Tessa uses a toy slingshot to launch a tennis ball across the park for her dog to fetch. For her first launch, she
uses 100 N of force. Her second launch uses 200 N of force, and her third launch uses 300 N. Which launch had
the greatest acceleration of the tennis ball?
Answer:
See the explanation below.
Explanation:
To solve this problem we must apply Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration and this force can be calculated by means of the following equation.
F = m*a
where:
F = force [N] (units of Newtons)
m = mass [kg]
a = acceleration [m/s²]
The mass of the tennis ball will always be the same therefore it will never change.
Now clearing a:
[tex]a=\frac{F}{m}[/tex]
If the mass of the ball remains the same:
[tex]a = \frac{100}{m} ; a = \frac{200}{m};a =\frac{300}{m}[/tex]
We see that for a force of 300 [N], the acceleration exerted on the ball must be greater. Therefore with the force of 300 [N] the greatest acceleration is achieved.
a sound has a frequency of 230 hz. what is the velocity of this sound if it has a wavelength of 46 cm?
Answer:
V=f×λ
Where,
V is the velocity of the wave measure using m/s.
f is the frequency of the wave measured using Hz.
λ is the wavelength of the wave measured using m.
transform 46 cm in m = 0.46m and there u have it
Explanation:
A ship has a constant velocity of 8.33 m/sec. How far does it travel in done day?
Distanced travelled by a ship with a constant velocity in a day is 719712m
VELOCITY is defined as rate of change of displacement in a given interval of time.
it is a vector quantity.
its unit is m/s
to calculate the distance of a ship travelled
distance = speed x time
d = s x t ----1.
velocity of ship = 8.33m/s
time taken = 1 day =86400 sec
now using the above values in equation 1 we get
d =8.33m/s x 86400 sec
d = 719712m
thus a ship travels 719712m in a day.
Distanced travelled by a ship with a constant velocity in a day is 719712m
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What is the force of gravity acting on a 2000 kg spacecraft when it is about the Earth at a in orbit distance of 8.375 x 106 m?
Answer:
W = 13.44 KN
Explanation:
First, we need to find the value of acceleration due to gravity at the specified height:
[tex]g' = g(1-2\frac{h}{R})[/tex]
where,
g' = value of acceleration due to gravity at given height = ?
g = value of acceleration due to gravity at surface of earth = 9.81 m/s²
h = height of space craft = 8.375 x 10⁶ m - 6.37 x 10⁶ m = 2.005 x 10⁶ m
R = Radius of Earth = 6.37 x 10⁶ m
Therefore,
[tex]g' = (9.81\ m/s^2)(1-2\frac{2.005\ x\ 10^6\ m}{6.37\ x\ 10^6\ m})\\\\g' = (9.81\ m/s^2)(1 - 0.315)\\g' = 6.72\ m/s^2[/tex]
Now, we can find the weight or force of gravity on space craft:
[tex]W = mg'[/tex]
where,
W = Force of gravity on space craft = ?
m = mass of space craft = 2000 kg
Therefore,
[tex]W = (2000 kg)(6.72\ m/s^2)[/tex]
W = 13.44 KN
what is gravity ?short ans
Answer:
Gravity, or gravitation, is a natural phenomenon by which all things with mass or energy—including planets, stars, galaxies, and even light—are brought toward one another. On Earth, gravity gives weight to physical objects, and the Moon's gravity causes the ocean tides.
At what seed is Jay traveling at 5 minutes into walk? At what time in his walk does he have the greatest speed?
The velocity of Jay after five minutes is 1 Km/min. The greatest speed occurred at 45 mins.
What is the velocity time graph?The velocity time graph is a representation of the movement of a body a body from one point to another. It is the plot of the velocity on the vertical axis against the time on the horizontal axis. Ideally this graph would be divided into three portions;
The region of uniform accelerationThe region of constant velocity The region of uniform decelerationWe can be able to read off from the graph, the instantaneous velocity of the object at any time. This could be done by tracing the time against the velocity axis and extrapolating according from the graph as shown.
