Answer:
i). (a) in the same direction as , (d) lower
ii). (f) opposite the direction of, (g) higher
Explanation:
An proton may be defined as a sub atomic particle and it has a positive electrical charge. Its mass is slightly less than that of a neutron. When a proton is placed in an electrical field that is uniformly charged, it is at rest. When the proton first moves out from rest from the uniform electric field, it will move in a direction which is same as that of the electric field and it will move to a region of higher potential.
An electron is defined as the subatomic particle having negative electric charge. When an electron is released form rest from an uniform electric field, it will move in the opposite direction of the uniform electric field and will move to the region of lower electric potential.
physical properties for soap
Answer:
there are germs in your hand and soap has chemicals that kill germs
Answer:
dino nuggies
Explanation:
cause whynot
A truck travels on a straight road at a velocity of 17 meters per second. Over 20
seconds, it accelerates uniformly to 27 meters per second. What distance did the truck
travel during this acceleration?
Answer:
Distance, S = 440 meters.
Explanation:
Given the following data;
Initial velocity, u = 17m/s
Time, t = 20 seconds
Final velocity, v = 27m/s
To find the distance;
First of all, we would determine the acceleration of the truck.
Acceleration = (v-u)/t
Substituting the given values into the equation, we have;
Acceleration = (27 - 17)/20
Acceleration = 10/20
Acceleration = 0.5m/s²
Now, we would use the second equation of motion to find the distance traveled.
S = ut + ½at²
S = 17*20 + ½*0.5*20²
S = 340 + 0.25*400
S = 340 + 100
S = 440m
The equations of motion can be used to obtain the distance covered as 440 m.
We have to use of the equations that are used for uniformly accelerated motion in solving the problem. The chosen equation must be;
v^2 = u^2 + 2as and v = u + at
v = final velocity
u = initial velocity
a = acceleration
s = distance
To obtain the acceleration;
27 = 17 + 20(a)
27 - 17 = 20a
a = 0.5 ms-2
Now, to obtain the distance;
v^2 = u^2 + 2as
v^2 - u^2/as = s
s = (27)^2 - (17)^2/2(0.5)
s = 440 m
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Which two substances are elements?
A. sand and air B. salt and sand
C. iron and helium
D. helium and water
A motorcycle passes over the top of a hill that has a radius of curvature of 100 m. The mass of the motorcycle plus rider is 300 kg. The motorcycle is moving at a speed of 30 m/s. The surface exerts a normal force of magnitude FN on the motorcycle.The motorcycle passes over the top of the hill again but now is moving at a speed of 33 m/s. How does the new normal force exerted on the motorcycle compare to FN
Answer:
[tex]F_N>F_N'[/tex]
Explanation:
From the question we are told that
Radius of curvature[tex]r=100m[/tex]
Mass [tex]M=300kg[/tex]
initial Speed of Motorcycle [tex]V_1=30m/s[/tex]
Final Speed of Motorcycle [tex]V_2=33m/s[/tex]
Generally the equation Force at initial speed is mathematically given as
[tex]F_N=mg-\frac{mv^2}{R}[/tex]
[tex]F_N=300*9.8-\frac{(300*30)^2}{100}[/tex]
[tex]F_N=240N[/tex]
Generally the equation Force at Final speed is mathematically given as
[tex]F_N'=mg-\frac{mv'^2}{R}[/tex]
[tex]F_N'=300*9.8-\frac{(300*33)^2}{100}[/tex]
[tex]F_N'=-327N[/tex]
Therefore
[tex]F_N>F_N'[/tex]
Following are the calculation to the given question:
Solution:
Using formula:
[tex]\to mg - F^{'}_N=\frac{mv^{2}}{R} \\[/tex]
Calculating the Initial value:
[tex]\to F_N = mg - \frac{mv^2}{R}\\\\[/tex]
[tex]= 300 \times (9.8 - \frac{(30)^2}{100}) \\\\= 300 \times (9.8 - \frac{900}{100}) \\\\= 300 \times (9.8 - 9) \\\\= 300 \times (0.8) \\\\ = 240\ N \\\\[/tex]
Calculating the Final value:
[tex]\to F^{'}_{N}= 300 (9.8 -\frac{33^2}{100})\\\\[/tex]
[tex]= 300 (9.8 -\frac{1089}{100})\\\\= 300 (9.8 - 10.89)\\\\= 300 (- 1.09)\\\\=-327[/tex]
Therefore, the answer is "the new normal force is less than [tex]F_{N}[/tex]" or [tex]\bold{F^{'}_{N}< F_{N}}[/tex].
