Answer:
(a) 490 N
(b) 490 N
(c) 540 N
Explanation:
mass, m = 50 kg
acceleration, a = 1 m/s2
(a) The weight is given by
W = m g = 50 x 9.8 = 490 N
(b) The normal force is
N = m g = 490 N
(c) Let the force required is F.
Use Newton's second law
F - m g = m a
F = m(g + a)
F = 50(9.8 + 1)
F = 540 N
Which of the following absorbs the energy required by photosynthesis?
Answer:
There are no options, so....
Explanation:
Chlorophyll a absorbs its energy from the Violet-Blue and Reddish orange-Red wavelengths, and little from the intermediate (Green-Yellow-Orange) wavelengths.
Calculate the energy in electron volts of X-rays that have a frequency of 6.00 x 1016 Hz.
207 eV
228 eV
249 eV
Answer:
228
Explanation:
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.
What is the voltage drop across an alarm clock that is connected to a circuit with a current of 1.10A and a resistance of 90Ω?
V = 99 volts
Explanation:
The voltage drop can be calculated using Ohm's law:
V = IR
= (1.10 A)(90 Ω)
= 99 volts
What are moana's hobbies
A spherical light bulb dissipates 100W and is of 5cm diameter. Assume the emissivity is 0.8 and the irradiation is negligible. What is the surface temperature of this spherical light bulb
Answer:
[tex]T=728.9K[/tex]
Explanation:
Power [tex]P=100W[/tex]
Diameter [tex]d=5[/tex]
Radius [tex]r=2.5cm=>2.5*10^{-2}m[/tex]
Emissivity [tex]e=0.8[/tex]
Generally the equation for Area of Spherical bulb is mathematically given by
[tex]A=4\pi r^2[/tex]
[tex]A=4\pi (2.5*10^{-2}m)^2[/tex]
[tex]A=7.85*10^{-3}m^2[/tex]
Generally the equation for Emissive Power bulb is mathematically given by
[tex]E=e\mu AT^4[/tex]
Where
[tex]\mu=Boltzmann constants\\\\\mu=5.67*10^{-8}[/tex]
Therefore
[tex]T^4=\frac{E}{e\mu A}[/tex]
[tex]T^4=\frac{100}{0.8*5.67*10^{-8}*7.85*10^{-3}m^2}[/tex]
[tex]T=^4\sqrt{2.80*10^{11}}[/tex]
[tex]T=728.9K[/tex]
The plum pudding model of the atom states that
Answer:
According to this model, the atom is a sphere of positive charge, and negatively charged electrons are embedded in it to balance the total positive charge.
Explanation:
Hope this helps you
Answer:
The plum pudding model of the atom states that had negatively-charged electrons embedded within a positively-charged "soup."
Explanation:
Thomson's plum pudding model of the atom had negatively-charged electrons embedded within a positively-charged "soup." Rutherford's gold foil experiment showed that the atom is mostly empty space with a tiny, dense, positively-charged nucleus. Based on these results, Rutherford proposed the nuclear model of the atom.
A 500-nm wavelength light in vacuum illuminates a soap film with an index of refraction of 1.33. Air (n=1.00) is on both sides of the film. If the light strikes the film nearly perpindicularly, what is the smallest film thickness such that the film appears bright?
ANS --> 94.0
Please show your work as to how to end up with this answer.
Answer:
Wavelength of light in film (let y = lambda)
y = 500 nm / (4/3) = 375 nm
There will be a phase change at the air/film interface (not the other side)
S = 4 t thickness of film = S/4 where S equals 1 wavelength
This is because of the phase change at one surface
375 nm = 4 * t
t = 93.8 nm
A 0.2-kg stone is attached to a string and swung in a circle of radius 0.6 m on a horizontal and frictionless surface. If the stone makes 150 revolutions per minute, the tension force of the string on the stone is:____.
a. 0.75 N.
b. 1.96 N.
c. 0.03 N.
d. 30 N.
e. 0.2 N.
