It sounds like the ball's velocity along the [tex]y[/tex]-axis should be taken to be constant, in which case the ball has velocity vector
[tex]\mathbf v(t)=\left(2.10\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\hat{\imath}+\left(1.30\,\dfrac{\mathrm m}{\mathrm s}\right)\,\hat{\jmath}[/tex]
We note that at [tex]t=t_0[/tex], we're told the ball is at the origin, so its position at this point is [tex]\mathbf r(t_0)=\mathbf 0[/tex]. We then integrate the velocity vector to find the position vector:
[tex]\mathbf r(t)=\mathbf r(t_0)+\displaystyle\int_{\tau=t_0}^{\tau=t}\mathbf v(\tau)\,\mathrm d\tau[/tex]
[tex]\mathbf r(t)=\displaystyle\int_{\tau=t_0}^{\tau=t}\left(2.10\,\dfrac{\mathrm m}{\mathrm s^2}\right)\tau\,\hat{\imath}+\left(1.30\,\dfrac{\mathrm m}{\mathrm s}\right)\,\hat{\jmath}\,\mathrm d\tau[/tex]
[tex]\mathbf r(t)=\left(1.05\,\dfrac{\mathrm m}{\mathrm s^2}\right)\tau^2\,\hat{\imath}+\left(1.30\,\dfrac{\mathrm m}{\mathrm s}\right)\tau\,\hat{\jmath}\bigg|_{\tau=t_0}^{\tau=t}[/tex]
[tex]\mathbf r(t)=\left(1.05\,\dfrac{\mathrm m}{\mathrm s^2}\right)(t^2-{t_0}^2)\,\hat{\imath}+\left(1.30\,\dfrac{\mathrm m}{\mathrm s}\right)(t-t_0)\,\hat{\jmath}[/tex]
For the sake of simplicity, I'll assume [tex]t_0=0[/tex], so that
[tex]\mathbf r(t)=\left(1.05\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\,\hat{\imath}+\left(1.30\,\dfrac{\mathrm m}{\mathrm s}\right)t\,\hat{\jmath}[/tex]
(a) Then at [tex]t=2.50\,\mathrm s[/tex], the ball's position in the plane is given by the vector
[tex]\mathbf r(2.50\,\mathrm m)=(6.56\,\mathrm m)\,\hat{\imath}+(3.25\,\mathrm m)\,\hat{\jmath}[/tex]
which is to say its position in the plane is at the point (6.56, 3.25).
(b) The ball's displacement is the same as its position, but in vector form (as above).
(c) The ball has velocity
[tex]\mathbf v(2.50\,\mathrm s)=\left(5.25\,\dfrac{\mathrm m}{\mathrm s}\right)\,\hat{\imath}+\left(1.3\,\dfrac{\mathrm m}{\mathrm s}\right)\,\hat{\jmath}[/tex]
That is, the ball has a speed of [tex]\|\mathbf v(2.50\,\mathrm s)\|=5.41\,\dfrac{\mathrm m}{\mathrm s}[/tex] and is traveling in a direction of about [tex]13.9^\circ[/tex] relative to the positive [tex]x[/tex]-axis.
At t=2.5s, the ball's position will be at coordinates (6.5625, 3.25), have a displacement of approximately 7.23 metres, and have a velocity of 5.25 m/s in the x direction and 1.3 m/s in the y direction.
Explanation:The subject of this question is physics, specifically kinematics, and the topic is motion in two dimensions. We will calculate the x and y coordinates, the displacement, and the velocity for the given conditions. First we will calculate the x-coordinate using the equation of motion x = ut + 0.5at^2, where 'u' is initial velocity, 'a' is acceleration and 't' is time. Here, the initial velocity along x-axis is zero, acceleration is 2.10 m/s^2, and time is 2.5 s. Substituting these values, we get x = 0 + 0.5 * 2.1 * (2.5)^2 = 6.5625 m.
Next, for the y-coordinate, since there is no acceleration along y-axis, its position at any time is given by y = ut, where 'u' is initial velocity along y-axis, and 't' is time. So, y = 1.30 * 2.50 = 3.25 m.
