Answer:
81.42 km/h
Explanation:
t = Time taken for the car to slow down
u = Initial velocity = 100 km/h
v = Final velocity
s = Displacement = 20 m = 0.02 km
a = Acceleration = -6.5 m/s² = -0.0065×60×60×60×60 = -84240 km/h²
Equation of motion
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -84240\times 0.02+100^2}\\\Rightarrow v=81.42\ km/h[/tex]
Speed of the truck at the end of this distance is 81.42 km/h
Using the equation of motion v^2 = u^2 + 2as, and converting the initial speed from km/hr to m/s, the final velocity is approximately 15.22 m/s when deceleration and distance are considered. Converted back to km/hr, this is approximately 54.8 km/hr.
Explanation:The final speed of the truck can be calculated using one of the equations of motion, specifically v^2 = u^2 + 2as, where 'v' is final velocity, 'u' is initial velocity, 'a' is acceleration (which in this case is -6.50 m/s^2 due to deceleration), and 's' is distance. Convert the initial speed from km/hr to m/s: 100 km/hr = 27.78 m/s.
Then, we can plug the values into the formula: v^2 = (27.78 m/s)^2 - 2(6.50 m/s^2)(20.0 m). Solving the equation gives a final velocity of approximately 15.22 m/s. This speed in km/h is then obtained by converting m/s back, 15.22 m/s * (3600 / 1000) = 54.8 km/h. So, the speed of the truck at the end of the distance is approximately 54.8 km/h.
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Displacement vector points due east and has a magnitude of 2.8 km. Displacement vector points due north and has a magnitude of 2.8 km. Displacement vector points due west and has a magnitude of 2.4 km. Displacement vector points due south and has a magnitude of 1 km. Find the magnitude and direction (relative to due east) of the resultant vector + + + .
Answer:
The magnitude of resultant vector and direction are 1.843 m and 77.47° east of north.
Explanation:
Given that,
Magnitude of displacement due to east = 2.8 km
Magnitude of displacement due to north = 2.8 km
Magnitude of displacement due to west = 2.4 km
Magnitude of displacement due to south = 1 km
We need to calculate the resultant of the displacement
[tex]D = d_{1}+d_{2}+d_{3}+d_{4}[/tex]
[tex]D=2.8\hat{i}+2.8\hat{j}-2.4\hat{i}-1\hat{j}[/tex]
[tex]D=0.4\hat{i}+1.8\hat{j}[/tex]
The magnitude of the resultant vector
[tex]D=\sqrt{(0.4)^2+(1.8)^2}[/tex]
[tex]D=1.843\ m[/tex]
We need to calculate the direction
Using formula of direction
[tex]\tan\theta=\dfrac{j}{i}[/tex]
Put the value into the formula
[tex]\tan\theta=\dfrac{1.8}{0.4}[/tex]
[tex]\theta=\tan^{-1}4.5[/tex]
[tex]\theta=77.47^{\circ}[/tex]
Hence, The magnitude of resultant vector and direction are 1.843 m and 77.47° east of north.
Assume Young’s modulus for bone is 1.50 3 1010 N/m2. The bone breaks if stress greater than 1.50 3 108 N/m2 is imposed on it. (a) What is the maximum force that can be exerted on the femur bone in the leg if it has a minimum effective diameter of 2.50 cm? (b) If this much force is applied compressively, by how much does the 25.0-cm-long bone shorten?
Answer:
[tex]F_{max}=2.95*10^{5}N[/tex]
[tex]\Delta l=0.25cm[/tex]
Explanation:
E=1.50x10^10 N/m2 Young's modulus of bone
σmax=1.50x10^8 N/m2 Max stress tolerated by the bone
Relation between stress and Force:
[tex]\sigma=\frac{F}{A}=\frac{F}{\pi*d^{2}/4}[/tex]
[tex]F_{max}=\sigma_{max}*\pi*d^{2}/4}=1.50*10^{8}*\pi*(2.5*10^{-2})^{2}=2.95*10^{5}N[/tex]
Relation between stress and strain:
Young's modulus is defined by the ratio of longitudinal stress σ , to the longitudinal strain ε:
[tex]E=\frac{\sigma}{\epsilon}[/tex]
[tex]\epsilon=\frac{\Delta l}{l}[/tex]
We solve these equations to find the bone compression:
[tex]\Delta l=l*\frac{\sigma}{E}=25*\frac{1.50*10^{8}}{1.50*10^{10}}=0.25cm[/tex]
Final answer:
To find the maximum force that can be exerted on the femur bone in the leg, we need to calculate the stress using Young's modulus. The maximum force is approximately 7.35 * 10^7 N. To find the amount by which the bone shortens, we need to calculate the strain using Young's modulus and the stress.
Explanation:
To find the maximum force that can be exerted on the femur bone in the leg, we need to calculate the stress. The stress is equal to the force divided by the cross-sectional area of the bone. The cross-sectional area can be determined using the diameter of the bone.
(a) The diameter of the bone is 2.50 cm, which is equal to 0.025 m. The cross-sectional area can be calculated using the formula for the area of a circle: π * (radius)^2. In this case, the radius is half of the diameter, so the area is approximately 0.4909 m2. To find the maximum force, we can use the formula for stress: stress = force / area. Rearranging the formula, we have: force = stress * area. Plugging in the values, we get: force = (1.50 * 108 N/m2) * (0.4909 m2). The maximum force that can be exerted on the femur bone is approximately 7.35 * 107 N.
(b) To find the amount by which the bone shortens, we need to calculate the strain. The strain is equal to the change in length divided by the original length of the bone. The change in length can be determined using the formula for strain: strain = change in length / original length. Rearranging the formula, we have: change in length = strain * original length. We can calculate the strain using the stress and Young's modulus: strain = stress / Young's modulus. Plugging in the values, we get: strain = (1.50 * 108 N/m2) / (1.50 * 1010 N/m2). The change in length can be calculated using the formula: change in length = strain * original length. Plugging in the values, we get: change in length = (1.50 * 108 N/m2) / (1.50 * 1010 N/m2) * 25.0 cm.
