Answer:
27.44 J
Explanation:
We can find the energy at the top of the slide by using the potential energy equation:
PE = mghAt the top of the slide, the swimmer has 0 kinetic energy and maximum potential energy.
The swimmer's mass is given as 7.00 kg.
The acceleration due to gravity is 9.8 m/s².
The (vertical) height of the water slide is 0.40 m.
Substitute these values into the potential energy equation:
PE = (7.00)(9.8)(0.40) PE = 27.44Since there is 0 kinetic energy at the top of the slide, the total energy present is the swimmer's potential energy.
Therefore, the answer is 27.44 J of energy when the swimmer is at the top of the slide.
An astronaut weighing 248 lbs on Earth is on a mission to the Moon and Mars. (a) What would he weigh in newtons when he is on the Moon
Answer:
[tex]W_m=183.495N[/tex]
Explanation:
From the question we are told that:
Weight [tex]W=248Ibs[/tex]
Mass of Weight [tex]m= 248*0.453[/tex]
[tex]m=112.344kg[/tex]
Generally the acceleration due to gravity on the Moon is one-sixth that on Earth.
Therefore
The equation for Weight on Moon is given as
[tex]W_m=m*g/6[/tex]
[tex]W_m=122.344*\frac{9.8}{6}[/tex]
[tex]W_m=183.495N[/tex]
Answer:
F ’= 40.9 lb
Explanation:
The weight of a body is the attraction of the Earth on the body
F = [tex]G \frac{m M_e}{R_e^2}[/tex]
F = mg
m = F / g
m = 248/32
m = 7.75 slug
The weight of the body on the moon is the attraction of the body for the satellite
F ’= [tex]G \frac{mM}{R^2}[/tex]
from the tables the mass and radius of the moon are M = 7.34 10²² kg and R = 1.74 10⁶ m
let's reduce the mass to the SI system
m = 7.75 slug (14.59 kg / 1 slug) = 113 kg
F ’= 6.67 10⁻¹¹ 113 7.34 10²² / (1.74 10⁶) ²
F ’= 1.82 10² N
F ’= 1.82 10² N (0.2248 lb / 1 N)
F ’= 40.9 lb
A position vector in the first quadrant has an x-component of 6 m and a magnitude of 10 m. What is the value of its y-component
Answer:
8m
Explanation:
The magnitude m of a vector (x, y) is given by
m = [tex]\sqrt{x^2 + y^2}[/tex] -------------------------(i)
where;
x and y are the x- and y- components of the vector.
From the question;
m = 10m
x = 6m
Substitute these values into equation (i) as follows;
10 = [tex]\sqrt{6^2 + y^2}[/tex]
Solve for y;
Find the square of both sides
10² = 6² + y²
100 = 36 + y²
y² = 100 - 36
y² = 64
y = √64
y = 8
Therefore, the y-component of the position vector is 8m
convert 37 microgram to Gigagram
Answer:
3.7e-14
Explanation:
mark me brainliest!!
Ftension = 120 N
10 kg
Fg
What is the weight (not mass) of the box?
O 10 kg
0 19 N
98 N
O 98 kg
PLZ HELP
Answer:
98N
Explanation:
El peso se mide en kg y la fuerza no afecta
A horizontal force of 40N is needed to pull a 60kg box across the horizontal floor at which coefficient of friction between floor and box? Determine it to three significant figures even through that's quite unrealistic. How much work is done in overcoming friction between the object and floor if the box slides 8m along horizontally on the floor?
Answer:
Coefficient of friction is [tex]0.068[/tex].
Work done is [tex]320~J[/tex].
Explanation:
Given:
Mass of the box ([tex]m[/tex]): [tex]60[/tex] kg
Force needed ([tex]F[/tex]): [tex]40[/tex] N
The formula to calculate the coefficient of friction between the floor and the box is given by
[tex]F=\mu mg...................(1)[/tex]
Here, [tex]\mu[/tex] is the coefficient of friction and [tex]g[/tex] is the acceleration due to gravity.
Substitute [tex]40[/tex] N for [tex]F[/tex], [tex]60[/tex] kg for [tex]m[/tex] and [tex]9.80[/tex] m/s² for [tex]g[/tex] into equation (1) and solve to calculate the value of the coefficient of friction.
[tex]40 N=\mu\times60 kg\times9.80 m/s^{2} \\~~~~~\mu=\frac{40 N}{60 kg\times9.80 m/s^{2}}\\~~~~~~~=0.068[/tex]
The formula to calculate the work done in overcoming the friction is given by
[tex]W=Fd..........................(2)[/tex]
Here, [tex]W[/tex] is the work done and [tex]d[/tex] is the distance travelled.
