Answer:
a) the cheetah's top speed is 64.4 miles/hr
b) time taken to reach top speed is 3.3 seconds
Explanation:
Given the data in the question;
Cheetahs can accelerate to a speed of 21.8 m/s in 2.50 s.
They can continue to accelerate to reach a top speed of 28.8 m/s.
a) Express the cheetah's top speed in mi/h. (Express your answer to three significant figures.)
The cheetah's top speed = 28.8 m/s = ( 28.8 × 2.237 ) miles/hr
= 64.4256 ≈ 64.4 miles/hr
Therefore, the cheetah's top speed is 64.4 miles/hr
b) Starting from a crouched position, how long does it take a cheetah to reach its top speed.
given that
v₁ = 21.8 m/s and t₁ = 2.50 s
let a represent the acceleration of the cheetah
From the First Equation of Motion;:
v = u + at
we substitute
21.8 = 0 + ( a × 2.50 )
21.8 = a × 2.50
a = 21.8 / 2.50
a = 8.72 m/s²
Now, let the time taken by cheetah to reach top speed ( 28.8 m/s ) be t
so from the first equation of motion;
v = u + at
we substitute
28.8 = 0 + ( 8.72 × t )
t = 28.8 / 8.72
t = 3.3 seconds
Therefore, time taken to reach top speed is 3.3 seconds
With what speed must a ball be thrown directly upward so that it remains in the air for 10 seconds?
a) What will be its speed when it hits the ground?
b) How high does the ball rise?
Answer:
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A steel ball is dropped onto a thick piece of foam. The ball is released 2.5 meters above the foam. The foam compresses 3.0 cm as the ball comes to rest. What is the magnitude of the ball's acceleration as it comes to rest on the foam
Answer:
the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²
Explanation:
Given the data in the question;
initial velocity; u = 0 m/s
height; h = 2.5 m
we find the velocity of the ball just before it touches the foam.
using the equation of motion;
v² = u² + 2gh
we know that acceleration due gravity g = 9.81 m/s²
so we substitute
v² = ( 0 )² + ( 2 × 9.81 × 2.5 )
v² = 49.05
v = √49.05
v = 7.00357 m/s
Now as the ball touches the foam
final velocity v₀ = 0 m/s
compresses S = 3 cm = 0.03 m
so
v₀² = v² + 2as
we substitute
( 0 )² = 49.05 + 0.06a
0.06a = -49.05
a = -49.05 / 0.06
a = -817.5 m/s²
Therefore, the magnitude of the ball's acceleration as it comes to rest on the foam is 817.5 m/s²
In addition to acceleration, what else will be a maximum at the amplitude for SHM?
A. Potential energy
B. Kinetic energy
C. Nuclear energy
D. Chemical energy
It is Potential energy's
If the distance between a neutral atom and a point charge is increased by a factor of six, by what factor does the force on the atom by the point charge change
Answer:
A neutral atom has no net charge and should not be affected by a point charge, and the distance of separation is not a factor.
The force will remain the same and is equal to zero.
We have a point charge and a neutral atom.
We have to investigate if the distance between a neutral atom and a point charge is increased by a factor of six, by what factor does the force on the atom by the point charge change.
State Coulomb's Law of Electrostatic force.The Coulomb's law states that the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. Mathematically -
[tex]$F=\frac{q _{1}\cdot q _{2}}{4\cdot \pi \cdot \varepsilon \cdot \varepsilon _{0}\cdot r^{2}}[/tex]
According to question, we have -
A point charge and a neutral atom.
If initially the distance between the point charge and neutral atom is r meters, then -
q(1) = Q (say)
q(2) = 0 ( Neutral atom has zero charge)
Using Coulomb's law -
[tex]$F=\frac{Q\times 0}{4\cdot \pi \cdot \varepsilon \cdot \varepsilon _{0}\cdot r^{2}}[/tex]
F = 0 Newtons.
Now, if you will increase the distance by the factor of six, still the force will remain zero as there is only one point charge and other is neutral atom. There is no electrostatic force between a charged and uncharged particle.