An electron enters a region of uniform electric field with an initial velocity of 50 km/s in the same direction as the electric field, which has magnitude E = 50 N/C, (a) what is the speed of the electron 1.5 ns after entering this region? (b) How far does the electron travel during the 1.5 ns interval?
Answer:
(a). The speed of the electron is [tex]3.68\times10^{4}\ m/s[/tex]
(b). The distance traveled by the electron is [tex]4.53\times10^{-5}\ m[/tex]
Explanation:
Given that,
Initial velocity = 50 km/s
Electric field = 50 N/C
Time = 1.5 ns
(a). We need to calculate the speed of the electron 1.5 n s after entering this region
Using newton's second law
[tex]F = ma[/tex].....(I)
Using formula of electric force
[tex]F = qE[/tex].....(II)
from equation (I) and (II)
[tex]-qE= ma[/tex]
[tex]a = \dfrac{-qE}{m}[/tex]
(a). We need to calculate the speed of the electron
Using equation of motion
[tex]v = u+at[/tex]
Put the value of a in the equation of motion
[tex]v = 50\times10^{3}-\dfrac{1.6\times10^{-19}\times50}{9.1\times10^{-31}}\times1.5\times10^{-9}[/tex]
[tex]v=36813.18\ m/s[/tex]
[tex]v =3.68\times10^{4}\ m/s[/tex]
(b). We need to calculate the distance traveled by the electron
Using formula of distance
[tex]s = ut+\dfrac{1}{2}at^2[/tex]
Put the value in the equation
[tex]s = 3.68\times10^{4}\times1.5\times10^{-9}-\dfrac{1}{2}\times\dfrac{1.6\times10^{-19}\times50}{9.1\times10^{-31}}\times(1.5\times10^{-9})^2[/tex]
[tex]s=0.0000453\ m[/tex]
[tex]s=4.53\times10^{-5}\ m[/tex]
Hence, (a). The speed of the electron is [tex]3.68\times10^{4}\ m/s[/tex]
(b). The distance traveled by the electron is [tex]4.53\times10^{-5}\ m[/tex]
How far from a -7.80 μC point charge must a 2.40 μC point charge be placed in order for the electric potential energy of the pair of charges to be -0.500 J ? (Take the energy to be zero when the charges are infinitely far apart.)
Answer:
Distance between two point charges, r = 0.336 meters
Explanation:
Given that,
Charge 1, [tex]q_1=-7.8\ \mu C=-7.8\times 10^{-6}\ C[/tex]
Charge 2, [tex]q_2=2.4\ \mu C=2.4\times 10^{-6}\ C[/tex]
Electric potential energy, U = -0.5 J
The electric potential energy at a point r is given by :
[tex]U=k\dfrac{q_1q_2}{r}[/tex]
[tex]r=k\dfrac{q_1q_2}{U}[/tex]
[tex]r=9\times 10^9\times \dfrac{-7.8\times 10^{-6}\times 2.4\times 10^{-6}}{-0.5}[/tex]
r = 0.336 meters
So, the distance between two point charges is 0.336 meters. Hence, this is the required solution.
The 2.40 μC point charge must be placed approximately 0.34 meters away from the -7.80 μC point charge for their electric potential energy to be -0.500 J.
To find the distance between a -7.80 μC point charge and a 2.40 μC point charge such that their electric potential energy is -0.500 J, we use the formula for electric potential energy:
U = k × (q₁ × q₂) / r
where U is the electric potential energy, k is Coulomb's constant (8.99 × 10⁹ Nm²/C²), q₁ and q₂ are the point charges, and r is the distance between them.
Given:
q₁ = -7.80 μC = -7.80 × 10⁻⁶ C
q₂ = 2.40 μC = 2.40 × 10⁻⁶ C
U = -0.500 J
Rearranging the formula to solve for r, we get:
r = k × (q₁ × q₂) / U
Substituting the given values:
r = (8.99 × 10⁹ Nm²/C²) × (-7.80 × 10⁻⁶ C × 2.40 × 10⁻⁶ C) / -0.500 J
r ≈ 0.34 meters
Therefore, the 2.40 μC point charge must be placed approximately 0.34 meters away from the -7.80 μC point charge for their electric potential energy to be -0.500 J.
Ships A and B leave port together. For the next two hours, ship A travels at 28 mph in a direction 32° west of north while ship B travels 24° east of north at 35 mph . -- What is the distance between the two ships two hours after they depart? -- What is the speed of ship A as seen by ship B?
Answer:
Explanation:
We shall write velocities in vector form
Ship A travels in the direction of 32 °west of north with velocity 28 mph
V₁ = - 28 Sin 32 i + 28 Cos 32 j
Ship B travels in the direction of 24 ° east of north with velocity 35 mph
V₂ = 35 Sin 24 i + 35 Cos 24 j
Their relative velocity
= V₁ -V₂ = - 28 Sin 32 i + 28 Cos 32 j - (35 Sin 24 i + 35 Cos 24 j )
-14.83 i - 14.23 i + 23.74 j - 31.97 j
= - 29.06 i - 8.23 j
Distance between them = relative velocity x time
- 29.06 i - 8.23 j x 2
= - 58.12 i - 16.46 j
magnitude² =( 58.12 ) ² + ( 16.46)² = 60.40²
magnitude = 60.40 km
Speed of ship A as seen by ship B
= Relative velocity of A wrt B
= - 28 Sin 32 i + 28 Cos 32 j - (35 Sin 24 i + 35 Cos 24 j )
= - 29.06 i - 8.23 j
Emergency Landing. A plane leaves the airport in Galisteo and flies 170 km at 68° east of north and then changes direction to fly 230 km at 48° south of east, after which it makes an immediate emergency landing in a pasture. When the airport sends out a rescue crew, in which direction and how far should this crew fly to go directly to this plane?
