given info is... Acceleration(a)=2.6m/s^2
final velocity(v)=26.8m/s
initial velocity(u)=24.6m/s
need to find.... time(t)=?
[tex]a=\frac{v-u}{t} \\2.6=\frac{26.8-24.6}{t} \\\\[/tex]
[tex]t=\frac{v-u}{a}[/tex]
[tex]t=\frac{26.8-24.6}{2.6}[/tex]
[tex]t=0.846s[/tex]
Explanation:
It takes 0.84 second her car to accelerate from 24.6m/s to 26.8m/s.
What is acceleration?Acceleration is the rate at which speed and direction of velocity vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates. Even if the speed is constant, motion on a circle accelerates because the direction is always shifting.
Given parameters:
Initial speed of the car: u = 24.6 m/s
Final speed of the car: v = 26.8 m/s.
Acceleration of the car: a = 2.6 m/s²
Time interval: t = ?
change is speed = final speed - initial speed
= 26.8 m/s - 24.6 m/s
= 2.2 m/s
From the definition of acceleration,
acceleration = change is speed / time interval
So, time interval = change is speed / acceleration
= 2.2 m/s/2.6 m/s²
= 0.84 second.
Hence, it takes 0.84 second her car to accelerate from 24.6m/s to 26.8m/s.
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Design a tension member and slip-critical splice to carry a factored load of 500 kips. Please use a wide-flange section for the tension member. Please use A572 Gr. 50 steel plates for the splice plates. Please use Group B, A490 bolts for the splice connection. The splice connection should be slip-critical, and have adequate strength after slip occurs as well. Please make any other assumptions you need in order to complete the problem. Provide detailed sketches and drawings for your design.
Answer:
Kindly check the explanation section.
Explanation:
For the design we are asked for in this question/problem there is the need for us to calculate or determine the strength in fracture and that of the yield. Also, we need to calculate for the block shear strength.
From the question, we have that the factored load = 500kips. Also, note that the tension splice must not slip.
Also, the shear force are resisted by friction, that is to say shear resistance = 1.13 × Tb × Ns.
Assuming our db = 3/4 inches, then the slip critical resistance to shear service load = 18ksi(refer to AISC manual for the table).
If db = 7/8 inches, then the shear force resistance for n bolt = 10.2kips, n > 49.6.
The yielding strength = 0.9 × Aj × Fhb= 736 kips > 500
The fracture strength = .75 × Ah × Fhb = 309 kips.
The bearing strength of 7/8 inches bolt at the edge hole and other holes = 46 kips and 102 kips.
What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?
Answer:
303 Ω
Explanation:
Given
Represent the resistors with R1, R2 and RT
R1 = 633
RT = 205
Required
Determine R2
Since it's a parallel connection, it can be solved using.
1/Rt = 1/R1 + 1/R2
Substitute values for R1 and RT
1/205 = 1/633 + 1/R2
Collect Like Terms
1/R2 = 1/205 - 1/633
Take LCM
1/R2 = (633 - 205)/(205 * 633)
1/R2 = 428/129765
Take reciprocal of both sides
R2 = 129765/428
R2 = 303 --- approximated
A 10-ohm resistor has a constant current. If 1200 C of charge flow through it in 4 minutes what
is the value of the current?
A. 3.0 A
B 5.0 A
C. 11 A
D. 15 A
E. 20A
Answer:
B 5.0 A .
Explanation:
Hello.
In this case, since we know the charge (1200 C), time (4 min =240 s) and resistance (10Ω) which is actually not needed here, we compute the current as follows:
[tex]I=\frac{Q}{t}[/tex]
Then, for the given data, we obtain:
[tex]I=\frac{1200C}{4min}*\frac{1min}{60s}\\\\I=5A[/tex]
Therefore, answer is B 5.0 A .
Best regards!
If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be Imax
The complete question is;
A person with body resistance between his hands of 10 kΩ accidentally grasps the terminals of a 16-kV power supply. What is the power dissipated in his body?
A) If the internal resistance of the power supply is 1600 Ω , what is the current through the person's body?
B) What is the power dissipated in his body?
C) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be I_max = 1.00mA or less?
