Answer:
she should write about how big is it and what the sun looks and how far away is it from earth.
I need help please will mark brainliest
Answer: IT B TRUST ME MANN OK
Explanation: OK BYE TRUST
3. A 10-centimeter diameter solid sphere made of a conducting material has 10 micro-Coulombs of charge placed upon it. What is the potential difference between a point on one side of the sphere to a point on the exact opposite side of the sphere
Answer:
zero
Explanation:
For a solid conducting sphere, charges are present on the surface of the sphere due to a phenomenon known as electrostatic sheilding. This affects the charge present in the body and makes it zero. However, the electrostatic potential appears to be equal to the whole present point that shows on the surface. The surface of a spherical conducting solid sphere is known as an equipotential surface. Thus, the potential difference between the two opposite points on the surface of the sphere will also be zero.
Which mode of kinetic energy contributes to temperature?
Answer:
Kinetic energy Temperature
Explanation:
What is the speed of a cyclist that rides west 88 km in 32 minutes?
Answer:
The speed of the cyclist is 2.75 km/min.
Explanation:
Given
The distance d = 88 km Time t = 32 minutesTo determine
We need to find the speed of a cyclist.
In order to determine the speed of a cyclist, all we need to do is to divide the distance covered by a cyclist by the time taken to cover the distance.
Using the formula involving speed, time, and distance
[tex]s=\frac{d}{t}[/tex]
where
s = speed d = distance covered t = time takensubstitute d = 88, and t = 32 in the formula
[tex]s=\frac{d}{t}[/tex]
[tex]s=\frac{88}{32}[/tex]
Cancel the common factor 8
[tex]s=\frac{11}{4}[/tex]
[tex]s=2.75[/tex] km/min
Therefore, the speed of the cyclist is 2.75 km/min.
A person stands on the ball of one foot. The normal force due to the ground pushing up on the ball of the foot has magnitude 750 N. Ignore the weight of the foot itself. The other significant forces acting on the foot are the tension in the Achilles tendon pulling up and the force of the tibia pushing down on the ankle joint. If the tension in the Achilles tendon is 2225 N, what is the force exerted on the foot by the tibia
Answer:
the force exerted on the foot by the tibia would be 2975 N
Explanation:
Given the data in the question;
To maintain equilibrium between the foot and the ball vertically, the addition normal normal force [tex]N^>[/tex] (750 N) and the tension in the Achilles tendon [tex]F^>_{Achilles}[/tex] (2225 N) must be equal to the force exerted on the foot by the tibia;
so
| [tex]N^>[/tex] | + |[tex]F^>_{Achilles}[/tex] | = | [tex]F^>_{Tibia}[/tex] |
so force exerted on the foot by the tibia will be;
| [tex]F^>_{Tibia}[/tex] | = |[tex]N^>[/tex] | + |[tex]F^>_{Achilles}[/tex] |
so we substitute IN OUR VALUES
| [tex]F^>_{Tibia}[/tex] | = 750 N + 2225 N
| [tex]F^>_{Tibia}[/tex] | = 2975 N
Therefore, the force exerted on the foot by the tibia would be 2975 N
Define position
i am not sure?
A ball is projected at an angle of 53º. If the initial velocity is 48 meters/second, what is the vertical component of the velocity with which it was
launched?
OA. 31 meters/second
OB. 38 meters/second
OC
44 meters/second
OD
55 meters/second
Answer: B
Explanation:
The vertical component of a vector such as velocity is the magnitude of the vector multiplied by the sine of the angle.
[tex]V_y=48*sin(53)=38.3m/s[/tex]
If the mass of the book is 50 sliding with acceleration 1.2 m/s ^ 2 then the friction force is
364N
185N
173N
73N
ANSWER AND I WILL GIVE YOU BRAINILIEST
Answer:
73N
Explanation:Just multiply 1.2^2 by 50
An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field towards the other end. If they move fast enough when they strike the positive electrode at the other end, they will give up their energy as X-Rays
(a) Through what potential difference should electrons be accelerated so that their speed is 1% of the speed of light?
(b) What potential difference would be needed to give the protons same kinetic energy as electrons?
(c) What speed would this potential difference give to the protons, both in m/s and as a % of the speed of light.
