Answer:
Look at explanation
Explanation:
a)Only force acting on the object is gravity, so a=-g (consider up to be positive)
use: v^2=v0^2+2a(y-y0)
plug in givens, at max height v=0
0=400-19.6(H)
Solve for H
H= 20.41m
b) Use: y=y0+v0t+1/2at^2
Plug in givens
0=0+20t-4.9t^2
solve for t
t=4.08 seconds
c) v=v0+at
v=20-39.984= -19.984m/s
d. On the afternoon of January 15, 1919, an unusually warm day in Boston, a 17.7-m-high, 27.4-m-diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded into the streets in a 5-m-deep stream, killing pedestrians and horses and knocking down buildings. The molasses had a density of 1600 kg>m3 . If the tank was full before the accident, what was the total outward force the molasses exerted on its sides
Answer:
F = 1.638 x 10⁸ N = 163.8 MN
Explanation:
The total force exerted by the molasses is given as:
F = PA
where,
F = Force exerted by the molasses = ?
P = Pressure = ρgh
ρ = density of molasses = 1600 kg/m³
g = acceleration due to gravity = 9.81 m/s²
h = height of tank = 17.7 m
A = cross-sectional area of tank = πr²
r = radius of tank = 27.4 m/2 = 13.7 m
Therefore,
[tex]F = \rho ghA = \rho gh(\pi r^2)\\\\F = (1600\ kg/m^3)(9.81\ m/s^2)(17.7\ m)(\pi)(13.7\ m)^2[/tex]
F = 1.638 x 10⁸ N = 163.8 MN
In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with zero initial speed and constant acceleration a with arrow of magnitude 0.35 m/s2. What angle between a with arrow and the positive direction of the y axis would result in a collision?
Answer:
59.26°
Explanation:
Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.
Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.
Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration = acosθ.
So, y = ut + 1/2a't²
y = 0 × t + 1/2(acosθ)t²
y = 0 + 1/2(acosθ)t²
y = 1/2(acosθ)t² (1)
Also, both particles must move the same horizontal distance to collide in time, t.
Let x be the horizontal distance,
x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision
Also, using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration = asinθ.
So, x = ut + 1/2a"t²
x = 0 × t + 1/2(ainsθ)t²
x = 0 + 1/2(asinθ)t²
x = 1/2(asinθ)t² (3)
Equating (2) and (3), we have
vt = 1/2(asinθ)t² (4)
From (1) t = √[2y/(acosθ)]
Substituting t into (4), we have
v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²
v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)
v√[2y/(acosθ)] = ytanθ
√[2y/(acosθ)] = ytanθ/v
squaring both sides, we have
(√[2y/(acosθ)])² = (ytanθ/v)²
2y/acosθ = (ytanθ/v)²
2y/acosθ = y²tan²θ/v²
2/acosθ = ytan²θ/v²
1/cosθ = aytan²θ/2v²
Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1
secθ = ay(sec²θ - 1)/2v²
2v²secθ = aysec²θ - ay
aysec²θ - 2v²secθ - ay = 0
Let secθ = p
ayp² - 2v²p - ay = 0
Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have
ayp² - 2v²p - ay = 0
0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0
10.85p² - 15.68p - 10.85 = 0
dividing through by 10.85, we have
p² - 1.445p - 1 = 0
Using the quadratic formula to find p,
[tex]p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125[/tex]
Since p = secθ
secθ = 1.95625 or secθ = -0.51125
cosθ = 1/1.95625 or cosθ = 1/-0.51125
cosθ = 0.5112 or cosθ = -1.9956
Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.
So, cosθ = 0.5112
θ = cos⁻¹(0.5112)
θ = 59.26°
So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.
You need to calculate the volume of berm that has a starting cross-sectional area of 118 SF, and an ending cross-sectional area of 245 SF. The berm is 300 ft long and is assumed to taper evenly between the two cross-sectional areas, what is the calculated volume of the berm in cubic feet
two resistors with resistance values 4.5 ohms and 2.3 ohms are connected in series or parallel across a potential difference of 30V to a light bulb find the current flowing through the light bulb in both cases
Answer:
Look at work
Explanation:
Series:
I is the same for all resistors so just find the value of Req. In series Req= R1+R2+...+Rn. So here it will be 4.5+2.3=6.8ohms. Ieq=Veq/Req=4.41A. And since current is the same across all resistors the current to the lightbulb is 4.41A.