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It takes 500 W of power to move an object 96 m in 12 s. What force is being applied to the object?
Answer:
Explanation:
Power, by definition, is the amount of work per unit of time. We arent given work in this question, but we can find it because work is how much force per unit of distance.
[tex]P=\frac{W}{t_f-t_i} =\frac{F*d}{t_f-t_i}[/tex]
Plug in all the values, and its algebra at this point
[tex]500=\frac{F*96}{12}[/tex]
6000 = 96F
F = 62.5 Newtons
The force which is being applied to the object with the power of 500 Watt is 62.5 N.
Power can be defined as the rate of doing work. It is the work done in unit time. The SI unit of power is Watt (W) which is equal to joules per second (J/s). Sometimes, the power of motor vehicles and other machines is given in terms of Horsepower (hp), This unit is approximately equal to 745.7 watts of power.
The power of a system can be calculated as the product of force applied and distance travelled by the system per unit time taken.
Therefore, the force of the object with power 500W can be calculated as:
P = f × d/ t
where, P = Power of the object,
f = Force applied,
d = distance travelled,
t = time taken to cover the distance
f = (P × t)/ d
f = (500 × 12) / 96
f = 6000/ 96
f = 62.5 N
Therefore, the force which is being applied to the object is 62.5N (Newton).
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Police driving with a velocity of 50 m/s decide to chase a speeder who is 3 km ahead and moving at 55 m/s. The police car accelerates at 2 m/s2. Instantly the speeder becomes aware that he is being chased and starts to accelerate at 1 m/s2. How much time (in s) passes until the police catch the speeder
Answer:
The time that passes until the police catch the speeder is 82.6204 seconds.
Explanation:
A body performs a uniformly accelerated rectilinear motion or uniformly varied rectilinear motion when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases its modulus in a uniform way.
The position is calculated by the expression:
x = x0 + v0*t + 1/2*a*t²
where:
x0 is the initial position. v0 is the initial velocity. a is the acceleration. t is the time interval in which the motion is studied.First, let’s look at the police car’s equations of motion. In this case:
x0= 0 v0= 50 m/s a= 2 m/s²So: x = 50 m/s*t + 1/2*2 m/s²*t²
Now for the speeder’s car’s equations of motion you know:
x0= 3 km= 3,000 m v0= 55 m/s a= 1 m/s²So: x = 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²
When the police catch the speeder they are both in the same position. So:
50 m/s*t + 1/2*2 m/s²*t²= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²
Solving:
0= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t² - 50 m/s*t - 1/2*2 m/s²*t²
0= 3,000 + 55 *t + 1/2*t² - 50*t - 1*t²
0= 3,000 + 55 *t - 50*t - 1*t² + 1/2*t²
0= 3,000 + 5*t - 1/2*t²
Applying the quadratic formula:
[tex]x1,x2=\frac{-5+-\sqrt{5^{2}-4*(-\frac{1}{2})*3000 } }{2*(-\frac{1}{2} )}[/tex]
x1= -72.6209
and x2= 82.6209
Since you are calculating the value of a time and it cannot be negative, then the time that passes until the police catch the speeder is 82.6204 seconds.