To locate the emergency landed plane, we add the vectors representing the plane's two separate legs, breaking them into components using trigonometry. Summing these vectors gives us the direct path for the rescue crew in terms of both distance and bearing from the airport to the plane.
Explanation:To assist in this emergency landing scenario, we need to compute the resulting position vector by analyzing the two separate motions of the plane. The first motion has the plane fly 170 km at 68° east of north, and the second has it flying 230 km at 48° south of east. By representing these movements as vectors and adding them, we find the direct path the rescue crew should take.
This vector addition can be done graphically or by using trigonometry to break each leg of the plane's journey into its horizontal (east-west) and vertical (north-south) components. After determining the components, we can find the direct distance and bearing from the airport to the plane's location. The past examples and explanations equip us with strategies to calculate the required velocity of the plane relative to the ground and the direction the pilot must head by accounting for the known wind velocities, when necessary.
In summary, to find the direction and distance for the rescue crew, we add the vectors representing the plane's path, utilizing trigonometry to solve the components and then applying vector sum principles to find the result. This procedure allows us to efficiently direct the rescue efforts.
If you fired a rifle straight upwards at 1000 m/s, how far up will the bullet get?
Answer:
h = 51020.40 meters
Explanation:
Speed of the rifle, v = 1000 m/s
Let h is the height gained by the bullet. It can be calculated using the conservation of energy as :
[tex]\dfrac{1}{2}mv^2=mgh[/tex]
[tex]h=\dfrac{v^2}{2g}[/tex]
[tex]h=\dfrac{(1000\ m/s)^2}{2\times 9.8\ m/s^2}[/tex]
h = 51020.40 meters
So, the bullet will get up to a height of 51020.40 meters. Hence, this is the required solution.
A piece of purple plastic is charged with 8.45 × 10^6 extra electrons compared to its neutral state. What is its net electric charge (including its sign) in coulombs.
Answer:
The net electric charge is [tex]-1.352\times10^{-12}\ C[/tex]
Explanation:
Given that,
No of electron [tex]n=8.45\times10^{6}[/tex]
We need to calculate the net electric charge in coulombs
Using formula of net electric charge
Net electric charge = number of electron X charge of one electron
[tex]Q=ne[/tex]
Put the value into the formula
[tex]Q=8.45\times10^{6}\times(-1.6\times10^{-19})[/tex]
[tex]Q=-1.352\times10^{-12}\ C[/tex]
Hence, The net electric charge is [tex]-1.352\times10^{-12}\ C[/tex]
An electric field of 790,000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -3.00 uC at this spot? (14C = 10 6C) Give your answer in Sl unit rounded to two decimal places.
Answer:
-2370000 N force acts on the charge particle
Explanation:
We have given electric field E = 790000 N/C
Charge [tex]q=-3\mu C=-3\times 10^{-6}C[/tex]
We know that force on any charge particle due to electric field is given by
[tex]F=qE[/tex], here q ia charge and E is electric field
So force [tex]F=-3\times 10^{-6}\times 790000=-2370000N[/tex]
So -2370000 N force acts on the charge particle
A small earthquake starts a lamppost vibrating back and forth. The amplitude of the vibration of the top of the lamppost is 6.4 cm at the moment the quake stops, and 8.8 s later it is 1.6 cm .What is the time constant for the damping of the oscillation? What was the amplitude of the oscillation 4.4 s after the quake stopped?
Answer:
[tex]\tau = 6.35 s[/tex]
A = 3.2 cm
Explanation:
given,
Amplitude of the vibration of the top of lamppost = 6.4 cm
after 8.8 s the amplitude is 1.6 cm
time constant for damping of oscillation = ?
Amplitude at 4.4 second= ?
using formula
[tex]A = A_0e^{-\dfrac{T}{\tau}}[/tex]
[tex]1.6 = 6.4\times e^{-\dfrac{8.8}{\tau}}[/tex]
taking ln both side
[tex]ln (1.6) = ln(6.4)-\dfrac{8.8}{\tau}[/tex]
[tex]\tau = 6.35 s[/tex]
[tex]A = A_0e^{-\dfrac{T}{\tau}}[/tex]
[tex]A =6.4\times e^{-\dfrac{4.4}{6.35}}[/tex]
A = 3.2 cm
The amplitude of the oscillation 4.4 s after the quake stopped is:
A= 3.2cm
This refers to the maximum length which an object can attain when it oscillates or vibrates.
Given that the amplitude from the top of the lamppost is given as:
6.4cm
We need to find the time constant for the damping of the oscillation.
We use the formula
[tex]A= Aoe[/tex]^-T/t
We would input the values
1.6 = 6.4 x [tex]e[/tex]^-8.8/t
Taking Ln on both sides
Ln(1.6) = ln(6.4)^-8.8/t
t=6.35₈
A= [tex]Aoe[/tex]^T/t
A= 3.2cm
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If a mile is 5280 ft long and a yard contains 3 ft, how many yards are there in a mile?
To determine how many yards are in a mile, knowing that a mile equals 5280 feet and a yard contains 3 feet, divide the total feet in a mile by the feet in a yard, resulting in 1760 yards in a mile.
Explanation:To find how many yards are there in a mile, given that a mile is 5280 feet long and a yard contains 3 feet, we can divide the total number of feet in a mile by the number of feet in a yard. Using the formula for conversion, we calculate:
Yards in a mile = Total feet in a mile ÷ Feet in a yard
By substituting the given values:
Yards in a mile = 5280 ft ÷ 3 ft
Yards in a mile = 1760
This calculation clearly shows that there are 1760 yards in a mile. This example emphasizes the importance of understanding unit conversions in mathematics, allowing us to easily switch between units of measurement.
