Answer:
C. At the highest point of flight.
Explanation:
Given data;
[tex]u_{x} = 4 m/s ,u_{y} = 0, h = 60 m[/tex]
Time of flight,
by third equation of motion ,
[tex]h=\frac{1}{2} gt^{2}[/tex]
[tex]t=\sqrt{\frac{2*60}{9.8} }[/tex]
[tex]t = 3.5 Sec[/tex]
so the range ,
by horizontal component of velocity,([tex]u_{x}[/tex])
R = 4X3.5
R = 14 m
Horizontal component of velocity remain constant in the entire motion.
Can someone write this question clearly and send it to me? Don't just say the answer. Draw and write clearly please
Explanation:
acceleration is weight*gravity
tension is the weight In Newtons
A bowling ball with a mass of 8 kg is moving at a speed of 5 m/s. What is its
kinetic energy?
Answer:
100J
Explanation:
Kinetic energy=1/2mv^2
Kinetic energy=(1/2 x 8)x5^2
Kinetic energy=4x25
Kinetic energy=100
100J
can someone help plz
Answer:
29.15 N
Explanation:
Applying,
Pythagoras theorem,
a² = b²+c²................ Equation 1
Where a = resultant of the two forces, b = first force, c = second force.
From the diagram,
Given: b = 15 N, c = 25 N
Substitute these values into equation 1
a² = 15²+25²
a² = 225+625
a² = 850
a = √850
a = 29.15 N
Hence the resultant of the two forces is 29.15 N
Convert Rev/min to rad/s x 2pie/60?
Anyone knows this please?
Answer:
Thus, [tex]\frac{1 rev}{min} =\frac{2\pi}{60} rad/s[/tex]
Explanation:
The angular speed is defined as the rate of change of angular velocity.
Its SI unit is rad/s and other units are rev/min or rev/s.
[tex]\frac{1 rev}{min } = \frac{1 rev}{60 sec}\\\\1 rev = 2\pi rad\\\\So\\\\\frac{1 rev}{min} = \frac{2\pi}{60} rad/s[/tex]
6. Which of these contain muscles that are not under the mind's control? (Select all that apply.)
Answer:
a c
Explanation:
edu 2021
Which of the following is acceleration toward the center of a circular motion? O A. Centripetal acceleration O B. Uniform circular motion O C. Centrifugal force D. Centripetal force
PLEASE HELP ASAP!!
We call the acceleration of an object moving in uniform circular motion— resulting from a net external force—the centripetal ...
A crucible (container) of molten metal has an open top with an area of 5.000 m^2. The molten metal acts as a blackbody radiator. The intensity spectrum of its radiation peaks at a wavelength of 320 nm. What is the temperature of that blackbody?
Answer:
T = 9056 K
Explanation:
In the exercise they indicate that the body can be approximated by a black body, for which we can use the Wien displacement relation
λ T = 2,898 10⁻³
where lam is the wavelength of the maximum emission
T = 2,898 10⁻³ /λ
let's calculate
T = 2,898 10⁻³ / 320 10⁻⁹
T = 9.056 10³ K
T = 9056 K
Two parallel circular plates with radius carrying equal-magnitude surface charge densities of are separated by a distance of How much stored energy do the plates have? A. 120 B. 360 C. 12 D. 37
Answer:
I guess it is A. I am not sure
A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.6 m and the horizontal range of the ball from the base of the platform is 20 m. What is the horizontal velocity of the ball just before it touches the ground?
Explanation:
the answer is in the above image
6)An electric field of 6 N/C points in the positive X direction. What is the electric flux through a surface that is 4 m2, if its surface normal isin the XY plane and along a line that isinclined at 60 degrees to the positive Y axisand 30 degrees to the positive X axis
Answer:
Flux is 21 Nm^2/C.