Hence, the force will remain the same and is equal to zero.
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If energy is transferred spontaneously as heat from a substance with a temperature of T1 to a substance with a temperature of T2, which of the following statements must be true?
1-T1 < T2
2-T1 = T2
3-T1 > T2
4-more information is needed
Answer: The statement [tex]T_{1} > T_{2}[/tex] must be true.
Explanation:
As it is given that heat is being transferred from a substance with temperature [tex]T_{1}[/tex] to a substance with temperature [tex]T_{2}[/tex].
It is known that heat will always being transferred from a higher temperature to a lower temperature. Because at higher temperature the molecules of a substance acquire more energy and when they lose energy then a decrease in temperature occurs.
Hence, in the given situation [tex]T_{1} > T_{2}[/tex].
Thus, we can conclude that the statement [tex]T_{1} > T_{2}[/tex] must be true.
A 30g bullet at v=900m/s strikes a 1kg soft iron target stopping inside the iron [c=490J/kg°C. How much will the temperature of the iron increase? Ignore the heat that will be shared with the bullet
A) 25°C
B) 24795°C
C) 826°C
D) 82653°C
show your full work please
Answer:
ΔT = 25°C
Explanation:
Given that.
The mass of a bullet, m₁ = 30 g = 0.03 kg
The speed of the bullet, v = 900 m/s
Mass of soft iron, m₂ = 1 k
The specific heat of iron, c=490J/kg°C
We need to find the increase in temperature of iron. using the conservation of energy,
Kinetic energy = heat absorbed
[tex]\dfrac{1}{2}m_1v^2=m_2c\Delta T\\\\\Delta T=\dfrac{\dfrac{1}{2}m_1v^2}{m_2c}\\\\\Delta T=\dfrac{\dfrac{1}{2}\times 0.03\times 900^2}{1\times 490}\\\\=24.79^{\circ} C\\\\or\\\\\Delta T=25^{\circ} C[/tex]
So, the correct option is (A).
which unit would be most suitable for its scale?
A mm
B
с
crn?
D
cm
[0625_504_9p_1].
8
A piece of cotton is measured between two points on a ruler.
1
coton
BAS
2
4
5
6
7
8
9
10
11
12
13
14
15 16
when the lenge of coton is wound closely around a pen, goes round six times.
pen
six turns of coton
दे-
What is the distance onde round the pen?
4 2.2 m
B 26 cm
с
13.2 cm
D 15.6 cm
Answer:
Mm, thats the answer trust me men
Diwn unscramble the word
Answer:
wind
Explanation:
just a possible answer.
Select the correct answer.
Which figure shows a correct pattern of field lines?
A. Figure A
B. Figure B
C. Figure C
D. Figure D
Which describes a characteristic of metallic bonds?
Answer:
arge number of electrons free to move between the charged ions in the lattice.
Explanation:
The metallic bond occurs when an atom with few electrons is united in its last level, therefore the best way to decrease the total energy of the system is to lose all its electrons to remain with the configuration of a noble gas. The electrons that it loses cannot be acquired by other atoms since they all have few electrons, thus leaving a large number of electrons free to move between the charged ions in the lattice.
Some important characteristics emerge from this description of the metallic bond:
* It has many free electrons therefore its electrical conductivity is high
* As the charged ions are fixed, the material can be malleable, bent without breaking since the free electrons create the bond that keeps the system stable.
* As the electrons are free when heating a part of the material, these electrons acquire energy and rapidly propagate it to the other side, giving a high thermal conductivity
* As the temperature increases, the electrons acquire more kinetic energy, which is why there are more collisions between them and consequently the resistivity of the material increases.