Answer:
A) I = 1.379 A
B) P = 19016.41 W
C) r = 15990000 Ω
Explanation:
A) We are given;
Internal resistance of the power supply; r = 1600 Ω
Body resistance between hands; R = 10kΩ = 10000 Ω
Power supply voltage; E =16 kV = 16000 V
Formula for the current through the person's body with internal resistance is given by;
I = E/(R + r)
Thus;
I = 16000/(10000 + 1600)
I = 1.379 A
B) Formula for power dissipated is;
P = I²R
P = 1.379² × 10000
P = 19016.41 W
C) Now, we are told that the maximum current should be I_max = 1.00mA or less. So, I_max = 0.001 A
Thus, from I = E/(R + r) and making r the subject, we have;
r = (E/I) - R
r = (16000/0.001) - 10000
r = 15990000 Ω
Take the regular compass and hold it so the case is vertical. Now use it to investigate the direction of the coil’s magnetic field at locations other than the central axis. What happens as you move away from the center axis toward the coil? What happens above the coil? Outside the coil? Below the coil?
Answer:
Please find the answer in the explanation
Explanation:
Take the regular compass and hold it so the case is vertical. Now use it to investigate the direction of the coil’s magnetic field at locations other than the central axis.
What happens as you move away from the center axis toward the coil? The direction of the magnetic compass needle will move in an opposite direction since the direction of the induced voltage is reversed.
What happens above the coil?
the needle on the magnetic compass will be deflected. Since compasses work by pointing along magnetic field lines
Outside the coil? The magnetic compass needle will experience no deflection. Since there is no induced voltage or current.
Below the coil?
The needle will move in an opposite direction.
An object moving 20 m/s
experiences an acceleration of 4 m/s' for 8
seconds. How far did it move in that time?
Variables:
Equation and Solve:
Answer:
We are given:
initial velocity (u) = 20m/s
acceleration (a) = 4 m/s²
time (t) = 8 seconds
displacement (s) = s m
Solving for Displacement:
From the seconds equation of motion:
s = ut + 1/2 * at²
replacing the variables
s = 20(8) + 1/2 * (4)*(8)*(8)
s = 160 + 128
s = 288 m
The Jamaican Bobsled Team is sliding down a hill in a toboggan at a rate of 5 m/s when he reaches an even steeper slope. If he accelerates at 2 m/s2 for the 5 m slope, how fast is he traveling when he reaches the bottom of the 5 m slope?
Answer:
6.7 m/s
Explanation:
Given:
Δx = 5 m
v₀ = 5 m/s
a = 2 m/s²
Find: v
v² = v₀² + 2aΔx
v² = (5 m/s)² + 2 (2 m/s²) (5 m)
v = 6.7 m/s
A mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. She weighs 555 N. As the drawing shows, she is closer to the left cliff than to the right cliff, with the result that the tensions in the left and right sides of the rope are not the same. Find the tension in the rope to the left of the mountain climber.
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The tension in the rope on the left of the mountain climber is [tex] T_a = 1106 \ N [/tex]
Explanation:
From the question we are told that
The weight of the mountain climber is m = 555 N
Generally from the diagram , the total amount of force acting on the rope along the vertical axis at equilibrium is mathematically represented as
[tex]T_a* cos 65 -555 + T_b * cos(85) = 0[/tex]
Here [tex]T_a, T_b[/tex] are the tension of the rope on the left and on the right hand side
So
[tex]0.423T_a + 0.0871T_b = 555[/tex]
=> [tex] 0.0871T_b = 555 - 0.423T_a[/tex]
=> [tex] T_b = \frac{555 - 0.423T_a}{0.0871}[/tex]
Generally from the diagram , the total amount of force acting on the rope along the horizontal axis at equilibrium is mathematically represented as
[tex]T_a* sin 65 - T_b * sin(85) = 0[/tex]
=> [tex] 0.9063T_a - 0.9962T_b = 0[/tex]
=> [tex] 0.9063T_a = 0.9962T_b [/tex]
=> [tex] 0.9063T_a = 0.9962[\frac{555 - 0.423T_a}{0.0871}] [/tex]
=> [tex] 0.9063T_a = [\frac{552.891 - 0.421T_a}{0.0871}] [/tex]
=> [tex] 0.0789T_a = [552.891 - 0.421T_a[/tex]
=> [tex] 0.4999T_a = 552.891 [/tex]
=> [tex] T_a = 1106 \ N [/tex]
The interaction between electrical energy and magnetism has been an important
topic in 20th century science, Which term describes this interaction?
Answer:
Maybe
Explanation:
I say maybe because it will help them still but not quite
Converting compound units
You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfortunately, you have the density of mercury in units of kilogram/meter3 and the density of silicon in other units: 2.33 gram/centimeter3. You decide to convert the density of silicon into units of kilogram/meter3 to perform the comparison. By which combination of conversion factors will you multiply 2.33 gram/centimeter3 to perform the unit conversion?