Through what potential difference should electrons be accelerated so that their speed is 1.0 % of the speed of light when they hit the target
Answer:
Explanation:
Considering non - relativistic approach : ----
Speed of electron = 1 % of speed of light
= .01 x 3 x 10⁸ m /s
= 3 x 10⁶ m /s
Kinetic energy of electron = 1/2 m v²
= .5 x 9.1 x 10⁻³¹ x ( 3 x 10⁶ )²
= 40.95 x 10⁻¹⁹ J
Kinetic energy in electron comes from lose of electrical energy equal to
Ve where V is potential difference under which electron is accelerated and e is electronic charge .
V x e = kinetic energy of electron
V x 1.6 x 10⁻¹⁹ = 40.95 x 10⁻¹⁹
V = 25.6 Volt .
What is sin(77)?
Α. 0.77
Β. 4.33
Ο Ο Ο
C. 0.22
Ο D. 0.97
Answer:
0.97 id the correct answer
Explanation:
what is the definition of power ? what are the units of power?
Answer:
power is the rate at which energy is transferred or converted
the unit of power is Watts
The body breaks down or converts most carbohydrates into the
sugar____?*
1.Fructose
2.Galactose
3.Lactose
4.Glucose
Green plants need light in order to survive. Structures in the leaves absorb light, which in turn, helps plants make
their own food
Under which color of light will plants be least likely to make food?
ted
blue
orange
Save and Exit
Answer:
I don't know lol good luck i guess
Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece's weight. Assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a 12.0 mg piece of tape held 1.00 cm above another.
Answer:
36.1 nC
Explanation:
The electrostatic force F = kqq'/r² since q = q', F = kq²/r² where q = charge on tape, r = distance between tapes = 1.00 cm = 1 × 10⁻² m and k = 9 × 10⁹ Nm²/C².
Given that F = weight of 12.0 mg piece of tape, F = mg where m = mass of tape = 12.0 mg = 12 × 10⁻³ kg and g = acceleration due to gravity = 9.8 m/s²
So, kq²/r² = mg
q²/r² = mg/k
q² = mgr²/k
taking square-root of both sides,
q = √(mg/k)r
So, substituting the values of the variables into the equation, we have
q = √(mg/k)r
q = √(12 × 10⁻³ kg × 9.8 m/s²/ 9 × 10⁹ Nm²/C²)1 × 10⁻² m
q = √(117.6 × 10⁻³ kgm/s²/9 × 10⁹ Nm²/C²)1 × 10⁻² m
q = √(13.07 × 10⁻¹² C²/m²)1 × 10⁻² m
q = 3.61 × 10⁻⁶ C/m × 1 × 10⁻² m
q = 3.61 × 10⁻⁸ C
q = 36.1 × 10⁻⁹ C
q = 36.1 nC
A piece of aluminium with mass 1 kg and density 2700 kg/m3 is suspended from a string and then completely immersed in a container of water. Calculate the tension in the string (a) before and (b) after the metal is immersed.
(with steps please)
A piece of aluminium with mass 1 kg and density 2700 kg/m3 is suspended from a string and then completely immersed in a container of water. Calculate the tension in the string (a) before and (b) after the metal is immersed.
ANSWER: 6.296NI am unable to add the workings out, however please do message me and I will be able to provide you with them :)
3. One effective method for coping with change is using:
Recall
Revisitation
Reminiscence
Reminding
Cope up with changes is significant in everyone's life. One of the suitable way is to remind ourselves that we can withstand with the changes.
What is reminder?Reminding a thing that is important to anyone or ourselves until we do it a better way. Thinking past time and wasting time is just ridiculous. Because we must cope up with the current scenario and build up our mental strength.
Our lives are constantly impacted by changes. Change is a constant, uncontrolled aspect of life that has an impact on our work ethics and routines.
Anything that changes may have major or little repercussions. As a result, since the extent of change's first consequences is uncertain, we must refrain from discriminating against it.
Organizations can effectively use change as a tool. Without change, people would be hesitant to form fresh perspectives and ideas. They wouldn't be able to respond to daily obstacles and would become inert in their work and their roles.