Substitute [tex]40[/tex] N for [tex]F[/tex] and [tex]8 m[/tex] for [tex]d[/tex] into equation (2) to calculate the work done.
[tex]W=40~N\times8~m\\~~~~= 320~J[/tex]
A proton moves across a magnetic field and feels a force. If an electron were to
move at the same speed in the same direction across the same magnetic field, the
force
O 1) would be smaller in the opposite direction
O2) would be the same magnitude in the same direction
3) would be the same magnitude in the opposite direction
4) would be larger in the opposite direction
5) would be smaller in the same direction
potang ina mooooooo bubu hayop kaaaaapestrng yawa
Explanation:
peste kakkkkaaaaaaaa bubu pesteng yawaaaayaka kaayu kaaaaaaa
Allison pulls a sled up a hill, which has an incline of 20 degrees to the horizontal. Of the sled has a mass of 20 kg, and the coefficient of kinetic friction between the sled and the ground is 0.1, what is the minimum force Allison must apply to the sled to keep it moving forward at a constant speed
Answer:
[tex]F=48.62N[/tex]
Explanation:
From the question we are told that:
Angle [tex]\theta= 20[/tex]
Mass [tex]m= 20kg[/tex]
Coefficient of kinetic friction [tex]\mu=0.1[/tex]
Generally the equation for Force Required to jeep sled moving is mathematically given by
[tex]F= mg sin \theta - \mu N[/tex]
Where N is normal force
[tex]F_N=Wcos\theta[/tex]
[tex]F_N=20*9.8*cos 20[/tex]
[tex]F_N=184.18N[/tex]
Therefore
[tex]F= (20*9.8) sin 20 - (0.1) (184.18)[/tex]
[tex]F=48.62N[/tex]
A solid piece of clear transparent materialhas an index of refraction of 1.61. If you place itinto a clear transparentsolution and it seems to disappear, approximately what is the index of refraction of the solution
Answer: The index of refraction of a solution is 1.61
Explanation:
Refractive index or index of refraction is defined as the measure of bending of light when passed through a medium. It is simply the ratio of the speed of light in a vacuum and the speed of light in the phase.
We are given:
Index of refraction of a material = 1.61
When the material is put in a solution, it disappears. This means that there is no bending of light to distinguish between the two phases (material and solution). Thus, the refractive index of the solution is equal to the refractive index of the material.
Index of refraction of a solution = Index of refraction of a material = 1.61
Hence, the index of refraction of a solution is 1.61
1. What did you observe about the magnitudes of the forces on the two charges? Were they the same or different? Does your answer depend on whether the charges were of the same magnitude or different? How does this relate to Newton’s 3rd law?
Answer:
Following are the solution to the given question:
Explanation:
Its strength from both charges is equivalent or identical. The power is equal. And it is passed down
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
Therefore, the extent doesn't rely on the fact that charges are the same or different. Newton's third law complies with Electrostatic Charges due to a couple of charges. They are similar in magnitude, and they're in the other way.
[tex]|F_{12}| = |F_{21}|[/tex]
An object travels down a ramp at a constant acceleration. The object experiences a force of friction and a gravitational force. Which of the following could be true about the motion of the object?
a. The force of friction acts in the same direction as the object travels.
b. The force of friction between the surface and the object is less than the component of the gravitational force that is parallel to the ramp.
c. If the object increased in mass, the object's acceleration would change.
d. If the object increased in mass, the normal force exerted on the object would remain the same.
Answer:
Option (b) is correct.
Explanation:
An object travels down a ramp at a constant acceleration.
Friction force and the gravitational force is there.
(a). It is false. As the friction acts in the opposite direction of motion of the object.
(b). It is true. As the object moves down the ramp so the force of friction is less the component of gravitational force along the ramp.
(c). It is false. the acceleration does not depends on the mass.
a = g sin A - u g cos A
where, A is the angle of inclination and u is the coefficient of friction.
(d). It is false. The normal force is given by
N = m g cos A
so it depends on the mass.
With the frequency set at the mid-point of the slider and the amplitude set at the mid-point of the slider, approximately how many grid marks is the wavelength of the wave (use the pause button and step button as you need to in order to get a good measure, and round to the nearest whole grid mark)?