Answer:
the tension force of the string on the stone is 30 N
Option d) 30 N is the correct answer.
Explanation:
Given the data in the question;
mass m = 0.2 kg
radius r = 0.6 m
θ = 150 revolutions = 300π rad
time t = 60 seconds
we know that; Angular speed ω = θ / t
we substitute
ω = 300π / 60
ω = 5π rad
Linear speed of stone u = ω × r
we substitute
u = 5π × 0.6
u = 3π m/s
The tension force of the string on the stone is equal to centripetal force, which aid it move in circle;
so
T = mv² / r
we substitute
T = [ 0.2 × (3π)² ] / 0.6
T = 17.7652879 / 0.6
T = 29.6 ≈ 30 N
Therefore, the tension force of the string on the stone is 30 N
Option d) 30 N is the correct answer.
What is the voltage drop across an alarm clock that is connected to a circuit with a current of 1.10A and a resistance of 90Ω?
Answer:
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Explanation:
Hannah tests her new sports car by racing with Sam, an experienced racer. Both start from rest, but Hannah leaves the starting line 1.00 s after Sam does. Sam moves with a constant acceleration of 3.50 m/s2, while Hannah maintains an acceleration of 4.90 m/s2. Find (a) the time at which Hannah overtakes Sam, (b) the distance she travels before she catches him, and (c) the speeds of both cars at the instant Hannah overtakes Sam.
Answer:
a) t = 6.46 s, b) x = 72.98 m, c) v₁ = 26.75 m / s, v₂ = 22.61 m / s
Explanation:
This is an exercise in kinematics, let's write the expressions for each person
Hanna
leaves a time t₀o = 1s after Sam's output, both with zero initial velocity and acceleration of a₁ = 4.90 m / s²
x₁ = 0 + ½ a₁ (t-t₀) ²
v₁ = 0 + a₁ (t-t₀)
Sam
with an acceleration of a₂ = 3.50 m / s² and with an initial velocity of zero
x₂ = 0+ ½ a₂ t²
v₂ = 0 + a₂ t
a) at the point where the position of the two is found is the same
x₁ = x₂
½ a₁ (t-t₀) ² = ½ a₂ t²
let's solve
t-t₀ = [tex]\sqrt{\frac{a_2}{a_1} }[/tex] t
t (1 - [tex]\sqrt{ \frac{a_2}{a_1} }[/tex]) = t₀
t = [tex]\frac{t_o}{ 1-\sqrt{ \frac{a_2}{a_1} } }[/tex]
let's calculate
t = [tex]\frac{ 1}{1- \sqrt{\frac{3.50}{4.90} } }[/tex]
t = [tex]\frac{1}{1- 0.845}[/tex] 1 / 1- 0.845
t = 6.46 s
b) the distance traveled is
x = ½ a₂ t²
x = ½ 3.5 6.46²
x = 72.98 m
c) Hanna's speed
v₁ = 4.9 (6.46 -1)
v₁ = 26.75 m / s
sam's speed
v₂ = a2 t
v₂ = 3.50 6.46²
v₂ = 22.61 m / s
If a 50 W light bulb and a 90 W light bulb operate from 120 V, which bulb has a greater current in it?
50 W
90 W
give you point but please i need help in physcis
Explanation:
Hi
Hi
Hi
Hi
BRAINILIEST PLEASE
What is 4+4+2+2
.The answer is 12
Hope this helps
What was the earliest energy source for humans?
A. coal
B. oil
C. natural gas
D. wood
Answer:
i think its oil
Explanation:
It was made & used as early as the fourth century BC.
(TCO 4) A signal consists of only two sinusoids, one of 65 Hz and one of 95 Hz. This signal is sampled at a rate of 245 Hz. Find the first six positive frequencies that will be present in the replicated spectrum.