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Two soccer players kick a soccer ball back and forth along a straight line. The first player kicks the ball 11 m to the right to the second player. The second player kicks the ball to the left weakly; it only moves 2.7 m before stopping. (Consider the right to be the positive direction. Where applicable, indicate the direction with the sign of your answer.)
(a) What is the total distance that the ball moved?
m
(b) At the end, what is the displacement of the ball (from the first player)?
m
Answer:
a) 13.7 m
b) 8.3 m
Explanation:
a)
The total distance traveled by the ball is just the sum of the distances traveled after being kicked by the first player (11 m.) plus the distance moved after being kicked by the second player (2.7 m), independent of the direction of the movement:[tex]d = d_{1} + d_{2} = 11.0 m + 2.7 m = 13.7 m (1)[/tex]
b)
The total displacement of the ball, is just the difference between the final and initial position, by definition.If we choose the origin to be in the place that the ball is kicked by the first player,[tex]x_{o} = 0 (2)[/tex]
The final position along the x-axis, is just the distance traveled to the right (11 m) minus the distance traveled to the left (considering the right to be the positive direction), 2.7 m:[tex]x_{f} = x_{1} - x_{2} = 11.0 m - 2.7 m = 8.3 m (3)[/tex]
⇒ Δx = xf - x₀ = 8.3 m - 0 m = 8.3 m
An electric field is represented by the force on which particle?
a neutron
a dipole
a negative test charge
a positive test charge
Answer: a positive test charge
Explanation:
An electric field is a vector measure. Its direction is determined as the direction that a positive test charge would experience a force when it is placed in the field.
A positive charge has a higher potential and a negative charge is lower potential.
Thus, an electric field is represented by the force on a positive test charge.
Hence, the correct option is "a positive test charge".
Answer:
D. a positive test charge.
Explanation:
Edge 2021
A blue ball is thrown upward with an initial speed of 24.1 m/s, from a height of 0.5 meters above the ground. 2.9 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 7.2 m/s from a height of 32 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2. Take up as the positive direction.
1. What is the speed of the blue ball when it reaches its maximum height?.
2. How long does it take the blue ball to reach its maximum height?.
3. What is the maximum height the blue ball reaches?.
4. What is the height of the red ball 3.77 seconds after the blue ball is thrown?.
5. How long after the blue ball is thrown are the two balls in the air at the same height?.
Answer:
1. Speed=0
2. 2.46 s
3.30.1 m
4. 22.0 m
5.1.004 s
Explanation:
We are given that
Initial speed of blue ball, u=24.1 m/s
Height of blue ball from ground y_0=0.5 m
Initial speed of red ball , u'=7.2 m/s
Height of red from ground=y'0=32 m
Gravity, g=[tex]9.81ms^{-2}[/tex]
1.When the ball reaches its maximum height then the speed of the blue ball is zero.
2.v=0
[tex]v=u+at[/tex]
Using the formula and substitute the values
[tex]0=24.1-9.81t[/tex]
Where g is negative because motion of ball is against gravity
[tex]24.1=9.81t[/tex]
[tex]t=\frac{24.1}{9.81}=2.46s[/tex]
3.[tex]y=y_0+ut+\frac{1}{2}at^2[/tex]
Using the formula
[tex]y=0.5+24.1(2.46)-\frac{1}{2}(9.81)(2.46)^2[/tex]
[tex]y=30.1 m[/tex]
4.Time of flight for red ball=3.77-2.9=0.87s
[tex]y'=32-7.2(0.87)-\frac{1}{2}(9.81)(0.87)^2[/tex]
[tex]y'=22.0m[/tex]
Hence, the height of red ball 3.77 s after the blue ball is 22.0 m.