A ball is thrown straight up in the air at an initial speed of 30 m/s. At the same time the ball is thrown, a person standing 70 m away begins to run toward the spot where the ball will land.How fast will the person have to run to catch the ball just before it hits the ground?Vperson= m/s
Answer:
Explanation:
Here's what we know and in which dimension:
y dimension:
[tex]v_0=30[/tex] m/s
v = 0 (I'll get to that injust a second)
a = -9.8 m/s/s
The final velocity of 0 is important because that's the velocity of the ball right at the very top of its travels. If we knew how long it takes to get to that max height, we can also use that to find out how long it will take to hit the ground. Therefore, we will find the time it takes to reach its max height and pick up with the investigation of what this means after.
x dimension:
Δx = 70 m
v = ??
Velocity is our unknown.
Solving for the time in the y dimension:
[tex]v=v_0+at[/tex] and filling in:
0 = 30 + (-9.8)t and
-30 = -9.8t so
t = 3.1 seconds
We know it takes 3.1 seconds to get to its max height. In order to determine how long it will take to hit the ground, just double the time. Therefore, it will take 6.2 seconds for the ball to come back to the ground, which is where the persom trying to catch the ball comes in. We will use that time in our x dimension now.
In the x dimension, the equation we need is just a glorified d = rt equation since the acceleration in this dimension is 0.
Δx = vt and
70 = v(6.2) so
v = 11.3 m/s
Answer the following using equations, number substitution and keep units. 1. What is the speed of an object that travels 5m in 10s. 2. What force is on a 10kg mass that accelerates at 3m/s/s. 3. What is the potential energy of a 7kg object 4m off the ground *
show all your work please
Explanation:
1. Distance, d = 5 m
Time, t = 10 s
Speed = distance/time
[tex]v=\dfrac{5}{10}=0.5\ m/s[/tex]
2. Mass, m = 10 kg
Acceleration, a = 3 m/s³
Force, F = mass (m) × acceleration (a)
F = 10 × 3
= 20 N
3. Mass, m = 7 kg
Height, h = 4 m
Potential energy, E = mgh
E = 7 × 9.8 × 4
E = 274.4 J
Hence, this is the required solution.
Điện tích trên một vật dẫn bất kỳ có giá trị bằng:
A. Tổng độ lớn các giá trị điện tích âm và điện tích dương có trên vật.
B. Tổng đại số các giá trị điện tích âm và điện tích dương có trên vật.
C. Không. Vì lúc nào số điện tích âm cũng bằng số điện tích dương.
D. Tất cả đều sai.
Answer:
A.
sửa cho tôi nếu tôi sai
a rock with the mass of 10 kg sits at the top of a hill 20 m high. what is the potential energy
Answer:
What is the potential energy? PE= mghPE= hwKE= 1/2mv2
Answer:1960J
Explanation:
Charge of uniform density (80 nC/m3) is distributed throughout a hollow cylindrical
region formed by two coaxial cylindrical surfaces of radii, 1.0 mm and 3.0 mm. Determine
the magnitude of the electric field at a point which is 4.0 mm from the symmetry axis.
Answer:
The electric field is given by 4.5 N/C.
Explanation:
Charge density = 80 nC/m3
inner radius, r' = 1 mm
outer radius, r'' = 3 mm
distance, r = 4 mm
The linear charge density is given by
[tex]\lambda =\rho \times\pi\times (r''^2 - r'^2)\\\\\lambda = 80\times 10^{-9}\times 3.14\times 10^{-6}\times(9-1)\\\\\lambda = 2\times 10^{-12}\\[/tex]
The electric field is given by
[tex]E = \frac{\lambda }{4\pi\varepsilon_or}\\E=\frac{9\times 10^9\times 2 \times 10^{-12}}{0.004}\\\\E=4.5 N/C[/tex]
The number 0.00325 × 10-8 cm can be expressed in millimeters as A) 3.25 × 10-11 mm. B) 3.25 × 10-10 mm. C) 3.25 × 10-12 mm. D) 3.25 × 10-9 mm.
Answer:
Option B. 3.25×10¯¹⁰ mm.
Explanation:
Measurement (cm) = 0.00325×10⁻⁸ cm
Measurement (mm) =?