Recall that the acceleration of the body is obtained as the slope of the velocity time graph at any point and the acceleration could be uniform or non uniform as judged from the velocity time graph.
Having said all these, it is clear that the velocity of Jay after five minutes is 1 Km/min.
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courtney dauwalter leaves her home at 7 a.m. and takes his usual path to the top of mount elbert, arriving at 7 p.m. the following morning, she starts at 7 a.m. at the top of mount elbert and takes the same path back, arriving at her home at 7 p.m. determine if there exists a point on the path the courtney will cross at exactly the same time of day on both days. justify your answer.
There exists a point on the path the Courtney will cross at exactly the same time of day on both days.
Given that Courtney first takes his usual path and travel between 7.a.m. and 7.p.m.
Next morning, she travel backs between the time 7 a.m. and 7 p.m.
The time taken by Courtney on each day = 12 hours
12/2 = 6 hours.
6 hours from 7 a.m. is 1 pm. So this 6 hours will divide the total journey into two equal parts.
At 1 p.m. on first day, she will be in the middle of her way from home to Mount Elbert.
At 1.p.m. of the second day also, she will be in the middle of her way from Mount Elbert to her home.
So she cross the exact middle point of the path at exactly 1.pm on both days.
Intermediate Value theorem states this result. Since the distance travelled is a continuous function, there is a particular value of time corresponding to the exact same point crossed by Courtney on both days.
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A proton having an initial velvocity of 20.0i Mm/s enters a uniform magnetic field of magnitude 0.300 T with a direction perpendicular to the proton's velocity. It leaves the field-filled region with velocity -20.0j Mm/s. Determine(a) the direction of the magnetic field.
The the direction of magnetic field is perpendicular to the velocity of proton and is directed towards the + z axis.
We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.
We have to determine the direction of magnetic field.
What is the magnitude of force on the charged particle moving in a uniform magnetic field?The magnitude of force on the charged particle moving in a uniform magnetic field is given by -
F = qvB sinθ
According to the question, we have -
Entering Velocity (v) = 20 i m/s.
Magnetic field intensity (B) = 0.3 T.
Leaving velocity (u) = - 20 j m/s.
Now -
F = q |v| |B| sinθ
F = 1.6 x [tex]10^{-19}[/tex] x 20 x 0.3 x sin(90)
F = 9.6 x [tex]10^{-19}[/tex] Newton.
Now, using the Fleming's left hand rule, the the direction of magnetic field is perpendicular to the velocity of proton and is directed towards the + z axis. The force is towards the '- y' axis. Therefore -
B = + 0.3 k
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can light cause plastic to melt? If yes does it matter what kind of light?
Answer:
plastics are susceptible to degradation due to ultraviolet (UV) light from the sun. With enough exposure, UV light can cause a chemical reaction in the plastic, which results in scission, or severing, of those big polymer molecules.
Explanation:
Answer:
the short answer is yes and sun or strong uv lights
Explanation:
LEDs cannot melt plastic fixtures as they just do not get that hot, not even at the base These new bulbs are based on field-induced polymer electroluminescent (FIPEL) technology, with a twist. Without heat as a continual stressor, the polymer should remain stable for years. Any kind of light bulbs, from fluorescent to incandescent to halogen, can cause fires if they are not used correctly. As more heat is put into the plastic, more bonds are broken and the material becomes more like a liquid and less like a solid. Semi-crystalline polymers have a melting temperature which is the point at which the secondary bonds in the crystalline regions break. At this point the mateiral is completely liquid. There is melt and there is soften. Certainly they soften and become weak in sunlight both by the heat and by the UV degredation of the plastic. Unfortunately, the plastics used for garbage bags have a low melting point but by design, they should not melt in strong sunlight. the underlined sentences are the most important
Look at the picture
Answer:
Your answer is d
Explanation:
Consider these pictures in answering the question below.