5.According to question
[tex]0.5+24.1(t+2.9)-\frac{1}{2}(9.81)(2.9+t)^2=32-7.2t-\frac{1}{2}(9.81)t^2[/tex]
[tex]0.5+24.1t+69.89-4.905(t^2+5.8t+8.41)=32-7.2t-4.905t^2[/tex]
[tex]0.5+24.1t+69.89-4.905t^2-28.449t-41.25105=32-7.2t-4.905t^2[/tex]
[tex]0.5+69.89-41.25105-32=-24.1t+28.449t-7.2t[/tex]
[tex]-2.86105=-2.851t[/tex]
[tex]t=\frac{2.86105}{2.851}=1.004 s[/tex]
Hence, 1.004 s after the blue ball is thrown are the two balls in the air at the same height.
A young gazelle is grazing in a beautifully sunlit corner of the savanna. Suddenly, the gazelle raises herhead and spots a lioness in the tall grass 173 m away, so she turns away running at roughly constantspeed. The lioness immediately chases the gazelle, with an explosive acceleration of 2.57 m/s2. Aninformed source tells us that this lioness is capable of enduring her maximum speed of 21.0 m/s for 25.0seconds at the longest. [Assume that both predator and prey never change their direction or motion inthis case, for the sake of simplicity.]
(a) On average, how fast (at least) does that gazelle need to run to survive? [Show all your steps.]
(b) Produce a qualitative graph of animal's position (vertical axis) versus time (horizontal axis), shared for the lioness's motion and as well as the gazelle's. Place graph labels in a way that fits the narration.
Answer:
a) v₂ = 13.20 m / s
Explanation:
To solve this exercise we will use the kinematic relations
Let's start with the Lioness. Let's find the time to reach top speed
v = v₀ + a t
as part of rest, its initial velocity is zero
t = v / a
t₁ = 21.0 / 2.57
t₁ = 8.17 s
the total time is the acceleration time plus the time (t₂ = 25 s) that the maximum speed can withstand
t = t₁ + t₂
t = 8.17 +25.0 = 33.17 s
Now let's find out what distance the lioness travels in these times
during acceleration
x = v₀ t + ½ a t²
x = ½ a t²
x₁ = ½ 2.57 8.17²
x₁ = 85.77 m
during constant speed part
x₂ = v t₂
x₂ = 21.0 25.0
x₂ = 525 m
therefore the total distance traveled is
x = x₁ + x₂
x = 85.77 + 525
x = 610.77 m
a) the average speed of the gazelle
this must be the distance that the lioness travels minus the initial distance that separates the two animals (xo = 173 m) between the time taken
v₂ = [tex]\frac{x -x_o}{t}[/tex]
v₂ = [tex]\frac{610.77 - 173}{33.17}[/tex]
v₂ = 13.20 m / s
b) in the attachment we can see a graph of the displacement of the two animals
A car traveling 85 km/h is 250 m behind a truck
traveling 73 km/h.
Time needed = t = 20.83 s
Further explanationGiven
car speed = 85 km/h
truck speed = 73 km/h
Required
the time it takes for the car to reach the truck
Solution
When the car reaches the truck, the distance between them will be the same
x car - 250 m = x truck
General formula for distance (d) :
d = v.t
So the equation becomes :
85t-250 = 73t
12t=250
t = 20.83 s
Conductivities are often measured by comparing the resistance of a cell filled with the sample to its resistance when filled with some standard solution,such as aqueous potassium chloride. The conductivity of water is 76 mS m^(-1) at 25 C and the conductivity of 0.100 mol dm^(-3) KCl (aq) is 1.1639 S m^(-1) A cell has a resistance of 33.21ohm when filled with 0.100 mol dm^(-3) KCl (aq) and 300.0 ohm when filled with 0.100 mol dm CHaCOOH (aq). What is the molar conductivity of acetic acid at that concentration and temperature?