Answer:
65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz
Explanation:
Given :
Frequencies of the sinusoids,
[tex]$f_{m_1}= 65 \ Hz$[/tex] , and
[tex]$f_{m_2}= 95 \ Hz$[/tex]
Sampling rate [tex]f_s = \ 245 \ Hz[/tex]
The positive frequencies at the output of the sampling system are :
[tex]$f_{o_1}=\pm f_{m_1} \pm nf_s, f_{o_2}=\pm f_{m_2} \pm nf_s $[/tex]
When n = 0,
[tex]$f_{o_1}= f_{m_1} = 65 \ Hz,\ \ f_{o_2}= f_{m_2} = 95 \ Hz $[/tex]
when n = 1,
[tex]$f_{o_1}=\pm f_{m_1} \pm f_s, \ \ f_{o_2}=\pm f_{m_2} \pm f_s $[/tex]
[tex]$f_{o_1}= \pm 65 \pm 245,\ \ f_{o_2}=\pm 95 \pm 245$[/tex]
[tex]$f_{o_1}= 180 \ Hz, 310 \ Hz,\ \ f_{o_2}= 150 \ Hz,340 \ Hz$[/tex]
When n = 2,
[tex]$f_{o_1}= \pm 65 \pm 2(245),\ \ f_{o_2}=\pm 95 \pm 2(245)$[/tex]
[tex]$f_{o_1}= 555 \ Hz, 425 \ Hz,\ \ f_{o_2}= 395 \ Hz,585 \ Hz$[/tex]
Therefore, the first six positive frequencies present in the replicated spectrum are :
65 Hz, 95 Hz, 150 Hz, 180 Hz, 310 Hz, 340 Hz
3. A 5 gm/100 ml solution of drug X is stored in a closed test tube
at 25°C. If the rate of degradation of the drug is 0.05 day-1,
calculate the time required for the initial concentration to
drop to (a) 50% (half-life) and (b) 90% (shelf-life) of its initial
value.
Answer:
See explanation
Explanation:
The degradation of the drug is a first order process;
Hence;
ln[A] = ln[A]o - kt
Where;
ln[A] = final concentration of the drug
ln[A]o= initial concentration of the drug = 5 gm/100
k= degradation constant = 0.05 day-1
t= time taken
When [A] =[ A]o - 0.5[A]o = 0.5[A]o
ln2.5 = ln5 - 0.05t
ln2.5- ln5 = - 0.05t
t= ln2.5- ln5/-0.05
t= 0.9162 - 1.6094/-0.05
t= 14 days
b) when [A] = [A]o - 0.9[A]o = 0.1[A]o
ln0.5 = ln5 -0.05t
t= ln0.5 - ln5/0.05
t= -0.693 - 1.6094/-0.05
t= 46 days
As a 2.0-kg object moves from (4.4 i + 5j) m to ( 11.6 i - 2j) m, the constant resultant force
acting on it is equal to (4i - 9 j) N. If the speed of the object at the initial position is 4.0 m/s,
what is its kinetic energy at its final position?
Answer:
Answer: [tex]107.8\ J[/tex]
Explanation:
Given
Initial position of object is (4.4 i+5 j)
Final position of object is (11.6 i -2 j)
Force acting (4i-9j)
Work done is given by
[tex]\Rightarrow W=F\cdot dx\\\Rightarrow W=(4i-9j)\cdot (11.6i-4.4i-2j-5j)\\\Rightarrow W=(4i-9j)\cdot (7.2i-7j)\\\Rightarrow W=28.8+63\\\Rightarrow W=91.8\ J[/tex]
Initial kinetic energy
[tex]K_i=\dfrac{1}{2}\times 2\times 4^2\\\\K_i=16\ J[/tex]
Change in kinetic energy is equal to work done by object
[tex]\Rightarrow K_f=K_i+W\\\Rightarrow K_f=16+91.8\\\Rightarrow K_f=107.8\ 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
Describe 3 Levers of Power and how they work.