Next, the displacement, which is the shortest distance from the initial to the final position, can be found using the Pythagorean theorem, d = sqrt(x^2 + y^2). Thus, d = sqrt((6.5625)^2 + (3.25)^2) = 7.23 m approximately.
Lastly, the velocity of the ball at any point is given by v = u + at, where 'a' is the acceleration along the x-axis, 't' is time and 'u' is the initial velocity (which is zero in this case along the x-axis). So, v = 0 + 2.1 * 2.5 = 5.25 m/s along the + x-direction. And, the velocity along the y-axis remains constant as 1.30 m/s because there's no acceleration in the y direction.
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A quantity of a gas has an absolute pressure of 400kPa and an absolute temperature of 110 degrees Kelvin. When the temperature of the gas is raised to 235 degrees Kelvin, what's the new pressure of the gas?
Intially, we have:
Pressure P1 = 400KPa
Temperature T1= 110Kelvin
When,
Temperature T2= 235 Kelvin
Pressure P2= ?
We have gas equation:
PV= nRT
P/T= nR/V
Considering nR/T as constant, we have:
P1/T1 = P2/T2
400/110= P2/235
P2= 854.5 KPa
So the new temperature will be 854.5 KPa
A block of wood 5cm x 4cm x 3cm. It has a mass of 13.4g. What is the density of the wood?
The volume of a cube is given by s^3. So the volume of this block is 3cm x 3cm x3 cm = 27 cm^3.
density = mass/volume =27 g / 27 cm^3 = 1 g/cm^3.
the answer is 1 g/cm^3. I hope this helps!
A mole runs 10 meters due East, then runs 9 meters due South, then runs 3 meters due North. The mole then burrows 2 meters down a hole. From its original position, what is the magnitude of the mole's displacement?
Let the directions are defined by vectors as following
East = + x direction
West = - x direction
North = + y direction
South = - y direction
upwards = + z direction
downwards = - z direction
So here we are given that it moves
1). 10 m East
2). 9 m South
3). 3 m North
4). 2 m down
So we can write it as
[tex]d_1 = 10 \hat i[/tex]
[tex]d_2 = 9 (-\hat j)[/tex]
[tex]d_3 = 3 \hat j[/tex]
[tex]d_4 = 2 (-\hat k)[/tex]
Now total displacement will be given as
[tex]d = d_1 + d_2 + d_3 +d_4[/tex]
[tex]d = 10 \hat i - 9 \hat j + 3 \hat j - 2 \hat k[/tex]
[tex]d = 10 \hat i - 6 \hat j - 2 \hat k[/tex]
now the magnitude of the displacement will be
[tex]d = \sqrt{10^2 + 6^2 + 2^2}[/tex]
[tex]d = 11.8 m[/tex]
so the magnitude of displacement will be 11.8 m
The magnitude of the mole's displacement from its original position is approximately 11.83 meters.
To find the magnitude of the mole's displacement, you can break down the motion into its components and use vector addition. Here's the step-by-step calculation:
1. The mole runs 10 meters due East, which we'll represent as +10 meters in the horizontal (X) direction.
2. Then, the mole runs 9 meters due South, which we'll represent as -9 meters in the vertical (Y) direction. This is because it's in the opposite direction of the positive Y-axis.
3. After that, the mole runs 3 meters due North, which we'll represent as +3 meters in the vertical (Y) direction.
4. Finally, the mole burrows 2 meters down a hole, which we'll represent as -2 meters in the depth (Z) direction. This is because it's in the opposite direction of the positive Z-axis.
Now, we can calculate the total displacement vector by adding the individual components:
Displacement in the X direction: 10 meters (East)
Displacement in the Y direction: (3 meters - 9 meters) = -6 meters (South)
Displacement in the Z direction: -2 meters (Down)
To find the magnitude of the displacement, we can use the Pythagorean theorem in three dimensions:
[tex]\[|D| = \sqrt{(Dx^2 + Dy^2 + Dz^2)}\][/tex]
[tex]\[|D| = \sqrt{(10^2 + (-6)^2 + (-2)^2)}\][/tex]
[tex]\[|D| = \sqrt{(100 + 36 + 4)}\][/tex]
[tex]\[|D| = \sqrt{140}\][/tex]
The magnitude of the mole's displacement from its original position is approximately 11.83 meters.
