Answer:
t = 3 [s]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f}=v_{o}-g*t[/tex]
where:
Vf = final velocity [m/s]
Vo = initial velocity = 15 [m/s]
g = gravity acceleration = 10 [m/s²]
t = time [s]
Now replacing we have:
[tex]0 = 15 -10*t\\10*t=15\\t= 1.5[s][/tex]
Note: In the equation above the gravity acceleration is negative, because the movement of the ball bearing is pointing againts the gravity acceleration.
The time calculated is only when the ball bearing reaches the highest elevation, and it will take the same time for descending, therefore the total time is:
t = 1.5 + 1.5 = 3 [s]
At a certain instant, a ball is thrown downward with a velocity of 8.0 m/s from a height of 40 m. At the same instant, another ball is thrown upward from ground level directly in line with the first ball with a velocity of 12 m/s. Find (a) the time when the balls collide and (b) the height at which they collide. Take g = 10 m/s2
Answer:
(a) The two balls collide [tex]2\; \rm s[/tex] after launch.
(b) The height of the collision is [tex]4\; \rm m[/tex].
(Assuming that air resistance is negligible.)
Explanation:
Let vector quantities (displacements, velocities, acceleration, etc.) that point upward be positive. Conversely, let vector quantities that point downward be negative.
The gravitational acceleration of the earth points dowards (towards the ground.) Therefore, the sign of [tex]g[/tex] should be negative. The question states that the magnitude of [tex]g\![/tex] is [tex]10\; \rm m \cdot s^{-2}[/tex]. Hence, the signed value of [tex]\! g[/tex] should be [tex]\left(-10\; \rm m \cdot s^{-2}\right)[/tex].
Similarly, the initial velocity of the ball thrown downwards should also be negative: [tex]\left(-8.0\; \rm m \cdot s^{-1}\right)[/tex].
On the other hand, the initial velocity of the ball thrown upwards should be positive: [tex]\left(12\; \rm m \cdot s^{-1}\right)[/tex].
Let [tex]v_0[/tex] and [tex]h_0[/tex] denote the initial velocity and height of one such ball. The following SUVAT equation gives the height of that ball at time [tex]t[/tex]:
[tex]\displaystyle h(t) = \frac{1}{2}\, g \cdot {t}^{2} + v_0 \cdot t + h_0[/tex].
For both balls, [tex]g = \left(-10\; \rm m \cdot s^{-2}\right)[/tex].
For the ball thrown downwards:
Initial velocity: [tex]v_0 = \left(-8.0\; \rm m \cdot s^{-1}\right)[/tex].Initial height: [tex]h_0 = 40\; \rm m[/tex].[tex]\displaystyle h(t) = -5\, t^{2} + (-8.0)\, t + 40[/tex] (where [tex]h[/tex] is in meters and [tex]t[/tex] is in seconds.)
Similarly, for the ball thrown upwards:
Initial velocity: [tex]v_0 = \left(12\; \rm m \cdot s^{-1}\right)[/tex].Initial height: [tex]h_0 = 0\; \rm m[/tex].[tex]\displaystyle h(t) = -5\, t^{2} + 12\, t[/tex] (where [tex]h[/tex] is in meters and [tex]t[/tex] is in seconds.)
Equate the two expressions and solve for [tex]t[/tex]:
[tex]-5\, t^{2} + (-8.0)\, t + 40 = -5\, t^{2} + 12\, t[/tex].
[tex]t = 2[/tex].
Therefore, the collision takes place [tex]2\, \rm s[/tex] after launch.
Substitute [tex]t = 2[/tex] into either of the two original expressions to find the height of collision:
[tex]h = -5\times 2^{2} + 12 \times 2 = 4\; \rm m[/tex].
In other words, the two balls collide when their height was [tex]4\; \rm m[/tex].
The time the two balls collide is 0.4 seconds while the height at which they collide is 4m
The given parameters are :
Initial Velocity U = 8m/s
Height H = 40m
For the second ball, the initial velocity = 12m/s
a.) For the first ball, the height attained at the point of collision will be
h = ut + 1/2gt^2
h = 8t + 1/2 x 10t^2 ........ (1)
For the second ball, the height attained at the point of collision will be
h = 12t - 1/2 x 10t^2 .........(2)
Since the height will be the same for the two balls, equate the two equations
8t + 10t^2 = 12t - 10t^2
Collect the like term
8t - 12t = -5t^2 - 5t^2
-4t = -10^2
10t = 4
t = 4/10
t = 0.4s
b.) Substitute time t in any of the equation to find the height
h = 12(0.4) - 0.5 x 10(0.4)^2
h = 4.8 - 0.8
h = 4m
Therefore, the time the two balls collide is 0.4 seconds while the height at which they collide is 4m
Learn more here : https://brainly.com/question/20709166
A person weighs 60kg. Calculate the amount of work done if he is raised 12m into the air.