DESCRIBE
Assume the following vehicles are all moving at the Sam speed .it would be harder to change the velocity of which vehicle . What law is it
Answer:what the choices
Explanation:
When the bell in a clock tower rings with a sound of 474 Hz, a pigeon roosting in the belfry flies directly away from the bell.
If the pigeon hears a frequency of 453Hz, what is its speed?
Answer:
15.44 m/s
Explanation:
Below is the given values:
The bell in rings with a sound = 474 Hz
If the frequency of pigeon hears = 453 Hz
Speed = ?
Use below formula:
Frequency = [(Vs - Vo) / Vs ] x fo
Vs = Speed of sound
Vo = Speed of observer
fo = Sound frequency
Frequency = [(Vs - Vo) / Vs ] x fo
453 = [(343 - Vo) / 343 ] x 474
453 / 474 = [(343 - Vo) / 343 ]
0.955 = (343 - Vo) / 343
0.955 x 343 = 343 - Vo
327.56 = 343 - Vo
Vo = 343 - 327.56
Vo = 15.44 m/s
dipole moment are used to calculate the
Answer:
Explanation:
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physics
i really need help with this physics question
this was due last week but i had no idea what to do
pls help
Answer: Water fills the bucket until its force/weight is greater than the block’s. The lever tilts over, causing the bucket to water the plant.
5
What is back-lash error and how it is avoided
Answer:
The correct answer is - it is an error in the observation.
Explanation:
Backlash error is the error in the motion that takes place during shifting the direction of gears. A backlash error is an error in the observation which occurs due to the wear and tear of threads of the screw observed that takes place at the time of reversing the direction of rotation of the thimble where the tip of the screw does not move in the opposite direction but remains stationary for a part of the rotation.
Avoid backlash error:
While taking measurements, the screw should be rotated in one direction only.
A car (m=1200kg) accelerates at 3m/s/s for d=10m. How much work has the engine done? *
A) 5400J
B) 36000J
C) 352800J
show your work please
This is the answer hope it helps
Select the correct answer.
Two charged objects, A and B, are exerting an electric force on each other. What will happen if the charge on A is increased?
A meterstick of negligible mass is placed on a fulcrum at the 0.4 m mark, with a 1 kg mass hung at the zero mark and a 0.5 kg mass hung at the 1.0 m mark. The meterstick is held horizontal and released. Immediately after release, the magnitude of the net torque on the meterstick about the fulcrum is most nearly:________
a. 1 Nm
b. 2 Nm.
c. 2.5 Nm.
d. 7 Nm
e. 7.5 Nm
Answer:
The net torque is 0.98 Nm.
Explanation:
The torque is given by
Torque = force x perpendicular distance
The clock wise torque is taken as negative while the counter clock wise torque is taken as positive.
Take the torques about the fulcrum.
Torque = 1 x 9.8 x 0.4 - 0.5 x 9.8 x 0.6
Torque = 3.92 - 2.94 = 0.98 Nm
Suppose a proton is moving with a speed of 10 m/s in a direction parallel to a uniform magnetic field of 3.0 T. What is the magnitude and direction of the magnetic force on the proton
Answer:
the magnetic force on the proton is zero.
Explanation:
Given;
speed of the proton, v = 10 m/s
magnitude of the magnetic field, B = 3 T
The magnitude of the magnetic force on the particle is calculated as;
F = qvBsinθ
where;
θ is the angle between the velocity of the particle and the magnetic field
Since the particle is moving parallel to the magnetic field, θ = 0
F = qvBsin(0)
F = 0
Therefore, the magnetic force on the proton is zero.
Answer:
This is a trick question, in that the numbers do not matter.
Study the dependence of the magnetic force on the direction of the magnetic field and the direction of motion of the particle. When is it a maximum? When is it a minimum?
Hint: In vector notation, this is often expressed as q v x B, where q is the electric charge of the particle, v is its velocity (a vector), and B is the magnetic field vector.