The measurement in mm can be obtained as follow:
1 cm = 10 mm
Therefore,
0.00325×10⁻⁸ cm = 0.00325×10⁻⁸ cm × 10 mm / 1 cm
0.00325×10⁻⁸ cm = 3.25×10¯¹⁰ mm
Thus, 0.00325×10⁻⁸ cm is equivalent to 3.25×10¯¹⁰ mm.
The conversion from centimeter to millimeter of the number 0.00325*10^-8cm is 3.25*10^-10mm
The number given is in standard form and can be written as 3.25*10^-11 cm.
To convert this from centimeter to millimeter, we have to multiply this value by 10.
Conversion Units1 cm - 10mm100cm = 1m1000m = 1kmSo, let's 3.25*10^-11 by 10 and get our value in mm
[tex]3.25*10^-^1^1 * 10 = 3.25*10^-^1^0[/tex]
From the calculation above, we can see that option B is the right answer since it carries [tex]3.25*10^-^1^0mm[/tex]
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Chlorine (chemical symbol Cl) is located in Group 17, Period 3. Which is
chlorine most likely to be?
A. A metalloid with properties of both metals and nonmetals
B. A gaseous, highly reactive nonmetal
C. A soft, shiny, highly reactive nonmetal
D. A soft, shiny, highly reactive metal
Answer:
Option B
Explanation:
A gaseous, highly reactive non-metal
Answer:B
Explanation:I just took the test
An iron nail floats in mercury and sinks in water. explain why?
Answer:
because density of iron is more than that of water but less then that of Mercury
hope it's helpful
Two small silver spheres, each of mass m=6.2 g, are separated by distance d=1.2 m. As a result of transfer of some fraction of electrons from one sphere to the other, there is an attractive force F=900 KN between the spheres. Calculate the fraction of electrons transferred from one of the spheres: __________
To evaluate the total number of electrons in a silver sphere, you will need to invoke Avogadro's number, the molar mass of silver equal to 107.87 g/mol and the fact that silver has 47 electrons per atom.
Answer:
4.60 × 10⁻⁸
Explanation:
From the given information;
Assuming that q charges are transferred, then:
[tex]F = \dfrac{kq^2}{d^2}[/tex]
where;
k = 9 ×10⁹
[tex]900000 = \dfrac{9*10^9 \times q^2}{1.2^2}[/tex]
[tex]q = \sqrt{\dfrac{900000\times 1.2^2 }{9*10^9}}[/tex]
q = 0.012 C
No of the electrons transferred is:
[tex]= \dfrac{0.012}{1.6\times 10^{-19}} C[/tex]
[tex]= 7.5 \times 10^{16} \ C[/tex]
Initial number of electrons = N × 47 × no of moles
here;
[tex]\text{ no of moles }= \dfrac{6.2}{107.87}[/tex]
no of moles = 0.0575 mol
∴
Initial number of electrons = [tex]6.023\times 10^{23} \times 47 \times 0.0575 mol[/tex]
= 1.63 × 10²⁴
The fraction of electrons transferred [tex]=\dfrac{7.5\times 10^{16} }{1.6 3\times 10^{24}}[/tex]
= 4.60 × 10⁻⁸
Jason throws a basketball straight downward, letting it bounce once before catching it. We can ignore air
resistance
What is true about the acceleration and velocity of the ball on its way up?
Answer:
The acceleration is a negative as the ball is now moving in the opposite direction.
The velocity would decrease as the ball moves upward
Acceleration remains constant and velocity is negative and decreasing as per the given scenario. The correct option is B.
What is acceleration?In mechanics, acceleration is defined as the rate of change of an object's velocity with respect to time.
Vector quantities are accelerations. The orientation of an object's acceleration is determined by the orientation of its net force.
Velocity is the directional speed of a moving object as an indication of its rate of change in position as observed from a specific frame of reference and measured by a specific time standard.
The rate of displacement of an entity known as its velocity. It is measured in meters per second. The rate of change in velocity of an object is defined as acceleration.