Answer: I belive the answer would be 720 Joules
Explanation:
Work can be calculated by multiplying an objects weight times the distance traveled. Work is measured in joules (J).
Formula: W x D = J
W= weight
D= Distance
J= joules of energy
Hope it helps
There are two different isotopes; X and Y, both contain the same number of radioactive substances. If sample X
has a longer half-life than Y, compare their rate of radioactive decay.
O A. Rate does not depend on half-life
B. Both of their rates are equal
O C. X has a smaller rate than Y
O D. X has a greater rate than Y
Answer:
Half life refers to the time for 1/2 of the radioactive atoms to decay.
Suppose that X has a half life of 10 days and Y has a half life of 20 days
If both start out with 1000 radioactive atoms then after 20 days
X would have 250 radioactive atoms and Y would have 500 atoms
The rate of decay is greater for the shorter half life:
In the example given X must have the smaller rate of decay because it has a longer half life.
John runs 3 km north then walks 2 km south. What is his total distance traveled and displacement?
Answer:
the total distance is 5km and the displacement is 1km
Explanation:
The total distance would be the addition of John running both ways so 3 km, 2 km.
However since he only walked back from a distance of 3 km to 2 km, he would be displaced 1 km because displacement is more like the position from the original point.
Think about 2 km as a positive value for the first part of the question and a negative value for the second part.
Assuming no friction, how does the initial gravitational potential energy of
the marble on a downward slope compare to the final kinetic energy?
a) they are the same
b) the initial gravitational potential energy is greater than the final kinetic energy
c) the initial gravitational potential energy is less then the final kinetic energy
Answer:
a) They are the same.
Explanation:
Assuming no friction, there should be no energy transfer and thus the Law of Conservation of Energy says:
[tex]PE=KE,\\mgh=\frac{1}{2}mv^2[/tex]
These types of problems also disregard any air resistance the surface of the object may cause. Therefore, no energy is transferred and from the Law of Conservation of Energy, [tex]100\%[/tex] of energy is preserved.
Which action will leave the dump trucks inertia unchanged?? PLEASE ANSWER FAST!!!
A. add gas
B. increase force applied to engine
Answer:
B.
Explanation:
You perform nine (identical) measurements of the acceleration of gravity (units of m/s2): 10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, and 9.90. The true value is 9.81. Calculate the standard error of your results to ONE significant digit.
Answer:
0.01
Explanation:
Given the data:
10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, 9.90
True value = 9.81
Mean value :
Σx / n
Sample size, n = 9
(10.1 + 9.87 + 9.76 + 9.91 + 9.75 + 9.88 + 9.69 + 9.83 + 9.90) / 9
= 88.69 / 9
= 9.854
Standard deviation (σ) :
Sqrt (Σ(X - m)² / n)
[(10.1 - 9.854)^2 + (9.87 - 9.854)^2 + (9.76 - 9.854)^2 + (9.91 - 9.854)^2 + (9.75 - 9.854)^2 + (9.88 - 9.854)^2 + (9.69 - 9.854)^2 + (9.83 - 9.854)^2 + (9.90 - 9.854)^2] / 9
Sqrt(0.113824 / 9)
Sqrt(0.0126471)
σ = 0.1124593
Standard Error = σ / sqrt(n)
Standard Error = 0.1124593 / 9
Standard Error = 0.0124954
Standard Error = 0.01 ( 1 significant digit)
A small rock is thrown vertically upward with a speed of 17.0m/s from the edge of the roof of a 26.0m tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected.
Part A
What is the speed of the rock just before it hits the street?
Express your answer with the appropriate units.
Part B
How much time elapses from when the rock is thrown until it hits the street?
Express your answer with the appropriate units.
