of blood is placed in a centrifuge of radius 12.0 cm. The mass of a red blood cell is 3.0 ✕ 10^−16 kg, and the magnitude of the force acting on it as it settles out of the plasma is 4.0 ✕ 10^−11 N. At how many revolutions per second should the centrifuge be operated?
Answer:
167.85 rev / s
Explanation:
r = 12 cm = 0.12 m, m = 3 x 10^-16 kg, F = 4 x 10^-11 N
F = m r w^2
where, w is the angular velocity.
4 x 10^-11 = 3 x 10^-16 x 0.12 x w^2
w = 1054.1 rad / s
w = 2 π f
f = w / 2 π = 1054.1 / (2 x 3.14) = 167.85 rev / s
The number of revolutions given by the calculated frequency value in which the centrifuge would be operated is 167.8 Hz.
Recall :
Frequency, f = ω/2πForce, F = mω²rMass, m = [tex] 3 \times 10^{-16}[/tex] Force, F = [tex] 4 \times 10^{-11} N[/tex] Radius, r = 12 cm = 12/100 = 0.12 mWe calculate the angular velocity, ω thus :
ω² = F/mr
ω² = [tex] \frac{4 \times 10^{-11}}{3 \times 10^{-16} \times 0.12 = 11.11 \times 10^{5}[/tex]
ω = [tex] \sqrt{1.11 \times 10^{6}} = 1053.56 rad/s[/tex]
Frequency = 1053.56 ÷ (2π)
Frequency = 167.68 Hz
Therefore, the Number of revolutions per seconds would be about 167.8 Hz
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