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If you drop a ball off a cliff oh, it starts out at 0m/s. After 1s, it will be traveling at about 10m/s. if air resistance is moved, what will happen in 2 s?
A.the ball will accelerate to 10m/s
B.the ball will still be moving at 10m/s
C.the ball will accelerate 20m/s
A tennis ball, a bowling ball, and a feather are dropped from the top of a tall building at the same time. Consider what you have learned about air resistance. Which object reaches the ground first? A.The tennis ball B.The bowling ball C.All three reach the ground at the same time. D.The feather
The answer is B because the bowling ball is the heaviest.
What happens if the value of VOq doubles
If Voq doubles, abN gets multiplied by 16 .
If Voq=3, abN=81 .
If the value of VOq doubles, the magnitude of the restoring force doubles.
The formula for the restoring force of an abductin pad is:
F = k * x
where:
F is the restoring force
k is the spring constant
x is the distance the spring is compressed or stretched
In this case, we are told that VOq doubles. VOq is the spring constant. So, if VOq doubles, k doubles. This means that the restoring force doubles.
The restoring force is the force that pushes the shell open when the muscles relax. If the restoring force doubles, the shell will be pushed open with twice the force. This means that the shell will open more quickly and easily.
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Maintenance __________ are needed in great numbers to service all sorts of technical equipment.
fill in the blank please
The blank should be filled with the word 'technicians,' reflecting the need for specialized workers who possess practical technical skills to maintain and repair equipment as technology evolves.
Maintenance technicians are needed in great numbers to service all sorts of technical equipment. Technicians are those individuals who have the practical and technical knowledge to build, operate, maintain, and repair various kinds of equipment and machinery. This demand for technicians is a reflection of the growing complexity and proliferation of technology in contemporary society. As new technologies emerge, they often require different skills than those needed for older technologies, creating a need for continuous training and updating of skills among workers.
Moreover, significant infrastructure projects, such as the expansion of electrical transmission systems, will necessitate an increase in specialized workers, including technicians, to maintain and repair these systems efficiently. Preventive maintenance schedules and activities further highlight the essential role that technicians play in ensuring the smooth operation of various systems and equipment in industries as diverse as manufacturing, transportation, and housing.
How much heat would be absorbed by 75.20 g of iron when heated from 22 C to 28 C
Mass x SH x °C (or K) ΔT
= 75g x 0.45J/g/K x 6.0 ΔT
= 202.5 Joules of heat absorbed.
(202.5J / 4.184J/cal = 48.4 calories).
I guess that is the answer
The amount of heat would be absorbed by 75.20 g of iron when from 22°C to 28°C is equal to 8834.5 J.
What is specific heat capacity?Specific heat capacity for any substance is the amount of heat energy required to raise the temperature by 1°C of a unit mass of that substance. The mathematically Specific heat capacity can be expressed as:
Q= m C ΔT
where Q is the amount of heat energy, ΔT is the change in temperature, and C is the specific heat capacity of the system.
Specific heat capacity is an intensive property as it is independent of the size or quantity of the matter.
Given, the Specific heat capacity of the iron, C = 0.44 J/gK
The mass of the iron, m = 75.20 g
The change in the temperature = ΔT = 28 - 22 = 6°C = 273-6 = 267 K
The heat absorbed by iron, Q = mCΔT = 75.20 × 0.44 × 267
Q = 8834.5 J
Therefore, the heat would be absorbed by 75.20 g of iron is 8834.5 Joules.
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A 70 kg boy is about to dive 5 meters into the water. How much kinetic energy will he have when he is 2 yards above the water?
initial height of the boy when he jump or dive is 5 meter
[tex]h_1 = 5 m[/tex]
now his final position is 2 yards above the surface
[tex]h_2 = 2 yards[/tex]
as we know that
[tex]1 yard = 0.9144 m[/tex]
[tex]2 yards = 1.83 m[/tex]
now by energy conservation we can say
change in potential energy = gain in kinetic energy
[tex]mg(h_1 - h_2) = \frac{1}{2} mv^2[/tex]
divide both sides by mass "m"
[tex]g*(5 - 1.83) = \frac{1}{2}*v^2[/tex]
[tex]v^2 = 2*9.8*(5 - 1.83)[/tex]
Now kinetic energy will be given as
[tex]KE = \frac{1}{2} mv^2[/tex]
[tex]KE = \frac{1}{2}*70 * 2*9.8*( 5 - 1.83)[/tex]
[tex]KE = 2175 J[/tex]
so his kinetic energy will be 2175 J
Suppose a vector is drawn at 3 cm long and represents an object traveling at 100 m/s. How fast would an object be traveling if the vector is drawn 6 cm long?