с
D
Which diagram might represent the initial vertical component of the velocity of a ball thrown upward at an angle?
a
Ob
CU O
c
Od
Оe
E
Answer:
the answer would be D because not only is it going up but it's going up as an angle
hello help me pls! i need serious help
Answer:c
Explanation:
S A nonuniform electric field is given by the expression→E = ay i^ + bz j^ + cxk^where a, b , and c are constants. Determine the electric flux through a rectangular surface in the x y plane, extending from x=0 to x=w and from y=0 to y=h .
The electric flux is 2chω² where the k^ term was eliminated since k. k^ =0
The electric force per unit charge is referred to as the electric field. It is assumed that the field's direction corresponds to the force it would apply to a positive test charge. From a positive point charge, the electric field radiates outward, and from a negative point charge, it radiates in.
We are given an electric field in the general form
E→ =ay i^+bz j^+cx k^
In the xy plane z=0 so that the electric field reduces to
E→ =ay i^+cx k^
To obtain the flux, we integrate
ϕ E=∫ E→⋅d A→ =∫ (ay i^ +cx k^) ⋅k^
dAϕE=ch∫
x=0w
xdx=2chω²
where the k^ term was eliminated since k. k^ =0
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a ping-pong ball is held submerged in a bucket of water by a string attached to the bucket's bottom. part a salt is now added to the water in the bucket, increasing the density of the liquid. what happens to the tension in the string ?
The tension force in the string will increase because the difference between the densities became increased.
What is the free body diagram?Free-body diagrams are utilized to display the relative direction and strength of all forces that are being applied to an item in a certain scenario. A unique illustration of the geometric diagrams that were covered in a previous lesson is the free-body diagram. We will make use of these graphics throughout the entire study of physics.
A thread fastened to the bucket's base is holding a ping-pong ball immersed in water. The water in the bucket is now supplemented with salt, making the liquid denser.
Before adding the salt let T be the tension in the string.
Due to the growing disparity in densities, the force applied to the rope will increase.
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A missile is moving 1350 m/s at a 25.0° angle. It needs to hit a target 23,500 m away in a 55.0° direction in 10.20 s. What is the direction of its final velocity? Answer in direction (deg)
Answer:
(3504 m/s, 66°)
The final velocity of the missile is approximately 3504 m/s at an angle of 66 degrees.
Explanation:
We are going to use the horizontal and vertical components of this object since it is in projectile motion.
We are given the initial velocity of the missile; 1350 m/s at an angle of 25 degrees. We are also given the displacement in the x-direction: 23,500 meters at an angle of 55 degrees. The total time of this projectile in motion is 10.20 seconds.
Using this information, we can break up the motion of the projectile into its horizontal and vertical components and solve for the unknown variables [tex]a[/tex] and [tex]v_f[/tex].
Horizontal direction (x):List out the known variables:
[tex]v_i=1350[/tex] [tex]v_f=?[/tex] [tex]\triangle x=23500[/tex] [tex]a=?[/tex] [tex]t=10.20[/tex]Since we have initial velocity, displacement, and time, we can use one of the kinematic equations for constant acceleration that contains these variables, including acceleration, and solve for a:
[tex]\triangle x=v_it + \frac{1}{2}at^2[/tex]We are solving for acceleration in the x-direction, so this equation should be in terms of the x-direction:
[tex]\triangle x_x=(v_i)_x t +\frac{1}{2}a_x t^2[/tex]Let's break up the displacement into its horizontal (x) component:
[tex]23500cos(55)[/tex]Let's break up the initial velocity into its horizontal component:
[tex]1350cos(25)[/tex]Plug these values into the equation and 10.20 seconds for t.
[tex]23500cos(55)=1350cos(25) \cdot t + \frac{1}{2}a_x (10.20)^2[/tex]Solve for [tex]a_x[/tex].
[tex]23500cos(55)=12479.85823+52.02a_x[/tex] [tex]999.1880243=52.02a_x[/tex] [tex]a_x=19.20776671[/tex]The acceleration in the x-direction is about 19.21 m/s².