Parallel:
V is the same for all resistors so start of by finding Req. In parallel, Ieq=I1+I2+...+In. So I1= 30/4.5= 6.67A and I2= 13.04A. Ieq= 6.67+13.04= 19.71A.
In 1.0 second, a battery charger moves 0.50 C of charge from the negative terminal to the positive terminal of a 1.5 V AA battery.
Part A:
How much work does the charger do? Answer is 0.75 J
Part B:
What is the power output of the charger in watts?
Answer:
W = Q * V work done on charge Q
A. W = .5 C * 1.5 V = .75 Joules
B. P = W / t = .75 J / 1 sec = .75 Watts
how do you calculate voltage drop
Answer:
Multiply current in amperes by the length of the circuit in feet to get ampere-feet. Circuit length is the distance from the point of origin to the load end of the circuit.
Divide by 100.
Multiply by proper voltage drop value in tables. The result is voltage drop.
Explanation:
what is the force of a body which have mass of 7 kg
Answer:
Force acting on a body of mass 7 kg which produces an accceleration of 10 m/s2 is 70 N
Answer:
10 m/s2 or 70 newtons.
Explanation:
............................
............
Help me with my physics, please
1.- Que distancia recorrió una carga de 2,5x10-6 coul, generando así un campo eléctrico de 55new/coul.
Answer:
r = 20.22 m
Explanation:
Given that,
Charge,[tex]q=2.5\times 10^{-6}\ C[/tex]
Electric field, [tex]E=55\ N/C[/tex]
We need to find the distance. We know that, the electric field a distance r is as follows :
[tex]E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}}\\\\r=\sqrt{\dfrac{9\times 10^9\times 2.5\times 10^{-6}}{55}}\\\\r=20.22\ m[/tex]
So, the required distance is 20.22 m.
a vessel with mass 10kg intially moving withthe velocicity 12m s along the x axis explodes into three exactly identical pieces Just after the explosion one piece moves with speed 10 m s along the x axis and asecond piece moves with speed 10 m s along the y axis What iis the magnitude of the component of velocity of the third piece along the y axiss
Answer:
Explanation:
Apply law of conservation of momentum along y-axis.
Initially there was no momentum along y-axis. So there will be nil momentum along y-axis again finally.
Let the mass of each piece after breaking be m .
Momentum of piece moving along positive y-axis
= m x 10 = 10m .
Let the component of velocity of third piece along y-axis be v .
Its momentum along the same direction = m v .
Total momentum along y -axis = 10 m + m v
According to law of conservation of momentum
10 m + mv = 0
v = - 10 m/s .
Component of velocity of the third piece along y-axis will be - 10 m/s .
In other words it will be along negative y-axis with speed of 10 m/s.
A 0.033-kg bullet is fired vertically at 222 m/s into a 0.15-kg baseball that is initially at rest. How high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the ball
Answer:
The maximum height risen by the bullet-baseball system after the collision is 81.76 m.
Explanation:
Given;
mass of the bullet, m₁ = 0.033 kg
mass of the baseball, m₂ = 0.15 kg
initial velocity of the bullet, u₁ = 222 m/s
initial velocity of the baseball, u₂ = 0
let the common final velocity of the system after collision = v
Apply the principle of conservation of linear momentum to determine the common final velocity.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
0.033 x 222 + 0.15 x 0 = v(0.033 + 0.15)
7.326 = v(0.183)
v = 7.326 / 0.183
v = 40.03 m/s
Let the height risen by the system after collision = h
Initial velocity of the system after collision = Vi = 40.03 m/s
At maximum height, the final velocity, Vf = 0
acceleration due to gravity for upward motion, g = -9.8 m/s²
[tex]v_f^2 = v_i^2 +2gh\\\\0 = 40.03^2 - (2\times 9.8)h\\\\19.6h = 1602.4\\\\h = \frac{1602.4}{19.6} \\\\h = 81.76 \ m[/tex]
Therefore, the maximum height risen by the bullet-baseball system after the collision is 81.76 m.