length of the vector is proportional to the magnitude of the vector
so length of vector will be in the proportional to the magnitude of the vector here
initial length of the vector is 3 cm which represent the velocity vector as 100 m/s
now it means 3 cm length is proportional to speed which is 100 m/s
now if we draw another vector of length 6 cm then here we can see that if length of vector is doubled then initial length
so here the magnitude of velocity will also be doubled
so here final speed must be double of initial speed
[tex]v_f = 2* v_i[/tex]
[tex]v_f = 2*100 = 200 m/s[/tex]
We timed how long it took for the ball to travel 1 meter several times, so we could calculate an “average” time to use in the velocity calculation. In the following set of times, (2.26s, 2.38s, 3.02s, 2.26s, 2.31s), the value 3.02s is considered an ___________ and is not used in the calculation.
We need to find the average speed of the ball during the motion of 1 m
In order to find that we took several reading and found following times to cover the distance of 1 m
t1 = 2.26 s
t2 = 2.38 s
t3 = 3.02 s
t4 = 2.26 s
t5 = 2.31 s
Now in order to find the average time we can write
[tex]T_{mean} = \frac{t_1 + t_2 + t_3 + t_4 + t_5}{5}[/tex]
[tex]T_{mean} = \frac{2.26 + 2.38 + 3.02 + 2.26 + 2.31}{5}[/tex]
[tex]T_{mean} = 2.45 s[/tex]
So average time to cover the distance of 1 m by ball will be 2.45 s
here 3.02 s is not the average time but we can say it is the median of the readings of all possible values which we can not use in our calculation as average time
A sled is pushed along an ice covered lake. It has some Initial velocity before coming to rest in 15m. It took 23 seconds before the sled and rider come to rest. If the rider and sled have a combined mass of 52.5kg, what is the magnitude and direction of the stopping force?
If we assume the acceleration that the sled undergoes is constant throughout its motion, then we have the average velocity [tex]\bar v[/tex] of the sled to be
[tex]\bar v=\dfrac{v_i+v_f}2=\dfrac{\Delta x}{23\,\mathrm s}[/tex]
where [tex]\Delta x[/tex] is the total displacement of the sled, and [tex]v_i[/tex] and [tex]v_f[/tex] are the sled's initial and final velocities, respectively. The sled eventually stops, so we take [tex]v_f=0[/tex] and solve for [tex]v_i[/tex]:
[tex]\dfrac{v_i}2=\dfrac{15\,\mathrm m}{23\,\mathrm s}\implies v_i=1.3\,\dfrac{\mathrm m}{\mathrm s}[/tex]
Now, take the sled's starting position to be the origin. The sled moves in one direction, which we take to be the positive direction. Then because it's slowing down, we expect its acceleration to be in the negative direction (and hence with negative sign). In particular, the sled's position [tex]x[/tex] at time [tex]t[/tex] is
[tex]x=x_i+v_it+\dfrac12at^2[/tex]
We have [tex]\Delta x=x-x_i=15\,\mathrm m[/tex], [tex]v_i=1.3\,\frac{\mathrm m}{\mathrm s}[/tex], and [tex]t=23\,\mathrm s[/tex], so we can solve for acceleration [tex]a[/tex]:
[tex]15\,\mathrm m=\left(1.3\,\dfrac{\mathrm m}{\mathrm s}\right)(23\,\mathrm s)+\dfrac12a(23\,\mathrm s)^2[/tex]
[tex]\implies a=-0.056\,\dfrac{\mathrm m}{\mathrm s^2}[/tex]
With a mass of [tex]m=52.5\,\mathrm{kg}[/tex], we find that the stopping force is
[tex]F=ma=-2.9\,\mathrm N[/tex]
which means the stopping force has magnitude [tex]2.4\,\mathrm N[/tex] in the negative direction (opposite the direction of the sled's initial velocity).