Explanation:
Electric field, E = 6 N/C along X axis
Electric filed vector, E = 6 i N/C
Area, A = 4 square meter
Area vector
[tex]\overrightarrow{A} = 4 (cos30 \widehat{i} + sin 30 \widehat{j})\\\\\overrightarrow{A} = 3.5 \widehat{i} + 2 \widehat{j}\\[/tex]
The flux is given by
[tex]\phi= \overrightarrow{E}.\overrightarrow{A}\\\\\phi = 6 \widehat{i} . \left (3.5 \widehat{i} + 2 \widehat{j} \right )\\\\\phi = 21 Nm^2/C[/tex]
For an object with a given mass on Earth, calculate the weight of the object with the mass equal in magnitude to the number representing the day given in part 3 in kilograms using the formula F=W=mg. On the surface of the Earth g=9.8m/s^2
Answer: The weight of the object is 29.4 N
Explanation:
To calculate the weight of the object, we use the equation:
[tex]W=m\times g[/tex]
where,
m = mass of the object = 3 kg
g = acceleration due to gravity = [tex]9.8m/s^2[/tex]
Putting values in above equation, we get:
[tex]W=3kg\times 9.8m/s^2\\\\W=29.4N[/tex]
Hence, the weight of the object is 29.4 N
Two astronauts, each having a mass of 88.0 kg, are connected by a 10.0-m rope of negligible mass. They are isolated in space, moving in circles around the point halfway between them at a speed of 5.40 m/s. Treating the astronauts as particles, calculate each of the following.
a. the magnitude of the angular momentum of the system
b. the rotational energy of the system
c. What is the new angular momentum of the system?
d. What are their new speeds?
e. What is the new rotational energy of the system
Answer:
a) L = 4.75 103 kg m² / s, b) K_total = 2.57 10³ J,
c) L₀ = L_f =4.75 103 kg m² / s, d) K = 1.03 10⁴ J, K = 1.03 10⁴ J
Explanation:
a) the angular momentum is the sum of the angular momentum of each astronaut
the distance is measured from the center of the circle r = 10/2 = 5.0 m
L = 2m v r
L = 2 88.0 5.40 5.0
L = 4.75 103 kg m² / s
b) rotational kinetic energy
K = ½ I w²
As there are two astronauts, the total energy is the sum of the energy of each no.
The moment of inertia of a point mass
I = m r²
I = 88 5²
I = 2.2 10³ kg m²
the angular velocity is given by
v = w r
w = v / r
w = 5.40 / 5
w = 1.08 rad / s
the kinetic energy of the system
K_total = 2 K
K_total = 2 (½ I w²)
K_total = 2.2 10³ 1.08²
K_total = 2.57 10³ J
c, d) as astronauts are isolated in space, these speeds do not change unless there is an interaction between them, for example they approach each other, suppose they reduce their distance by half
r = 2.5 m
I = 88 2.5²
I = 5.5 10² kg m²
for the change in angular velocity let us use the conservation of moment
L₀ = L_f
2Io wo = 2 I w
w = Io / I wo
w = 2.2 10³ / 5.5 10² 1.08
w = 4.32 rad / s
linear velocity is
v = w r
v = 4.32 2.5
K = 1.03 10⁴ J
the kinetic energy of the system is
K = 5.5 10² 4.32²
K = 1.03 10⁴ J
China's GDP passed one trillion USD between Between 2005 and 2010, China's GDP increased by approximately
Answer:
1995 and 2000 , 4 trillions
Explanation:
Answer:
1995 and 2000.4 trillion
Explanation:
Edge 2021
Steve pushes a crate 20 m across a level floor at a constant speed with a force of 200 N, this time on a frictionless floor. The velocity of the crate is in the direction of the force Steve is applying to the crate. What is the net work done on the crate
Answer:
The correct answer is "4000 J".
Explanation:
Given that,
Force,
= 200 N
Displacement,
= 20 m
Now,
The work done will be:
⇒ [tex]Work=Force\times displacement[/tex]
By putting the values, we get
[tex]=200\times 20[/tex]
[tex]=4000 \ J[/tex]
If you warm up the volume of a balloon but keep the pressure the same, you would be using which gas law?
Answer:
Charles law
Explanation:
Charle's law states that the volume (V) of a given gas is directly proportional to the absolute temperature (T) at a constant pressure.