Answer:
A) v = 28.3 m/s
B) t = 4.64 s
Explanation:
A)
Assuming no other forces acting on the rock, since the accelerarion due to gravity close to the surface to the Earth can be taken as constant, we can use one of the kinematic equations in order to get first the maximum height (over the roof level) that the ball reaches:[tex]v_{f}^{2} - v_{o}^{2} = 2* g* \Delta h (1)[/tex]
Taking into account that at this point, the speed of the rock is just zero, this means vf=0 in (1), so replacing by the givens and solving for Δh, we get:[tex]\Delta h = \frac{-v_{o} ^{2}}{2*g} = \frac{-(17.0m/s)^{2} }{2*(-9.8m/s2)} = 14.8 m (2)[/tex]
So, we can use now the same equation, taking into account that the initial speed is zero (when it starts falling from the maximum height) and that the total vertical displacement is the distance between the roof level and the ground (26.0 m) plus the maximum height that we have just found in (2) , 14.8m:Δh = 26.0 m + 14. 8 m = 40.8 m (3)Replacing now in (1), we can solve for vf, as follows:[tex]v_{f} =\sqrt{2*g*\Delta h} = \sqrt{2*9.8m/s2*40.8m} = 28.3 m/s (4)[/tex]
B)
In order to find the total elapsed from when the rock is thrown until it hits the street, we can divide this time in two parts:1) Time elapsed from the the rock is thrown, until it reaches to its maximum height, when vf =02) Time elapsed from this point until it hits the street, with vo=0.For the first part, we can simply use the definition of acceleration (g in this case), making vf =0, as follows:[tex]v_{f} = v_{o} + a*\Delta t = v_{o} - g*\Delta t = 0 (5)[/tex]
Replacing by the givens in (5) and solving for Δt, we get:[tex]\Delta t = \frac{v_{o}}{g} = \frac{17.0m/s}{9.8m/s2} = 1.74 s (6)[/tex]
For the second part, since we know the total vertical displacement from (3), and that vo = 0 since it starts to fall, we can use the kinematic equation for displacement, as follows:[tex]\Delta h = \frac{1}{2} * g * t^{2} (7)[/tex]
Replacing by the givens and solving for t in (7), we get:[tex]t_{fall} =\sqrt{\frac{2*\Delta h}{g}} = \sqrt{\frac{2*40.8m}{9.8m/s2} } = 2.9 s (8)[/tex]
So, total time is just the sum of (6) and (8):t = 2.9 s + 1.74 s = 4.64 sExplain how momentum is determined and conserved.
ASAP!!
Explanation:
Momentum is conserved in the collision. Momentum is conserved for any interaction between two objects occurring in an isolated system.
What is the strength of an electric field that will put a force of
1.28 x 10-15 N on a proton?
Answer: E = 7,490.6 N/C
Explanation:
If we have a field E, and a particle with a charge q, the force that the particle experiences is:
F = E*q
In this case, we know that the force is:
F = 1.2*10^(-15) N
And we know that the particle is a proton, where the charge of a proton is:
q = 1.602*10^(-19) C
Then we can replace these two values in the equation to get:
1.2*10^(-15) N = E*1.602*10^(-19) C
We just need to isolate E.
(1.2*10^(-15) N)/(1.602*10^(-19) C) = E
7,490.6 N/C = E
That is the strength of the electric field.
A receiver catches a football on the 50.0 yard line and is tackled 5.42 seconds later on the 12 yard line. What
was the runner's average speed?
Answer:
7.01yard/sec
Explanation:
Given parameters:
Initial position = 50yard
Final position = 12yard
Time = 5.42s
Unknown:
Average speed of runner = ?
Solution:
To solve this problem;
Speed = [tex]\frac{distance}{time}[/tex]
Distance covered = Initial position - final position = 50 - 12 = 38yards
So;
Speed = [tex]\frac{38}{5.42}[/tex] = 7.01yard/sec
A 12-kg object is moving rightward with a constant velocity of 4 m/s. How much net force is required to keep the object moving with
the same speed and in the same direction?
How many miles per day can you walk at a MODERATE Intensity level and your heart rate is 170?
Answer:
Not enough detail as it is very defendant on the person and a bunch of factors in health, but overall your heart rate shouldn't reach 170 as an adult walking at a moderate intensity level, that would be closer to extreme intensity.
Explanation:
help me help me help me
what are ribosomes?
I'm tired. But I have insomnia. Big ugh moment. <.<.
Answer:
Ribosomes are organelles the make protein for the cell.
Plz help this is so confusing
Answer:
5 Km/h
Explanation:
From the question given above, the following data were obtained:
Distance travelled = 10 Km
Time = 2 hours
Speed =?