If a weight lifter holds a 200 kg barbell in place over his head, he _______
A:does no work because the force he applies does not cause the barbell to move.
B: does work because the barbell is heavy to hold.
We have that from the Question, it can be said that
does no work because the force he applies does not cause the barbell to move. Option AFrom the Question we are told
If a weight lifter holds a 200 kg barbell in place over his head, he _______
A:does no work because the force he applies does not cause the barbell to move.
B: does work because the barbell is heavy to hold.
Generally the equation for work is mathematically given as
w=f*d
Therefore
For an instance where distance traveled is zero and force exerted is a 1000N the work done will still be zero because
1000*0 =0
Therefore
does no work because the force he applies does not cause the barbell to move.
Option A
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Can anyone fill in the blanks for the potential and kentic energy? Also, is this showing energy transformation? Thank you so much!
Potential Kinetic Kinetic Potential Potential
What is the eccentricity of a planet's orbit if the focal distance is 8 and the major axis is 32?
Focal distance of an ellipse is given by the formula
[tex]f = ae[/tex]
here a = length of semi major axis
e = eccentricity of the path
now here we know that
length of major axis for the path of planet is given as 32 units
so here we can say
[tex]2a = 32 units[/tex]
[tex]a = 16 units[/tex]
so length of semi major axis is 16 units
focal distance for the planet path is given as 8 units
now from the above formula we can write
[tex]f = a*e[/tex]
[tex]8 = 16*e[/tex]
[tex]e = \frac{8}{16}[/tex]
[tex]e = \frac{1}{2} = 0.5[/tex]
so eccentricity for the path of planet will be 0.5
How long will it take an airplane to go 800 miles if it is traveling at an average speed of 650 miles per hour?
1 h 13 min and 50 s it will take.
Why do you have to stand during an lab experiment?
answer: you stand in a lab experiment because if you sit during the lab, you have much reach of the materials on the table. and also, you might have a risk on some chemical spill on your clothes. the chemical might be flammable and it might set your clothes on fire.
so that's why you have to stand during lab experiments.
hope this helps! ❤ from peachimin
Standing during a lab experiment is important for safety and accuracy. It allows better control over equipment and materials and minimizes the risk of hazards.
Explanation:When conducting a lab experiment, it is important to stand to ensure safety and accuracy of the experiment. Standing allows you to have better control and stability over the equipment and materials you are working with. Additionally, it helps minimize the risk of accidental spills, breakage, or other hazards.
For example, if you are working with chemicals or glassware, standing can prevent the equipment from tipping over and causing injury. It also allows you to observe the reaction or process more closely, making it easier to record accurate data and observations.
Overall, standing during a lab experiment promotes safety, precision, and optimal results.
a flower pot is theown out of a window with a horizontal velocity of 8 m/s. If the window is 1.5 m off the ground, how far from the window does it land?
T= The time it takes for the flower pot to pass the top of my window.
V= The velocity of the flower pot at the moment it is passing the top of my window.
X= The height above the top of my window that the flower pot was dropped.
h = Lw + X
Lw = (1/2)*g*t^2 + V*t
V*t = Lw - (1/2)*g*t^2
V= Lw/t - (1/2)*g*t , On the other hand we know : V=gT.
Therefore we will have: Tg= Lw/t - (1/2)*g*t
T= Lw/(tg) - t/2
Now substitute for T in the following equation: X = (1/2)*g*T^2
X= (1/2)*g*(Lw/(tg) - t/2)^2
Now substitute for X in the very first equation I mentioned: h = Lw + X
h = Lw + (1/2)*g*(Lw/(tg) - t/2)^2
In case you wanted the answer to be simplified, then:
h= (Lw^2)/(2*g*t^2) + (g*t^2)/8 + Lw/2
Answer:
3.26m
Explanation:
Using one of the equation of motion to get the distance of the pot from the window and the ground;
v² = u²+2as where
v is the final velocity = 8m/s
u is the initial velocity = 0m/s
a =+g = acceleration due to gravity (this acceleration is positive since the body is falling downwards)
g = 9.81m/s
s is the distance between the object and the window from which it dropped.