Answer:
Explanation:
Given the density of silicon as 2.33g/cm³
We are to convert this to kg/cm³
We will be using the following conversion factors
1000g = 1kg
2.33g = x
Cross multiply
1000x = 2.33
x = 2.33/1000
x = 0.00233kg
Also we need to convert 1cm³ to 1m³
1cm = 0.01m
1cm³ = 0.01×0.01×0.01
1cm³ = 0.000001m³
Substituting into the density value of silicon
2.33g/cm³ = 0.00233kg/0.000001m³
= 2330kg/m³
A vector of components (−23, −22) is multiplied by the scalar value of −6. What is the magnitude and direction of the resultant vector?
Answer:
(1,)
Explanation:
Answer:
magnitude: 21.6; direction: 33.7°
Explanation:
The steam from a boiling pot of water is
A: conduction
B: Convection
C: radiation
D: Radiant energy
A car traveling at 27 m/s slams on its brakes to come to a stop. It decelerates at a rate of 8 m/s2 . What is the stopping distance of the car?
v² - u² = 2 a ∆x
where u = initial velocity (27 m/s), v = final velocity (0), a = acceleration (-8 m/s², taken to be negative because we take direction of movement to be positive), and ∆x = stopping distance.
So
0² - (27 m/s)² = 2 (-8 m/s²) ∆x
∆x = (27 m/s)² / (16 m/s²)
∆x ≈ 45.6 m
The stopping distance of car achieved during the braking is of 45.56 m.
Given data:
The initial speed of car is, u = 27 m/s.
The final speed of car is, v = 0 m/s. (Because car comes to stop finally)
The magnitude of deacceleration is, [tex]a = 8\;\rm m/s^{2}[/tex].
In order to find the stopping distance of the car, we need to use the third kinematic equation of motion. Third kinematic equation of motion is the relation between the initial speed, final speed, acceleration and distance covered.
Therefore,
[tex]v^{2}=u^{2}+2(-a)s[/tex]
Here, s is the stopping distance.
Solving as,
[tex]0^{2}=27^{2}+2(-8)s\\\\s = 45.56 \;\rm m[/tex]
Thus, we can conclude that the stopping distance of car achieved during the braking is of 45.56 m.
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I WILL MARK YOU AS BRAINLIEST IF RIGHT
What is the magnitude of the net force acting on this object? And what direction?
Answer:
The magnitude of the net force acting on an object is equal to the mass. and the direction is in 20N
Explanation:
You measure the radius of a sphere as (6.45 ± 0.30) cm, and you measure its mass as (1.79 ± 0.08) kg. What is the density and uncertainty in the density of the sphere, in kilograms per cubic meter?
Answer:
[tex](1630.13\pm 300.10)\ kg/m^3[/tex]
Explanation:
Given that,
The radius of a sphere is (6.45 ± 0.30) cm
Mass of the sphere is (1.79 ± 0.08) kg
Density = mass/volume
For sphere,
[tex]d=\dfrac{m}{V}\\\\d=\dfrac{m}{\dfrac{4}{3}\pi r^3}\\\\d=\dfrac{1.79\ kg}{\dfrac{4}{3}\pi (6.4\times 10^{-2}\ m)^3}\\\\d=1630.13\ kg/m^3[/tex]
We can find the uncertainty in volume as follows :
[tex]\dfrac{\delta V}{V}=3\dfrac{\delta r}{r}\\\\=3\times \dfrac{0.3\times 10^{-2}}{6.45\times 10^{-2}}\\\\=0.1395[/tex]
Uncertainty in mass,
[tex]\dfrac{\delta m}{m}=\dfrac{0.08}{1.79}\\\\=0.0446[/tex]
Now, the uncertainty in density of sphere is given by :
[tex]\dfrac{\delta d}{d}=\dfrac{\delta m}{m}+\dfrac{\delta V}V}\\\\=0.0446+0.1395\\\\\dfrac{\delta d}{d}=0.1841\\\\\delta d=0.1841\times d\\\\\delta d=0.1841\times 1630.13\\\\\delta d = 300.10\ kg/m^3[/tex]
Hence, the density pf the sphere is [tex](1630.13\pm 300.10)\ kg/m^3[/tex]
Which object will require the greatest amount of force to change its motion?