Speed is simply defined as the distance travelled per unit time. Mathematically, it can be represented as:
Speed = distance travelled /time.
With the above formula, we can obtain the speed at which the duck is travelling as follow:
Distance travelled = 10 Km
Time = 2 hours
Speed =?
Speed = distance travelled /time.
Speed = 10 / 2
Speed = 5 Km/h
Thus, the duck is travelling at a speed of 5 Km/h
a student lift a 25kg mass at vertical distance of 1.6m in a time of 2.0 seconds. a. Find the force needed to lift the mass (in N ). b. Find the work done by the student (in J). c. Find the power exerted by the student (in W)
Answer:
a. F = 245 Newton.
b. Workdone = 392 Joules.
c. Power = 196 Watts
Explanation:
Given the following data;
Mass = 25kg
Distance = 1.6m
Time = 2secs
a. To find the force needed to lift the mass (in N );
Force = mass * acceleration
We know that acceleration due to gravity is equal to 9.8
F = 25*9.8
F = 245N
b. To find the work done by the student (in J);
Workdone = force * distance
Workdone = 245 * 1.6
Workdone = 392 Joules.
c. To find the power exerted by the student (in W);
Power = workdone/time
Power = 392/2
Power = 196 Watts.
find the volume of an object with a density of 3.2 g/mL and a mass of 12 g.
The driver of a car wishes to pass a truck that is traveling at a constant speed of 19.0 m/s. Initially, the car is also traveling at a speed 19.0 m/s and its front bumper is a distance 23.1 m behind the truck's rear bumper. The car begins accelerating at a constant acceleration 0.570 m/s^2, then pulls back into the truck's lane when the rear of the car is a distance 25.0 m ahead of the front of the truck. The car is of length 5.00 m and the truck is of length 21.3 m.
1. How much time is required for the car to pass the truck?
2. What distance does the car travel during this time?
3. What is the final speed of the car?
Answer:
16.16 s
381.5 m
28.21 m/s
Explanation:
Acceleration of the car = [tex]0.57\ \text{m/s}^2[/tex]
The distance the car moves is [tex]23.1+25+21.3+5=74.4\ \text{m}[/tex]
Initial position of the car is [tex]23.1+21.3+5=49.4\ \text{m}[/tex]
Position of the truck is given by the equation
[tex]x_r(t)=49.4+19t[/tex]
Position of the car is given by the equation
[tex]x_c(t)=19t+\dfrac{1}{2}\times0.57t^2[/tex]
Difference in their positions is 25 m
[tex]x_c(t)-x_r(t)=25\\\Rightarrow 19t+\dfrac{1}{2}\times0.57t^2-(49.4+19t)=25\\\Rightarrow 0.285t^2-74.4=0\\\Rightarrow t=\sqrt{\dfrac{74.4}{0.285}}\\\Rightarrow t=16.16\ \text{s}[/tex]
Time required for the car to pass the truck is 16.16 s
[tex]x_c(16.16)=19\times 16.16+\dfrac{1}{2}\times0.57\times 16.16^2=381.5\ \text{m}[/tex]
The distance the car traveled during this time is 381.5 m
[tex]v_c(16.16)=19+0.57\times 16.16=28.21\ \text{m}[/tex]
The final speed of the car is 28.21 m/s
Q1. A man wants to install a surveillance mirror in his shop, which mirror should he use?(1)
a) Convex mirror
b) Concave mirror
c) Plane mirror
d) Both (a) and (b)
answer is convex mirror
Explanation:
A
Because convex mirror will provide maximum view
Heather drives her Super-Beetle around a turn on a circular track which has a radius of 200 m. The Super-Beetle has a mass of 1500 kg and the coefficient of static friction between the road and tires is 0.6.
a. What is the force of static friction the road can apply batore the car starts to selon (use Ft= uFn).
b. What is the maximum speed the car can travel before it would start to slide?
Answer:
a) The force of static friction the road can apply before the car starts to move is 8826.3 newtons.
b) The maximum speed that a car can travel before it would start to slide is approximately 34.305 meters per second.
Explanation:
a) Let suppose that the car is on a horizontal ground and travels at constant speed. The vehicle experiments a centripetal acceleration due to friction, which can be seen in the Free Body Diagram (please see image attached for further details). By Newton's Laws, we construct the following equations of equilibrium:
[tex]\Sigma F_{x} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}[/tex] (1)
[tex]\Sigma F_{y} = N -m\cdot g = 0[/tex] (2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.