Consider the cylindrical weir of diameter 3 m and length 6m. If the fluid on the left has a specific gravity of 0.8, find the magnitude and direction of the resultant force
This question is incomplete, the complete question is;
Consider the cylindrical weir of diameter 3m and length 6m. If the fluid on the left has a specific gravity of 1.6 and on the right has a specific gravity of 0.8, Find the magnitude and direction of the resultant force.
Answer:
- the magnitude of the resultant force is 557.32 kN
- the direction of resultant force is 48.29°
Explanation:
Given the data in the question and the diagram below,
First we work on the force on the left hand side.
Left Horizontal
[tex]F_{LH[/tex] = βgAr
here, h = 3/2 = 1.5 m, β = 1.6, g = 9.81 m/s², A = 3 m × 6 m = 18 m²
we substitute
[tex]F_{LH[/tex] = βgAh = ( 1.6 × 1000 ) × 9.81 × 18 × 1.5 = 423792 N
Left Vertical
[tex]F_{LV[/tex] = ( βgπh² / 2 ) × W
we substitute
[tex]F_{LV[/tex] = [ ( ( 1.6 × 1000 ) × 9.81 × π(1.5)² ) / 2 ] × 6 = 332845.458 N
Now we go to the right hand side
Right Horizontal
[tex]F_{RH[/tex] = βgAh
here, h' = 1.5/2 = 0.75 m, β = 0.8, g = 9.81 m/s², A = 1.5 m × 6 m = 9 m²
we substitute
[tex]F_{RH[/tex] = ( 0.8 × 1000 ) × 9.81 × 9 × 0.75 ) = 52974 N
Right Vertical
[tex]F_{RV[/tex] = ( βgπh² / 4 ) × W
we substitute
[tex]F_{RV[/tex] = [ ( ( 0.8 × 1000 ) × 9.81 × π(1.5)² ) / 4 ] × 6 = 83211.36 N
Hence
Fx = [tex]F_{LH[/tex] - [tex]F_{RH[/tex] = 52974 N - 423792 N = 370818 N
Fy = [tex]F_{LV[/tex] + [tex]F_{RV[/tex] = 332845.458 N + 83211.36 N = 416056.818 N
R = √( Fx² + Fy² ) = √[ (370818 N)² + (416056.818 N)² ] = 557323.3 N
R = 557.32 kN
Therefore, the magnitude of the resultant force is 557.32 kN
Direction of resultant force;
tanθ = Fy / Fx
we substitute
tanθ = 416056.818 N / 370818 N
tanθ = 1.121997
θ = tan⁻¹( 1.121997 )
θ = 48.29°
Therefore, the direction of resultant force is 48.29°
When a person sits erect, increasing the vertical position of their brain by 38.6 cm, the heart must continue to pump blood to the brain at the same rate. (a) What is the gain in gravitational potential energy (in J) for 110 mL of blood raised 38.6 cm
Answer:
the gain in gravitational potential energy is 0.4369 J
Explanation:
Given the data in the question;
from the definition of density, we know that;
ρ = m / V
where ρ is density, m is the mass and V is the volume.
Now, lets make mass the subject of the formula
m = ρV ------ let this be equation 1
Now, we know that potential energy PE = mgh ------- let this be equation 2
where m is mass, g is acceleration due gravity, and h is the height.
substitute equation 1 into 2
PE = ρVgh
given that; V = 110 mL = 110 × 10⁻⁶ m³, h = 38.6 cm = 38.6 × 10⁻² m, g = 9.8 m/s², ρ = 1.05 × 10³ kg/m³
we substitute
PE = (1.05 × 10³ kg/m³) × (110 × 10⁻⁶ m³) × 9.8 m/s² × 38.6 × 10⁻² m
PE = 0.4369 J
Therefore, the gain in gravitational potential energy is 0.4369 J