Answer:
1200 Sm^2mol^-1
Explanation:
Given data :
conductivity of water ( kwater ) = 76 mS m^-1 = 0.076 Sm^-1
conductivity of kcl (aq)( Kkcl ) = 1.1639 Sm^-1
Kkcl = 1.1639 - 0.076 = 1.0879 Sm^-1
Resistance = 33.21 Ω
where conductivity can be expressed as = [tex]\frac{Cell constant}{Resistance }[/tex]
hence cell constant = conductivity * Resistance
= 1.0879 * 33.21 = 36.13m^-1
conductivity of CH3COOH ( kCH3COOH ) = 36.13 / 300
= 0.120 Sm^-1
Determine the molar conductivity of acetic acid
= ( kCH3COOH * 1000 ) / C
C = 0.1 mol dm
= (0.120 * 1000) / 0.1 = 1200 Sm^2mol^-1
Suppose that a ball is dropped from the upper observation deck of the CN Tower in Toronto, 450 m above the ground. Find the velocity of the ball after 2 seconds. Through experiments carried out four centuries ago, Galileo discovered that the distance fallen by any freely falling body is proportional to the square of the time it has been falling. (This model for free fall neglects air resistance.) If the distance fallen after t seconds is denoted by s ( t ) and measured in meters, then Galileo's law is expressed by the equation
Answer:
v = -19.6 m / s, y =y₀ + v₀ t - ½ g t²
Explanation:
This is an exercise in kinetics in one dimension, let's take the upward direction as positive
v = v₀ - g t
in this case as the body is released its initial velocity is zero and the acceleration is -g the sign indicates that it is directed downwards
v = 0 -g t
v = - 9.8 2
v = -19.6 m / s
the sign indicates that the speed is down.
Galileo's equation is
y =y₀ + v₀ t - ½ g t²
where i is the initial height, v₀ the initial velocity and -g the acceleration of the body
A boat moves up and down as water waves pass under the boat. If the amplitude of the wave gets bigger, then
A)
the boat will rise up higher.
B)
the boat will not rise up as high.
C)
the boat will go up and down more often.
D)
the boat will continue to move the same way.
Answer: The Boat will rise
Explanation: Because high amplitude means high in heights.
A student sits on a rotating stool holding two 1 kg objects. When his arms are extended horizontally, the objects are 1.1 m from the axis of rotation, and he rotates with angular speed of 0.64 rad/sec. The moment of inertia of the student plus the stool is 4 kg m2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.33 m from the rotation axis.
Required:
a. Find the new angular speed of the student.
b. Find the kinetic energy of the student before and after the objects are pulled in.
Answer:
a) ω₁ = 0.97 rad/sec
b) K₀ = 1.31 J
Kf = 1.99 J
Explanation:
a)
Assuming no external torques acting on the student, total angular momentum must be conserved, as follows:[tex]L_{o} = L_{f} (1)[/tex]
The angular momentum of a rigid body rotating respect an axis of rotation can be written as follows:[tex]L = I*\omega (2)[/tex]
In order to get ωf, since ω₀ = 0.64 rad/sec, we need to find the values of the initial moment of inertia, I₀, and the final one, If:I₀ = 4 kg*m² + 1kg*(1.1 m)² + 1kg*(1.1m)² = 6.42 kg*m² (3)If = 4 kg*m² + 1kg*(0.33m)² + 1kg*(0.33m)² = 4.22 kg*m² (4)Replacing (3), (4) in (1) we can solve for ωf:[tex]\omega_{f} = \frac{I_{o} *\omega_{o} }{I_{f} } = \frac{6.42kgm2*0.64rad/sec}{4.22kgm2} = 0.97 rad/sec (5)[/tex]
b)
Since the student is not translating but he is only rotating, all his kinetic energy is rotational kinetic energy.The expression for the kinetic energy of a rotating rigid body, around an axis of rotation is as follows:[tex]K_{rot} = \frac{1}{2} * I * \omega^{2} (6)[/tex]
The initial kinetic energy of the student, before the objects are pulled in, is as follows:[tex]K_{roto} = \frac{1}{2} *I_{o} * \omega_{o} ^{2} = \frac{1}{2}* 6.42kgm2*(0.64rad/sec)^{2} = 1.32 J (7)[/tex]
The final kinetic energy is given by the following expression:[tex]K_{rotf} = \frac{1}{2} *I_{f} * \omega_{f} ^{2} = \frac{1}{2}* 4.22kg*m2*(0.97rad/sec)^{2} = 1.99 J (8)[/tex]A bald eagle is flying to the left with a speed of 34 meters
per second when a gust of wind blows back against the
eagle causing it to slow down with a constant acceleration
of a magnitude 8 meters per second squared.