If the amplitude is increased the wavelength:___________.
a. decreases
b. stays the same
c. increases
Answer:
correct answer is b
Explanation:
The frequency of a wave depends on the properties of medium density and the elasticity properties change the amplitude depends on the energy carried by the wave, that is, the amplitude is proportional to the height of the wave (oscillation).
Consequently the amplitude of independent of the frequency because it depends on different factors.
Therefore when changing the amplitude the wavelength remains constant
the correct answer is b
A particle of mass M moves along a straight line with initial speed vi. A force of magnitude F pushes the particle a distance D along the direction of its motion.
Required:
By what multiplicative factor RK does the initial kinetic energy increase, and by what multiplicative factor RWdoes the work done by the force increase (with respect to the case when the particle had a mass M)?
Solution :
From the Newton's second law of motion :
F = ma
[tex]a=\frac{F}{m}[/tex]
[tex]$\frac{dv}{dt}=\frac{F}{m}$[/tex]
[tex]$\left(\frac{dv}{ds} \times \frac{ds}{dt}\right)=\frac{F}{m}$[/tex]
[tex]$v \frac{dv}{ds} = \frac{F}{m}$[/tex]
[tex]$v dv =\frac{F}{m}\ ds$[/tex]
Integrating above the expression by applying the limits :
[tex]$\int_{v_i}^{v_f} v \ dv= \frac{F}{m} \int_0^s ds$[/tex]
Here the diameter is s= D
[tex]$\frac{v_f^2 - v_i^2}{2}=\frac{FD}{m}$[/tex]
The final speed of the particle after travelling distance D is
[tex]$v_f = \sqrt{v_i^2 + \frac{2FD}{m}}$[/tex]
The kinetic energy of the particle of mass M is :
[tex]$K_1=\frac{1}{2}Mv^2$[/tex]
For M = 3M
[tex]$K_2=\frac{1}{2}(3M)v^2$[/tex]
[tex]$=3(K_1)$[/tex]
Thus the kinetic energy increases by a factor of 3.
The work done depends on the factor and the displacement of the body. Thus, the work done remains same even though the mass increases. Hence the work down increases by factor 1.
which equation describes the sum of the vectors plotted below
Answer:
i think it could be d
Explanation:
I'm not sure
The sum of the vectors plotted below describes by 2x i + 4y j. Hence, option (B) is correct.
What is vector?Physical quantities that exhibit both magnitude and direction are referred to as vector quantities. As an illustration, consider displacement, force, torque, momentum, acceleration, and velocity.
It is more complicated to add vectors than it is to add scalars. It is impossible to simply add two vectors together to get their sum since vectors include both magnitude and direction. The Triangle law is used to calculate the vector addition.
According to the figure:
the sum of the vectors is position vector of second vector, that is, r = 2x i + 4y j.
Learn more about vector here:
https://brainly.com/question/29740341
#SPJ2
which wave characteristic is the same for all electromagnetic waves traveling through a vacuum?
Answer:
All waves travel at the same speed c.
How much power is required to light a lightbulb at 100V of voltage when the lightbulb has a resistance of 500 Ohms?
Answer:
20 Watt
Explanation:
Applying
P = V²/R................... Equation 1
Where V = Voltage, P = Power, R = resistance
From the question,
Given: V = 100V, R = 500 ohms
Substitute these values into equation 1
P = (100²)/500
P = 10000/500
P = 20 Watt.
Hence the power required to light the bulb is 20 W
in a Mercury thermometer the level of Mercury Rises when its bulb comes in contact with a hot object what is the reason for this rise in the level of Mercury
Answer:
because thermometric liquid readily expands on heating or contracts on cooling even for a small difference in the temperature of the body.
Find the acceleration due to gravity on the surface of a planet with a mass of 3.5 * 10^24 kg and an average radius of 4.5 * 10^6 m.
A truck is driving over a scale at a weight station. When the front wheels drive over the scale, the scale reads 5800 N. When the rear wheels drive over the scale, it reads 6500 N. The distance between the front and rear wheels is 3.20 m Determine the distance between the front wheels and the truck's center of gravity.
Answer:
[tex]x_2=1.60m[/tex]
Explanation:
From the Question We are told that
Initial Force [tex]F_1=5800N[/tex]
Final Force [tex]F_2=6500N[/tex]
Distance between the front and rear wheels \triangle x=3.20 m
Since
[tex]\triangle x=3.20 m[/tex]
Therefore
[tex]x_1+x_2=3.20[/tex]
[tex]x_1=3.20-x_2[/tex]
Generally the equation for The center of mass is at x_2 is mathematically
given by
[tex]x_2 =\frac{(F_1x_1+F_2x_2)}{(F_1+F_2)}[/tex]
[tex]x_2=3.20F_1-\frac{x_2F_1+F_2x_2}{(F_1+F_2)}[/tex]
[tex]2*F_1*x_2 =3.20F_1[/tex]
[tex]x_2=1.60m[/tex]
Center of gravity of a body is the sum of its moments divided by the overall weight of the object. The distance between the front wheels and the truck's center of gravity is 1.6 meters.