According to the scenario, acceleration remains constant while velocity decreases and is negative.
Thus, the correct option is B.
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Your question seems incomplete, the missing options are:
Acceleration increases and velocity is negative and decreasing.Acceleration remains constant and velocity is negative and decreasing,Acceleration decreases and velocity is positive and increasing.Acceleration remains constant and velocity is negative and increasing.a system absorb 500 J of heat and the same time 400J of work is done one the system find change in internal enery ?
Answer:
+ 900 J
Explanation:
Since the total energy change ΔE = internal energy change ΔU since there is no change in kinetic and potential energy,
ΔE = ΔU
ΔE = Q - W where Q = heat absorbed by system and W = work done by system
Now since the system absorbs 500 J of heat, Q = + 500 J and work of 400 J is done on the system, W = -400 J
So, the values of the variables into the equation, we have
ΔE = Q - W
ΔE = + 500 J - (-400 J)
ΔE = + 500 J + 400 J
ΔE = + 900 J
So, the internal energy change, ΔE = + 900 J
The total mass of the wheelbarrow and the road is 80 kg calculate the weight of the wheelbarrow and the road
Force required to lift the wheelbarrow
Answer:
The weight of the wheelbarrow and the road is 784 N and the force required to lift the wheelbarrow is 784 N.
Explanation:
Given that,
The total mass of the wheelbarrow and the road is 80 kg.
The weight of an object is given by :
W = mg
where
g is acceleration due to gravity
So,
W = 80 × 9.8
= 784 N
So, the force required to lift the wheelbarrow is equal to its weight i.e. 784 N.
A rock is thrown from the edge of the top of a 51 m tall building at some unknown angle above the horizontal. The rock strikes the ground a horizontal distance of 74 m from the base of the building 8 s after being thrown. Assume that the ground is level and that the side of the building is vertical. Determine the speed with which the rock was thrown.
Answer:
The speed of projection is 34 m/s.
Explanation:
Height of building, h = 51 m
horizontal distance, d = 74 m
time, t = 8 s
Let the angle is A and the speed is u.
d = u cos A x t
74 = u cos A x 8
u cos A = 9.25 .... (1)
Use second equation of motion
[tex]h = u sin A t - 0.5 gt^2\\\\-51 = u sinA \times 8 - 0.5\times 9.8\times8\times 8\\\\u sin A = 32.8 .... (2)[/tex]
Squaring and adding both the equations
[tex]u^2 = 9.25^2 + 32.8^2 \\\\u = 34 m/s[/tex]
Choose the FALSE statements. In Simple harmonic motion,
I. The velocity of the object does not change at all position
II. The acceleration of the object does not change at all position.
Ill. When velocity is zero, acceleration is also zero
IV. The velocity has maximum magnitude at the equilibrium position.
V. When the net force is maximum, the velocity is zero.
A. I and II
B. III and IV
C. IV and V
D. I, II and III
E. I, II, III and IV
F. I, II, III and V
G. All the above statements are false.
Answer:
b is correct.
Explanation:
because of the question you have given
The coefficent of static friction between the floor of a truck and a box resting on it is 0.38. The truck is traveling at 87.9 km/hr. What is the least distance in which the truck can stop and ensure that the box does not slide?
Answer:
[tex]d=79.9m[/tex]
Explanation:
From the question we are told that:
coefficient of static friction [tex]\mu=0.38[/tex]
Velocity [tex]v=87.9=>24.41667m/s[/tex]
Generally the equation for Conservation of energy is mathematically given by
[tex]\mu*mgd = 0.5 m v^2[/tex]
[tex]d=\frac{0.5*24.42^2}{0.38*9.8}[/tex]
[tex]d=79.9m[/tex]
An object 1.00cm high is placed 18.0cm from a converging lens, forms a real Image 2.00cm high Calculate the forcal length of to the lens
Answer:
focal length=12cm
Explanation:
object size is equal to 1.00cm
object distance = 18cm
heigh of image = 2.00cm
image distance = ??
but magnification is given by;
M = 2.00/1.00 = 2
but u/v = M
u/18 = 2
u = 36
1/f = 1/u+1/v
1/f = 1/18+ 1/36
1/f = 1/12
f = 12cm
The human ear can respond to an extremely large range of intensities - the quietest sound the ear can hear is smaller than 10-20 times the threshold which causes damage after brief exposure. If you could measure distances over the same range with a single instrument, and the smallest distance you could measure was 1 mm, what would the largest be, in kilometers?