That is;
: V ∝ T
: V/T = K
According to this question, the volume of a balloon is warmed up but the pressure is kept the same. Charles law will be used because it shows the relationship between the volume (V) and the temperature (heat) at a constant pressure (P).
A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and speed (in m/s) when t = 5. f(t) = 18 + 48/t + 1
Answer:
The velocity of the particle = -1.92 m/s
The speed of the particle = 5.72 m/s
Explanation:
Given equation of motion;
[tex]f(t) = 18 \ + \ \frac{48}{t} \ + \ 1[/tex]
Velocity is defined as the change in displacement with time.
[tex]V = \frac{df(t)}{dt} = -\frac{48}{t^2} \\\\at \ t = 5 \ s\\\\V = -\frac{48}{5^2} = \frac{-48}{25} = - 1.92 \ m/s[/tex]
The distance traveled by the particle in 5 s:
[tex]s = f(5) = 18 + \frac{48}{5} + 1\\\\s= 28.6 \ m[/tex]
The speed of the particle when t = 5s
[tex]Speed = \frac{28.6}{5} = 5.72 \ m/s[/tex]
Một chất điểm chuyển động tròn đều, sau 5 giây nó quay được 20 vòng. Chu kì quay của chất điểm là:
Answer:
T=0,25s
Explanation
5s>20vong 1s>4vong
omega= 8pi
omega=2pi/T
A car of mass M traveling with velocity v strikes a car of mass M that is at rest. The two cars’ bodies mesh in the collision. The loss of the kinetic energy the moving car undergo in the collision is
a) a quarter of the initial kinetic energy.
b) half of the initial kinetic energy.
c) all the initial kinetic energy.
d) zero.
Answer:
the correct answer is B
Explanation:
Let's propose the solution of the problem, for this we form a system formed by the two cars, so that the forces during the collision are internal, the momentum is conserved
instantly starts. Before the crash
p₀ = M v +0
final instant. After the crash
m_f = (M + M) v_f
the moment is preserved
p₀ = p_f
M v = 2 M v_f
v_f = v / 2
let's look for kinetic energy
before the crash
K₀ = ½ M v²
after the crash
K_f = ½ 2M (v_f)²
K_f = ½ 2M (v/2)²
K_f = (½ M v²) ½
K_f = K₀ / 2
therefore the correct answer is B
Five air-filled parallel-plate capacitors have the plate areas and plate separations listed below, where A and d are constants. The capacitors are each connected to the same potential difference. Which capacitor stores the greatest amount of energy?
a.
Area: 2A
Separation : d/2
b.
Area: 2A
Separation : 2d
c.
Area: A
Separation : d
d.
Area: A/2
Separation : d/2
e.
Area: A/2
Separation : 2d
Answer:
The answer is "Option A"
Explanation:
Energy stored in capcitor[tex]=\frac{1}{2}\ cv^2[/tex]
For point A:
[tex]C_A=\frac{\varepsilon_0 2A}{\frac{d}{2}}=\frac{4\ \varepsilon_0 A}{d}\\\\[/tex]
For point B:
[tex]C_B=\frac{\varepsilon_0 2A}{2d}=\frac{\varepsilon_0 A}{d}\\\\[/tex]
For point C:
[tex]C_c=\frac{\varepsilon_0 A}{d}\\\\[/tex]
For point D:
[tex]C_D=\frac{\varepsilon_0 A}{2 \frac{d}{2}}=\frac{\varepsilon_0 A}{d}\\\\[/tex]
For point E: [tex]C_E=\frac{\varepsilon_0 A}{2 \times 2d}=\frac{\varepsilon_0 A}{4d}\\\\[/tex]
therefore C_A has the maximum capacitance and max energy same energy is dir proportional to C for the same J
Ocean waves of wavelength 22 m are moving directly toward a concrete barrier wall at 4.6 m/s . The waves reflect from the wall, and the incoming and reflected waves overlap to make a lovely standing wave with an antinode at the wall. (Such waves are a common occurrence in certain places.) A kayaker is bobbing up and down with the water at the first antinode out from the wall.
