Answer:
Photosynthesis, the process by which green plants and certain other organisms transform light energy into chemical energy. ... During photosynthesis in green plants, light energy is captured and used to convert water, carbon dioxide, and minerals into oxygen and energy-rich organic compounds.
Explanation:
thank me later
Which statement is true?
a particle of violet light has less energy than a particle of red light
a particle of violet light has more energy than a particle of red light
a particle of violet light has exactly the same energy as a particle of red light
particles of light do not have any energy, regardless of what color the light is
a particle of violet light has exactly the same energy as a particle of red light
If the child has a mass of 13.9 kg, calculate the magnitude of the force in newtons the mother exerts on the child under the following conditions. (b) The elevator accelerates upward at 0.898 m/s2. 148.702 N
The elevator accelerates upward at an acceleration, then the magnitude of the force is 148.84 N.
What is Force?The force is the action of push or pull which makes an object to move or stop.
Given the mass of child m =13.9 kg, acceleration a =0.898 m/s², then the force will be given by
F = m(g-a)
F = 13.9 x (9.81 - (-0.898))
F = 148.84 N
Thus, the magnitude of the force is 148.84 N.
Learn more about force.
https://brainly.com/question/13191643
#SPJ2
In an experiment, a student brings up the rotational speed of a piece of laboratory apparatus to 24 rpm. She then allows the apparatus to slow down uniformly on its own, and counts 236 revolutions before the apparatus comes to a stop. The moment of inertia of the apparatus is known to be 0.076 kg m2. What is the magnitude of the torque on the apparatus
Answer:
T = 6.43 x 10⁻⁵ N.m
Explanation:
First, we will calculate the deceleration of the apparatus by using the third equation of motion:
[tex]2\alpha \theta = \omega_f^2-\omega_i^2[/tex]
where,
α = angular decelration = ?
θ = angular displacement = (236 rev)(2π rad/rev) = 1482.83 rad
ωi = initial angular speed = (24 rpm)(2π rad/1 rev)(1 min/ 60 s) = 2.51 rad/s
ωf = final angular speed = 0 rad/s
Therefore,
[tex]2\alpha(1482.83\ rad) = (0\ rad/s)^2-(2.51\ rad/s)^2\\\\\alpha = -\frac{(2.51\ rad/s)^2}{2965.66\ rad} \\\\\alpha = - 8.46\ x\ 10^{-4}\ rad/s^2[/tex]
negative sign shows deceleration
Now, for torque:
T = Iα
where,
T = Torque = ?
I = moment of inertia = 0.076 kg.m²
Therefore,
T = (0.076 kg.m²)(8.46 x 10⁻⁴ N.m)
T = 6.43 x 10⁻⁵ N.m
compare the time period of two pendulums of length 4m and 9m
area= length × length
area = 4m × 9m
ans 36
Find the value of T1 if 1 = 30°, 2 = 60°, and the weight of the object is 139.3 newtons.
A.
69.58 newtons
B.
45.05 newtons
C.
25 newtons
D.
98.26 newtons
Answer:
Option A (69.56 newtons) is the appropriate solution.
Explanation:
According to the question,
On the X-axis,
⇒ [tex]T_1Cos30^{\circ}-T_2Cos60^{\circ}=0[/tex]
or,
[tex]T_1Cos 30^{\circ}=T_2Cos60^{\circ}[/tex]
On substituting the values, we get
[tex]T_1\times \frac{\sqrt{3} }{2}=T_2\times \frac{1}{2}[/tex]
[tex]T_1\times \sqrt{3} =T_2[/tex]....(equation 1)
On the Y-axis,
⇒ [tex]T_1Sin30^{\circ}+T_2Sin60^{\circ}=139.3 \ N[/tex]
[tex]\frac{T_1}{2} +\frac{\sqrt{3} }{2} =139.2 \ N[/tex]
[tex]T_1+\sqrt{3}T_2=139.2\times 2[/tex]
From equation 1, we get
[tex]T_1+\sqrt{3}\times \sqrt{3}T_1 =278.4 \ N[/tex]
[tex]T_1+3T_1=278.4 \ N[/tex]
[tex]4T_1=278.4 \ N[/tex]
[tex]T_1=\frac{278.4}{4}[/tex]
[tex]=69.6 \ N[/tex]
Answer:
69.58
Explanation:
5N
5 N
19 N
19 N
Pls help look at the pic
Answer:
b. is the correct answer ....
You want to produce a magnetic field of magnitude 5.50 x 10¹ T at a distance of 0.0 6 m from a long, straight wire's center. (a) What current is required to produce this field? (b) With the current found in part (a), how strong is the magnetic field 8.00 cm from the wire's center?
Answer:
(a) I = 1650000 A
(b) 4.125 T
Explanation:
Magnetic field, B = 5.5 T
distance, r = 0.06 m
(a) Let the current is I.
The magnetic field due to a long wire is given by
[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\5.5= 10^{-7}\times \frac{2\times I}{0.06}\\I =1650000 A[/tex]
(b) Let the magnetic field is B' at distance r = 0.08 m.
[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\B = 10^{-7}\times \frac{2\times 1650000}{0.08}\\B'= 4.125 T[/tex]