Answer:
the largest distance we can measure is 10¹⁴ km
Explanation:
Given the data in the question;
Threshold hearing = 10⁻²⁰
smallest distance measured = 1 mm
Largest distance measured will be;
⇒ ( threshold hearing )⁻¹ × smallest distance
= ( 1 / 10⁻²⁰ ) × 1 mm
= 10²⁰ × 1mm
= 10²⁰ mm
we know that; 1000 mm = 10⁶ km
Largest distance = ( 10²⁰ / 10⁶ ) km
= 10¹⁴ km
Therefore, the largest distance we can measure is 10¹⁴ km
Find the uncertainty in a calculated electrical potential difference from the measurements of current and resistance. Electric potential difference depends on current and resistance according to this function V(I,R) = IR. Your measured current and resistance have the following values and uncertainties I = 5.9 Amps, delta I space equals space 0.4 Amps and R = 42.7 Ohms and delta R space equals space 0.6 Ohms. What is the uncertainty in the , delta V ? Units are not needed in your answer.
Answer:
ΔV = 2 10¹ V
Explanation:
The calculation of the uncertainty or error in an expression is given by
ΔV = [tex]\frac{dV}{di}[/tex] |Δi| + [tex]\frac{dV}{dR}[/tex] |ΔR |
V = i R
let's make the derivatives
[tex]\frac{dV}{di}[/tex] = R
[tex]\frac{dV}{dR}[/tex] = i
we substitute
ΔV = R | Δi | + i | ΔR |
in the exercise give the values
i = (5.9 ± 0.4) A
R = (42.7 ± 0.6) Ω
we calculate
ΔV = 42.7 0.4 + 5.9 0.6
ΔV = 20.6 V
ΔV = 2 10¹ V
the voltage is
V = i R
V = 5.9 42.7
V = 251.9 V
the result is
V = (25 ± 2) 10¹ V
Two resistors of 10 and 15 n are connected. What is their combined resistance if they are connected: a) in series b) in parallel?
Explanation:
Given that,
Two resistors of 10 ohms and 15 ohms are connected.
In series combination, the equivalent resistance is given by :
[tex]R_s=R_1+R_2\\\\R_s=10+15\\\\R_s=25\ \Omega[/tex]
In parallel combination, the equivalent resistance is given by :
[tex]\dfrac{1}{R_p}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\\\\\dfrac{1}{R_p}=\dfrac{1}{10}+\dfrac{1}{15}\\\\R_p=6\ \Omega[/tex]
Hence, this is the required solution.
A factory worker pushes a 32.0 kg crate a distance of 7.0 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and floor is 0.26.
Required:
a. What magnitude of force must the worker apply?
b. How much work is done on the crate by this force?
c. How much work is done on the crate by friction?
d. How much work is done on the crate by the normal force? By gravity?
e. What is the total work done on the crate?
Answer:
(a) 81.54 N
(b) 570.75 J
(c) - 570.75 J
(d) 0 J, 0 J
(e) 0 J
Explanation:
mass of crate, m = 32 kg
distance, s = 7 m
coefficient of friction = 0.26
(a) As it is moving with constant velocity so the force applied is equal to the friction force.
F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N
(b) The work done on the crate
W = F x s = 81.54 x 7 = 570.75 J
(c) Work done by the friction
W' = - W = - 570.75 J
(d) Work done by the normal force
W'' = m g cos 90 = 0 J
Work done by the gravity
Wg = m g cos 90 = 0 J
(e) The total work done is
Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J