Required:
a. How far from the wall is she?
b. What is the period of her up-and-down motion?
Answer:
a) [tex]d=11m[/tex]
b) [tex]T=4.68s[/tex]
Explanation:
From the question we are told that:
Wavelength [tex]\lambda=22m[/tex]
Velocity [tex]v=4.6m/s[/tex]
a)
Generally the equation for distance between her and the wall d is mathematically given by
Since
The First Anti node distance is [tex]\frac{\lambda}{2}[/tex]
Therefore
[tex]d= \frac{\22}{2}[/tex]
[tex]d=11m[/tex]
b)
Generally the equation for her up-and-down motion is mathematically given by
[tex]T=\frac{22}{4.7}[/tex]
[tex]T=4.68s[/tex]
two groups of students carry out experiments to investigate the relationships between force and extenstion
Answer:
They investigated Hooke's law.
Explanation:
[tex]{ \tt{force \: \alpha \: extension }} \\ { \boxed{ \bf{F = kx}}} \\ x \: is \: extension \\ F \: is \: force \\ k \: is \: constant : k = 1[/tex]
a soap bubble was slowly enlarged from radius 4cm to 6cm and amount of work necessary for enlargement is 1.5 *10 calculate the surface tension of soap bubble joules
Answer:
The surface tension is 190.2 N/m.
Explanation:
Initial radius, r = 4 cm
final radius, r' = 6 cm
Work doen, W = 15 J
Let the surface tension is T.
The work done is given by
W = Surface Tension x change in surface area
[tex]15 = T \times 4\pi^2(r'^2 - r^2)\\\\15 = T \times 4 \times 3.14\times 3.14 (0.06^2- 0.04^2)\\\\15 = T\times 0.0788\\\\T = 190.2 N/m[/tex]
You are packing for a trip to another star, and on your journey you will be traveling at a speed of 0.99c. Can you sleep in a smaller cabin than usual, because you will be shorter when you lie down? Explain your answer.
Explanation:
No. From your own reference frame, nothing will change. Everything will look the same to you even if you're traveling at 99% speed of light. However, an outside observer will see you shrink to about 14% of your original length along the direction of your motion according to the Lorentz contraction predicted by special relativity:
[tex]L=L_{0} \sqrt{1 - \frac{v^2 }{c^2 }}[/tex]
What is 1 second….???? Give a meaningful answer…..
Explanation:
.....................................
Answer:
1 second is defined as 1/86400th part of a mean solar day.
In a 2-dimensional Cartesian coordinate system the y-component of a given vector is equal to that vector's magnitude multiplied by which trigonometric function, with respect to the angle between vector and y-axis?
a. sine
b. cosine
c. tangent
d. cotangent
Answer:
Option b, cosine.
Explanation:
Below you can see an image that illustrates this situation.
Remember that for a triangle rectangle with a given angle θ, we have:
Cos(θ) = (adjacent cathetus)/(hypotenuse)
In the image, you can see a vector of magnitude M, and the angle θ defined between the vector and the positive y-axis.
In this case, the y-component is the adjacent cathetus and the hypotenuse is the magnitude of the vector.
Then we will have:
Cos(θ) = (adjacent cathetus)/(hypotenuse) = y/M
solving that for y, we get:
y = Cos(θ)*M
Then the y-component is the vector's magnitude multiplied by the cosine of the angle between the vector and the y-axis.
The correct option is b.
Answer:
(b) cosine
Explanation:
In a 2-dimensional Cartesian coordinate system, a vector has a x-component and/or a y-component. To get these components, the magnitude of the vector is resolved with respect to the x-axis and the y-axis by multiplying it (the magnitude) by some trigonometric function with respect to the angle between the vector and the x or y axis.