Given-
Scale reading value when the front wheels drive over the scale [tex]m_{1}[/tex] is 5800 N.
Scale reading value when the rear wheels drive over the scale [tex]m_{2}[/tex] is 6500 N
Distance between the front and rear wheel [tex]\bigtriangleup x[/tex] is 3.20 meters.
Let, the distance between the front wheels and the truck's center of gravity is [tex]x_{2}[/tex].
Since sum of the distance between front wheel to truck's center of gravity [tex]x_{1}[/tex], and rear wheel to truck's center of gravity [tex]x_{2}[/tex], is equal to the distance between the front and rear wheel [tex]\bigtriangleup x[/tex]. Therefore,
[tex]\bigtriangleup x=x_{1} +x_{2}[/tex]
[tex]3.20=x_{1} +x_{2}[/tex]
[tex]x_{1} =3.20-x_{2}[/tex]
For the distance between the front wheels and the truck's center of gravity is the formula of center of gravity can be written as,
[tex]x_{2} =\dfrac{m_{1}x_{1}+m_{2} x_{2} }{m_{1} +m_{2} }[/tex]
[tex]x_{2} =\dfrac{5800\times (3.20- x_{2})+6500\times x_{2} }{5800 +6500 }[/tex]
[tex]1230 x_{2} ={18560-5800 x_{2}+6500 x_{2} }[/tex]
[tex]x_{2}= 1.6[/tex]
Hence, the distance between the front wheels and the truck's center of gravity is 1.6 meters.
For more about the center of gravity, follow the link below-
https://brainly.com/question/20662119
150 0.0000
2. Use the distance from the previous problem to calculate how long it takes for light to travel from the
Sun to Earth.
300,000,000X 15000,000,
Answer:
t = 5 10² s
Explanation:
Light is an electromagnetic wave that travels at a constant speed of 3 10⁸ m/s in vacuum.
The distance from the Sun to the Earth is 150 10⁶ km
Let's reduce the distance to the SI system
x = 150 10⁶ km (10³ m / 1 km) = 150 10⁹ m
Since the velocity is constant, we can use the uniform motion relations
v = x / t
t = x / v
let's calculate
t = 150 10⁹ / 3 10⁸
t = 5 10² s
An ideal spring is used to stop blocks as they slide along a table without friction. A 0.80 kg block traveling at a speed of 2.2 m/s can be stopped over a distance of 0.11 m once it makes contact with the spring.
A rectangular block on a level surface moves at velocity v to the right towards a spring that rests on the surface and is attached to a fixed mount on the right.
What distance would a 1.40 kg block travel after making contact with the spring if the block is traveling at a speed of 3.0 m/s before it makes contact with the spring?
Use the work-energy theorem: the total work done on the 0.80 kg block by the spring to make it come to a stop is equal to the change in the block's kinetic energy.
If we take the block's initial direction of motion to be positive, then the spring does negative work on the block, and
-1/2 k (0.11 m)² = 0 - 1/2 (0.80 kg) (2.2 m/s)²
Solve for the spring constant k :
k = (1/2 (0.80 kg) (2.2 m/s)²) / (1/2 (0.11 m)²) = 320 N/m
We can use the same equation as above to find the distance the 1.40 kg block would travel as it is slowed down by the same spring:
-1/2 (320 N/m) x ² = 0 - 1/2 (1.40 kg) (3.0 m/s)²
Solve for the displacement x :
x = √((1/2 (1.40 kg) (3.0 m/s)²) / (1/2 (320 N/m))) ≈ 0.20 m
Which statement describe. good hypothesis ? Check all that apply
How do we get heat on Earth? Does thermal energy travel directly from the sun?
A wave with a frequency of 60 hertz would generate 60 wave crests every
Answer:
A wave with a frequency of 60 hertz will have 60 successive wave crests occurring every second
Explanation:
A crest is the highest point the medium rises to in a wave. That means that the crest is a point on the wave where the displacement is at a maximum.
Frequency is the number of oscillations of a wave in one second. Hence, frequency of a wave is the number of successive crests occurring in a second.