Substituting this values to get the distance s we have;
8² = 0²+2(9.81)s
64 = 19.62s
s = 64/19.62
S = 3.26m
Reaction time is the lapse of time between muscular movement and external stimuli. True or false
Answer:
true
Explanation:
If there is a loss of 4.36 × 10-5 g of mass in a nuclear reaction, how much energy would be released? Recall that c = 3 × 108 m/s.
A.
1.45 × 10-13 kJ
B.
3.92 × 106 kJ
C.
3.92 × 109 kJ
D.
3.92 × 1012 kJ
As E= mc²
E = (4.36 × 10^ -5) ×( 3 × 108 m/s.)²
E= 3.924×10^12J
E= (3.924×10^12)KJ/ 1000
E =3.92 × 109 kJ
Answer:
Released energy, [tex]E=3.92\times 10^6\ kJ[/tex]
Explanation:
It is given that,
Loss in mass in a nuclear reaction, [tex]m=4.36\times 10^{-5}\ g=4.36\times 10^{-8}\ g[/tex]
The relation between the mass and energy is given by Einstein mass energy equivalence equation :
[tex]E=mc^2[/tex]
c is the speed of light
So, [tex]E=4.36\times 10^{-8}\times (3\times 10^8)^2[/tex]
[tex]E=3.92\times 10^9\ J[/tex]
[tex]E=3.92\times 10^6\ kJ[/tex]
The energy released in a nuclear reaction is [tex]3.92\times 10^6\ kJ[/tex]. Hence, the correct option is (B)
The sensor in the torso of a crash test dummy records the magnitude and direction of the net force acting on the dummy.If the dummy is thrown forward with a force of 130.0N while simultandously being hit from the side wigh a force of 4500.0N,what force will the sensor report
Explanation:
Given that,
The dummy is thrown forward with a force of 130 N, [tex]F_1=130\ N[/tex]
Side force acting on the dummy, [tex]F_2=4500\ N[/tex]
We need to find the force acting on the sensor report. It can be calculated using Pythagoras theorem as :
[tex]F_{net}=\sqrt{F_1^2+F_2^2}[/tex]
[tex]F_{net}=\sqrt{130^2+4500^2}[/tex]
[tex]F_{net}=4501.87\ N[/tex]
So, the net force acting on the sensor report is 4501.87 N. Hence, this is the required solution.
The net force acting on the dummy is 4502 N.
Force is a vector, the resultant force (net force) is that single force that has the same effect in magnitude and direction as two or more forces acting together. The resultant of a vector must take into cognizance, the geometry of the problem.
The dummy is thrown forward with a force of 130.0N while simultaneously being hit from the side with a force of 4500.0N. The net force must now be obtained by Pythagoras theorem.
Fnet^2 = F1^2 + F2^2
F1 = 130.0N
F2 = 4500.0N
Fnet = √(130.0N)^2 + (4500.0N)^2
Fnet = 4502 N
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A solid is 5 cm tall, 3 cm wide and 2 cm thick. It has a mass of 129 g. What is the density
Rain falling vertically will make vertical streaks on a car's side window. However,if the car is moving, the streaks are slanted. If the streaks feom na vertically falling rain make 45 degree angle streaks, how fast is the car moving compared with the apeed of the falling rain?
When car is at rest the steaks makes makes vertical lines
which means the rain is falling in vertically downward direction
Now when car is moving with some speed v
Now the steaks makes and an angle 45 degree
So here we can say that relative velocity of rain with car is 45 degree
Now this is the resultant speed of rain in car frame
[tex]V_{rc} = V_r - V_c[/tex]
now if relative velocity makes 45 degree angle so this vector must have same components in vertical and horizontal direction
Since we know that relative velocity is resultant of rain velocity and car velocity so we can say here its two components are rain velocity and car velocity
So these two components must be of same magnitude
as it makes 45 degree
because when two vector are of same magnitude then the resultant vector always makes 45 degree with them if these two vectors are perpendicular to each other
car is moving at same speed as the speed of rain
Final answer:
When the rain makes a 45-degree angle streak on a moving car's window, the speed of the car is equal to the speed of the falling rain.