A. A 148 kg object moving 131 m/s
B. A 153 kg object moving 127 m/s
C. A 160 kg object moving 126 m/s
O D. A 162 kg object moving 124 m/s
Answer: D 160kg object moving 126 m/s
Explanation:
An object having a mass of 162 kg and moving with a velocity of 124 m/sec will require the greatest amount of force to change its motion. The correct option is D.
What is force?Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body.
If the object has to stop, the final velocity must be zero. If the time is constant, the amount of force only depends on the mass and the velocity at which the body is moving.
The amount of force on the object depends on the momentum of the body.
The momentum of the body is;
P = mv
Object D will require the greatest amount of force to change its motion. Because the momentum of the body for option D is the greatest.
Hence, the correct option is D.
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Calculate the work WC done by the gas during the isothermal expansion. Express WC in terms of p0, V0, and Rv.
Complete Question
The complete question is shown on the first and second uploaded image
Answer:
The expression is [tex]W_c = P_o V_o ln (R_v)[/tex]
Explanation:
Generally smallest workdone done by a gas is mathematically represented as
[tex]dW = PdV[/tex]
Generally for an isothermal process
[tex]PV = nRT = constant [/tex]
=> [tex]P = \frac{nRT}{V}[/tex]
Generally the total workdone is mathematically represented as
[tex]W_c = \int\limits^{v_f}_{V_o} {\frac{nRT}{V} } \, dV[/tex]
=> [tex]W_c = nRT \int\limits^{V_f}_{V_o} {\frac{1}{V} } \, dV[/tex]
=> [tex]nRT [lnV] | \left \ {V_f}} \atop {V_o}} \right.[/tex]
=> [tex]W_c = nRT [ln(V_f) - ln(V_o)][/tex]
=> [tex]W_c = nRT ln \frac{V_f}{V_o}[/tex]
From the question [tex]\frac{V_f}{V_o } = R_v[/tex]
=> [tex]W_c = P Vln (R_v)[/tex]
at initial state
[tex]W_c = P_o V_o ln (R_v)[/tex]
If a rock is skipped into a lake at 24 m/s2, with that what force was the rock thrown if it was 1.75kg?
Answer: f= M×A
1.75kg×24= 42N
Explanation:
Because to find force you do Mass times acceleration so I did 1.75 kg times 24 would equal 42 Newtons!
The feeling of weightlessness occurs because _____________________.
there is no supporting force under your mass.
there is no gravity present.
there is only a small amount of gravity present.
Answer:
there is only a small amount of gravity present.
Explanation:
this is because the only force acting upon your body during free fall is the force of gravity which is a non contact force.
PLEASE HELP EASY MULTIPLE CHOICE!!!!!!!!!!!
Answer:
options C is correct
Explanation:
asking questions is super in this education life
Answer:
option c should be the answer
Chris races his Audi north down a road for 1000 meters in 20 seconds, what is his velocity?
Answer:
I think it would be 50 I am not really sure
Explanation:
I think you would have to divid 1000 by 20 Again I'm not sure
A car is traveling south at 8.77 m/s. It then begins a uniform acceleration until it reaches a velocity of 47.8 m/s over a period of 3.84s. What is the car's acceleration?
Please help !
Answer:
The acceleration of the car is 10.16m/s²
Explanation:
Given parameters:
Initial velocity = 8.77m/s
Final velocity = 47.8m/s
Time duration = 3.84s
Unknown:
Acceleration of the car = ?
Solution:
To find the acceleration, we must bear in mind that this physical quantity is the change in velocity with time;
Acceleration = [tex]\frac{V - U}{T}[/tex]
V is the final velocity
U is the initial velocity
T is the time taken
Input the parameters and solve for acceleration;
Acceleration = [tex]\frac{47.8 - 8.77}{3.84}[/tex] = 10.16m/s²
The acceleration of the car is 10.16m/s²
How much would a 15.0 kg object weigh on that planet? Round the answer to the nearest whole number.
Answer:
168
Explanation:
Answer: a 15 kg object would weigh the most on Neptune
168 N
A 1870 kg car traveling at 13.5 m/s collides with a 2970 kg car that is initally at rest at a stoplight. The cars stick together and move 1.93 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.