The block in the drawing has dimensions L0×2L0×3L0,where L0 =0.2 m. The block has a thermal conductivity of 150 J/(s·m·C˚). In drawings A, B, and C, heat is conducted through the block in three different directions; in each case the temperature of the warmer surface is 35 ˚C and that of the cooler surface is 16 ˚C Determine the heat that flows in 6 s for each case.
Answer:
1140 J, 6840 J, 10260 J
Explanation:
Lo x 2 Lo x 3 Lo, Lo = 0.2 m, K = 150 J/(s · m · C˚) , T = 35 ˚C, T' = 16 ˚C,
time, t = 6 s
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 1140 J[/tex]
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{3\times 0.2}\\\\H = 6840 J[/tex]
The heat conducted is
[tex]H = \frac{K A (T - T') t}{d}\\\\H = \frac{150\times 3\times 0.2\times 2\times0.2\times (35-16) \times 6}{2\times 0.2}\\\\H = 10260 J[/tex]
If the source moves, the wavelength of the sound in front of the direction of motion is____than the wavelength behind the direction of motion.
a. the same.
b. smaller than.
c. unrealted to.
d. larger then.
Answer:
B. Smaller than
Explanation:
This question is from the Doppler effect. As the object which is in motion goes off from the other, there's a reduction in the frequency. This is due to the fact that successive soundwave get to be longer. So that the pitch will then be lowered. When the person observing moves towards what is making the sound, each soundwave that follows gets faster than the previous.
A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 N force on the ball.
What is the magnitude of the velocity after it hits the ground?
Answer:
9.25 m/s
Explanation:
A 100-W light bulb is left on for 20.0 hours. Over this period of time, how much energy did the bulb use?
Answer:
Power = Energy/time
Energy = Power xtime.
Time= 20hrs
Power = 100Watt =0.1Kw
Energy = 0.1 x 20 = 2Kwhr.
This Answer is in Kilowatt-hour ...
If the one given to you is in Joules
You'd have to Change your time to seconds
Then Multiply it by the power of 100Watts.
The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the two boxes is:__________.
A) 0 m/s
B) 1 m/s
C) 2 m/s
D) 3 m/s
E) Not enough info
Answer:
The correct option is (E).
Explanation:
Given that,
Mass of object 1, m₁ = 1 kg
Mass of object 2, m₂ = 2 kg
They collides after the collision. We need to find the speed of the two boxes after the collision.
The initial speeds of both boxes is not given. So, we can't put the values of their speeds in the momentum conservation equation.
So, the information is not enough.
prove mathematically :
1. v = u + at
2. s = ut+1*2 at
Answer:
a.v=u+v/2
a.v=s/t
combining two equation we get,
u+v/2=s/t
(u+v)t/2=s
(u+v)t/2=s
{u+(u+at)}t/2=s
(u+u+at)t/2=s
(2u+at)t/2=s
2ut+at^2/2=s
2ut/2+at^2/2=s
UT +1/2at^2=s
proved
a=v-u/t
at=v-u
u+at=v
A grade 12 Physics student shoots a basketball
from the ground at a hoop which is 2.0 m above
her release. The shot was at a velocity of 10 m/s
and at an angle of 80° to the ground.
a. Determine the vertical velocity of the ball
when it is at the level of the net. You
should get two answers.
Please show ALL steps
Answer:
7.84 m/s
Explanation:
Height, h = 2 m
Initial velocity, u = 10 m/s
Angle, A = 80°
(a) Let the time taken to go to the net is t.
Use second equation of motion
[tex]h = u t + 0.5 at^2\\\\- 2 = - 10 sin 80 t - 4.9 t^2\\\\4.9 t^2 + 9.8 t - 2 = 0 \\\\t= \frac{- 9.8\pm\sqrt{9.8^2 + 4\times 4.9\times 2}}{9.8}\\\\t = \frac{- 9.8 \pm 11.6}{9.8}\\\\t = - 2.2 s , 0.2 s[/tex]
Time cannot be negative.