What will the speed of the bald eagle be after the wind has
blown for 3 seconds?
Answer:
the speed after 3 seconds is 10 m/s
Explanation:
The computation of the speed is shown below:
As we know that
V = U + at
Here,
U = 34 m/s
a = - 8 m/s²
t = 3 Sec
V = velocity after 3 sec
V = 34 + (-8)3
= 34 - 24
V = 10 m/s
Hence, the speed after 3 seconds is 10 m/s
The other name of eureka can
3. What is the acceleration of a cart with a F = 80 N and m = 32 kg? *
2.0 m/s^2
O 2.5 m/s^2
2.8 m/s^2
2.7 m/s^2
Answer:
2.5 m/s^2
Explanation:
Given the following data;
Force = 80N
Mass = 32kg
To find acceleration;
Force is given by the multiplication of mass and acceleration.
Mathematically, Force is;
[tex] F = ma[/tex]
Where;
F represents force.
m represents the mass of an object.
a represents acceleration.
Making acceleration (a) the subject, we have;
[tex]Acceleration (a) = \frac{F}{m}[/tex]
Substituting into the equation;
[tex]Acceleration (a) = \frac{80}{32}[/tex]
Acceleration = 2.5m/s²
Elizabeth has always believed that people's thoughts can help heal them. She wants to help people use positive thinking to positively affect their
illnesses. What type of psychology would be MOST appropriate for Elizabeth to study?
Answer: Family
Explanation:
A sports car of mass m has the same kinetic energy as an SUV with a mass 3m as each is driven along the same road. Which vehicle, if either, has the larger momentum and what is the difference in their momenta, if any
Answer:
Explanation:
Kinetic energy ( KE ) = 1/2 m v²
= m²v² / 2 m = p² / 2m where p is momentum
KE = p² / 2m
p² = 2m KE
KE is constant
p is proportional to mass
So car having higher mass will have higher momentum .
p₁ = √ ( 2 m x KE )
p₂ = √ ( 6 m x KE )
p₂ - p₁ = √ ( 6 m x KE ) - √ ( 2 m x KE )
= √KE m ( √6 - √2 )
Kinetic energy ( K.E )
[tex]= \frac{1}{2} m v^2\\\\= \frac{m^2 v^2}{2 m} \\\\= \frac{p^2}{2m}[/tex]
where p is momentum
[tex]K.E =\frac{p^2}{2m}\\\\p^2 = 2m. KE[/tex]
KE is constant
p is proportional to mass
So car having higher mass will have higher momentum .
[tex]p_1 =\sqrt{(2m*K.E)}\\\\p_2 = \sqrt{(6m*K.E)} \\\\p_2 - p_1 = \sqrt{(6m*K.E)} -\sqrt{(2M*K.E} \\\\p_2 - p_1 = \sqrt{K.E m(\sqrt{6}-\sqrt{2}) }[/tex]
The difference is shown above.
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brainly.com/question/20658056
A child on a sled slides (starting from rest) down an icy slope that makes an angle of 15◦ with the horizontal. After sliding 20 m down the slope, the child enters a flat, slushy region, where she slides for 2.0 s with a constant negative acceleration of −1.5 m/s2 with respect to her direction of motion. She then slides up another icy slope that makes a 20◦ angle with the horizontal.
A) How fast was the child going when she reached the bottom of the first slope? How long did it take her to get there?B) How long was the flat stretch at the bottom?C) How fast was the child going as she started up the second slope?D) How far up the second slope did she slide?