Answer:
a) ΔV = 25.59 V, b) ΔV = 25.59 V, c) v = 7 10⁴ m / s, v/c= 2.33 10⁻⁴ ,
v/c% = 2.33 10⁻²
Explanation:
a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy
starting point. Where the electrons come out
Em₀ = U = e DV
final point. Where they hit the target
Em_f = K = ½ m v2
energy is conserved
Em₀ = Em_f
e ΔV = ½ m v²
ΔV = [tex]\frac{1}{2}[/tex] mv²/e (1)
If the speed of light is c and this is 100% then 1% is
v = 1% c = c / 100
v = 3 10⁸/100 = 3 10⁶6 m/ s
let's calculate
ΔV = [tex]\frac{1}{2} \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }[/tex]
ΔV = 25.59 V
b) Ask for the potential difference for protons with the same kinetic energy as electrons
[tex]K_e = K_p[/tex]
K_p = ½ m v_e²
K_p = [tex]\frac{1}{2}[/tex] 9.1 10⁻³¹ (3 10⁶)²
K_p = 40.95 10⁻¹⁹ J
we substitute in equation 1
ΔV = Kp / M
ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹
ΔV = 25.59 V
notice that these protons go much slower than electrons because their mass is greater
c) The speed of the protons is
e ΔV = ½ M v²
v² = 2 e ΔV / M
v² = [tex]\frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }[/tex]
v² = 49,035 10⁸
v = 7 10⁴ m / s
Relation
v/c = [tex]\frac{7 \ 10^4 }{ 3 \ 10^8}[/tex]
v/c= 2.33 10⁻⁴
Please Help!!!!
When energy is transferred in a system, the total amount of energy before the transfer is _____________ after the transformation is complete, just in different forms.
Group of answer choices
different
lost
transformed into light
the same
A square plate is produced by welding together four smaller square plates,
each of side
a. The weight of each of the four plates is
shown in the figure.
Find the x-coordinate of the center of gravity (as a multiple of a).
Answer in units of a.
(PICTURED)
PART TWO
Find the y-coordinate of the center of gravity
(as a multiple of a).
Answer in units of a
Answer:
eExplanation:
Daffy Duck is standing 6.8 m away from Minnie Duck. The attractive gravitational force between them is 5.4x10-8 N. If Daffy Duck has a mass of 86.5 kg, What is Minnie Duck's mass?'
Answer:
432.78 Kg
Explanation:
From the question given above, the following data were obtained:
Distance apart (r) = 6.8 m
Force of attraction (F) = 5.4×10¯⁸ N
Mass of Daffy Duck (M₁) = 86.5 kg
Mass of Minnie Duck (M₂) =?
NOTE: Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
The mass of Minnie Duck can be obtained as follow:
F = GM₁M₂ / r²
5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 6.8²
5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 46.24
Cross multiply
6.67×10¯¹¹ × 86.5 × M₂ =5.4×10¯⁸ × 46.24
Divide both side by 6.67×10¯¹¹ × 86.5
M₂ = 5.4×10¯⁸ × 46.24 / 6.67×10¯¹¹ × 86.5
M₂ = 432.78 Kg
Therefore, the mass of Minnie Duck is 432.78 Kg
Henrietta is going off to her physics class, jogging down the sidewalk at a speed of 3.05 m/sm/s . Her husband Bruce suddenly realizes that she left in such a hurry that she forgot her lunch of bagels, so he runs to the window of their apartment, which is a height 40.8 mm above the street level and directly above the sidewalk, to throw them to her. Bruce throws them horizontally at a time 7.00 ss after Henrietta has passed below the window, and she catches them on the run. You can ignore air resistance.
Required:
a. With what initial speed must Bruce throw the bagels so Henrietta can catch them just before they hit the ground?
b. Where is Henrietta when she catches the bagels?
Answer:
v = 10.46 m/s
x = 30.134 m from house
Explanation:
given data
speed = 3.05 m/s
time = 7 s
height = 40.8 mm
solution
we get here first time required to fall that is t
t = [tex]\sqrt{\frac{2\times 40.8}{9.8}}[/tex] ..................1
t = 2.88 s
now we take here initial speed that is v
so v × t = 3.05 × ( t+ 7)
v = [tex]\frac{3.05\times (2.88 + 7)}{2.88}[/tex]
v = 10.46 m/s
and
when she catch the bagel henrietta was at
x = 3.05 × ( 2.88 + 7)
x = 30.134 m from house
Along the remote Racetrack Playa in Death valley, California, stones sometimes gouge out prominent trails in the desert floor, as if they had been migrating. For years curiosity mounted about why the stones moved. One explanation was that strong winds during the occasional rainstorms would drag the rough stones over ground softened by rain. When the desert dried out, the trails behind the stones were hard-baked in place. According to measurements, the coefficient of kinetic friction between the stones and the wet playa ground is about 0.80. What horizontal force is needed on a 25 kg stone (a typical mass) to maintain the stone's motion once a gust has started it moving
Answer:
F = 196 N
Explanation:
For this exercise we will use Newton's second law, we define a reference system with the x axis in the direction of movement of the stones and the y axis vertically
Y axis
N- W = 0
N = mg
X axis
F -fr = ma
In this case, they ask us for the force to keep moving, so the stones go at constant speed, which implies that the acceleration is zero.