So, t = 0.2 s
The vertical velocity at t = 0.2 s is
v = u + at
v = 10 sin 80 - 9.8 x0.2
v = 9.8 - 1.96 = 7.84 m/s
I need help with this problem can anybody help me please , it’s physics 2 course
Answer:
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Explanation:
The 2 capacitors in the middle are connected in parallel so simply add their capacitance together:
[tex]5.0\:\mu\text{F} + 8.0\:\mu\text{F} = 13.0\:\mu \text{F}[/tex]
Now we have 3 capacitors connected in series so their equivalent capacitance [tex]C_{eq}[/tex] is
[tex]\dfrac{1}{C_{eq}} = \dfrac{1}{10.0\:\mu \text{F}} + \dfrac{1}{13.0\:\mu \text{F}} + \dfrac{1}{9.0\:\ mu \text{F}} [/tex]
or
[tex]C_{eq} = 3.5\:\mu \text{F}[/tex]
The outer surface of a spacecraft in space has an emissivity of 0.44 and a solar absorptivity of 0.3. If solar radiation is incident on the spacecraft at a rate of 950 W/m2, determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed.
Answer:
[tex]T=326.928K[/tex]
Explanation:
From the question we are told that:
Emissivity [tex]e=0.44[/tex]
Absorptivity [tex]\alpha =0.3[/tex]
Rate of solar Radiation [tex]R=0.3[/tex]
Generally the equation for Surface absorbed energy is mathematically given by
[tex]E=\alpha R[/tex]
[tex]E=0.3*950[/tex]
[tex]E=285W/m^2[/tex]
Generally the equation for Emitted Radiation is mathematically given by
[tex]\mu=e(\sigmaT^4)[/tex]
Where
T=Temperature
[tex]\sigma=5.67*10^8Wm^{-2}K_{-4}[/tex]
Therefore
[tex]\alpha*E=e \sigma T^4[/tex]
[tex]0.3*(950)=0.44(5.67*10^-8)T^4[/tex]
[tex]T=326.928K[/tex]
A cylindrical tank with radius 7 m is being filled with water at a rate of 2 m3/min. How fast is the height of the water increasing (in m/min)?
Answer:
0.013 m/min
Explanation:
Applying,
dV/dt = (dh/dt)(dV/dh)............. Equation 1
Where
V = πr²h................ Equation 2
Where V = volume of the tank, r = radius, h = height.
dV/dh = πr²............ Equation 3
Substitute equation 3 into equation 1
dV/dt = πr²(dh/dt)
From the question,
Given: dV/dt = 2 m³/min, r = 7 m, π = 3.14
Substitute these values into equation 3
2 = (3.14)(7²)(dh/dt)
dh/dt = 2/(3.14×7²)
dh/dt = 0.013 m/min
A block slides down a frictionless plane that makes an angle of 24.0° with the horizontal. What is the
acceleration of the block?
Answer:
F = m g sin theta force accelerating block
m a = m g sin theta
a = 9.8 sin 24 = 3.99 m/sec^2
Question 9 of 10
According to the law of conservation of momentum, the total initial
momentum equals the total final momentum in a(n)
A. Interacting system
B. System interacting with one other system
C. Isolated system
D. System of balanced forces
Answer:
The answer is C. Isolated System
Answer:
C. Isolated system
Explanation :
∵According to law of conservation of momentum ,In an isolated system ,the total momentum remains conserved.
A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by y = (0.10 m)(sin πx/2) sin 12πt, where x = 0 at one end of the rope, x is in meters, and t is in seconds.
What are:
a. the length of the rope.
b. the speed of the waves on the rope
c. the mass of the rope
d. If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation.