A standard 1 kilogram weight is a cylinder 55.0 mm in height and 46.0 mm in diameter. What is the density of the material?
Answer:
10945.9 kg/m^3
Explanation:
mass of cylinder, m = 1 kg
Height of cylinder, h = 55 mm = 0.055 m
Diameter of cylinder = 46 mm
Radius of cylinder, r = 23 mm = 0.023 m
The formula of the volume of the cylinder is given by
[tex]V = \pi r^{2}h[/tex]
V = 3.14 x 0.023 x 0.023 x 0.055
V = 9.136 x 10^-5 m^3
Density is defined as mass per unit volume .
[tex]Density = \frac{1}{9.136 \times 10^{-5}}[/tex]
Density = 10945.9 kg/m^3
Ship A is located 3.90 km north and 2.50 km east of ship B. Ship A has a velocity of 21.0 km/h toward the south and ship B has a velocity of 40.0 km/h in a direction 37.0° north of east. What are the (a) x-component and (b) y-component of the velocity of A relative to B? (Axis directions are determined by the unit vectors i and j, where i is toward the east.) (c) At what time is the separation between the ships least? (d) What is that least separation?
Answer:
a) Vx = -31.95 km/h b) Vy = -45.07 km/h
c) t = 0.083 h d) d = 0.22 km
Explanation:
First we have to express these values as vectors:
ra = (2.5, 3.9) km rb = (0,0)km
Va = (0, - 21) km/h Vb = (31.95, 24.07) km/h
Now we can calculate relative velocity:
[tex]V_{A/B} = V_{A} - V_{B} = (0, -21) - (31.95, 24.07) = (-31.95, -45.07) km/h[/tex]
For parts (c) and (d) we need the position of A relative to B and the module of the position will be de distance.
[tex]r_{A/B} = (2.5, 3.9) + (-31.95, -45.07) * t[/tex]
[tex]d = |r_{A/B}| = \sqrt{(2.5 -31.95*t)^{2}+(3.9-45.07*t)^{2}}[/tex]
In order to find out the minimum distance we have to derive and find t where it equals zero:
[tex]d' = \frac{-2*(2.5-31.95*t)*(-31.95)-2*(3.9-45.07*t)*(-45.07)}{2*\sqrt{(2.5 -31.95*t)^{2}+(3.9-45.07*t)^{2}}} =0[/tex]
Solving for t we find:
t = 0.083 h
Replacing this value into equation for d:
d = 0.22 km
An aluminum wing on a passenger jet is 25 m long when its temperature is 21°C. At what temperature would the wing be 6 cm (0.06 m) shorter?
Answer:
The temperature at which the wing would be shorter is [tex]- 80.69^{\circ}C[/tex]
Solution:
The original length of the Aluminium wing, [tex]l_{w} = 25 m[/tex]
Temperature, T = [tex]21^{\circ}C[/tex]
Change in the wing's length, [tex]\Delta l_{w} = 0.06 m[/tex]
Also, for Aluminium, at temperature between [tex]20^{\circ}C[/tex] to [tex]100^{\circ}C[/tex], the linear expansion coefficient, [tex]\alpha = 23.6\times 10^{- 6}/^{\circ}C[/tex]
Now, Change in length is given by:
[tex]\Delta l_{w} = l_{w}\alpha \Delta T[/tex]
[tex]0.06 = 25\times 23.6\times 10^{- 6}\(T - T')[/tex]
[tex]\frac{0.06}{5.9\times 10^{- 4}} = 21^{\circ}C _ T'[/tex]
[tex]T' = - 80.69^{\circ}C[/tex]
-Final answer:
To find the temperature at which the aluminum wing would be 6 cm shorter, use the thermal expansion formula. Given the initial length, final length, and coefficient of linear expansion for aluminum, the required temperature will be -285.8°C.
Explanation:
Thermal Expansion Formula: ΔL = αL0ΔT
To find the temperature at which the aluminum wing would be 6 cm shorter, we can use the thermal expansion formula. Given initial length L0 = 25m, final length = 25m - 0.06m = 24.94m, and coefficient of linear expansion for aluminum α = 22 × 10-6 °C. Rearranging the formula gives us the change in temperature ΔT = ΔL / (αL0). Substituting the values results in ΔT ≈ 306.8°C.
Now, ΔT = Temperature for 25meter length - Temperature for 24.94meter length.
Thus, Temperature for 24.94meter length = -285.8°C
Hydrogen protons are used in MRI because of their abundance.
Answer:
Because of that but also because of its magnetic properties.
Explanation:
But its abundance in water and fat, since Magnetic Resonance Imaging uses our own body's magnetic properties to produce the images, and for this is much better to use something that is abundant in our own body and has magnetic properties, that is, and hydrogen nucleus ( single proton), being useful since it behaves like a small magnet.
A cricket ball has mass 0.155 kg. If the velocity of a bowled ball has a magnitude of 35.0 m/s and the batted ball's velocity is 65.0 m/s in the opposite direction, find the magnitude of the change in momentum of the ball.
Find the magnitude of the impulse applied to it by the bat.
If the ball remains in contact with the bat for 2.00 ms , find the magnitude of the average force applied by the bat.