F- fr = 0
F = fr
the friction force has the equation
fr = μ N
fr = μ mg
we substitute
F = μ mg
let's calculate
F = 0.80 9.8 25
F = 196 N
How would you compare the acceleration between the unbalanced net force of 100 N and of 50 N
Answer:
The acceleration produced by the 100 N net force will be two times greater than the acceleration produced by 50 N net force.
Explanation:
Given;
first net force, F₁ = 100 N
second net force, F₂ = 50 N
If we consider equal mass for the two net forces, and apply Newton's second law of motion, the acceleration produced by the 100 N net force will be two times greater than the acceleration produced by 50 N net force.
Let a₁ be the acceleration produced by the first net force
then, a₂ be the acceleration produced by the second net force
Thus, a₁ = 2a₂
For this assignment, you should mathematically solve and record a video testing your solution for the following prompt: Two rolls of toilet paper, of equal mass and radius, are dropped from different heights so that they hit the ground at the same time. One roll of toilet paper is dropped normally while the other is dropped while a person holds onto a sheet of toilet paper such that the roll unravels as it descends. Determine the ratio of heights h1/h2, where h1 represents the height of the toilet paper dropped normally and h2 represents the height of the toilet paper that unravels, so that both rolls hit the ground at the same time.
Answer:
h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]
Explanation:
Using two rolls of tissue paper : One roll dropped normally while the other drops as some holds onto a sheet of the toilet paper ( I.e. the tissue paper drops rotating about its axis )
Determine the ratio of heights h1/h2
mass of tissues = same
radius of tissues = same
h1 = height of tissue 1
h2 = height of tissue 2
For the first tissue ( Tissue that dropped manually )
potential energy = kinetic energy
mgh = 1/2 mv^2
therefore the final velocity ( v^2 ) = 2gH ----- ( 1 )
second tissue ( Tissue that dropped while rotating )
gh = [tex]\frac{v^2}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] ) ------ ( 2 )
To determine the ratio of heights we will equate equations 1 and 2
hence :
gh = [tex]\frac{2gH}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] )
∴ h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]
a car travels 10 miles east in 30 minutes. what is its velocity in miles per hour. what is its velocity in miles per hour?
Answer:
popu
Explanation:
2u2uwju2i2je82jei
You push a box across the floor at a constant speed of 1 m/s, applying a horizontal constant force of magnitude 20 N. Your friend pushes the same box across the same floor at a constant speed of 2 m/s, applying a horizontal force. What is the magnitude of the force that your friend applies to the box
Answer:
the force your friend applied on the box is 40 N.
Explanation:
Given;
speed of the box, v₁ = 1 m/s
force applied to the box, F₁ = 20 N
the speed of the box when your friend pushes it, v₂ = 2 m/s
then your friends applied force, F₂ = ?
Assuming the time, t, through which both forces were applied and mass of the box, m, to be constant;
[tex]F_1 = \frac{mv_1}{t} \\\\\frac{m}{t} = \frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]
[tex]F_2 = \frac{F_1v_2}{v_1} \\\\F_2 = \frac{20\times 2}{1} \\\\F_2 = 40 \ N[/tex]
Therefore, the force your friend applied on the box is 40 N.
a device that spreads light into different wavelengths is a what?
maybe a spectrograph ?
What does m/s/s mean?
Explanation:
There are two answers
m/(s/s)=m
or
(m/s)/s=m/s²
A cylindrical bar that us well insulated around its sides connects hot and cold reservoirs and conducts heat at a rate of 10.0 J/s under steady state conditions. If all of its linear dimensions (diameter and length) are reduced by half, the rate at which it will now conduct heat between the same reservoirs is closest to
Answer: the at which the bar conducts now is 5 Js⁻¹
Explanation:
Given the data in the question;
we know that; Heat transfer by conduction is given by;
Q ∝ [tex]A / l[/tex]
such that,
[tex]Q_{1}/Q_{2}[/tex] = [tex]A_{1}l_{2} / A_{2}l_{1}[/tex]
[tex]Q_{2} = Q_{1} A_{2}l_{1}/A_{1}l_{2}[/tex]
so
[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × π([tex]\frac{r}{2}[/tex])² × [tex]l[/tex] ) / (πr² × [tex]\frac{l}{2}[/tex])
[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × π([tex]\frac{r^{2} }{4}[/tex]) × [tex]l[/tex] ) / (πr² × [tex]\frac{l}{2}[/tex])
[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × πr² × 1/4 × [tex]l[/tex] ) / (πr² × 1/2 × [tex]l[/tex] )
[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × 1/4) / ( 1/2)
[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × 1/4) / ( 1/2)
[tex]Q_{2} =[/tex] 5 Js⁻¹
Therefore, the at which the bar conducts now is 5 Js⁻¹
I need help please will mark brainliest
Answer: 30 to 40 s
Explanation:
two objects are moving in the xy plane. no external forces are acting on the objects. object a has a mass of 3.2 kg and has a velocity of v= (2.3m/s)i + (4.2m/s)j and object b has a mass of 2.9kg and has a velocity of v=(-1.8m/s)i = (2.7 m/s)j. sometime later object a is seen to have a velocity va=(1.7m/s)i+(3.5m/s)j what is the velocity of object b at that instant
Answer:
58.469 kg.m/s are moving in the xy plane
A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antinode is 2.20 mm and the speed of propagation of transverse waves on the string is 260 m/s. The string extends along the x-axis, with one of the fixed ends at x= 0, so that there is a node at x =0. The smallest value of x where there is an antinode is x= 0.150m.