Answer:
The value is [tex] \mu = 0.72 [/tex]
Explanation:
From the question we are told that
The mass of the first car is [tex]m_1 = 1870\ kg[/tex]
the initial speed of the car is [tex]u = 13.5 \ m/s[/tex]
The mass of the second car is [tex]m_2 = 2970\ kg[/tex]
The distance move by both cars is s = 1.93 m
Generally from the law of momentum conservation
[tex]m_1 * u_1 + m_2 * u_2 = (m_1 + m_2 ) * v_f[/tex]
Here [tex]u_2 = 0[/tex] because the second car is at rest
and [tex]v_f[/tex] is the final velocity of the the two car
So
[tex]1870* 13.5+ 0= ( 1870 + 2970 ) * v_f[/tex]
=> [tex]v_f = 5.22\ m/s[/tex]
Generally from kinematic equation
[tex]v_f^2 = u_2^2 + 2as[/tex]
here a is the deceleration
So
[tex]5.22^2 = 0 + 2 *a * 1.93[/tex]
=> [tex]a = 7.06 \ m/s^2 [/tex]
Generally the frictional force is equal to the force propelling the car , this can be mathematically represented as
[tex]F_f = F[/tex]
Here F is mathematically represented as
[tex]F = (m_1 + m_2) * a[/tex]
[tex]F = (1870 + 2970) * 7.06 [/tex]
[tex]F =34170.4 \ N[/tex]
and
[tex]F_f = \mu * (m_1 + m_2 ) * g[/tex]
[tex]F_f = 47432 * \mu [/tex]
So
[tex] 47432 * \mu = 34170.4 [/tex]
=> [tex] 47432 * \mu = 34170.4 [/tex]
=> [tex] \mu = 0.72 [/tex]
"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"
This question is incomplete, the complete question is;
The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.
"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"
Answer:
the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
Explanation:
Given that;
P₁ = 1.00 atm
P₂ = ?
V₁ = 1 L
V₂ = 1.60 L
the temperature of the gas is kept constant
we know that;
P₁V₁ = P₂V₂
so we substitute
1 × 1 = P₂ × 1.60
P₂ = 1 / 1.60
P₂ = 0.625 atm
Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
g A child bounces a 50 g super ball on the sidewalk. The velocity change of the super bowl is from 27 m/s downward to 17 m/s upward. If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk
Answer:
The average force exerted on the superball by the sidewalk is 0.00122 N.
Explanation:
Given;
mass of the super ball, m = 50 g = 0.05 kg
initial velocity of the super bowl, u = -27 m/s (assuming downward motion to be negative)
final velocity of the super bowl, u = 17 m/s (assuming upward motion to be positive)
time of motion, t = 1800 s
The average force exerted on the superball by the sidewalk is given by;
[tex]F = ma\\\\F = \frac{m(v-u)}{t} \\\\F = \frac{0.05(17-(-27))}{1800}\\\\ F = \frac{0.05(44)}{1800}\\\\F = 0.00122 \ N[/tex]
Therefore, the average force exerted on the superball by the sidewalk is 0.00122 N.
Write a haiku
poem
explaining
why graphing
is useful.
If you are
able, share
your poem
with others.
Answer:
Explanation:
graphing is helpful
helps visualize the line
of your equation
21. A toy car starts from rest and begins to
accelerate at 11.0 m/s2. What is the toy car's
final velocity after 6.0 seconds?
Answer:
v = 66 m/s
Explanation:
Given that,
The initial velocity of a car, u = 0
Acceleration of the car, a = 11 m/s²
We need to find the final velocity of the toy after 6 seconds.
Let v is the final velocity. It can be calculated using first equation of motion. It is given by :
v = u +at
v = 0 + 11 m/s² × 6 s
v = 66 m/s
So, the final velocity of the car is 66 m/s.
A uniform magnetic field of magnitude 0.72 T is directed perpendicular to the plane of a rectangular loop having dimensions 8.2 cm by 14 cm. Find the magnetic flux through the loop.
Answer:
Explanation:
Magnetic flux is expressed as the product of magnetic field and cross sectional area.
Φ = BAsintheta
Given
B = 0.72T
A = 8.2cm×14cm
A = 0.082m × 0.14m
Area = 0.01148m²
Theta = 90°
Substitute into the formula
Φ = BAsintheta
Φ = 0.72(0.01148)sin90°
Φ = 0.72(0.01148)(1)
Φ = 0.0082656
Hence the magnetic flux through the loop is 8.2656 × 10^-3 Weber
A particle is moved along the x-axis by a force that measures 10/(1+x)^2 pounds at a point x feet from the origin. Find the work (in ft-lb) done in moving the particle from the origin to a distance of 9 feet.
Answer:
9 ft*lb
Explanation:
super simple but you just have to understand that the integral is going with the curve
work = integral a to b of f(x)dx = integral 0 to 9 of 10/(1+x)^2dx = 9ft*lb