Now, we can use this acceleration and solve for the final velocity in the x-direction using this constant acceleration kinematic equation:
[tex]v_f=v_i+at[/tex]Use this equation in terms of the x-direction:
[tex](v_f)_x=(v_i)_x+a_xt[/tex]Plug the known values into the equation and solve for [tex](v_f)_x[/tex].
[tex](v_f)_x=1350cos(55)+19.20776671 \cdot 10.20[/tex] [tex](v_f)_x=1419.434733[/tex]The final velocity in the x-direction is about 1419.43 m/s.
Vertical direction (y):This process will be the same as solving for acceleration and final velocity in the x-direction, except this time we will be using the vertical (y) components.
[tex]\triangle x_y=(v_i)_y t +\frac{1}{2}a_y t^2[/tex]Vertical component of initial velocity:
[tex]1350sin(25)[/tex]Vertical component of displacement:
[tex]23500sin(55)[/tex]Plug known values into the equation and solve for [tex]a_y[/tex].
[tex]23500sin(55)=1350sin(25) \cdot (10.20) + \frac{1}{2}a_y (10.20)^2[/tex] [tex]23500sin(55)=5819.453464 +52.02a_y[/tex] [tex]13430.61958=52.02a_y[/tex] [tex]a_y=258.181845[/tex]The acceleration in the y-direction is about 258.18 m/s².
Now let's use the same equation we used previously to solve for the final velocity in the y-direction.
[tex]v_f=v_i+at[/tex]Use this equation in terms of the y-direction:
[tex](v_f)_y=(v_i)_y+a_y t[/tex]Plug known values into the equation and solve for [tex](v_f)_y[/tex].
[tex](v_f)_y=1350sin(25)+258.181845 \cdot 10.20[/tex] [tex](v_f)_y=3203.989472[/tex]The final velocity in the y-direction is about 3203.99 m/s.
Final velocity and direction:The magnitude of the final velocity is the square root of the final velocity in the horizontal and vertical directions squared and added together.
[tex]|v|=\sqrt{(v_f)^2_x + (v_f)^2_y}[/tex]Plug the final velocity in the x and y-directions into the equation.
[tex]|v|=\sqrt{(1419.434733)^2+(3203.989472)^2}[/tex] [tex]|v|=\sqrt{2014794.961+10265548.54}[/tex] [tex]|v|=\sqrt{12280343.5}[/tex] [tex]|v|=3504.332105[/tex]The magnitude of the final velocity is about 3504 m/s. We can solve for the direction of the final velocity by using arccos to find the angle that is formed with the x-axis.
[tex]\displaystyle\theta = cos^-^1 \big{(} \frac{v_x}{|v|}\big{)} }[/tex]Plug in the x-component of the initial velocity and the magnitude of the final velocity of the projectile into the equation.
[tex]\displaystyle \theta =cos^-^1 \big{(}\frac{1419.434733}{3504.332105}\big{)}}[/tex] [tex]\displaystyle \theta = cos^-^1 \big{(} .4050514308 \big{)}}[/tex] [tex]\displaystyle \theta = 66.1056499[/tex]The direction of the final velocity is about 66 degrees.
Question 10 While looking in the yard, Jill noticed that the areas directly under the trees had less grass than the other areas in the yard. Which of the following is the BEST explanation for this observation? A The trees are blocking the sunlight. B The grass is receiving too much fertilizer. 2 C The areas under the trees are too cool for grass to grow. D The areas under the trees are receiving less water.
Answer:
The trees are blocking the sunlight.
Explanation:
B is obviously not the answer because fertilizer wouldn't have that effect. For C, shade doesn't change the temperature that much that it would inhibit plant growth. For D, even if the tree blocks the water, the surrounding ground will still be wet and therefore the grass should still have a sufficient amount of water to grow. Sunlight can't be gotten from somewhere else, unlike water, so this would inhibit plant growth.
A 0.5-kg ball accelerated at 50 m/s 2 . What force was applied?
Answer:
25N
Explanation:
soln:
given,
mass(m)=0.5kg
acceleration(a)=50m/s^2
force(F)=?
we know,
F=m*a
or,F=0.5*50
•°•F=25N
help me please!!!!!!!!!!!!!!!