Required:
a. What is the maximum transverse speed of a point on the string at an antinode?
b. What is the maximum transverse speed of a point on the string at x = 0.075 m?
Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so [tex]V_{max1}[/tex] = A × ω
[tex]V_{max1}[/tex] = 2.2×10⁻³ × 2722.69 m/s
[tex]V_{max1}[/tex] = 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
[tex]V_{max2}[/tex] = A' × ω
[tex]V_{max2}[/tex] = 1.555×10⁻³ × 2722.69
[tex]V_{max2}[/tex] = 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Which hormone do ovaries release?
A. estrogen
B. glucagon
C. insulin
D. testosterone
Answer:
A. estrogen
Explanation:
This is released in the female reproductive organ.
A spring with a spring constant of 22 N/m is stretched from equilibrium to 2.9 m. How much work is done in the process?
O A. 186 )
OB. 47 J
O C. 933
OD. 121 )
what is the answer ?
Answer:
using W=1/2kW2
k=22N/m w=2.9
w=1/2×22×2.9×2.9
w=92.51Joules
Approximately 93J answer is C
A skier of mass 72 kg is pulled up a slope by a motor-driven cable. a. how much work is required to pull him 75 m up a 30 degree slope (assumed frictionless) at a constant speed of 3.4 m/s
Answer: [tex]26.460\times 10^3\ J[/tex]
Explanation:
Given the mass of skier m=72 kg
distance traveled d=75 m
constant speed v=3.4 m/s
If speed is constant then there must no force acting in the direction of motion
i.e. tension force must be equal to the component of weight
[tex]T=mg\sin 30^{\circ}[/tex]
Work done is given by
[tex]\Rightarrow W=F\cdot d=Td\cos 0^{\circ}\\\Rightarrow W=mg\sin 30^{\circ} d=72\times 9.8\times 0.5\times 75\\\Rightarrow W=26.460\times 10^3\ J[/tex]
What will the reading of the voltmeter be at the instant the switch returns to position a if the inertia of the d'Arsonval movement is negligible
Answer:
hello your question is incomplete attached below is the complete question
answer :
20.16 v
Explanation:
The reading of the voltmeter at the instant the switch returns to position a
L = 5H
i ( current through inductor ) = 1/L ∫ V(t) d(t) + Vo
= 1/5 ∫ 3*10^-3 d(t) + 0 = 0.6 * 10^-3 t
iL ( 1.6 s ) = 0.6 * 10^-3 * 1.6 = 0.96 mA
Rm ( resistance ) = 21 * 1000 = 21 kΩ
The reading of the voltmeter ( V )
V = IR
= 0.96 mA * 21 k Ω = 20.16 v
QUICK SOMEONE PLEASE HELP!!!! I’LL MARK BRAINLIEST!!!
Two small charged objects repel each other with a force F when separated by a distance d. If the charge on each object is reduced to 1/ 14 of its original value and the distance between them is reduced to d/ 24 the force becomes Group of answer choices F F * 24 / 196 F * 576 / 14 F * 576 / 196 F * 196 / 576
Answer:
Ff = F₀ *(576/196)
Explanation:
Assuming that both charges are equal each other, we can express the repulsion force between the charges (assuming that we can treat them as point charges) using Coulomb's Law, as follows:[tex]F_{o} = \frac{k*q^{2} }{d^{2}} (1)[/tex]
Now, if q reduces to q/14, and d is reduced to d/24, the new value of the force will be:[tex]F_{f} = \frac{k*(q/14)^{2} }{(d/24)^{2}} = \frac{k*q^{2}}{d^{2}} * \frac{(24)^{2}}{(14)^{2}} (2)[/tex]
⇒[tex]F_{f} =\frac{k*q^{2}}{d^{2}} * \frac{(24)^{2}}{(14)^{2}} = F_{o} * \frac{576}{196} (3)[/tex]