Answer:
A) v₁ = 10.1 m/s t₁= 4.0 s
B) x₂= 17.2 m
C) v₂=7.1 m/s
D) x₂=7.5 m
Explanation:
A)
Assuming no friction, total mechanical energy must keep constant, so the following is always true:[tex]\Delta K + \Delta U = (K_{f} - K_{o}) +( U_{f} - U_{o}) = 0 (1)[/tex]
Choosing the ground level as our zero reference level, Uf =0.Since the child starts from rest, K₀ = 0.From (1), ΔU becomes:[tex]\Delta U = 0- m*g*h = -m*g*h (2)[/tex] In the same way, ΔK becomes:[tex]\Delta K = \frac{1}{2}*m*v_{1}^{2} (3)[/tex] Replacing (2) and (3) in (1), and simplifying, we get:[tex]\frac{1}{2}*v_{1}^{2} = g*h (4)[/tex]
In order to find v₁, we need first to find h, the height of the slide.From the definition of sine of an angle, taking the slide as a right triangle, we can find the height h, knowing the distance that the child slides down the slope, x₁, as follows:[tex]h = x_{1} * sin \theta_{1} = 20.0 m * sin 15 = 5.2 m (5)[/tex]
Replacing (5) in (4) and solving for v₁, we get:
[tex]v_{1} = \sqrt{2*g*h} = \sqrt{2*9.8m/s2*5.2m} = 10.1 m/s (6)[/tex]
As this speed is achieved when all the energy is kinetic, i.e. at the bottom of the first slide, this is the answer we were looking for.Now, in order to finish A) we need to find the time that the child used to reach to that point, since she started to slide at the its top.We can do this in more than one way, but a very simple one is using kinematic equations.If we assume that the acceleration is constant (which is true due the child is only accelerated by gravity), we can use the following equation:[tex]v_{1}^{2} - v_{o}^{2} = 2*a* x_{1} (7)[/tex]
Since v₀ = 0 (the child starts from rest) we can solve for a:[tex]a = \frac{v_{1}^{2}}{2*x_{1} } = \frac{(10.1m/s)^{2}}{2* 20.0m} = 2.6 m/s2 (8)[/tex]
Since v₀ = 0, applying the definition of acceleration, if we choose t₀=0, we can find t as follows:[tex]t_{1} =\frac{v_{1} }{a} =\frac{10.1m/s}{2.6m/s2} = 4.0 s (9)[/tex]
B)
Since we know the initial speed for this part, the acceleration, and the time, we can use the kinematic equation for displacement, as follows:[tex]x_{2} = v_{1} * t_{2} + \frac{1}{2} *a_{2}*t_{2}^{2} (10)[/tex]
Replacing the values of v₁ = 10.1 m/s, t₂= 2.0s and a₂=-1.5m/s2 in (10):[tex]x_{2} = 10.1m/s * 2.0s + \frac{1}{2} *(-1.5m/s2)*(2.0s)^{2} = 17.2 m (11)[/tex]
C)
From (6) and (8), applying the definition for acceleration, we can find the speed of the child whem she started up the second slope, as follows:[tex]v_{2} = v_{1} + a_{2} *t_{2} = 10.1m/s - 1.5m/s2*2.0s = 7.1 m/s (12)[/tex]
D)
Assuming no friction, all the kinetic energy when she started to go up the second slope, becomes gravitational potential energy when she reaches to the maximum height (her speed becomes zero at that point), so we can write the following equation:[tex]\frac{1}{2}*v_{2}^{2} = g*h_{2} (13)[/tex]
Replacing from (12) in (13), we can solve for h₂:[tex]h_{2} =\frac{v_{2} ^{2}}{2*g} = \frac{(7.1m/s) ^{2}}{2*9.8m/s2} = 2.57 m (14)[/tex]
Since we know that the slide makes an angle of 20º with the horizontal, we can find the distance traveled up the slope applying the definition of sine of an angle, as follows:[tex]x_{3} = \frac{h_{2} }{sin 20} = \frac{2.57m}{0.342} = 7.5 m (15)[/tex]
what is efficiency?
Explanation:
the state or quality of being efficient or able to accomplish something