Therefore a wave with a frequency of 60 hertz will have 60 successive wave crests occurring every second.
How could extreme heat (resulting from Climate Change) affect human and
animal life?
Answer: See explanation
Explanation:
Climate change, is also referred to as global warming, and it simply means the rise in Earth's average surface temperature.
Effects of climate change include rising sea levels, heat waves, drought, storms, etc.
Extreme heat events is dangerous to the health of both animals and humans. For human beings, it can bring about increase in heat- related illness, weakness, heat stroke and excessive water consumption.
For animals, it can lead to struggling of the animals in losing their excess body heat by evaporation. Other effects include panting, loss of appetite, increased drinking, difficulty breathing, anxious behaviour, and weakness.
a 250 mH coil of negligible resistance is connected to an AC circuit in which as effective current of 5 mA is flowing. if the frequency is 850Hz, find the inductive reactance
Answer:
the inductive reactance of the coil is 1335.35 Ω
Explanation:
Given;
inductance of the coil, L = 250 mH = 0.25 H
effective current through the coil, I = 5 mA
frequency of the coil, f = 850 Hz
The inductive reactance of the coil is calculated as;
[tex]X_l = \omega L = 2\pi f L\\\\X_l = 2\pi \times 850 \times 0.25\\\\X_l = 1335.35 \ ohms[/tex]
Therefore, the inductive reactance of the coil is 1335.35 Ω
They realize there is a thin film of oil on the surface of the puddle. If the index of refraction of the oil is 1.81, and they observe the reflected light in the air to be orange (wavelength of 600 nm), what is the thickness of the film of oil in nanometers? Give your answer with one decimal place please.
Answer:
The right solution is "165.8 nm".
Explanation:
Given:
Index of refraction,
n = 1.81
Wavelength,
λ = 600 nm
We know that,
⇒ [tex]t=\frac{\lambda}{2\times n}[/tex]
By putting the values, we get
[tex]=\frac{600}{2\times 1.81}[/tex]
[tex]=165.8 \ nm[/tex]
14) A magnetic force acts most strongly on a current carrying wire when it
A) carries a very large current.
B) is perpendicular to the magnetic field.
C) either or both of these
D) none of the above
omring wire the wire
Answer:
B) is perpendicular to the magnetic field
A concert loudspeaker suspended high off the ground emits 26.0 W of sound power. A small microphone with a 1.00 cm2 area is 53.0 m from the speaker. Part A What is the sound intensity at the position of the microphone
Answer:
the sound intensity at the position of the microphone is 7.4 × 10⁻⁴ W/m²
Explanation:
Given the data in the question;
Sound power P = 26.0 W
Area of microphone A = 1.00 cm²
Radius r = 53.0 m
sound intensity at the position of the microphone = ?
Now, intensity at the position of the microphone can be determined using the following expression;
[tex]I[/tex] = P / 4πr²
We substitute
[tex]I[/tex] = 26.0 / ( 4 × π × (53.0 )² )
[tex]I[/tex] = 26.0 / ( 4 × π × 2809 )
[tex]I[/tex] = 26.0 / 35298.935
[tex]I[/tex] = 26.0 / ( 4 × π × (53.0 )² )
[tex]I[/tex] = 0.000736566
[tex]I[/tex] = 7.4 × 10⁻⁴ W/m²
Therefore, the sound intensity at the position of the microphone is 7.4 × 10⁻⁴ W/m²
acceleration of a refrigerator 30s after a person begins pushing it at a force of 400 N
Question: Find acceleration of a refrigerator 30s after a person begins pushing it at a force of 400 N, If the mass of the refrigerator is 10 kg.
Answer:
40 m/s²
Explanation:
Applying,
F = ma................Equation 1
Where F = Force applied to the refrigerator, m = mass of the refrigerator, a = acceleration of the refrigerator.
make a the subject of the equation
a = F/m............ Equation 2
From the question,
Given: F = 400 N, m = 10 kg
Substitute these values into equation 2
a = 400/10
a = 40 m/s²
A thermodynamic system consists of an ideal gas at a volume of 3.50 L and initial pressure of 6.2 × 104 Pa. As the volume is held constant, the pressure is increased to 8.2 × 104 Pa. What work is involved in this process?
Answer:
0 J
Explanation:
Since work done W = PΔV where P = pressure and ΔV = change in volume.
Since the volume is constant, ΔV = 0
So, Work done, W = PΔV = P × 0 = 0 J
So, the work done is 0 J.