Answer:
Magnitude of change in momentum = 4.65 kg.m/s
Magnitude of impulse = 4.65 kg.m/s
Magnitude of the average force applied by the bat = 1550 N
Explanation:
Mass of the cricket ball, m = 0.155 kg
Initial velocity of the ball, u = 35.0 m/s
final velocity of the ball after hitting the bat, v = 65.0 m/s
Time of contact, t = 2.00 ms = 2.00 × 10⁻³ s
Now,
Magnitude of change in momentum = Final momentum - Initial momentum
or
Magnitude of change in momentum = ( m × v ) - ( m × u )
or
Magnitude of change in momentum = ( 0.155 × 65 ) - ( 0.155 × 35 )
or
Magnitude of change in momentum = 10.075 - 5.425 = 4.65 kg.m/s
Now, Magnitude of impulse = change in momentum
thus,
Magnitude of impulse = 4.65 kg.m/s
Now,
magnitude of the average force applied by the bat = [tex]\frac{\textup{Impulse}}{\textup{Time}}[/tex]
or
magnitude of the average force applied by the bat = [tex]\frac{\textup{4.65}}{\textup{3}\times\textup{10}^{-3}}[/tex]
or
Magnitude of the average force applied by the bat = 1550 N
A satellite is held in orbit by a 2000- N gravitational
force.Each time the satellite completes an orbit of circumference
80000km, the work done on it by gravity is?
Answer:
The work done by gravity is zero.
Explanation:
Given that,
Gravitational force = 2000 N
Circumference = 80000 km
We need to calculate the work done by gravity
Using formula of work done
[tex]W=F\cdot d\cos\theta[/tex]
Here, [tex]\cos\theta[/tex] = 0
Because, for a circular motion is always perpendicular to the force.
Where, F = force
d = distance
Put the value into the formula
[tex]W=2000\times80000\times0[/tex]
[tex]W=0[/tex]
Hence, The work done by gravity is zero.
The work done on the satellite by gravity is 160,000,000,000 joules.
Explanation:The work done on a satellite by gravity can be calculated using the formula: work = force x distance. In this case, the force is the gravitational force of 2000 N and the distance is the circumference of the orbit, which is 80000 km. To convert km to meters, we multiply by 1000, so the distance is 80000 x 1000 = 80,000,000 meters. Plugging these values into the formula, the work done on the satellite by gravity is 2000 x 80,000,000 = 160,000,000,000 J (joules).
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A player kicks a ball with an initial vertical velocity of 12 m/s and horizontal velocity of 16 m/s. (a) At what speed does the ball hit the ground? (b) For how long does the ball remain in the air? (c)What maximum height is attained by the ball?
Answer:
a)Vg=13.42m/s :Speed with which the ball hits the ground
b) t₁= 2.74s : Time the ball remains in the air
c)h=9.19m: Maximum height reached by the ball
Explanation:
We apply the kinematic equations of parabolic motion:
a) Vg= Vo
Vg:speed with which the ball hits the ground
Vo: initial speed
Initial Speed Calculation
[tex]v_{o} =\sqrt{v_{ox}^{2} +v_{oy} ^{2} }[/tex]
[tex]v_{o} =\sqrt{16^{2} +12^{2} }[/tex]
Vo=13.42m/s
Vg=13.42m/s
b)Calculation of the time the ball remains in the air
t₁=2*t₂
t₁;time the ball remains in the air
t₂ time when the ball reaches the maximum height
Vf=Vo-g*t₂ : When the ball reaches the maximum height Vf = 0
0=13.42-9.8*t₂
9.8*t₂=13.42
t₂=13.42 ÷9.8
t₂=1.37s
t₁=2*1.37s
t₁= 2.74s
c)Calculation of the maximum height reached by the ball
When the ball reaches the maximum height Vf = 0
Vf²=V₀²-2*g*h
0= V₀²-2*g*h
2*g*h= V₀²
h= V₀² ÷ 2*g
h= 13.42² ÷2*9.8
h=9.19m
In a vacuum, two particles have charges of q1 and q2, where q1 = +3.3C. They are separated by a distance of 0.24 m, and particle 1 experiences an attractive force of 4.1 N. What is the value of q2, with its sign?
Final answer:
The value of q2, with its sign, can be found using Coulomb's Law. By plugging in the given values for q1, the distance, and the force experienced, we can calculate the value of q2 as -2.25C. The negative sign indicates that q2 is a negative charge.
Explanation:
In order to find the value of q2, we can use Coulomb's Law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Here, we are given the charge of particle 1 (q1 = +3.3C), the distance between the particles (d = 0.24m), and the force experienced by particle 1 (F = 4.1N). Let's denote the charge of particle 2 as q2.
Using Coulomb's Law, we can write:
F = k(q1 * q2) / d^2
Plugging in the given values, we have:
4.1N = (9 x 10^9 N m^2/C^2)(3.3C * q2) / (0.24m)^2
Simplifying the equation, we can solve for q2:
q2 = (4.1N * (0.24m)^2) / (9 x 10^9 N m^2/C^2 * 3.3C)
Calculating this equation gives us the value of q2 as +2.25C. Since the force experienced by particle 1 is attractive, with a positive charge (+3.3C), the value of q2 must be negative to create an attractive force. Therefore, the value of q2 is -2.25C.
Which quantity does not change when an ice skater pulls in her arms during a spin?
A. angular momentum
B. angular velocity
C. moment of inertia
D. mass
Answer:
The Answer is Letter A :)
Explanation:
When the ice skater sticks out her hands, she spins slower. Then she rotates really fast when she pulls her arms to her sides. This is an example of a fundamental law in physics called Conservation of Angular Momentum. This law relates two observable quantities: the speed of rotation and the shape.
I Hope It's Helpful
Hint The Brainliest :)
Answer:
Option (A)
Explanation:
Angular momentum is usually defined as a vector quantity that controls the rotational momentum of an object, body or a system. It is the product of the three quantities namely the mass, radius, and velocity of the rotating body.
The given question is based on the conservation of the angular momentum, where an ice skater when pulls in her arms during the time of spinning, the angular momentum remains conserved. It does not change.
Thus, the correct answer is option (A).
The diameter of a sphere is measured to be 5.36 in. Find (a) the radius of the sphere in centimeters, (b) the surface area of the sphere in square centimeters, and (c) the volume of the sphere in cubic centimeters.