Answer:
10
Explanation:
would
Heat flows from __________ energy to __________ energy.
A. higher, lower
B. lower, higher
Answer:
A. higher, lower
Explanation:
Hope this helps! :3
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7. A student is rotating an object on a rope that is 4.5m long. If we increase the length
of the rope so that the student rotates in uniform circular motion with the mass
rotating 9.0 m away from the center of the rotation (2x farther), what will be the
effect on the tangential velocity for the athlete if we keep the time for 1 revolution
the same?
It will be 16 times greater
0
O
It will be 2 times greater
0
о
It will be 4 times greater
0
It will be the same
CLEAR ALL
7 08 09 10 0 11
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REVIEW & SUBMIT
The tangential velocity is proportional to the length of the rope hence when the length of the rope is doubled, the tangential velocity becomes two times greater.
We know that the tangential velocity of an object moving along a circular path depends on its radius and it's angular velocity.
In this case, we have been told that the motion of the object is uniform meaning that the angular velocity is held constant and the length of the rope which is the radius was doubled.
Since the tangential velocity is directly proportional to the length of the rope, the tangential velocity will be two times greater when the length of the rope is doubled.
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Phase conductors 6 awg and smaller ____ permitted to be re-identified as equipment ground conductors using green paint, tape, or other effective means
Phase conductors 6 awg and smaller are not permitted to be re-identified as equipment ground conductors using green paint, tape, or other effective means.
Depending on the type of system offered, the grounded conductor of a service is typically a neutral conductor, though it can also be a phase conductor. A grounded phase conductor and no grounded neutral conductor are present in a corner-grounded delta system, for instance. The unbalanced load is returned to the source by the neutral. The conductor that has been purposefully grounded is the grounded conductor. The neutral is the conductor that is purposefully grounded in the majority of wiring systems used in industrial facilities, businesses, and homes. For safety reasons, the grounding conductor is employed. The grounding wire does not carry current under typical circumstances. The grounding wire offers a low resistance channel for the current in a fault situation.
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Please help! Didn’t have right subject so i chose one that may be similar.
Tides are caused by
the rotation of the Earth
the orbit of the Moon
gravitational pull of just the Sun
gravitational pull of the Sun and Moon
Answer:
Moon's Orbit
Explanation:
Answer:
saaaaaaaaaaaay it please
Explanation:
Hi im really confused:
How much energy would be needed to raise the temperature of the air in a room by 5.0°C if the room measures 4.0m x 4.0m x 3.0m? (Density of air = 1.0kg/m³)
Assume that the room has no furniture and that the walls gain no thermal energy.
The energy required to raise the temperature of the air in a room by 5.0°C is 336 kJ
U = [tex]c_{p}[/tex] m ΔT
U = Energy
[tex]c_{p}[/tex] = Specific heat
m = Mass
ΔT = Change in temperature
ρ = Density
V = Volume
ρ = 1000 g / m³ (Dry air )
= 1 J / g K
ΔT = 5 °C
V = 4 * 4 * 3
V = 48 m³
m = ρ V
m = 1000 * 48
m = 48000 g
U = 1 * 48000 * 7
U = 336000 J
U = 336 kJ
Therefore, the energy required to raise the temperature of the air in a room by 5.0°C is 336 kJ
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Force is measured in what unit? What is this unit equal to in other units?
Force is measured in Newton ( kg.m. /s²)
The movie "The Gods Must Be Crazy" begins with a pilot dropping a bottle out of an airplane. A surprised native below, who thinks it is a message from the gods, recovers it. If the plane from which the bottle was dropped was flying at a height of 500 m, and the bottle lands 400 m horizontally from the initial dropping point, how fast was the plane flying when the bottle was released? Draw a 2-dimensional motion map for the velocities and another for the acceleration.