[tex]N[/tex] - Normal force from ground to the car, measured in newtons.
[tex]v[/tex] - Maximum speed of the car, measured in meters per second.
[tex]R[/tex] - Radius of the circular track, measured in meters.
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
By applying (2) in (1):
[tex]\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}[/tex] (3)
The force of static friction the road can apply in the car ([tex]f[/tex]), measured in newtons, is: ([tex]\mu_{s} = 0.6[/tex], [tex]m = 1500\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex])
[tex]f = \mu_{s}\cdot m \cdot g[/tex]
[tex]f = (0.6)\cdot (1500\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]f = 8826.3\,N[/tex]
The force of static friction the road can apply before the car starts to move is 8826.3 newtons.
b) Then, we calculate the maximum speed of the car by (3):
[tex]\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}[/tex]
[tex]\mu_{s}\cdot g = \frac{v^{2}}{R}[/tex]
[tex]v = \sqrt{\mu_{s}\cdot g\cdot R}[/tex]
If we know that [tex]\mu_{s} = 0.6[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]R = 200\,m[/tex], then the maximum speed of the car can travel before it would start to slide is:
[tex]v =\sqrt{(0.6)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (200\,m)}[/tex]
[tex]v \approx 34.305\,\frac{m}{s}[/tex]
The maximum speed that a car can travel before it would start to slide is approximately 34.305 meters per second.
3. As the mass of an object increases, the force of gravity
Answer:
As the mass of an object increases, the force of gravity increases as well.
Explanation:
Objects with more mass have more gravity. They work together.
explain resolution of Force
Answer:
it is defined as splitting up the given force into a number of components, without changing its effects on the body is called resolution of forces. A force is generally resolved along with two mutually perpendicular directions.
Explanation:
What relationship must exist between an applied force and the velocity of a moving object if uniform circular motion is to result?
Answer:
See explanation
Explanation:
Centripetal force is defined as the inward force required to keep an object moving with a constant speed in a circular path.
The magnitude of this force depends on the mass of the object, radius of the object and the velocity of the body.
So we can write;
F = mv^2/r
an airplane normally flying at 80 km/h north encounters a wind from the west of 10km/h at right angle to its foward motion a crosswind what will its resultant velocity be ?
Answer:
v = 80.62 km/h
Explanation:
Given that,
An airplane normally flying at 80 km/h north encounters a wind from the west of 10km/h at right angle to its forward motion a crosswind.
We need to find the resultant velocity of the airplane. It can be calculated as follows :
[tex]v=\sqrt{80^2+10^2} \\\\v=80.62\ km/h[/tex]
So, the required velocity is 80.62 km/h.
Kiara starts at 4, walks 6 blocks left and 2 blocks right. What is her displacement?
2) Given R = 3 ohms and R, = 1 ohm and V = 12 volts
I
a) Find the total resistance.
b) Find the current in the circuit:
c) Find the voltage drop in each resistor:
Answer:
a) because this is in series, we have:
the total resistance is 3 + 1 = 4 (ohm)
b) the curren in the circuit is 12/4 = 3 (A)
c) the voltage in R = 3 ohm is 3.3 = 9 (V)
the voltage in R = 1 ohm is 12 - 9 = 3 (V)
The force of gravity acting on an object is directed through this
center of gravity and toward the center of the
Answer:
Earth.
Explanation:
Center of gravity can be defined as the specific point where all of the weight of an object is concentrated.
Generally, all the objects found around the world all have a center of gravity.
When an object is balanced so that a displacement lowers its center of gravity, the object is said to be in stable equilibrium.
Hence, the force of gravity acting on an object is directed through this center of gravity and toward the center of the earth.
Weight can be defined as the force acting on a body or an object as a result of gravity.
Mathematically, weight is given by the formula;
[tex] Weight, W = mg [/tex]
Where;
m is the mass of an object.
g is acceleration due to gravity.
On a slope where does a marble have to most kinetic energy?
a) it is always the same
b) at the initial position
c) at the final position
d) somewhere between the initial and the final position
Answer:
C
Explanation:
Kinetic energy is the energy of motion. It has the most potential energy at the top but the most kinetic at the bottom after it's accelerated fully down the slope.
Can someone help me out please I got it wrong
Answer:
3 maybe since protons=atomic