For example, given a vector A of magnitude A which makes an angle α with the x-axis and an angle β with the y-axis, the x and y components of the vector A can be found as follows;
i. x-component is given by [tex]A_{x}[/tex]
[tex]A_{x}[/tex] = A cos α (with respect to the angle between A and the x-axis) or
[tex]A_{x}[/tex] = A sin β (with respect to the angle between A and the y-axis)
ii. y-component is given by [tex]A_{y}[/tex]
[tex]A_{y}[/tex] = A sin α (with respect to the angle between A and the x-axis) or
[tex]A_{y}[/tex] = A cos β (with respect to the angle between A and the y-axis)
Therefore, the y-component of a vector in a 2-dimensional Cartesian coordinate is given by the product of the magnitude of the vector and the cosine of the angle between the vector and the y-axis.
A puck moves 2.35 m/s in a -22.0 direction. A hockey stick pushes it for 0.215 s, changing its velocity to 6.42 m/s in a 50 degree direction. What was the direction of the acceleration?
Answer:
48.9 is the answer I think !
Answer:
28.4
Explanation:
A conducting sphere of radius R carries an excess positive charge and is very far from any other charges. Draw the graphs that best illustrates the potential (relative to infinity) produced by this sphere as a function of the distance r from the center of the sphere?
Answer:
See annex
Explanation:
By convention potential at ∞ V(∞ ) = 0
As the distance from the sphere decreases the potential increases up to the point d = R ( R is the radius of the sphere. That potential remains constant while d = R and becomes 0 inside the sphere where there is not free charges and therefore the electric field is 0 and so is the potential.
I am sorry I could not make a better graph
The graph that best illustrates the potential (relative to infinity) produced by this sphere as a function of the distance r from the center of the sphere is attached as an image below
[tex]V = \frac{KQ}{R}[/tex]
for r <= R
[tex]V = \frac{KQ}{r}[/tex]
for r > R
Therefore the graph will be
For more information on potentials as function of distance
https://brainly.com/question/24146175?referrer=searchResults
A 50-kg copper block initially at 140°C is dropped into an insulated tank that contains 90 L of water at 10°C. Determine the final equilibrium tempera
Answer:
16.33°C
Explanation:
Applying,
Heat lost by copper = heat gained by water
cm(t₁-t₃) = c'm'(t₃-t₂).............. Equation 1
Where c = specific heat capacity of copper, m = mass of copper, c' = specific heat capacity of water, m' = mass of water, t₁ = initial temperature of copper, t₂ = initial temperature of water, t₃ = final equilibrium temperature.
From the question,
Given: m = 50 kg, t₁ = 140°C, m' = 90 L = 90 kg, t₂ = 10°C
Constant: c = 385 J/kg°C, c' = 4200J/kg°C
Substitute these values into equation 1
50(385)(140-t₃) = 90(4200)(t₃-10)
(140-t₃) = 378000(t₃-10)/19250
(140-t₃) = 19.64(t₃-10)
140-t₃ = 19.64t₃-196.6
19.64t₃+t₃ = 196.4+140
20.64t₃ = 336,4
t₃ = 336.4/20.6
t₃ = 16.33°C
A 1000-kg car rolling on a smooth horizontal surface ( no friction) has speed of 20 m/s when it strikes a horizontal spring and is brought to rest in a distance of 2 m What is the spring’s stiffness constant?
Explanation:
kinetic energy was converted to potential energy in the spring.
the answer is in the above image
A 1.50-V battery supplies 0.414 W of power to a small flashlight. If the battery moves 4.93 1020 electrons between its terminals during the time the flashlight is in operation, how long was the flashlight used?
Answer:
2.86×10⁻¹⁸ seconds
Explanation:
Applying,
P = VI................ Equation 1
Where P = Power, V = Voltage, I = Current.
make I the subject of the equation
I = P/V................ Equation 2
From the question,
Given: P = 0.414 W, V = 1.50 V
Substitute into equation 2
I = 0.414/1.50
I = 0.276 A
Also,
Q = It............... Equation 3
Where Q = amount of charge, t = time
make t the subject of the equation
t = Q/I.................. Equation 4
From the question,
4.931020 electrons has a charge of (4.931020×1.6020×10⁻¹⁹) coulombs
Q = 7.899×10⁻¹⁹ C
Substitute these value into equation 4
t = 7.899×10⁻¹⁹/0.276
t = 2.86×10⁻¹⁸ seconds