Answer:
1) The Speed of the plane is approximately 39.596 m/s
2) Please find attached the required velocity and acceleration graphs
Explanation:
1) The height at which the plane was flying, h = 500 m
The location the bottle lands from the initial dropping point = 400 m
The time it takes the bottle to land is given by the equation for free fall as follows;
h = 1/2·g·t²
Where;
g = The acceleration due to gravity = 9.8 m/s²
t = √(h/(1/2·g) = √(500/(1/2×9.8) ≈ 10.102
The Speed of the plane = The horizontal velocity of the bottle = 400/10.102 ≈ 39.596
The Speed of the plane ≈ 39.596 m/s
2) Please find attached the velocity graph for the vertical and horizontal velocities of the bottle and the acceleration graph for the vertical motion of the bottle, created with Microsoft Excel
Aluminium is a non magnetic metal. However, it can become magnetised under special conditions. Explain how. (4 marks)
Answer:
Exposing it to a very strong magnetic field
Explanation:
Aluminium is a non magnetic metal thus it is not magnetic under normal conditions. This is because it interacts with other magnets. The only condition that magnetism can be induced in aluminium is exposing it to a very strong magnetic field.
Supposing d(t) is known to have value D,
what procedure will find the time t at which
this happens?
2. Set d(t) equal to v^2/a
6. Set a to zero
In the year 2000, a company made $4.7 million in profit. For each consecutive year after that, their profit increased by 15%. How much would the company's profit be in the year 2004, to the nearest tenth of a million dollars?
Answer: $8.2 million.
Explanation:
If we have a quantity A, and we have an increase of x%, this can be written as:
A + (x%/100%)*A
Now, for this particular case we have:
In year 2000 (we can define this year as the year zero, y = 0) the initial value is $4.7 million.
The next year, y = 1, there is an increase of 15%, then we will have a profit of:
P = $4.7 million + (15%/100%)*$4.7 million = $4.7 million + 0.15*$4.7 million
P = $4.7 million*(1 + 0.15) = $4.7 million*(1.15)
in the next year, y = 2, the profit will be:
P = $4.7 million*(1.15) + (15%/100%)* $4.7 million*(1.15)
= $4.7 million*(1.15) + 0.15* $4.7 million*(1.15)
= $4.7 million*(1.15)^2
We already can see the pattern, the profit in the year y will be:
P(y) = $4.7 million*(1.15)^y
In particular, in the year 2004 we have y = 4, then the profit that year will be:
P(4) = $4.7 million*(1.15)^4 = $8.2 million.
How much work is needed to move an object 20 meters with 40 newtons of force?
a) 800 J
b) 850 N
c) 875 W
d) 825 m
Answer:
a) 800 J
Explanation:
Work is said to be done when a force induces movement over a distance in an object. The amount of work done (W) is calculated by multiplying the force by the distance traveled.
That is;
W = F × d
Where;
W = work done (J or N/m)
F = force (N)
d = distance (m)
In the information given in this question, F = 40N, d = 20m
Hence,
Work done = 40N × 20m
Work done = 800N/m or 800 J
C: Calculate the kinetic energy of a car with a mass of 1563 kilograms that is traveling at 42 kilometers per hour.
HELP PLEASEE
Hello!
C. KE ≈ 9120.105 J
D. m = 151.45 kg
Question C:
Calculate kinetic energy using the formula:
[tex]KE = \frac{1}{2}mv^{2}[/tex]
Substitute in the given mass. We have to convert kilometers to meters in order to solve.
[tex]\frac{42km}{1hr} * \frac{1hr}{3600sec} *\frac{1000m}{1km} = 11.67 m/s[/tex]
Use the equation above:
KE = 1/2(1563)(11.67)
KE ≈ 9120.105 J
Question D:
Plug in the given Kinetic Energy and velocity to solve. Convert kilometers/hour to meters/second:
[tex]\frac{7km}{1hr} * \frac{1hr}{3600sec} *\frac{1000m}{1km} = 1.94 m/s[/tex]
[tex]KE = \frac{1}{2}mv^{2}[/tex]
285 = 1/2(m)(1.94²)
570 = (1.94²)m
151.45kg = m