Answer:
a) r = 6.81 cm : radius of the sphere
b) A = 582.78 cm² : surface area of the sphere
c) V = 1322.91 cm³ : volume of the sphere
Explanation:
Formula to calculate the surface area of the sphere:
A = 4×π×r² Formula (1)
Formula to calculate the volume of the sphere:
V = (4/3)×π×r³ Formula (2)
Problem development
a)
d = 5.36 in
1 in = 2.54 cm
[tex]d = 5.36 in * \frac{2.54cm}{1in} = 13.614 cm[/tex]
r = d/2
Where:
r: sphere radius (cm)
d: sphere diameter (cm)
r = 13.614/2 = 6.81 cm
b)
We replace in formula (1)
A = 4×π×(6.81)² = 582.78 cm²
c)
We replace in formula (2)
V = (4/3)×π×(6.81)³ = 1322.91 cm³
The direction of a natural process is indicated by which of the following? A. conservation of energy. B. change in entropy C. thermal efficiency D. specific heat E. expansion coefficient
Answer:
The correct answer is option'B': Change in entropy
Explanation:
We know from the second law of thermodynamics for any spontaneous process the total entropy of the system and it's surroundings will increase.
Meaning that any unaided process will move in a direction in which the entropy of the system will increase.It is because the system will always want to increase it's randomness
When we throw an object vertically upwards its initial velocity: a. It will be less than the final
b. It will be greater than the final
c. It will be equivalent to the final
d. It remains constant until reaching its maximum height
Answer:
Its initial velocity will be greater than final velocity so option (b) will be correct option
Explanation:
As we throw the any object vertically the motion of the object will be opposes by the gravity.
And as the velocity of object is opposes by gravity, the final velocity goes on decreasing and finally it becomes zero.
So the initial velocity is always greater than final velocity when the object is thrown vertically upward.
So option (b) will be the correct option
A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone. The acceleration of gravity is 9.8 m/s 2 . How long does it take for the ball to rotate once around the axis?
Answer:
Time taken, [tex]T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}[/tex]
Explanation:
It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.
From the figure,
The sum of forces in y direction is :
[tex]T\ cos\theta-mg=0[/tex]
[tex]T=\dfrac{mg}{cos\theta}[/tex]
Sum of forces in x direction,
[tex]T\ sin\theta=\dfrac{mv^2}{r}[/tex]
[tex]mg\ tan\theta=\dfrac{mv^2}{r}[/tex].............(1)
Also, [tex]r=l\ sin\theta[/tex]
Equation (1) becomes :
[tex]mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}[/tex]
[tex]v=\sqrt{gl\ tan\theta.sin\theta}[/tex]...............(2)
Let t is the time taken for the ball to rotate once around the axis. It is given by :
[tex]T=\dfrac{2\pi r}{v}[/tex]
Put the value of T from equation (2) to the above expression:
[tex]T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}[/tex]
[tex]T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}[/tex]
On solving above equation :
[tex]T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}[/tex]
Hence, this is the required solution.
The time is taken by the ball to rotate once around the axis is [tex]2\pi \sqrt{\dfrac{l\cos\theta}{g}}[/tex].
How long does it take for the ball to rotate once around the axis?
As it is given to us that the small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone.
Now since the ball is moving in circular as well as vertical motion, therefore, the sum of vertical forces at any given moment of time can be written as,
[tex]\rm T\ cos \theta = mg\\\\T = \dfrac{mg}{Cos\theta}[/tex]
Also, the sum of the forces in the x-direction,
[tex]\rm T\ sin\theta= \dfrac{mv^2}{r}[/tex]
Substitute the value of T,
[tex]\rm \dfrac{mg}{cos\theta}\ sin\theta= \dfrac{mv^2}{r}[/tex]
[tex]\rm \dfrac{mg}{cos\theta}\ sin\theta= \dfrac{mv^2}{l\ sin\theta}[/tex]
[tex]v=\sqrt{gl\ tan\theta \cdot sin\theta}[/tex]
We know that the time is taken by the ball to circulate around the axis can be given by the formula,
[tex]T = \dfrac{2\pi r}{v}[/tex]
Substitute the value of v,
[tex]T = \dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta\cdot sin\theta}}[/tex]
[tex]T=2\pi \sqrt{\dfrac{l\cos\theta}{g}}[/tex]
Hence, the time is taken by the ball to rotate once around the axis is [tex]2\pi \sqrt{\dfrac{l\cos\theta}{g}}[/tex].
Learn more about Motion:
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Three forces with magnitudes of 66 pounds, 110 pounds, and 138 pounds act on an object at angles of 30°, 45°, and 120° respectively, with the positive x-axis. Find the direction and magnitude of the resultant force. (Round your answers to one decimal place.)
Answer:
magnitude = 239.5 lbfdirection ( angle to the x axis) = 74.0Explanation:
We just need to sum the forces, we can do this easily in their Cartesian form.
Knowing the magnitude and angle with the positive x axis, we can find Cartesian representation of the vectors using the formula
[tex]\vec{A}= |\vec{A}| \ ( \ cos(\theta) \ , \ sin (\theta) \ )[/tex]
where [tex]|\vec{A}|[/tex] its the magnitude of the vector and θ the angle with the positive x axis.
So, for our forces we got:
[tex]\vec{F}_1 \ = \ 66 \ lbf \ * \ ( \ cos (30\°)\ , \ sin(30\°) \ )[/tex]
[tex]\vec{F}_2 \ = \ 110 \ lbf \ * \ ( \ cos (45\°)\ , \ sin(45\°) \ )[/tex]
[tex]\vec{F}_3 \ = \ 138 \ lbf \ * \ ( \ cos (120\°)\ , \ sin(120\°) \ )[/tex]
this will give us:
[tex]\vec{F}_1 \ = ( \ 57.157 \ lbf \ , \ 33 \ lbf \ )[/tex]
[tex]\vec{F}_2 \ = ( \ 77.782 \ lbf \ , \ 77.782 \ lbf \ )[/tex]
[tex]\vec{F}_3 \ = ( \ - 69 \ lbf \ , \ 119.511 \ lbf \ )[/tex]
Now, we just sum the forces:
[tex]\vec{F}_{net} \ = \ \vec{F}_1 \ + \ \vec{F}_2 \ + \ \vec{F}_3[/tex]
[tex]\vec{F}_{net} \ = ( \ 57.157 \ lbf \ , \ 33 \ lbf \ ) + ( \ 77.782 \ lbf \ , \ 77.782 \ lbf \ ) + (\ - 69 \ lbf \ , \ 119.511 \ lbf \ )[/tex]
[tex]\vec{F}_{net} \ = ( \ 57.157 \ lbf \ + \ 77.782 \ lbf \ - \ 69 \ lbf \, \ 33 \ lbf \ + \ 77.782 \ lbf \ + \ 119.511 \ lbf \ )[/tex]
[tex]\vec{F}_{net} \ = ( \ 65.939 \ lbf \, \ 230.293 lbf \ )[/tex]
This is the net force, to obtain the magnitude, we just need to find the length of the vector, using the Pythagorean formula:
[tex]|\vec{F}_{net}| = \sqrt{(F_{net_x})^2+(F_{net_y})^2}[/tex]
[tex]|\vec{F}_{net}| = \sqrt{(65.939 \ lbf)^2+(230.293 lbf)^2}[/tex]
[tex]|\vec{F}_{net}| = \ 239.547 \ lbf[/tex]
To obtain the angle with the positive x-axis we can use the formula:
[tex]\theta \ = \ arctan( \frac{F_y}{F_y})[/tex]
[tex]\theta \ = \ arctan( \frac{230.293 lbf}{65.939 \ lbf})[/tex]
[tex]\theta \ = \ arctan( 3.492)[/tex]
[tex]\theta \ = \ 74.02[/tex]
So, the answer its
[tex]magnitude = \ 239.547 \ lbf[/tex]
[tex]angle_{ (to the x axis)} = \ 74.02 [/tex]
Rounding up:
[tex]magnitude = \ 239.5 \ lbf[/tex]
[tex]angle_{ (to the x axis)} = \ 74.0 [/tex]
What is the magnitude (in N/C) and direction of an electric field that exerts a 3.50 ✕ 10−5 N upward force on a −1.55 µC charge?
Answer:
The magnitude of electric field is 22.58 N/C
Solution:
Given:
Force exerted in upward direction, [tex]\vec{F_{up}} = 3.50\times 10^{- 5} N[/tex]
Charge, Q = [tex] - 1.55\micro C = - 1.55\times 10^{- 6} C[/tex]
Now, we know by Coulomb's law,
[tex]F_{e} = \frac{1}{4\pi\epsilon_{o}\frac{Qq}{R^{2}}[/tex]
Also,
Electric field, [tex]E = \frac{1}{4\pi\epsilon_{o}\frac{q}{R^{2}}[/tex]
Thus from these two relations, we can deduce:
F = QE
Therefore, in the question:
[tex]\vec{E} = \frac{\vec F_{up}}{Q}[/tex]
[tex]\vec{E} = \frac{3.50\times 10^{- 5}}{- 1.55\times 10^{- 6}}[/tex]
[tex]\vec{E} = - 22.58 N/C[/tex]
Here, the negative side is indicative of the Electric field acting in the opposite direction, i.e., downward direction.
The magnitude of the electric field is:
[tex]|\vec{E}| = 22.58\ N/C[/tex]
The magnitude of the electric field is 22.58 × 10³ N/C and the direction is downward, which is opposite to the upward force applied.
Explanation:The question is asking to find the magnitude and direction of an electric field that applies a force on a charged particle. The formula used to calculate the electric field (E) is given by E = F/q, where F is the force applied to the charge and q is the magnitude of the charge.
Given a force (F) of 3.50 × 10⁻⁵ N upward and a charge (q) of -1.55 μC (microcoulombs), which is equivalent to -1.55 × 10⁻⁶ C (coulombs), first, we convert the charge into coulombs by multiplying the microcoulombs by 10⁻⁶. Next, we substitute the values into the formula, yielding:
E = (3.50 × 10⁻⁵ N) / (-1.55 × 10⁻⁶ C), which simplifies to E = -22.58 × 10³ N/C. The negative sign indicates that the direction of the electric field is opposite to the direction of the force, so if the force is upward, the electric field is downward.
The magnitude of the electric field is thus 22.58 × 10³ N/C and the direction is downward
Which of the following combinations of variables results in the greatest period for a pendulum? length = L, mass = M, and maximum angular displacement = degrees length = L, mass = M, and maximum angular displacement = 3 degrees length = 2L, mass = M/2, and maximum angular displacement = 1 degree length = 1.5L, mass = 2M, and maximum angular displacement = 2 degrees length = L, mass = 4M, and maximum angular displacement = 4 degrees
Answer:
length = 2L, mass = M/2, and maximum angular displacement = 1 degree
Explanation:
We consider only small amplitude oscillations (like in this case), so that the angle θ is always small enough. Under these conditions recall that the equation of motion of the pendulum is:
[tex]\ddot{\theta}=\frac{g}{l}\theta[/tex]
And its solution is:
[tex]\theta=Asin(\omega t + \phi)[/tex]
Where [tex]\omega=\sqrt\frac{g}{l}[/tex] are the angular frequency of the oscillations, from which we determine their period:
[tex]T=\frac{2\pi}{\omega}\\T=2\pi\sqrt\frac{l}{g}[/tex]
Therefore the period of a pendulum will only depend on its length, not on its mass or angle, for angles small enough. So, the answer is the one with the greater length.