Answer:
sup qwertyasdfghjk
Explanation:
What is the energy equivalent of an object with a mass of 2.5 kg? 5.5 × 108 J 7.5 × 108 J 3.6 × 1016 J 2.25 × 1017 J
Answer:
E = m c^2 = 2.5 * (3 * 10E8)^2 = 2.25 * 10E17 Joules
Answer:
The answer is D. 2.25 × 1017 J
Explanation:
got it right on edge 2021
convert 2.4 milimetre into metre
Answer is 0.0024
Explanation
divide the length value by 1000.
Two spheres are rolling without slipping on a horizontal floor. They are made of different materials, but each has mass 5.00 kg and radius 0.120 m. For each the translational speed of the center of mass is 4.00 m/s. Sphere A is a uniform solid sphere and sphere B is a thin-walled, hollow sphere. Part B How much work, in joules, must be done on the solid sphere to bring it to rest? Express your answer in joules. VO AE4D ? J WA Request Answer Submit Part C How much work, in joules, must be done on the hollow sphere to bring it to rest? Express your answer in joules. Wa Request
Answer:
Explanation:
Moment of inertia of solid sphere = 2/5 m R²
m is mass and R is radius of sphere.
Putting the values
Moment of inertia of solid sphere I₁
Moment of inertia of hollow sphere I₂
Kinetic energy of solid sphere ( both linear and rotational )
= 1/2 ( m v² + I₁ ω²) [ ω is angular velocity of rotation ]
= 1/2 ( m v² + 2/5 m R² ω²)
= 1/2 ( m v² + 2/5 m v²)
=1/2 x 7 / 5 m v²
= 0.7 x 5 x 4² = 56 J .
This will be equal to work to be done to stop it.
Kinetic energy of hollow sphere ( both linear and rotational )
= 1/2 ( m v² + I₂ ω²) [ ω is angular velocity of rotation ]
= 1/2 ( m v² + 2/3 m R² ω²)
= 1/2 ( m v² + 2/3 m v²)
=1/2 x 5 / 3 m v²
= 0.833 x 5 x 4² = 66.64 J .
This will be equal to work to be done to stop it.
During a practice shot put throw, the 7.9-kg shot left world champion C. J. Hunter's hand at speed 16 m/s. While making the throw, his hand pushed the shot a distance of 1.4 m. Assume the acceleration was constant during the throw.
Required:
a. Determine the acceleration of the shot.
b. Determine the time it takes to accelerate the shot.
c, Determine the horizontal component of the force exerted on the shot by hand.
Answer:
a) a = 91.4 m / s², b) t = 0.175 s, c)
Explanation:
a) This is a kinematics exercise
v² = vox ² + 2a (x-xo)
a = v² - 0/2 (x-0)
let's calculate
a = 16² / 2 1.4
a = 91.4 m / s²
b) the shooting time
v = vox + a t
t = v-vox / a
t = 16 / 91.4
t = 0.175 s
c) let's use Newton's second law
F = ma
F = 7.9 91.4
F = 733 N
Find the volume of cuboid of side 4cm. Convert it in SI form
Answer:
0.000064 cubic meters.
Explanation:
Given the following data;
Length of side = 4 centimeters
Conversion:
100 centimeters = 1 meters
4 cm = 4/100 = 0.04 meters
To find the volume of cuboid;
Mathematically, the volume of a cuboid is given by the formula;
Volume of cuboid = length * width * height
However, when all the sides are equal the formula is;
Volume of cuboid = L³
Volume of cuboid = 0.04³
Volume of cuboid = 0.000064 cubic meters.
Question 8 a-e plz
Answer:
(a) t = 0 s
(b) t = 0 s, 30 s, 55 s
(c) t = 40 s to t = 60 s
(d) t = 10 s to t = 15 s
(e) a = 6 m/s^2
Explanation:
(a) The car is at starting position at t = 0 s and v = 0 m/s.
(b) The velocity of car is zero when the time is t = 0 s, 30 s and 55 s.
(c) from t = 40 s to 60 s the car is moving in the negative direction.
(d) The fastest speed is 60m/s from t = 10 s to t = 15 s.
(e) The slope of the velocity time graph gives acceleration.
a = (60 - 0) / (10 - 0) = 6 m/s^2