Explanation:
The student's question involves understanding relative velocities and their relationship to observed angles. Specifically, if rain makes 45-degree streaks on a moving vehicle's window, we want to know how fast the car is moving compared with the speed of the falling rain.
To solve this problem, we make use of trigonometry, particularly the tangent function which relates opposite and adjacent sides in a right-angled triangle. If the streaks are at a 45-degree angle, the vertical and horizontal speeds (i.e., speed of the rain and speed of the car) are equal. Therefore, in this scenario, the car is moving at the same speed as the rain is falling. Using the formula tan(θ) = vhorizontal / vvertical, where θ is the angle of the rain relative to the vertical, we find that at 45 degrees, tan(45) = 1 which implies the car's speed (vhorizontal) is equal to the rain's speed (vvertical).
James gently releases a ball at the top of a slope, but does not push the ball. The ball rolls down the slope. Which force caused the ball to move downhill?
A.
applied force
B.
drag force
C.
friction force
D.
gravitational force
E.
normal force
D. Gravitational force.
Answer:
Gravitational force
Explanation:
When ball is placed on the slope and slightly disturbed then ball will roll down the slope with increasing speed.
This increasing speed or acceleration of ball is due to the component of weight of ball which is along the inclined plane.
As shown in the FBD the component of weight perpendicular to inclined plane is used to counterbalance the normal force while other component of weight parallel to the inclined plane is accelerating the ball down the plane
So here we can write
[tex]mgsin\theta = ma[/tex]
so here
[tex]a = gsin\theta[/tex]
so ball is accelerating due to gravitational force
Average velocity is different than average speed because calculating average velocity involves
Average velocity differs from average speed in that it considers displacement over elapsed time and has direction, while average speed is total distance over time without direction.
Explanation:Average velocity is different than average speed because calculating average velocity involves taking into consideration not just the total distance traveled but the displacement (the change in position) of an object, and it is a vector quantity which means it has both magnitude and direction. On the other hand, average speed is simply the total distance traveled divided by the elapsed time and is a scalar quantity, which means it does not have a direction associated with it.
For example, if you were to drive to a store, then turn around and drive back home, making the starting and ending points the same, your displacement would be zero, therefore your average velocity would be zero. However, your average speed would be the total distance for the trip divided by the time taken to travel that distance. This illustrates that average speed can be greater than the magnitude of average velocity due to changes in direction.
To calculate average speed, one would use the formula: total distance traveled ÷ elapsed time. Meanwhile, average velocity is calculated using the formula: displacement ÷ travel time. Therefore, if you are only interested in how fast you are going regardless of the direction, you would look at average speed. But if you need to know your speed in a particular direction, you would consider average velocity.
How much farther will a car skid if it locks its brakes at 60 mph as compared to a skid from 15 mph? Assume the braking force that will eventually stop the cars is provided by the road and that it acts with equal magnitude in both situations. Distance increases by a factor of 4 times 16 times impossible to tell because the mass of the car is not known impossible to tell because frictional force is not known
v₀ = initial velocity of the car before brakes are applied
v = final velocity of the car after it comes to a stop = 0 mph
d = stopping distance
a = acceleration caused due to braking force
Using the kinematics equation
v² = v²₀ - 2 a d
0² = v²₀ - 2 a d
d = v²₀ /(2a)
Since the acceleration is same , the stopping distance is directly proportional to the square of the initial speed of car before brakes are applied
hence
d₁/d₂ = v²₀₁/v²₀₂
Given that : v₀₁ = 60 mph and v₀₂ = 15 mph
inserting the values
d₁/d₂ = (60)²/(15)²
d₁ = 16 d₂
hence distance increases by a factor of 16 times.
A car skidding from 60 mph will skid 16 times farther than a car from 15 mph because the kinetic energy is proportional to the square of the velocity, and the work done by friction to stop the car is equal to its initial kinetic energy.