The period of a pendulum is only determined by its length and the acceleration due to gravity, and is independent of other factors such as mass and maximum displacement.
Explanation:The period of a simple pendulum depends on its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass and the maximum displacement. Therefore, none of the combinations of variables given will result in a greater period for the pendulum. The period is only determined by the length and the value of acceleration due to gravity.
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A model rocket rises with constant acceleration to a height of 4.2 m, at which point its speed is 27.0 m/s. How much time does it take for the rocket to reach this height? What was the magnitude of the rocket's acceleration? Find the height of the rocket 0.20 s after launch. Find the speed of the rocket 0.20 s after launch.
Answers:
a) [tex]t=0.311 s[/tex]
b) [tex]a=86.847 m/s^{2}[/tex]
c) [tex]y=1.736 m[/tex]
d) [tex]V=17.369 m/s[/tex]
Explanation:
For this situation we will use the following equations:
[tex]y=y_{o}+V_{o}t+\frac{1}{2}at^{2}[/tex] (1)
[tex]V=V_{o} + at[/tex] (2)
Where:
[tex]y[/tex] is the height of the model rocket at a given time
[tex]y_{o}=0[/tex] is the initial height of the model rocket
[tex]V_{o}=0[/tex] is the initial velocity of the model rocket since it started from rest
[tex]V[/tex] is the velocity of the rocket at a given height and time
[tex]t[/tex] is the time it takes to the model rocket to reach a certain height
[tex]a[/tex] is the constant acceleration due gravity and the rocket's thrust
a) Time it takes for the rocket to reach the height=4.2 m
The average velocity of a body moving at a constant acceleration is:
[tex]V=\frac{V_{1}+V_{2}}{2}[/tex] (3)
For this rocket is:
[tex]V=\frac{27 m/s}{2}=13.5 m/s[/tex] (4)
Time is determined by:
[tex]t=\frac{y}{V}[/tex] (5)
[tex]t=\frac{4.2 m}{13.5 m/s}[/tex] (6)
Hence:
[tex]t=0.311 s[/tex] (7)
b) Magnitude of the rocket's acceleration
Using equation (1), with initial height and velocity equal to zero:
[tex]y=\frac{1}{2}at^{2}[/tex] (8)
We will use [tex]y=4.2 m[/tex] :
[tex]4.2 m=\frac{1}{2}a(0.311)^{2}[/tex] (9)
Finding [tex]a[/tex]:
[tex]a=86.847 m/s^{2}[/tex] (10)
c) Height of the rocket 0.20 s after launch
Using again [tex]y=\frac{1}{2}at^{2}[/tex] but for [tex]t=0.2 s[/tex]:
[tex]y=\frac{1}{2}(86.847 m/s^{2})(0.2 s)^{2}[/tex] (11)
[tex]y=1.736 m[/tex] (12)
d) Speed of the rocket 0.20 s after launchWe will use equation (2) remembering the rocket startted from rest:
[tex]V= at[/tex] (13)
[tex]V= (86.847 m/s^{2})(0.2 s)[/tex] (14)
Finally:
[tex]V=17.369 m/s[/tex] (15)
The sun is 21° above the horizon. It makes a 54 m -long shadow of a tall tree. How high is the tree? Express your answer in meters.
Answer:
Height of the tree, h = 20.72 meters
Explanation:
Given that,
The sun is 21° above the horizontal, [tex]\theta=21^{\circ}[/tex]
Length of the shadow, d = 54 m
Let h is the height of the tree. It can be calculated using trigonometry as :
[tex]tan\theta=\dfrac{perpendicular}{base}[/tex]
Here, perpendicular is h and base is 54 meters.
[tex]tan(21)=\dfrac{h}{54}[/tex]
[tex]h=tan(21)\times 54[/tex]
h = 20.72 meters
So, the height of the tree is 20.72 meters. Hence, this is the required solution.
You kick a soccer ball at vo = 7 m/s horizontally off of the top of the Great Pyramid of Giza. The faces are slanted 40 degrees from the vertical. How far down the pyramid face does the ball hit the pyramid?
Answer:
The ball hit the pyramid 18.52m down the pyramid face.
Explanation:
As you can see in the image below, the will hit the pyramid at at point where the distance travelled vertically divided by the distance travelled horizontally is equal to tan(50), since it's at this moment where the path of the ball will coincide with the walls of the pyramid.
The horizontal vellocity of the ball will remain constant at a value of 7m/s along the whole journey. This is because there is no horizontal acceleration that can affect the horizontal velocity. On the contrary, the vertical velocity will start at 0m/s and will increase because of gravity.
The distance travelled horizontally will be:
[tex]x = v_x*t = 7t[/tex]
The distance travelled vertically will be:
[tex]y = \frac{1}{2}gt^2+v_o_yt+y_o | v_o_y = 0m/s, y_o=0m\\ y = \frac{1}{2}gt^2[/tex]
So, then:
[tex]\frac{y}{x} = tan(50)\\\frac{\frac{1}{2}gt^2}{v_xt} =tan(50)\\\frac{1}{2}gt = v_xtan(50)\\t= \frac{2v_xtan(50)}{g} = \frac{2*7m/s*tan(50)}{9.81m/s^2}=1.7s[/tex]
At time = 1.7s:
[tex]x = v_x*t = 7t = 7m/s*1.7s = 11.9m[/tex]
[tex]y=\frac{1}{2}gt^2 =\frac{1}{2}*9.81m/s^2*(1.7s)^2=14.19m [/tex]
Using Pythagorean theorem, we can find the distance:
[tex]d = \sqrt{x^2+y^2}=\sqrt{(11.9m)^2+(14.19m)^2} = 18.52m[/tex]