The question asks how much farther a car will skid if it locks its brakes at 60 mph compared to a skid from 15 mph. When a car locks its brakes, the distance it skids before coming to a stop is directly related to the kinetic energy it had when the brakes were applied. Kinetic energy is given by the formula [tex]KE = \frac{1}{2}mv^2[/tex]. Because the kinetic energy depends on the square of the velocity, a car moving at 60 mph (which is four times faster than 15 mph) will have 16 times the kinetic energy ([tex]4^2[/tex]).
Assuming that the frictional force acting on the car is the same in both situations, we know that the work done by the frictional force to stop the car is equal to the car's initial kinetic energy. Hence, the car skidding from 60 mph will skid 16 times farther than the car skidding from 15 mph, because work is directly proportional to distance when force is constant (Work = Force × Distance). Therefore, the distance increases by a factor of 16 times.
what is the velocity of a dropped object after it had fallen for 3.0s?
A 4-kg ball is 2 cm away from one 1-kg ball and 6 cm away from another 1-kg ball. Use the relationships among the balls to describe two factors that affect gravity. Also explain why the balls do not move toward each other unless acted upon by another force.
Force of gravitation between two balls is given by the formula
[tex]F = \frac{Gm_1 m_2}{r^2}[/tex]
here we know that
[tex]m_1[/tex] = mass of ball 1
[tex]m_2[/tex] = mass of ball 2
[tex]r[/tex] = distance between two balls
So here the two factors that will affect the force of gravitation is
1. Distance between two balls
2. mass of two balls
Here balls do not move due to gravitation attraction force because here the force of gravitation is very small as it is compared with other forces like frictional force between balls and ground.
So this weak gravitational force is balanced by frictional force on balls
SO all balls remains at rest
Gravitational force is directly affected by the masses of the objects involved and the distance between them. Despite this, objects do not simply move towards each other due to gravity because other forces such as friction and air resistance often counteract this attraction.
Explanation:The question essentially asks about two factors that affect gravity and why the balls do not move towards each other unless acted upon by an external force. Here's what we know from Newton's Law of Universal Gravitation: the gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. In simple terms, this means that the larger the masses of the objects, and the closer they are to each other, the stronger the gravitational pull between them.
Now, onto the second part of your question. In a perfect vacuum where no other forces exist, the balls would indeed move towards each other. However, in the real world, there are other forces at play, such as friction and air resistance, that counteract the gravitational pull. Therefore, unless these balls are acted upon by a stronger force (like a push or a pull), they will not move towards each other.
Learn more about Gravitation here:https://brainly.com/question/3009841
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Which of the following statements is true?
The more massive an object, the faster its acceleration of gravity.
Heavier objects will free-fall more quickly than lighter objects.
Objects in free-fall do not experience air resistance.
Mass and weight are the same.
Answer:
Objects in free-fall do not experience air resistance.
Explanation:
This comes from the definition of free-fall: an object in free fall is an object which is falling and the only force acting on it is gravity. Therefore, air resistance is not present for an object in free-fall. As a result, the acceleration of any object in free-fall is always equal to g (gravitational acceleration), and so all the objects fall with the same velocity and same time, regardless of their mass.
If the object is massive then the acceleration of gravity is faster, such objects will free fall more quickly than lighter objects. The objects in free fall do not experience air resistance are all true statements because the force of gravity which depends upon weight.
It is directly proportional to the weight. Yet the mass and weight are not same because mass is only value in kilograms but the weight is the value in which mass and gravitational acceleration are included. However, its unit is kilograms - 2 which is equal to Newton law of formula which is also unit of force.
The following quantities describe location or its change:position, distance, and a displacement. Which are vectors?
As per the question there are three physical quantities named as position,distance and displacement.
Before coming into a conclusion first we have to understand a vector and a scalar.
A scalar quantity is a quantity which has only magnitude for it's complete specifications.
A vector is a quantity which has magnitude as well as direction and at the same time it is in accordance with the paraellogram law of vector addition.
Out of the three options displacement and position are vector quantities.It is because it is the minimum distance between two points .It has magnitude as well as direction.
Distance can not be considered as a vector quantity as it has only magnitude.There is no specific directions of distance travelled.
Position vector is a vector which provides location of an object in a plane or space.It is nothing else except the point which has x,y,z coordinates with origin is taken as the reference point.
Hence position and displacement are vectors
Answer:
position and displacement
Explanation: